task_url stringlengths 30 116 | task_name stringlengths 2 86 | task_description stringlengths 0 14.4k | language_url stringlengths 2 53 | language_name stringlengths 1 52 | code stringlengths 0 61.9k |
|---|---|---|---|---|---|
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Java | Java | public class McNuggets {
public static void main(String... args) {
int[] SIZES = new int[] { 6, 9, 20 };
int MAX_TOTAL = 100;
// Works like Sieve of Eratosthenes
int numSizes = SIZES.length;
int[] counts = new int[numSizes];
int maxFound = MAX_TOTAL + 1;
boolean[] found = new boolean[maxFound];
int numFound = 0;
int total = 0;
boolean advancedState = false;
do {
if (!found[total]) {
found[total] = true;
numFound++;
}
// Advance state
advancedState = false;
for (int i = 0; i < numSizes; i++) {
int curSize = SIZES[i];
if ((total + curSize) > MAX_TOTAL) {
// Reset to zero and go to the next box size
total -= counts[i] * curSize;
counts[i] = 0;
}
else {
// Adding a box of this size still keeps the total at or below the maximum
counts[i]++;
total += curSize;
advancedState = true;
break;
}
}
} while ((numFound < maxFound) && advancedState);
if (numFound < maxFound) {
// Did not find all counts within the search space
for (int i = MAX_TOTAL; i >= 0; i--) {
if (!found[i]) {
System.out.println("Largest non-McNugget number in the search space is " + i);
break;
}
}
}
else {
System.out.println("All numbers in the search space are McNugget numbers");
}
return;
}
} |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #BASIC | BASIC | 10 DEFINT A-Z: DIM L$(5)
15 FOR I=0 TO 5: READ L$(I): NEXT I
20 LINE INPUT I$
30 M=LEN(I$): DIM D(M)
40 FOR I=1 TO M: D(I)=VAL(MID$(I$,I,1)): NEXT I
50 FOR J=M-1 TO 1 STEP -1
60 FOR I=1 TO J
70 D(I+1) = D(I+1) + 10*(D(I) AND 1)
80 D(I) = D(I)\2
90 NEXT I,J
100 S=1
110 IF D(S)=0 AND S<M THEN S=S+1: GOTO 110
120 FOR I=S TO M: PRINT "+----";: NEXT I: PRINT "+"
130 FOR L=3 TO 0 STEP -1
140 FOR I=S TO M
150 IF (D(I) OR L)=0 THEN PRINT "| @ ";: GOTO 180
160 N=D(I)-5*L: IF N>5 THEN N=5 ELSE IF N<0 THEN N=0
170 PRINT "|";L$(N);
180 NEXT I
190 PRINT "|"
200 NEXT L
210 FOR I=S TO M: PRINT "+----";: NEXT I: PRINT "+"
220 END
230 DATA " "," . "," .. ","... ","....","----" |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #BCPL | BCPL | get "libhdr"
let reads(v) be
$( v%0 := 0
$( let ch = rdch()
if ch = '*N' | ch = endstreamch break
v%0 := v%0 + 1
v%(v%0) := ch
$) repeat
$)
let digits(v) = valof
$( for i=1 to v%0
$( unless '0' <= v%i <= '9' resultis false
v%i := v%i - '0'
$)
resultis true
$)
let base20(v) be
$( let i = ?
for j=v%0-1 to 1 by -1
for i=1 to j
$( v%(i+1) := v%(i+1) + 10*(v%i & 1)
v%i := v%i >> 1
$)
i := 1
while v%i=0 & i<v%0 do i := i+1
for j=i to v%0 do v%(j-i+1) := v%j
v%0 := v%0-i+1
$)
let mayan(v) be
$( let border(n) be
$( for i=1 to n do writes("+----")
writes("+*N")
$)
let part(num, line) be
test num=0 do
writes(line=0 -> " @ ", " ")
or
$( num := num - line*5
writes(num<=0 -> " ",
num =1 -> " . ",
num =2 -> " .. ",
num =3 -> "... ",
num =4 -> "....",
"----")
$)
border(v%0)
for l=3 to 0 by -1
$( for d=1 to v%0
$( wrch('|')
part(v%d, l)
$)
writes("|*N")
$)
border(v%0)
$)
let start() be
$( let v = vec 1+255/BYTESPERWORD
$( writes("Number? ")
reads(v)
if v%0=0 finish
unless digits(v) loop
base20(v)
mayan(v)
$) repeat
$) |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #D | D | import std.stdio, std.random, std.string, std.array, std.algorithm,
std.file, std.conv;
enum int cx = 4, cy = 2; // Cell size x and y.
enum int cx2 = cx / 2, cy2 = cy / 2;
enum pathSymbol = '.';
struct V2 { int x, y; }
bool solveMaze(char[][] maze, in V2 s, in V2 end) pure nothrow @safe @nogc {
if (s == end)
return true;
foreach (immutable d; [V2(0, -cy), V2(+cx, 0), V2(0, +cy), V2(-cx, 0)])
if (maze[s.y + (d.y / 2)][s.x + (d.x / 2)] == ' ' &&
maze[s.y + d.y][s.x + d.x] == ' ') {
//Would this help?
// maze[s.y + (d.y / 2)][s.x + (d.x / 2)] = pathSymbol;
maze[s.y + d.y][s.x + d.x] = pathSymbol;
if (solveMaze(maze, V2(s.x + d.x, s.y + d.y), end))
return true;
maze[s.y + d.y][s.x + d.x] = ' ';
}
return false;
}
void main() {
auto maze = "maze.txt".File.byLine.map!(r => r.chomp.dup).array;
immutable h = (maze.length.signed - 1) / cy;
assert (h > 0);
immutable w = (maze[0].length.signed - 1) / cx;
immutable start = V2(cx2 + cx * uniform(0, w), cy2 + cy * uniform(0, h));
immutable end = V2(cx2 + cx * uniform(0, w), cy2 + cy * uniform(0, h));
maze[start.y][start.x] = pathSymbol;
if (!solveMaze(maze, start, end))
return "No solution path found.".writeln;
maze[start.y][start.x] = 'S';
maze[end.y][end.x] = 'E';
writefln("%-(%s\n%)", maze);
} |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #C.23 | C# |
using System;
namespace RosetaCode
{
class MainClass
{
public static void Main (string[] args)
{
int[,] list = new int[18,19];
string input = @"55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93";
var charArray = input.Split ('\n');
for (int i=0; i < charArray.Length; i++) {
var numArr = charArray[i].Trim().Split(' ');
for (int j = 0; j<numArr.Length; j++)
{
int number = Convert.ToInt32 (numArr[j]);
list [i, j] = number;
}
}
for (int i = 16; i >= 0; i--) {
for (int j = 0; j < 18; j++) {
list[i,j] = Math.Max(list[i, j] + list[i+1, j], list[i,j] + list[i+1, j+1]);
}
}
Console.WriteLine (string.Format("Maximum total: {0}", list [0, 0]));
}
}
}
|
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Java | Java | import org.bouncycastle.crypto.digests.MD4Digest;
import org.bouncycastle.util.encoders.Hex;
public class RosettaMD4
{
public static void main (String[] argv) throws Exception
{
byte[] r = "Rosetta Code".getBytes("US-ASCII");
MD4Digest d = new MD4Digest();
d.update (r, 0, r.length);
byte[] o = new byte[d.getDigestSize()];
d.doFinal (o, 0);
Hex.encode (o, System.out);
System.out.println();
}
} |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #REXX | REXX | /* REXX ***************************************************************
* 03.02.2013 Walter Pachl
* 19.04.2013 mid 3 is now a function returning the middle 3 digits
* or an error indication
**********************************************************************/
sl='123 12345 1234567 987654321 10001 -10001 -123 -100 100 -12345',
'2 -1 -10 2002 -2002 0 abc 1e3 -17e-3'
Do While sl<>'' /* loop through test values */
Parse Var sl s sl /* pick next value */
Say left(s,12) '->' mid3(s) /* test it */
End
Exit
mid3: Procedure
Parse arg d /* take the argument */
Select /* first test for valid input */
When datatype(d)<>'NUM' Then Return 'not a number'
When pos('E',translate(d))>0 Then Return 'not just digits'
When length(abs(d))<3 Then Return 'less than 3 digits'
When length(abs(d))//2<>1 Then Return 'not an odd number of digits'
Otherwise Do /* input is ok */
dx=abs(d) /* get rid of optional sign */
ld=length(dx) /* length of digit string */
z=(ld-3)/2 /* number of digits to cut */
res=substr(dx,z+1,3) /* get middle 3 digits */
End
End
Return res |
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #BASIC256 | BASIC256 | print MD5("")
print MD5("a")
print MD5("abc")
print MD5("message digest")
print MD5("abcdefghijklmnopqrstuvwxyz")
print MD5("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789")
print MD5("12345678901234567890123456789012345678901234567890123456789012345678901234567890")
end |
http://rosettacode.org/wiki/Menu | Menu | Task
Given a prompt and a list containing a number of strings of which one is to be selected, create a function that:
prints a textual menu formatted as an index value followed by its corresponding string for each item in the list;
prompts the user to enter a number;
returns the string corresponding to the selected index number.
The function should reject input that is not an integer or is out of range by redisplaying the whole menu before asking again for a number. The function should return an empty string if called with an empty list.
For test purposes use the following four phrases in a list:
fee fie
huff and puff
mirror mirror
tick tock
Note
This task is fashioned after the action of the Bash select statement.
| #Ursa | Ursa | def _menu (string<> items)
for (decl int i) (< i (size items)) (inc i)
out " " i ") " items<i> endl console
end for
end _menu
def _ok (string reply, int itemcount)
try
decl int n
set n (int reply)
return (and (or (> n 0) (= n 0)) (< n itemcount))
catch
return false
end try
end _ok
def selector (string<> items, string prompt)
# Prompt to select an item from the items
if (= (size items) 0)
return ""
end if
decl int itemcount reply
set reply -1
set itemcount (size items)
while (not (_ok reply itemcount))
_menu items
out prompt console
set reply (in int console)
end while
return items<(int reply)>
end selector
decl string<> items
append "fee fie" "huff and puff" "mirror mirror" "tick tock" items
decl string item
set item (selector items "Which is from the three pigs: ")
out "You chose: " item endl console |
http://rosettacode.org/wiki/Menu | Menu | Task
Given a prompt and a list containing a number of strings of which one is to be selected, create a function that:
prints a textual menu formatted as an index value followed by its corresponding string for each item in the list;
prompts the user to enter a number;
returns the string corresponding to the selected index number.
The function should reject input that is not an integer or is out of range by redisplaying the whole menu before asking again for a number. The function should return an empty string if called with an empty list.
For test purposes use the following four phrases in a list:
fee fie
huff and puff
mirror mirror
tick tock
Note
This task is fashioned after the action of the Bash select statement.
| #VBScript | VBScript | 'The Function
Function Menu(ArrList, Prompt)
Select Case False 'Non-standard usage hahaha
Case IsArray(ArrList)
Menu = "" 'Empty output string for non-arrays
Exit Function
Case UBound(ArrList) >= LBound(ArrList)
Menu = "" 'Empty output string for empty arrays
Exit Function
End Select
'Display menu and prompt
Do While True
For i = LBound(ArrList) To UBound(ArrList)
WScript.StdOut.WriteLine((i + 1) & ". " & ArrList(i))
Next
WScript.StdOut.Write(Prompt)
Choice = WScript.StdIn.ReadLine
'Check if input is valid
If IsNumeric(Choice) Then 'Check for numeric-ness
If CStr(CLng(Choice)) = Choice Then 'Check for integer-ness (no decimal part)
Index = Choice - 1 'Arrays are 0-based
'Check if Index is in array
If LBound(ArrList) <= Index And Index <= UBound(ArrList) Then
Exit Do
End If
End If
End If
WScript.StdOut.WriteLine("Invalid choice.")
Loop
Menu = ArrList(Index)
End Function
'The Main Thing
Sample = Array("fee fie", "huff and puff", "mirror mirror", "tick tock")
InputText = "Which is from the three pigs: "
WScript.StdOut.WriteLine("Output: " & Menu(Sample, InputText)) |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #JavaScript | JavaScript | (() => {
'use strict';
// main :: IO ()
const main = () => {
const
size = n => enumFromTo(0)(
quot(100, n)
),
nuggets = new Set(
size(6).flatMap(
x => size(9).flatMap(
y => size(20).flatMap(
z => {
const v = sum([6 * x, 9 * y, 20 * z]);
return 101 > v ? (
[v]
) : [];
}
),
)
)
),
xs = dropWhile(
x => nuggets.has(x),
enumFromThenTo(100, 99, 1)
);
return 0 < xs.length ? (
xs[0]
) : 'No unreachable quantities found in this range';
};
// GENERIC FUNCTIONS ----------------------------------
// dropWhile :: (a -> Bool) -> [a] -> [a]
const dropWhile = (p, xs) => {
const lng = xs.length;
return 0 < lng ? xs.slice(
until(
i => i === lng || !p(xs[i]),
i => 1 + i,
0
)
) : [];
};
// enumFromThenTo :: Int -> Int -> Int -> [Int]
const enumFromThenTo = (x1, x2, y) => {
const d = x2 - x1;
return Array.from({
length: Math.floor(y - x2) / d + 2
}, (_, i) => x1 + (d * i));
};
// ft :: Int -> Int -> [Int]
const enumFromTo = m => n =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// quot :: Int -> Int -> Int
const quot = (n, m) => Math.floor(n / m);
// sum :: [Num] -> Num
const sum = xs => xs.reduce((a, x) => a + x, 0);
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};
// MAIN ---
return console.log(
main()
);
})(); |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #C | C | #include <stdio.h>
#include <string.h>
#include <stdint.h>
#include <stdlib.h>
#define MAX(x,y) ((x) > (y) ? (x) : (y))
#define MIN(x,y) ((x) < (y) ? (x) : (y))
/* Find base-20 digits. */
size_t base20(unsigned int n, uint8_t *out) {
/* generate digits */
uint8_t *start = out;
do {*out++ = n % 20;} while (n /= 20);
size_t length = out - start;
/* put digits in high-endian order */
while (out > start) {
uint8_t x = *--out;
*out = *start;
*start++ = x;
}
return length;
}
/* Write a Mayan digit */
void make_digit(int n, char *place, size_t line_length) {
static const char *parts[] = {" "," . "," .. ","... ","....","----"};
int i;
/* write 4-part digit */
for (i=4; i>0; i--, n -= 5)
memcpy(place + i*line_length, parts[MAX(0, MIN(5, n))], 4);
/* if digit was 0 we should put '@' in 2nd position of last line */
if (n == -20) place[4 * line_length + 1] = '@';
}
/* Make a Mayan numeral */
char *mayan(unsigned int n) {
if (n == 0) return NULL;
uint8_t digits[15]; /* 2**64 is 15 Mayan digits long */
size_t n_digits = base20(n, digits);
/* a digit is 4 chars wide, plus N+1 divider lines, plus a newline
makes for a length of 5*n+2 */
size_t line_length = n_digits*5 + 2;
/* we need 6 lines - four for the digits, plus top and bottom row */
char *str = malloc(line_length * 6 + 1);
if (str == NULL) return NULL;
str[line_length * 6] = 0;
/* make the cartouche divider lines */
char *ptr;
unsigned int i;
/* top and bottom row */
for (ptr=str, i=0; i<line_length; i+=5, ptr+=5)
memcpy(ptr, "+----", 5);
memcpy(ptr-5, "+\n", 2);
memcpy(str+5*line_length, str, line_length);
/* middle rows */
for (ptr=str+line_length, i=0; i<line_length; i+=5, ptr+=5)
memcpy(ptr, "| ", 5);
memcpy(ptr-5, "|\n", 2);
memcpy(str+2*line_length, str+line_length, line_length);
memcpy(str+3*line_length, str+line_length, 2*line_length);
/* copy in the digits */
for (i=0; i<n_digits; i++)
make_digit(digits[i], str+1+5*i, line_length);
return str;
}
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: mayan <number>\n");
return 1;
}
int i = atoi(argv[1]);
if (i <= 0) {
fprintf(stderr, "number must be positive\n");
return 1;
}
char *m = mayan(i);
printf("%s",m);
free(m);
return 0;
} |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #Delphi | Delphi | procedure SolveMaze(var AMaze: TMaze; const S, E: TPoint);
var
Route : TRoute;
Position : TPoint;
V : TPoint; // delta vector
begin
ClearVisited(AMaze);
Position := S;
Route := TStack<TPoint>.Create;
with Position do
try
AMaze[x, y].Visited := True;
repeat
if (y > 0) and not AMaze[x, y-1].Visited and AMaze[x, y].PassTop then V := Point(0, -1) else
if (x < mwidth-1) and not AMaze[x+1, y].Visited and AMaze[x+1, y].PassLeft then V := Point(1, 0) else
if (y < mheight-1) and not AMaze[x, y+1].Visited and AMaze[x, y+1].PassTop then V := Point(0, 1) else
if (x > 0) and not AMaze[x-1, y].Visited and AMaze[x, y].PassLeft then V := Point(-1, 0) else
begin
if Route.Count = 0 then Exit; // we are back at start so no way found
Position := Route.Pop; // step back
Continue;
end;
Route.Push(Position); // save current position to route
Offset(V); // move forward
AMaze[x, y].Visited := True;
until Position = E; // solved
ClearVisited(AMaze);
while Route.Count > 0 do // Route to Maze
with Route.Pop do
AMaze[x, y].Visited := True;
finally
Route.Free;
end;
end;
procedure Main;
var
Maze: TMaze;
S, E: TPoint;
begin
Randomize;
PrepareMaze(Maze);
S := Point(Random(mwidth), Random(mheight));
E := Point(Random(mwidth), Random(mheight));
SolveMaze(Maze, S, E);
Write(MazeToString(Maze, S, E));
ReadLn;
end;
begin
Main;
end. |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #C.2B.2B | C++ |
/* Algorithm complexity: n*log(n) */
#include <iostream>
int main( int argc, char* argv[] )
{
int triangle[] =
{
55,
94, 48,
95, 30, 96,
77, 71, 26, 67,
97, 13, 76, 38, 45,
7, 36, 79, 16, 37, 68,
48, 7, 9, 18, 70, 26, 6,
18, 72, 79, 46, 59, 79, 29, 90,
20, 76, 87, 11, 32, 7, 7, 49, 18,
27, 83, 58, 35, 71, 11, 25, 57, 29, 85,
14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55,
2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23,
92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42,
56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 55, 72,
44, 25, 67, 84, 71, 67, 11, 61, 40, 57, 58, 89, 40, 56, 36,
85, 32, 25, 85, 57, 48, 84, 35, 47, 62, 17, 1, 1, 99, 89, 52,
6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15,
27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93
};
const int size = sizeof( triangle ) / sizeof( int );
const int tn = static_cast<int>(sqrt(2.0 * size));
assert(tn * (tn + 1) == 2 * size); // size should be a triangular number
// walk backward by rows, replacing each element with max attainable therefrom
for (int n = tn - 1; n > 0; --n) // n is size of row, note we do not process last row
for (int k = (n * (n-1)) / 2; k < (n * (n+1)) / 2; ++k) // from the start to the end of row
triangle[k] += std::max(triangle[k + n], triangle[k + n + 1]);
std::cout << "Maximum total: " << triangle[0] << "\n\n";
}
|
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #JavaScript | JavaScript | const md4func = () => {
const hexcase = 0; /* hex output format. 0 - lowercase; 1 - uppercase */
const b64pad = ""; /* base-64 pad character. "=" for strict RFC compliance */
const chrsz = 8; /* bits per input character. 8 - ASCII; 16 - Unicode */
const tab = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const hex_tab = hexcase ? "0123456789ABCDEF" : "0123456789abcdef";
/**
* Add integers, wrapping at 2^32. This uses 16-bit operations internally
* to work around bugs in some JS interpreters.
*/
const safe_add = (x, y) => {
const lsw = (x & 0xFFFF) + (y & 0xFFFF);
const msw = (x >> 16) + (y >> 16) + (lsw >> 16);
return (msw << 16) | (lsw & 0xFFFF);
};
/**
* Bitwise rotate a 32-bit number to the left.
*/
const rol = (num, cnt) => (num << cnt) | (num >>> (32 - cnt));
/**
* Convert a string to an array of little-endian words
* If chrsz is ASCII, characters >255 have their hi-byte silently ignored.
*/
const str2binl = str => {
const bin = Array();
const mask = (1 << chrsz) - 1;
for (let i = 0; i < str.length * chrsz; i += chrsz)
bin[i >> 5] |= (str.charCodeAt(i / chrsz) & mask) << (i % 32);
return bin;
};
/**
* Convert an array of little-endian words to a string
*/
const binl2str = bin => {
let str = "";
const mask = (1 << chrsz) - 1;
for (let i = 0; i < bin.length * 32; i += chrsz)
str += String.fromCharCode((bin[i >> 5] >>> (i % 32)) & mask);
return str;
};
/**
* Convert an array of little-endian words to a hex string.
*/
const binl2hex = binarray => {
let str = "";
for (let i = 0; i < binarray.length * 4; i++) {
str += hex_tab.charAt((binarray[i >> 2] >> ((i % 4) * 8 + 4)) & 0xF) +
hex_tab.charAt((binarray[i >> 2] >> ((i % 4) * 8)) & 0xF);
}
return str;
};
/**
* Convert an array of little-endian words to a base-64 string
*/
const binl2b64 = binarray => {
let str = "";
for (let i = 0; i < binarray.length * 4; i += 3) {
const triplet = (((binarray[i >> 2] >> 8 * (i % 4)) & 0xFF) << 16)
| (((binarray[i + 1 >> 2] >> 8 * ((i + 1) % 4)) & 0xFF) << 8)
| ((binarray[i + 2 >> 2] >> 8 * ((i + 2) % 4)) & 0xFF);
for (let j = 0; j < 4; j++) {
if (i * 8 + j * 6 > binarray.length * 32) str += b64pad;
else str += tab.charAt((triplet >> 6 * (3 - j)) & 0x3F);
}
}
return str;
};
/**
* Calculate the MD4 of an array of little-endian words, and a bit length
*/
const core_md4 = (x, len) => {
x[len >> 5] |= 0x80 << (len % 32);
x[(((len + 64) >>> 9) << 4) + 14] = len;
let a = 1732584193;
let b = -271733879;
let c = -1732584194;
let d = 271733878;
for (let i = 0; i < x.length; i += 16) {
const olda = a;
const oldb = b;
const oldc = c;
const oldd = d;
a = md4_ff(a, b, c, d, x[i], 3);
d = md4_ff(d, a, b, c, x[i + 1], 7);
c = md4_ff(c, d, a, b, x[i + 2], 11);
b = md4_ff(b, c, d, a, x[i + 3], 19);
a = md4_ff(a, b, c, d, x[i + 4], 3);
d = md4_ff(d, a, b, c, x[i + 5], 7);
c = md4_ff(c, d, a, b, x[i + 6], 11);
b = md4_ff(b, c, d, a, x[i + 7], 19);
a = md4_ff(a, b, c, d, x[i + 8], 3);
d = md4_ff(d, a, b, c, x[i + 9], 7);
c = md4_ff(c, d, a, b, x[i + 10], 11);
b = md4_ff(b, c, d, a, x[i + 11], 19);
a = md4_ff(a, b, c, d, x[i + 12], 3);
d = md4_ff(d, a, b, c, x[i + 13], 7);
c = md4_ff(c, d, a, b, x[i + 14], 11);
b = md4_ff(b, c, d, a, x[i + 15], 19);
a = md4_gg(a, b, c, d, x[i], 3);
d = md4_gg(d, a, b, c, x[i + 4], 5);
c = md4_gg(c, d, a, b, x[i + 8], 9);
b = md4_gg(b, c, d, a, x[i + 12], 13);
a = md4_gg(a, b, c, d, x[i + 1], 3);
d = md4_gg(d, a, b, c, x[i + 5], 5);
c = md4_gg(c, d, a, b, x[i + 9], 9);
b = md4_gg(b, c, d, a, x[i + 13], 13);
a = md4_gg(a, b, c, d, x[i + 2], 3);
d = md4_gg(d, a, b, c, x[i + 6], 5);
c = md4_gg(c, d, a, b, x[i + 10], 9);
b = md4_gg(b, c, d, a, x[i + 14], 13);
a = md4_gg(a, b, c, d, x[i + 3], 3);
d = md4_gg(d, a, b, c, x[i + 7], 5);
c = md4_gg(c, d, a, b, x[i + 11], 9);
b = md4_gg(b, c, d, a, x[i + 15], 13);
a = md4_hh(a, b, c, d, x[i], 3);
d = md4_hh(d, a, b, c, x[i + 8], 9);
c = md4_hh(c, d, a, b, x[i + 4], 11);
b = md4_hh(b, c, d, a, x[i + 12], 15);
a = md4_hh(a, b, c, d, x[i + 2], 3);
d = md4_hh(d, a, b, c, x[i + 10], 9);
c = md4_hh(c, d, a, b, x[i + 6], 11);
b = md4_hh(b, c, d, a, x[i + 14], 15);
a = md4_hh(a, b, c, d, x[i + 1], 3);
d = md4_hh(d, a, b, c, x[i + 9], 9);
c = md4_hh(c, d, a, b, x[i + 5], 11);
b = md4_hh(b, c, d, a, x[i + 13], 15);
a = md4_hh(a, b, c, d, x[i + 3], 3);
d = md4_hh(d, a, b, c, x[i + 11], 9);
c = md4_hh(c, d, a, b, x[i + 7], 11);
b = md4_hh(b, c, d, a, x[i + 15], 15);
a = safe_add(a, olda);
b = safe_add(b, oldb);
c = safe_add(c, oldc);
d = safe_add(d, oldd);
}
return Array(a, b, c, d);
};
/**
* These functions implement the basic operation for each round of the
* algorithm.
*/
const md4_cmn = (q, a, b, x, s, t) => safe_add(
rol(safe_add(safe_add(a, q), safe_add(x, t)), s), b);
const md4_ff = (a, b, c, d, x, s) => md4_cmn(
(b & c) | ((~b) & d), a, 0, x, s, 0);
const md4_gg = (a, b, c, d, x, s) => md4_cmn(
(b & c) | (b & d) | (c & d), a, 0, x, s, 1518500249);
const md4_hh = (a, b, c, d, x, s) => md4_cmn(
b ^ c ^ d, a, 0, x, s, 1859775393);
/**
* Calculate the HMAC-MD4, of a key and some data
*/
const core_hmac_md4 = (key, data) => {
let bkey = str2binl(key);
if (bkey.length > 16) {
bkey = core_md4(bkey, key.length * chrsz)
}
const ipad = Array(16);
const opad = Array(16);
for (let i = 0; i < 16; i++) {
ipad[i] = bkey[i] ^ 0x36363636;
opad[i] = bkey[i] ^ 0x5C5C5C5C;
}
const hash = core_md4(
ipad.concat(str2binl(data)), 512 + data.length * chrsz);
return core_md4(opad.concat(hash), 512 + 128);
};
/**
* These are the functions you'll usually want to call
*/
return {
hex_md4: s => binl2hex(core_md4(str2binl(s), s.length * chrsz)),
b64_md4: s => binl2b64(core_md4(str2binl(s), s.length * chrsz)),
str_md4: s => binl2str(core_md4(str2binl(s), s.length * chrsz)),
hex_hmac_md4: (key, data) => binl2hex(core_hmac_md4(key, data)),
b64_hmac_md4: (key, data) => binl2b64(core_hmac_md4(key, data)),
str_hmac_md4: (key, data) => binl2str(core_hmac_md4(key, data)),
};
};
const md4 = md4func();
console.log(md4.hex_md4('Rosetta Code')); |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #Ring | Ring |
n = 1234567
middle(n)
func middle nr
mnr = floor(len(string(nr))/2)
lennr = len(string(nr))
if lennr = 3 see "" + nr + nl
but lennr < 3 see "Number must have at least three digits"
but lennr%2=0 see "Number must have an odd number of digits"
else cnr = substr(string(nr),mnr,3) see cnr + nl ok
|
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #BBC_BASIC | BBC BASIC | PRINT FN_MD5("")
PRINT FN_MD5("a")
PRINT FN_MD5("abc")
PRINT FN_MD5("message digest")
PRINT FN_MD5("abcdefghijklmnopqrstuvwxyz")
PRINT FN_MD5("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789")
PRINT FN_MD5(STRING$(8,"1234567890"))
END
DEF FN_MD5(message$)
LOCAL I%, MD5$, MD5_CTX{}
DIM MD5_CTX{i%(1), buf%(3), in&(63), digest&(15)}
SYS "MD5Init", MD5_CTX{}
SYS "MD5Update", MD5_CTX{}, message$, LEN(message$)
SYS "MD5Final", MD5_CTX{}
FOR I% = 0 TO 15
MD5$ += RIGHT$("0"+STR$~(MD5_CTX.digest&(I%)),2)
NEXT
= MD5$ |
http://rosettacode.org/wiki/Menu | Menu | Task
Given a prompt and a list containing a number of strings of which one is to be selected, create a function that:
prints a textual menu formatted as an index value followed by its corresponding string for each item in the list;
prompts the user to enter a number;
returns the string corresponding to the selected index number.
The function should reject input that is not an integer or is out of range by redisplaying the whole menu before asking again for a number. The function should return an empty string if called with an empty list.
For test purposes use the following four phrases in a list:
fee fie
huff and puff
mirror mirror
tick tock
Note
This task is fashioned after the action of the Bash select statement.
| #Wren | Wren | import "/ioutil" for Input
var menu = Fn.new { |list|
var n = list.count
if (n == 0) return ""
var prompt = "\n M E N U\n\n"
for (i in 0...n) prompt = prompt + "%(i+1). %(list[i])\n"
prompt = prompt + "\nEnter your choice (1 - %(n)): "
var index = Input.integer(prompt, 1, n)
return list[index-1]
}
var list = ["fee fie", "huff and puff", "mirror mirror", "tick tock"]
var choice = menu.call(list)
System.print("\nYou chose : %(choice)") |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #jq | jq | [
[range(18) as $n6 |
range(13) as $n9 |
range(6) as $n20 |
($n6 * 6 + $n9 * 9 + $n20 * 20)] |
unique |
. as $possible |
range(101) |
. as $n |
select($possible|contains([$n])|not)
] |
max |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #CLU | CLU | % This program must be linked with PCLU's "misc.lib" and "useful.lib"
base20 = proc (n: bigint) returns (sequence[int])
own zero: bigint := bigint$i2bi(0)
own twenty: bigint := bigint$i2bi(20)
if n=zero then return(sequence[int]$[0]) end
digits: array[int] := array[int]$[]
while n>zero do
array[int]$addl(digits, bigint$bi2i(n//twenty))
n := n/twenty
end
return(sequence[int]$a2s(digits))
end base20
mayan = proc (digits: sequence[int]) returns (string)
own parts: array[string] := array[string]$[0:
" ", " . ", " .. ", "... ", "....", "----"
]
% generate edges
edge: stream := stream$create_output()
for i: int in int$from_to(1, sequence[int]$size(digits)) do
stream$puts(edge, "+----")
end
stream$putl(edge, "+")
% generate digits
lines: stream := stream$create_output()
for i: int in int$from_to_by(15, 0, -5) do
for d: int in sequence[int]$elements(digits) do
p: int := d-i
if p<0 then p:=0 end
if p>5 then p:=5 end
if i=0 & p=0
then stream$puts(lines, "| @ ")
else stream$puts(lines, "|" || parts[p])
end
end
stream$putl(lines, "|")
end
s_edge: string := stream$get_contents(edge)
return(s_edge || stream$get_contents(lines) || s_edge)
end mayan
start_up = proc ()
po: stream := stream$primary_output()
n: bigint := bigint$parse(sequence[string]$bottom(get_argv()))
stream$puts(po, mayan(base20(n)))
end start_up |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #Cowgol | Cowgol | include "cowgol.coh";
include "argv.coh";
# convert number to base 20
sub base20(n: uint32, out: [uint8]): (n_digits: uint8) is
n_digits := 0;
loop
[out] := (n % 20) as uint8;
n := n / 20;
out := @next out;
n_digits := n_digits + 1;
if n == 0 then break; end if;
end loop;
end sub;
# get the N'th line (from the top) for a Mayan digit
sub digit_line(n: uint8, line: uint8): (s: [uint8]) is
var parts: [uint8][] := {" "," . "," .. ","... ","....","----"};
if n == 0 then
if line == 3 then s := " @ ";
else s := parts[0];
end if;
else
var nn := n - 5*(3-line);
if nn > 128 then s := parts[0];
elseif nn > 5 then s := parts[5];
else s := parts[nn];
end if;
end if;
end sub;
# print Mayan number
sub print_mayan(n: uint32) is
sub edge(n: uint8) is
while n>0 loop
print("+----");
n := n-1;
end loop;
print_char('+');
print_nl();
end sub;
var digits: uint8[8]; # 8 digits is enough for 2**32
var size := base20(n, &digits[0]);
edge(size);
var line: uint8 := 0;
while line < 4 loop
var d: uint8 := size-1;
loop
print_char('|');
print(digit_line(digits[d], line));
if d==0 then break; end if;
d := d - 1;
end loop;
print_char('|');
print_nl();
line := line + 1;
end loop;
edge(size);
end sub;
sub Error() is
print("usage: mayan <number>. number must be positive");
print_nl();
ExitWithError();
end sub;
# read number from command line
ArgvInit();
var arg := ArgvNext();
if arg == 0 as [uint8] then Error(); end if;
var n: int32;
(n, arg) := AToI(arg);
if n <= 0 then Error(); end if;
print_mayan(n as uint32); |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #EasyLang | EasyLang | size = 20
n = 2 * size + 1
endpos = n * n - 2
startpos = n + 1
#
f = 100 / (n - 0.5)
len m[] n * n
#
background 000
func show_maze . .
clear
for i range len m[]
if m[i] = 0
x = i mod n
y = i div n
color 777
move x * f - f / 2 y * f - f / 2
rect f * 1.5 f * 1.5
.
.
sleep 0.001
.
offs[] = [ 1 n -1 (-n) ]
brdc[] = [ n - 2 -1 1 -1 ]
brdr[] = [ -1 n - 2 -1 1 ]
#
func m_maze pos . .
m[pos] = 0
call show_maze
d[] = [ 0 1 2 3 ]
for i = 3 downto 0
d = random (i + 1)
dir = d[d]
d[d] = d[i]
r = pos div n
c = pos mod n
posn = pos + 2 * offs[dir]
if c <> brdc[dir] and r <> brdr[dir] and m[posn] <> 0
posn = pos + 2 * offs[dir]
m[(pos + posn) div 2] = 0
call m_maze posn
.
.
.
func make_maze . .
for i range len m[]
m[i] = 1
.
call m_maze startpos
m[endpos] = 0
.
call make_maze
call show_maze
#
func mark pos col . .
x = pos mod n
y = pos div n
color col
move x * f + f / 4 y * f + f / 4
circle f / 4
.
func solve dir0 pos . found .
call mark pos 900
sleep 0.05
if pos = endpos
found = 1
else
for dir range 4
posn = pos + offs[dir]
if dir <> dir0 and m[posn] = 0 and found = 0
call solve (dir + 2) mod 4 posn found
if found = 0
call mark posn 777
sleep 0.05
.
.
.
.
.
sleep 3
call solve -1 startpos found |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #Clojure | Clojure |
(ns clojure.examples.rosetta
(:gen-class)
(:require [clojure.string :as string]))
(def rosetta "55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93")
;; The technique is described here in more detail http://mishadoff.com/blog/clojure-euler-problem-018/
;; Most of the code converts the string data to a nested array of integers.
;; The code to calculate the max sum is then only a single line
;; First convert string data to nested list
;; with each inner list containing one row of the triangle
;; [[55] [94 48] [95 30 96] ... [...10 69 93]
(defn parse-int [s]
" Convert digits to a number (finds digits when could be surrounded by non-digits"
(Integer. (re-find #"\d+" s)))
(defn data-int-array [s]
" Convert string to integer array"
(map parse-int (string/split (string/trim s) #"\s+")))
(defn nested-triangle [s]
" Convert triangle to nested vector, with each inner vector containing one triangle row"
(loop [lst s n 1 newlist nil]
(if (empty? lst) (reverse newlist)
(recur (drop n lst) (inc n) (cons (take n lst) newlist)))))
; Create nested list
(def nested-list (nested-triangle (data-int-array rosetta)))
;; Function to compute maximum path sum
(defn max-sum [s]
" Compute maximum path sum using a technique described here: http://mishadoff.com/blog/clojure-euler-problem-018/"
(reduce (fn [a b] (map + b (map max a (rest a)))) (reverse s)))
; Print result
(println (max-sum nested-list))
|
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Julia | Julia |
using Nettle
msg = "Rosetta Code"
h = HashState(MD4)
update!(h, msg)
h = hexdigest!(h)
println("\"", msg, "\" => ", h)
|
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #Ruby | Ruby | def middle_three_digits(num)
# minus sign doesn't factor into digit count,
# and calling #abs acts as a duck-type assertion
num = num.abs
# convert to string and find length
length = (str = num.to_s).length
# check validity
raise ArgumentError, "Number must have at least three digits" if length < 3
raise ArgumentError, "Number must have an odd number of digits" if length.even?
return str[length/2 - 1, 3].to_i
end |
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <openssl/md5.h>
const char *string = "The quick brown fox jumped over the lazy dog's back";
int main()
{
int i;
unsigned char result[MD5_DIGEST_LENGTH];
MD5(string, strlen(string), result);
// output
for(i = 0; i < MD5_DIGEST_LENGTH; i++)
printf("%02x", result[i]);
printf("\n");
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/Menu | Menu | Task
Given a prompt and a list containing a number of strings of which one is to be selected, create a function that:
prints a textual menu formatted as an index value followed by its corresponding string for each item in the list;
prompts the user to enter a number;
returns the string corresponding to the selected index number.
The function should reject input that is not an integer or is out of range by redisplaying the whole menu before asking again for a number. The function should return an empty string if called with an empty list.
For test purposes use the following four phrases in a list:
fee fie
huff and puff
mirror mirror
tick tock
Note
This task is fashioned after the action of the Bash select statement.
| #XPL0 | XPL0 | string 0;
func Menu(List);
int List;
int Size, I, C;
[Size:= List(0);
if Size < 1 then return List(0); \if empty list, return pointer to 0
for I:= 1 to Size-1 do
[IntOut(0, I); Text(0, ": ");
Text(0, List(I)); CrLf(0);
];
CrLf(0);
Text(0, List(Size)); \display prompt
loop [C:= ChIn(0); \buffered keyboard requires Enter key
if C>=^1 & C<=Size-1+^0 then return List(C-^0);
Text(0, "Please enter 1 thru "); IntOut(0, Size-1);
Text(0, ": ");
];
];
Text(0, Menu([5, "fee fie", "huff and puff", "mirror mirror", "tick tock",
"Which phrase is from the Three Little Pigs? "])) |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Julia | Julia | function mcnuggets(max)
b = BitSet(1:max)
for i in 0:6:max, j in 0:9:max, k in 0:20:max
delete!(b, i + j + k)
end
maximum(b)
end
println(mcnuggets(100))
|
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Kotlin | Kotlin | // Version 1.2.71
fun mcnugget(limit: Int) {
val sv = BooleanArray(limit + 1) // all false by default
for (s in 0..limit step 6)
for (n in s..limit step 9)
for (t in n..limit step 20) sv[t] = true
for (i in limit downTo 0) {
if (!sv[i]) {
println("Maximum non-McNuggets number is $i")
return
}
}
}
fun main(args: Array<String>) {
mcnugget(100)
} |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #F.23 | F# |
// Mayan numerals. Nigel Galloway: February 19th., 2021
let N=[|"│ ";"│. ";"│.. ";"│... ";"│....";"│~~~~"|]
let fN g=(fun(n)->if g=0 && n=0 then "│ Θ " else N.[let g=g-5*n in if g>4 then 5 else if g<0 then 0 else g])
let rec fG n g=match n/20L,n%20L with (0L,0L)->(g,List.length g) |(i,n)->fG i ((fN(int n))::g)
let mayan n=let n,g=fG n []
printf "┌────"; for _ in 2..g do printf "┬────"
printfn "┐"; for g in 3.. -1 ..0 do n|>List.iter(fun n->printf "%s" (n(g))); printfn "│"
printf "└────"; for _ in 2..g do printf "┴────"
printfn "┘"
[4005L;8017L;326205L;886205L]|>List.iter(fun n->printfn "%d" n; mayan n)
|
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #11l | 11l | F matrix_mul(m1, m2)
assert(m1[0].len == m2.len)
V r = [[0] * m2[0].len] * m1.len
L(j) 0 .< m1.len
L(i) 0 .< m2[0].len
V s = 0
L(k) 0 .< m2.len
s += m1[j][k] * m2[k][i]
r[j][i] = s
R r
F identity(size)
V rsize = 0 .< size
R rsize.map(j -> @rsize.map(i -> Int(i == @j)))
F matrixExp(m, pow)
assert(pow >= 0 & Int(pow) == pow, ‘Only non-negative, integer powers allowed’)
V accumulator = identity(m.len)
L(i) 0 .< pow
accumulator = matrix_mul(accumulator, m)
R accumulator
F printtable(data)
L(row) data
print(row.map(cell -> ‘#<5’.format(cell)).join(‘ ’))
V m = [[3, 2], [2, 1]]
L(i) 5
print("\n#.:".format(i))
printtable(matrixExp(m, i))
print("\n10:")
printtable(matrixExp(m, 10)) |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #EGL | EGL | program MazeGenAndSolve
// First and last columns/rows are "dead" cells. Makes generating
// a maze with border walls much easier. Therefore, a visible
// 20x20 maze has a maze size of 22.
mazeSize int = 22;
south boolean[][];
west boolean[][];
visited boolean[][];
// Solution variables
solution Dictionary;
done boolean;
startingRow, startingCol, endingRow, endingCol int;
function main()
initMaze();
generateMaze();
drawMaze(false); // Draw maze without solution
solveMaze();
drawMaze(true); // Draw maze with solution
end
private function initMaze()
visited = createBooleanArray(mazeSize, mazeSize, false);
// Initialize border cells as already visited
for(col int from 1 to mazeSize)
visited[col][1] = true;
visited[col][mazeSize] = true;
end
for(row int from 1 to mazeSize)
visited[1][row] = true;
visited[mazeSize][row] = true;
end
// Initialize all walls as present
south = createBooleanArray(mazeSize, mazeSize, true);
west = createBooleanArray(mazeSize, mazeSize, true);
end
private function createBooleanArray(col int in, row int in, initialState boolean in) returns(boolean[][])
newArray boolean[][] = new boolean[0][0];
for(i int from 1 to col)
innerArray boolean[] = new boolean[0];
for(j int from 1 to row)
innerArray.appendElement(initialState);
end
newArray.appendElement(innerArray);
end
return(newArray);
end
private function createIntegerArray(col int in, row int in, initialValue int in) returns(int[][])
newArray int[][] = new int[0][0];
for(i int from 1 to col)
innerArray int[] = new int[0];
for(j int from 1 to row)
innerArray.appendElement(initialValue);
end
newArray.appendElement(innerArray);
end
return(newArray);
end
private function generate(col int in, row int in)
// Mark cell as visited
visited[col][row] = true;
// Keep going as long as there is an unvisited neighbor
while(!visited[col][row + 1] || !visited[col + 1][row] ||
!visited[col][row - 1] || !visited[col - 1][row])
while(true)
r float = MathLib.random(); // Choose a random direction
case
when(r < 0.25 && !visited[col][row + 1]) // Go south
south[col][row] = false; // South wall down
generate(col, row + 1);
exit while;
when(r >= 0.25 && r < 0.50 && !visited[col + 1][row]) // Go east
west[col + 1][row] = false; // West wall of neighbor to the east down
generate(col + 1, row);
exit while;
when(r >= 0.5 && r < 0.75 && !visited[col][row - 1]) // Go north
south[col][row - 1] = false; // South wall of neighbor to the north down
generate(col, row - 1);
exit while;
when(r >= 0.75 && r < 1.00 && !visited[col - 1][row]) // Go west
west[col][row] = false; // West wall down
generate(col - 1, row);
exit while;
end
end
end
end
private function generateMaze()
// Pick random start position (within the visible maze space)
randomStartCol int = MathLib.floor((MathLib.random() *(mazeSize - 2)) + 2);
randomStartRow int = MathLib.floor((MathLib.random() *(mazeSize - 2)) + 2);
generate(randomStartCol, randomStartRow);
end
private function drawMaze(solve boolean in)
line string;
// Iterate over wall arrays (skipping dead border cells as required).
// Construct a row at a time and output to console.
for(row int from 1 to mazeSize - 1)
if(row > 1)
line = "";
for(col int from 2 to mazeSize)
if(west[col][row])
line ::= cellTest(col, row, solve);
else
line ::= cellTest(col, row, solve);
end
end
Syslib.writeStdout(line);
end
line = "";
for(col int from 2 to mazeSize - 1)
if(south[col][row])
line ::= "+---";
else
line ::= "+ ";
end
end
line ::= "+";
SysLib.writeStdout(line);
end
end
private function cellTest(col int in, row int in, solve boolean in) returns(string)
wall string;
// Determine cell wall structure. If in solve mode, show start, end and
// solution markers.
if(!solve)
if(west[col][row])
wall = "| ";
else
wall = " ";
end
else
if(west[col][row])
case
when(col == startingCol and row == startingRow)
wall = "| S ";
when(col == endingCol and row == endingRow)
wall = "| E ";
when(solution.containsKey("x=" + col + "y=" + row))
wall = "| * ";
otherwise
wall = "| ";
end
else
case
when(col == startingCol and row == startingRow)
wall = " S ";
when(col == endingCol and row == endingRow)
wall = " E ";
when(solution.containsKey("x=" + col + "y=" + row))
wall = " * ";
otherwise
wall = " ";
end
end
end
return(wall);
end
private function solve(col int in, row int in)
if(col == 1 || row == 1 || col == mazeSize || row == mazeSize)
return;
end
if(done || visited[col][row])
return;
end
visited[col][row] = true;
solution["x=" + col + "y=" + row] = true;
// Reached the end point
if(col == endingCol && row == endingRow)
done = true;
end
if(!south[col][row]) // Go South
solve(col, row + 1);
end
if(!west[col + 1][row]) // Go East
solve(col + 1, row);
end
if(!south[col][row - 1]) // Go North
solve(col, row - 1);
end
if(!west[col][row]) // Go West
solve(col - 1, row);
end
if(done)
return;
end
solution.removeElement("x=" + col + "y=" + row);
end
private function solveMaze()
for(col int from 1 to mazeSize)
for(row int from 1 to mazeSize)
visited[col][row] = false;
end
end
solution = new Dictionary(false, OrderingKind.byInsertion);
done = false;
// Pick random start position on first visible row
startingCol = MathLib.floor((MathLib.random() *(mazeSize - 2)) + 2);
startingRow = 2;
// Pick random end position on last visible row
endingCol = MathLib.floor((MathLib.random() *(mazeSize - 2)) + 2);
endingRow = mazeSize - 1;
solve(startingCol, startingRow);
end
end |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #Common_Lisp | Common Lisp | (defun find-max-path-sum (s)
(let ((triangle (loop for line = (read-line s NIL NIL)
while line
collect (with-input-from-string (str line)
(loop for n = (read str NIL NIL)
while n
collect n)))))
(flet ((get-max-of-pairs (xs)
(maplist (lambda (ys)
(and (cdr ys) (max (car ys) (cadr ys))))
xs)))
(car (reduce (lambda (xs ys)
(mapcar #'+ (get-max-of-pairs xs) ys))
(reverse triangle))))))
(defparameter *small-triangle*
" 55
94 48
95 30 96
77 71 26 67")
(format T "~a~%" (with-input-from-string (s *small-triangle*)
(find-max-path-sum s)))
(format T "~a~%" (with-open-file (f "triangle.txt")
(find-max-path-sum f))) |
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Kotlin | Kotlin | // version 1.0.6
import java.security.MessageDigest
class MD4() : MessageDigest("MD4"), Cloneable {
private val blockLength = 64
private var context = IntArray(4)
private var count = 0L
private var buffer = ByteArray(blockLength)
private var x = IntArray(16)
init {
engineReset()
}
private constructor(md: MD4): this() {
context = md.context.clone()
buffer = md.buffer.clone()
count = md.count
}
override fun clone(): Any = MD4(this)
override fun engineReset() {
context[0] = 0x67452301
context[1] = 0xefcdab89.toInt()
context[2] = 0x98badcfe.toInt()
context[3] = 0x10325476
count = 0L
for (i in 0 until blockLength) buffer[i] = 0
}
override fun engineUpdate(b: Byte) {
val i = (count % blockLength).toInt()
count++
buffer[i] = b
if (i == blockLength - 1) transform(buffer, 0)
}
override fun engineUpdate(input: ByteArray, offset: Int, len: Int) {
if (offset < 0 || len < 0 || offset.toLong() + len > input.size.toLong())
throw ArrayIndexOutOfBoundsException()
var bufferNdx = (count % blockLength).toInt()
count += len
val partLen = blockLength - bufferNdx
var i = 0
if (len >= partLen) {
System.arraycopy(input, offset, buffer, bufferNdx, partLen)
transform(buffer, 0)
i = partLen
while (i + blockLength - 1 < len) {
transform(input, offset + i)
i += blockLength
}
bufferNdx = 0
}
if (i < len) System.arraycopy(input, offset + i, buffer, bufferNdx, len - i)
}
override fun engineDigest(): ByteArray {
val bufferNdx = (count % blockLength).toInt()
val padLen = if (bufferNdx < 56) 56 - bufferNdx else 120 - bufferNdx
val tail = ByteArray(padLen + 8)
tail[0] = 0x80.toByte()
for (i in 0..7) tail[padLen + i] = ((count * 8) ushr (8 * i)).toByte()
engineUpdate(tail, 0, tail.size)
val result = ByteArray(16)
for (i in 0..3)
for (j in 0..3)
result[i * 4 + j] = (context[i] ushr (8 * j)).toByte()
engineReset()
return result
}
private fun transform (block: ByteArray, offset: Int) {
var offset2 = offset
for (i in 0..15)
x[i] = ((block[offset2++].toInt() and 0xff) ) or
((block[offset2++].toInt() and 0xff) shl 8 ) or
((block[offset2++].toInt() and 0xff) shl 16) or
((block[offset2++].toInt() and 0xff) shl 24)
var a = context[0]
var b = context[1]
var c = context[2]
var d = context[3]
a = ff(a, b, c, d, x[ 0], 3)
d = ff(d, a, b, c, x[ 1], 7)
c = ff(c, d, a, b, x[ 2], 11)
b = ff(b, c, d, a, x[ 3], 19)
a = ff(a, b, c, d, x[ 4], 3)
d = ff(d, a, b, c, x[ 5], 7)
c = ff(c, d, a, b, x[ 6], 11)
b = ff(b, c, d, a, x[ 7], 19)
a = ff(a, b, c, d, x[ 8], 3)
d = ff(d, a, b, c, x[ 9], 7)
c = ff(c, d, a, b, x[10], 11)
b = ff(b, c, d, a, x[11], 19)
a = ff(a, b, c, d, x[12], 3)
d = ff(d, a, b, c, x[13], 7)
c = ff(c, d, a, b, x[14], 11)
b = ff(b, c, d, a, x[15], 19)
a = gg(a, b, c, d, x[ 0], 3)
d = gg(d, a, b, c, x[ 4], 5)
c = gg(c, d, a, b, x[ 8], 9)
b = gg(b, c, d, a, x[12], 13)
a = gg(a, b, c, d, x[ 1], 3)
d = gg(d, a, b, c, x[ 5], 5)
c = gg(c, d, a, b, x[ 9], 9)
b = gg(b, c, d, a, x[13], 13)
a = gg(a, b, c, d, x[ 2], 3)
d = gg(d, a, b, c, x[ 6], 5)
c = gg(c, d, a, b, x[10], 9)
b = gg(b, c, d, a, x[14], 13)
a = gg(a, b, c, d, x[ 3], 3)
d = gg(d, a, b, c, x[ 7], 5)
c = gg(c, d, a, b, x[11], 9)
b = gg(b, c, d, a, x[15], 13)
a = hh(a, b, c, d, x[ 0], 3)
d = hh(d, a, b, c, x[ 8], 9)
c = hh(c, d, a, b, x[ 4], 11)
b = hh(b, c, d, a, x[12], 15)
a = hh(a, b, c, d, x[ 2], 3)
d = hh(d, a, b, c, x[10], 9)
c = hh(c, d, a, b, x[ 6], 11)
b = hh(b, c, d, a, x[14], 15)
a = hh(a, b, c, d, x[ 1], 3)
d = hh(d, a, b, c, x[ 9], 9)
c = hh(c, d, a, b, x[ 5], 11)
b = hh(b, c, d, a, x[13], 15)
a = hh(a, b, c, d, x[ 3], 3)
d = hh(d, a, b, c, x[11], 9)
c = hh(c, d, a, b, x[ 7], 11)
b = hh(b, c, d, a, x[15], 15)
context[0] += a
context[1] += b
context[2] += c
context[3] += d
}
private fun ff(a: Int, b: Int, c: Int, d: Int, x: Int, s: Int): Int {
val t = a + ((b and c) or (b.inv() and d)) + x
return (t shl s) or (t ushr (32 - s))
}
private fun gg(a: Int, b: Int, c: Int, d: Int, x: Int, s: Int): Int {
val t = a + ((b and (c or d)) or (c and d)) + x + 0x5a827999
return (t shl s) or (t ushr (32 - s))
}
private fun hh(a: Int, b: Int, c: Int, d: Int, x: Int, s: Int): Int {
val t = a + (b xor c xor d) + x + 0x6ed9eba1
return (t shl s) or (t ushr (32 - s))
}
}
fun main(args: Array<String>) {
val text = "Rosetta Code"
val bytes = text.toByteArray(Charsets.US_ASCII)
val md: MessageDigest = MD4()
val digest = md.digest(bytes)
for (byte in digest) print("%02x".format(byte))
println()
} |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #Rust | Rust | fn middle_three_digits(x: i32) -> Result<String, String> {
let s: String = x.abs().to_string();
let len = s.len();
if len < 3 {
Err("Too short".into())
} else if len % 2 == 0 {
Err("Even number of digits".into())
} else {
Ok(s[len/2 - 1 .. len/2 + 2].to_owned())
}
}
fn print_result(x: i32) {
print!("middle_three_digits({}) returned: ", x);
match middle_three_digits(x) {
Ok(s) => println!("Success, {}", s),
Err(s) => println!("Failure, {}", s)
}
}
fn main() {
let passing = [123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345];
let failing = [1, 2, -1, -10, 2002, -2002, 0];
for i in passing.iter() {
print_result(*i);
}
for i in failing.iter() {
print_result(*i);
}
} |
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #C.23 | C# | using System.Text;
using System.Security.Cryptography;
byte[] data = Encoding.ASCII.GetBytes("The quick brown fox jumped over the lazy dog's back");
byte[] hash = MD5.Create().ComputeHash(data);
Console.WriteLine(BitConverter.ToString(hash).Replace("-", "").ToLower()); |
http://rosettacode.org/wiki/Menu | Menu | Task
Given a prompt and a list containing a number of strings of which one is to be selected, create a function that:
prints a textual menu formatted as an index value followed by its corresponding string for each item in the list;
prompts the user to enter a number;
returns the string corresponding to the selected index number.
The function should reject input that is not an integer or is out of range by redisplaying the whole menu before asking again for a number. The function should return an empty string if called with an empty list.
For test purposes use the following four phrases in a list:
fee fie
huff and puff
mirror mirror
tick tock
Note
This task is fashioned after the action of the Bash select statement.
| #Yabasic | Yabasic | // Rosetta Code problem: https://www.rosettacode.org/wiki/Menu
// by Jjuanhdez, 06/2022
dim choose$(5)
restore menudata
for a = 0 to 5 : read choose$(a) : next a
print menu$(choose$())
end
sub menu$(m$())
clear screen
repeat
print color("green","black") at(0,0) "Menu selection"
vc = 0
b = arraysize(m$(),1)
while vc < 1 or vc > b
for i = 1 to b-1
print i, " ", choose$(i)
next i
print choose$(b)
print
input "Your choice: " c
print at(0,7) "Your choice: "
if c > 0 and c < 6 then
vc = c
print color("yellow","black") at(0,8) choose$(vc)
else
print color("red","black") at(0,8) choose$(0)
break
fi
wend
until vc = 5
end sub
label menudata
data "Ack, not good", "fee fie ", "huff and puff"
data "mirror mirror", "tick tock ", "exit " |
http://rosettacode.org/wiki/Menu | Menu | Task
Given a prompt and a list containing a number of strings of which one is to be selected, create a function that:
prints a textual menu formatted as an index value followed by its corresponding string for each item in the list;
prompts the user to enter a number;
returns the string corresponding to the selected index number.
The function should reject input that is not an integer or is out of range by redisplaying the whole menu before asking again for a number. The function should return an empty string if called with an empty list.
For test purposes use the following four phrases in a list:
fee fie
huff and puff
mirror mirror
tick tock
Note
This task is fashioned after the action of the Bash select statement.
| #zkl | zkl | fcn teleprompter(options){
os:=T("exit").extend(vm.arglist); N:=os.len();
if(N==1) return("");
while(1){
Utils.zipWith(fcn(n,o){"%d) %s".fmt(n,o).println()},[0..],os);
a:=ask("Your choice: ");
try{ n:=a.toInt(); if(0<=n<N) return(os[n]); } catch{}
println("Ack, not good");
}
}
teleprompter("fee fie" , "huff and puff" , "mirror mirror" , "tick tock")
.println(); |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Locomotive_Basic | Locomotive Basic | 100 CLEAR
110 DIM a(100)
120 FOR a=0 TO 100/6
130 FOR b=0 TO 100/9
140 FOR c=0 TO 100/20
150 n=a*6+b*9+c*20
160 IF n<=100 THEN a(n)=1
170 NEXT c
180 NEXT b
190 NEXT a
200 FOR n=0 TO 100
210 IF a(n)=0 THEN l=n
220 NEXT n
230 PRINT"The Largest non McNugget number is:";l
240 END |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #Factor | Factor | USING: arrays formatting io kernel make math math.extras
sequences ;
IN: rosetta-code.mayan-numerals
: mayan-digit ( n -- m pair ) 20 /mod 5 /mod swap 2array ;
: integer>mayan ( n -- seq )
[ [ mayan-digit , ] until-zero ] { } make reverse ;
: ones ( n -- str ) [ CHAR: ● ] "" replicate-as ;
: fives ( n -- str ) [ "——" ] replicate "<br>" join ;
: numeral ( pair -- str )
dup sum zero? [ drop "Θ" ]
[ first2 [ ones ] [ fives ] bi* 2array harvest "<br>" join ]
if ;
: .table-row ( pair -- )
"|style=\"width:3em;vertical-align:bottom;\"|" write numeral
print ;
: .table-head ( -- )
"class=\"wikitable\" style=\"text-align:center;\"" print ;
: .table-body ( n -- ) integer>mayan [ .table-row ] each ;
: .mayan ( n -- )
[ "Mayan %d:\n" printf ]
[ "{|" write .table-head .table-body "|}" print ] bi ;
: mayan-numerals ( -- )
{ 4005 8017 326205 886205 } [ .mayan ] each ;
MAIN: mayan-numerals |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #F.C5.8Drmul.C3.A6 | Fōrmulæ | package main
import (
"fmt"
"strconv"
)
const (
ul = "╔"
uc = "╦"
ur = "╗"
ll = "╚"
lc = "╩"
lr = "╝"
hb = "═"
vb = "║"
)
var mayan = [5]string{
" ",
" ∙ ",
" ∙∙ ",
"∙∙∙ ",
"∙∙∙∙",
}
const (
m0 = " Θ "
m5 = "────"
)
func dec2vig(n uint64) []uint64 {
vig := strconv.FormatUint(n, 20)
res := make([]uint64, len(vig))
for i, d := range vig {
res[i], _ = strconv.ParseUint(string(d), 20, 64)
}
return res
}
func vig2quin(n uint64) [4]string {
if n >= 20 {
panic("Cant't convert a number >= 20")
}
res := [4]string{mayan[0], mayan[0], mayan[0], mayan[0]}
if n == 0 {
res[3] = m0
return res
}
fives := n / 5
rem := n % 5
res[3-fives] = mayan[rem]
for i := 3; i > 3-int(fives); i-- {
res[i] = m5
}
return res
}
func draw(mayans [][4]string) {
lm := len(mayans)
fmt.Print(ul)
for i := 0; i < lm; i++ {
for j := 0; j < 4; j++ {
fmt.Print(hb)
}
if i < lm-1 {
fmt.Print(uc)
} else {
fmt.Println(ur)
}
}
for i := 1; i < 5; i++ {
fmt.Print(vb)
for j := 0; j < lm; j++ {
fmt.Print(mayans[j][i-1])
fmt.Print(vb)
}
fmt.Println()
}
fmt.Print(ll)
for i := 0; i < lm; i++ {
for j := 0; j < 4; j++ {
fmt.Print(hb)
}
if i < lm-1 {
fmt.Print(lc)
} else {
fmt.Println(lr)
}
}
}
func main() {
numbers := []uint64{4005, 8017, 326205, 886205, 1081439556}
for _, n := range numbers {
fmt.Printf("Converting %d to Mayan:\n", n)
vigs := dec2vig(n)
lv := len(vigs)
mayans := make([][4]string, lv)
for i, vig := range vigs {
mayans[i] = vig2quin(vig)
}
draw(mayans)
fmt.Println()
}
} |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Matrix is
generic
type Element is private;
Zero : Element;
One : Element;
with function "+" (A, B : Element) return Element is <>;
with function "*" (A, B : Element) return Element is <>;
with function Image (X : Element) return String is <>;
package Matrices is
type Matrix is array (Integer range <>, Integer range <>) of Element;
function "*" (A, B : Matrix) return Matrix;
function "**" (A : Matrix; Power : Natural) return Matrix;
procedure Put (A : Matrix);
end Matrices;
package body Matrices is
function "*" (A, B : Matrix) return Matrix is
R : Matrix (A'Range (1), B'Range (2));
Sum : Element := Zero;
begin
for I in R'Range (1) loop
for J in R'Range (2) loop
Sum := Zero;
for K in A'Range (2) loop
Sum := Sum + A (I, K) * B (K, J);
end loop;
R (I, J) := Sum;
end loop;
end loop;
return R;
end "*";
function "**" (A : Matrix; Power : Natural) return Matrix is
begin
if Power = 1 then
return A;
end if;
declare
R : Matrix (A'Range (1), A'Range (2)) := (others => (others => Zero));
P : Matrix := A;
E : Natural := Power;
begin
for I in P'Range (1) loop -- R is identity matrix
R (I, I) := One;
end loop;
if E = 0 then
return R;
end if;
loop
if E mod 2 /= 0 then
R := R * P;
end if;
E := E / 2;
exit when E = 0;
P := P * P;
end loop;
return R;
end;
end "**";
procedure Put (A : Matrix) is
begin
for I in A'Range (1) loop
for J in A'Range (1) loop
Put (Image (A (I, J)));
end loop;
New_Line;
end loop;
end Put;
end Matrices;
package Integer_Matrices is new Matrices (Integer, 0, 1, Image => Integer'Image);
use Integer_Matrices;
M : Matrix (1..2, 1..2) := ((3,2),(2,1));
begin
Put_Line ("M ="); Put (M);
Put_Line ("M**0 ="); Put (M**0);
Put_Line ("M**1 ="); Put (M**1);
Put_Line ("M**2 ="); Put (M**2);
Put_Line ("M*M ="); Put (M*M);
Put_Line ("M**3 ="); Put (M**3);
Put_Line ("M*M*M ="); Put (M*M*M);
Put_Line ("M**4 ="); Put (M**4);
Put_Line ("M*M*M*M ="); Put (M*M*M*M);
Put_Line ("M**10 ="); Put (M**10);
Put_Line ("M*M*M*M*M*M*M*M*M*M ="); Put (M*M*M*M*M*M*M*M*M*M);
end Test_Matrix; |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #Emacs_Lisp | Emacs Lisp | (require 'cl-lib)
(cl-defstruct maze rows cols data)
(defmacro maze-pt (w r c)
`(+ (* (mod ,r (maze-rows ,w)) (maze-cols ,w))
(mod ,c (maze-cols ,w))))
(defmacro maze-ref (w r c)
`(aref (maze-data ,w) (maze-pt ,w ,r ,c)))
(defun new-maze (rows cols)
(setq rows (1+ rows)
cols (1+ cols))
(let ((m (make-maze :rows rows :cols cols :data (make-vector (* rows cols) nil))))
(dotimes (r rows)
(dotimes (c cols)
(setf (maze-ref m r c) (copy-sequence '(wall ceiling)))))
(dotimes (r rows)
(maze-set m r (1- cols) 'visited))
(dotimes (c cols)
(maze-set m (1- rows) c 'visited))
(maze-unset m 0 0 'ceiling) ;; Maze Entrance
(maze-unset m (1- rows) (- cols 2) 'ceiling) ;; Maze Exit
m))
(defun maze-is-set (maze r c v)
(member v (maze-ref maze r c)))
(defun maze-set (maze r c v)
(let ((cell (maze-ref maze r c)))
(when (not (member v cell))
(setf (maze-ref maze r c) (cons v cell)))))
(defun maze-unset (maze r c v)
(setf (maze-ref maze r c) (delete v (maze-ref maze r c))))
(defun print-maze (maze &optional marks)
(dotimes (r (1- (maze-rows maze)))
(dotimes (c (1- (maze-cols maze)))
(princ (if (maze-is-set maze r c 'ceiling) "+---" "+ ")))
(princ "+")
(terpri)
(dotimes (c (1- (maze-cols maze)))
(princ (if (maze-is-set maze r c 'wall) "|" " "))
(princ (if (member (cons r c) marks) " * " " ")))
(princ "|")
(terpri))
(dotimes (c (1- (maze-cols maze)))
(princ (if (maze-is-set maze (1- (maze-rows maze)) c 'ceiling) "+---" "+ ")))
(princ "+")
(terpri))
(defun shuffle (lst)
(sort lst (lambda (a b) (= 1 (random 2)))))
(defun to-visit (maze row col)
(let (unvisited)
(dolist (p '((0 . +1) (0 . -1) (+1 . 0) (-1 . 0)))
(let ((r (+ row (car p)))
(c (+ col (cdr p))))
(unless (maze-is-set maze r c 'visited)
(push (cons r c) unvisited))))
unvisited))
(defun make-passage (maze r1 c1 r2 c2)
(if (= r1 r2)
(if (< c1 c2)
(maze-unset maze r2 c2 'wall) ; right
(maze-unset maze r1 c1 'wall)) ; left
(if (< r1 r2)
(maze-unset maze r2 c2 'ceiling) ; up
(maze-unset maze r1 c1 'ceiling)))) ; down
(defun dig-maze (maze row col)
(let (backup
(run 0))
(maze-set maze row col 'visited)
(push (cons row col) backup)
(while backup
(setq run (1+ run))
(when (> run (/ (+ row col) 3))
(setq run 0)
(setq backup (shuffle backup)))
(setq row (caar backup)
col (cdar backup))
(let ((p (shuffle (to-visit maze row col))))
(if p
(let ((r (caar p))
(c (cdar p)))
(make-passage maze row col r c)
(maze-set maze r c 'visited)
(push (cons r c) backup))
(pop backup)
(setq backup (shuffle backup))
(setq run 0))))))
(defun generate (rows cols)
(let* ((m (new-maze rows cols)))
(dig-maze m (random rows) (random cols))
(print-maze m)))
(defun parse-ceilings (line)
(let (rtn
(i 1))
(while (< i (length line))
(push (eq ?- (elt line i)) rtn)
(setq i (+ i 4)))
(nreverse rtn)))
(defun parse-walls (line)
(let (rtn
(i 0))
(while (< i (length line))
(push (eq ?| (elt line i)) rtn)
(setq i (+ i 4)))
(nreverse rtn)))
(defun parse-maze (file-name)
(let ((rtn)
(lines (with-temp-buffer
(insert-file-contents-literally file-name)
(split-string (buffer-string) "\n" t))))
(while lines
(push (parse-ceilings (pop lines)) rtn)
(push (parse-walls (pop lines)) rtn))
(nreverse rtn)))
(defun read-maze (file-name)
(let* ((raw (parse-maze file-name))
(rows (1- (/ (length raw) 2)))
(cols (length (car raw)))
(maze (new-maze rows cols)))
(dotimes (r rows)
(let ((ceilings (pop raw)))
(dotimes (c cols)
(unless (pop ceilings)
(maze-unset maze r c 'ceiling))))
(let ((walls (pop raw)))
(dotimes (c cols)
(unless (pop walls)
(maze-unset maze r c 'wall)))))
maze))
(defun find-exits (maze row col)
(let (exits)
(dolist (p '((0 . +1) (0 . -1) (-1 . 0) (+1 . 0)))
(let ((r (+ row (car p)))
(c (+ col (cdr p))))
(unless
(cond
((equal p '(0 . +1)) (maze-is-set maze r c 'wall))
((equal p '(0 . -1)) (maze-is-set maze row col 'wall))
((equal p '(+1 . 0)) (maze-is-set maze r c 'ceiling))
((equal p '(-1 . 0)) (maze-is-set maze row col 'ceiling)))
(push (cons r c) exits))))
exits))
(defun drop-visited (maze points)
(let (not-visited)
(while points
(unless (maze-is-set maze (caar points) (cdar points) 'visited)
(push (car points) not-visited))
(pop points))
not-visited))
(defun solve-maze (maze)
(let (solution
(exit (cons (- (maze-rows maze) 2) (- (maze-cols maze) 2)))
(pt (cons 0 0)))
(while (not (equal pt exit))
(maze-set maze (car pt) (cdr pt) 'visited)
(let ((exits (drop-visited maze (find-exits maze (car pt) (cdr pt)))))
(if (null exits)
(setq pt (pop solution))
(push pt solution)
(setq pt (pop exits)))))
(push pt solution)))
(defun solve (file-name)
(let* ((maze (read-maze file-name))
(solution (solve-maze maze)))
(print-maze maze solution)))
(solve "maze.txt") |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #D | D | void main() {
import std.stdio, std.algorithm, std.range, std.file, std.conv;
"triangle.txt".File.byLine.map!split.map!(to!(int[])).array.retro
.reduce!((x, y) => zip(y, x, x.dropOne)
.map!(t => t[0] + t[1 .. $].max)
.array)[0]
.writeln;
} |
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Lasso | Lasso | cipher_digest('Rosetta Code', -digest='MD4')->encodeHex->asString |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #S-lang | S-lang | Educational Notes:
1) Array-style string manipulations (e.g. as = as[[1:]] ) always use bytes,
not UTF-8 characters. As digits and '-' are all ASCII, all is copacetic.
2) The task doesn't require 64-bit integer support, but I added it, dependant
on whether the S-Lang library compile supports them, which it generally
does. [Main exception would be the current downloadable/installable MS-Win
version of the Jed editor.]
3) S-Lang functions peel arguments off right-to-left, opposite of many
languages. To align the order with other solutions, I enclosed the
parameters in a list. One alternative would have been for m3() to call
_stk_reverse(_NARGS), then pop each arg directly off the stack. Or of
course just list the numbers backwards.
|
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #C.2B.2B | C++ | #include <string>
#include <iostream>
#include "Poco/MD5Engine.h"
#include "Poco/DigestStream.h"
using Poco::DigestEngine ;
using Poco::MD5Engine ;
using Poco::DigestOutputStream ;
int main( ) {
std::string myphrase ( "The quick brown fox jumped over the lazy dog's back" ) ;
MD5Engine md5 ;
DigestOutputStream outstr( md5 ) ;
outstr << myphrase ;
outstr.flush( ) ; //to pass everything to the digest engine
const DigestEngine::Digest& digest = md5.digest( ) ;
std::cout << myphrase << " as a MD5 digest :\n" << DigestEngine::digestToHex( digest )
<< " !" << std::endl ;
return 0 ;
} |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Lua | Lua |
function range(A,B)
return function()
return coroutine.wrap(function()
for i = A, B do coroutine.yield(i) end
end)
end
end
function filter(stream, f)
return function()
return coroutine.wrap(function()
for i in stream() do
if f(i) then coroutine.yield(i) end
end
end)
end
end
function triple(s1, s2, s3)
return function()
return coroutine.wrap(function()
for x in s1() do
for y in s2() do
for z in s3() do
coroutine.yield{x,y,z}
end
end
end
end)
end
end
function apply(f, stream)
return function()
return coroutine.wrap(function()
for T in stream() do
coroutine.yield(f(table.unpack(T)))
end
end)
end
end
function exclude(s1, s2)
local exlusions = {} for x in s1() do exlusions[x] = true end
return function()
return coroutine.wrap(function()
for x in s2() do
if not exlusions[x] then
coroutine.yield(x)
end
end
end)
end
end
function maximum(stream)
local M = math.mininteger
for x in stream() do
M = math.max(M, x)
end
return M
end
local N = 100
local A, B, C = 6, 9, 20
local Xs = filter(range(0, N), function(x) return x % A == 0 end)
local Ys = filter(range(0, N), function(x) return x % B == 0 end)
local Zs = filter(range(0, N), function(x) return x % C == 0 end)
local sum = filter(apply(function(x, y, z) return x + y + z end, triple(Xs, Ys, Zs)), function(x) return x <= N end)
print(maximum(exclude(sum, range(1, N))))
|
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #Go | Go | package main
import (
"fmt"
"strconv"
)
const (
ul = "╔"
uc = "╦"
ur = "╗"
ll = "╚"
lc = "╩"
lr = "╝"
hb = "═"
vb = "║"
)
var mayan = [5]string{
" ",
" ∙ ",
" ∙∙ ",
"∙∙∙ ",
"∙∙∙∙",
}
const (
m0 = " Θ "
m5 = "────"
)
func dec2vig(n uint64) []uint64 {
vig := strconv.FormatUint(n, 20)
res := make([]uint64, len(vig))
for i, d := range vig {
res[i], _ = strconv.ParseUint(string(d), 20, 64)
}
return res
}
func vig2quin(n uint64) [4]string {
if n >= 20 {
panic("Cant't convert a number >= 20")
}
res := [4]string{mayan[0], mayan[0], mayan[0], mayan[0]}
if n == 0 {
res[3] = m0
return res
}
fives := n / 5
rem := n % 5
res[3-fives] = mayan[rem]
for i := 3; i > 3-int(fives); i-- {
res[i] = m5
}
return res
}
func draw(mayans [][4]string) {
lm := len(mayans)
fmt.Print(ul)
for i := 0; i < lm; i++ {
for j := 0; j < 4; j++ {
fmt.Print(hb)
}
if i < lm-1 {
fmt.Print(uc)
} else {
fmt.Println(ur)
}
}
for i := 1; i < 5; i++ {
fmt.Print(vb)
for j := 0; j < lm; j++ {
fmt.Print(mayans[j][i-1])
fmt.Print(vb)
}
fmt.Println()
}
fmt.Print(ll)
for i := 0; i < lm; i++ {
for j := 0; j < 4; j++ {
fmt.Print(hb)
}
if i < lm-1 {
fmt.Print(lc)
} else {
fmt.Println(lr)
}
}
}
func main() {
numbers := []uint64{4005, 8017, 326205, 886205, 1081439556}
for _, n := range numbers {
fmt.Printf("Converting %d to Mayan:\n", n)
vigs := dec2vig(n)
lv := len(vigs)
mayans := make([][4]string, lv)
for i, vig := range vigs {
mayans[i] = vig2quin(vig)
}
draw(mayans)
fmt.Println()
}
} |
http://rosettacode.org/wiki/Maze_generation | Maze generation |
This page uses content from Wikipedia. The original article was at Maze generation algorithm. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Generate and show a maze, using the simple Depth-first search algorithm.
Start at a random cell.
Mark the current cell as visited, and get a list of its neighbors. For each neighbor, starting with a randomly selected neighbor:
If that neighbor hasn't been visited, remove the wall between this cell and that neighbor, and then recurse with that neighbor as the current cell.
Related tasks
Maze solving.
| #11l | 11l | F make_maze(w = 16, h = 8)
V vis = [[0] * w [+] [1]] * h [+] [[1] * (w + 1)]
V ver = [[‘| ’] * w [+] [String(‘|’)]] * h [+] [[String]()]
V hor = [[‘+--’] * w [+] [String(‘+’)]] * (h + 1)
F walk(Int x, Int y) -> N
@vis[y][x] = 1
V d = [(x - 1, y), (x, y + 1), (x + 1, y), (x, y - 1)]
random:shuffle(&d)
L(=xx, =yy) d
I yy == -1
yy = @vis.len - 1
I xx == -1
xx = @vis[0].len - 1
I @vis[yy][xx]
L.continue
I xx == x
@hor[max(y, yy)][x] = ‘+ ’
I yy == y
@ver[y][max(x, xx)] = ‘ ’
@walk(xx, yy)
walk(random:(w), random:(h))
V s = ‘’
L(a, b) zip(hor, ver)
s ‘’= (a [+] [String("\n")] + b [+] [String("\n")]).join(‘’)
R s
print(make_maze()) |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #ALGOL_68 | ALGOL 68 | INT default upb=3;
MODE VEC = [default upb]COSCAL;
MODE MAT = [default upb,default upb]COSCAL;
OP * = (VEC a,b)COSCAL: (
COSCAL result:=0;
FOR i FROM LWB a TO UPB a DO result+:= a[i]*b[i] OD;
result
);
OP * = (VEC a, MAT b)VEC: ( # overload vec times matrix #
[2 LWB b:2 UPB b]COSCAL result;
FOR j FROM 2 LWB b TO 2 UPB b DO result[j]:=a*b[,j] OD;
result
);
OP * = (MAT a, b)MAT: ( # overload matrix times matrix #
[LWB a:UPB a, 2 LWB b:2 UPB b]COSCAL result;
FOR k FROM LWB result TO UPB result DO result[k,]:=a[k,]*b OD;
result
);
OP IDENTITY = (INT upb)MAT:(
[upb,upb] COSCAL out;
FOR i TO upb DO
FOR j TO upb DO
out[i,j]:= ( i=j |1|0)
OD
OD;
out
); |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #Erlang | Erlang |
-module( maze_solving ).
-export( [task/0] ).
cells( {Start_x, Start_y}, {Stop_x, Stop_y}, Maze ) ->
Start_pid = maze:cell_pid( Start_x, Start_y, Maze ),
Stop_pid = maze:cell_pid( Stop_x, Stop_y, Maze ),
{ok, Cells} = loop( Start_pid, Stop_pid, maze:cell_accessible_neighbours(Start_pid), [Start_pid] ),
Cells.
task() ->
Max_x = 16,
Max_y = 8,
Maze = maze:generation( Max_x, Max_y ),
Start_x = random:uniform( Max_x ),
Start_y = random:uniform( Max_y ),
Stop_x = random:uniform( Max_x ),
Stop_y = random:uniform( Max_y ),
Cells = cells( {Start_x, Start_y}, {Stop_x, Stop_y}, Maze ),
[maze:cell_content_set(X, ".") || X <- Cells],
maze:cell_content_set( maze:cell_pid(Start_x, Start_y, Maze), "S" ),
maze:cell_content_set( maze:cell_pid(Stop_x, Stop_y, Maze), "G" ),
maze:display( Maze ),
maze:stop( Maze ).
loop( _Start, _Stop, [], _Acc) -> {error, dead_end};
loop( _Start, Stop, [Stop], Acc ) -> {ok, lists:reverse( [Stop | Acc] )};
loop( Start, Stop, [Next], [Previous | _T]=Acc ) -> loop( Start, Stop, lists:delete(Previous, maze:cell_accessible_neighbours(Next)), [Next | Acc] );
loop( Start, Stop, Nexts, Acc ) -> loop_stop( lists:member(Stop, Nexts), Start, Stop, Nexts, Acc ).
loop_stop( true, _Start, Stop, _Nexts, Acc ) -> {ok, lists:reverse( [Stop | Acc] )};
loop_stop( false, Start, Stop, Nexts, Acc ) ->
My_pid = erlang:self(),
[erlang:spawn( fun() -> My_pid ! loop( Start, Stop, [X], Acc ) end ) || X <- Nexts],
receive
{ok, Cells} -> {ok, Cells}
end.
|
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #Elena | Elena | import system'routines;
import extensions;
import extensions'math;
string input = "55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93";
public program()
{
var list := IntMatrix.allocate(18,19);
int i := 0;
int j := 0;
input.splitBy(forward newLine).forEach:(string line)
{
j := 0;
line.trim().splitBy(" ").forEach:(string num)
{
list[i][j] := num.toInt();
j += 1
};
i += 1
};
for(int i := 16, i >= 0, i-=1)
{
for(int j := 0, j < 18, j += 1)
{
list[i][j] := max(list[i][j] + list[i+1][j], list[i][j] + list[i+1][j+1])
}
};
console.printLine("Maximum total: ", list[0][0])
} |
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Lua | Lua | #!/usr/bin/lua
require "crypto"
print(crypto.digest("MD4", "Rosetta Code")) |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #Scala | Scala | /**
* Optionally return the middle three digits of an integer.
*
* @example List(123,12345,-789,1234,12) flatMap (middleThree(_)), returns: List(123, 234, 789)
*/
def middleThree( s:Int ) : Option[Int] = s.abs.toString match {
case v if v.length % 2 == 0 => None // Middle three is undefined for even lengths
case v if v.length < 3 => None
case v =>
val i = (v.length / 2) - 1
Some( v.substring(i,i+3).toInt )
}
// A little test...
val intVals = List(123,12345,1234567,987654321,10001,-10001,-123,-100,100,-12345,1,2,-1,-10,2002,-2002,0)
intVals map (middleThree(_)) map {
case None => "No middle three"
case Some(v) => "%03d".format(v) // Format the value, force leading zeroes
} mkString("\n")
|
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #Cach.C3.A9_ObjectScript | Caché ObjectScript | USER>set hash=$System.Encryption.MD5Hash("The quick brown fox jumped over the lazy dog's back")
USER>zzdump hash
0000: E3 8C A1 D9 20 C4 B8 B8 D3 94 6B 2C 72 F0 16 80 |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #MAD | MAD | NORMAL MODE IS INTEGER
BOOLEAN NUGGET
DIMENSION NUGGET(101)
THROUGH CLEAR, FOR I=0, 1, I.G.100
CLEAR NUGGET(I) = 0B
THROUGH MARK, FOR A=0, 6, A.G.100
THROUGH MARK, FOR B=A, 9, B.G.100
THROUGH MARK, FOR C=B, 20, C.G.100
MARK NUGGET(C) = 1B
SEARCH THROUGH SEARCH, FOR I=100, -1, .NOT.NUGGET(I)
PRINT FORMAT F, I
VECTOR VALUES F = $29HMAXIMUM NON-MCNUGGET NUMBER: ,I2*$
END OF PROGRAM |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | FrobeniusNumber[{6, 9, 20}] |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #Haskell | Haskell | import Data.Bool (bool)
import Data.List (intercalate, transpose)
import qualified Data.Map.Strict as M
import Data.Maybe (maybe)
--------------------------- MAIN -------------------------
main :: IO ()
main =
(putStrLn . unlines) $
mayanFramed
<$> [ 4005,
8017,
326205,
886205,
1081439556,
1000000,
1000000000
]
---------------------- MAYAN NUMBERS ---------------------
mayanGlyph :: Int -> [[String]]
mayanGlyph =
filter (any (not . null))
. transpose
. leftPadded
. flip (showIntAtBaseS 20 mayanDigit) []
mayanDigit :: Int -> [String]
mayanDigit n
| 0 /= n =
replicate (rem n 5) mayaOne :
concat
( replicate (quot n 5) [mayaFive]
)
| otherwise = [[mayaZero]]
mayanFramed :: Int -> String
mayanFramed =
("Mayan " <>)
. ( (<>) <$> show
<*> ( (":\n\n" <>)
. wikiTable
( M.fromList
[ ( "style",
concat
[ "text-align:center;",
"background-color:#F0EDDE;",
"color:#605B4B;",
"border:2px solid silver;"
]
),
("colwidth", "3em;")
]
)
. mayanGlyph
)
)
mayaZero, mayaOne :: Char
mayaZero = '\920'
mayaOne = '\9679'
mayaFive :: String
mayaFive = "\9473\9473"
---------------------- NUMERIC BASES ---------------------
-- Based on the Prelude showIntAtBase but uses an
-- (Int -> [String]) (rather than Int -> Char) function
-- as its second argument.
--
-- Shows a /non-negative/ 'Integral' number using the base
-- specified by the first argument, and the **String**
-- representation specified by the second.
showIntAtBaseS ::
Integral a =>
a ->
(Int -> [String]) ->
a ->
[[String]] ->
[[String]]
showIntAtBaseS base toStr n0 r0 =
let go (n, d) r =
seq s $
case n of
0 -> r_
_ -> go (quotRem n base) r_
where
s = toStr (fromIntegral d)
r_ = s : r
in go (quotRem n0 base) r0
------------------------- DISPLAY ------------------------
wikiTable :: M.Map String String -> [[String]] -> String
wikiTable opts rows
| null rows = []
| otherwise =
"{| "
<> foldr
( \k a ->
maybe
a
( ((a <> k <> "=\"") <>)
. ( <> "\" "
)
)
(M.lookup k opts)
)
[]
["class", "style"]
<> ( '\n' :
intercalate
"|-\n"
( zipWith renderedRow rows [0 ..]
)
)
<> "|}\n\n"
where
renderedRow row i =
unlines
( fmap
( ( bool
[]
( maybe
"|"
(("|style=\"width:" <>) . (<> "\""))
(M.lookup "colwidth" opts)
)
(0 == i)
<>
)
. ('|' :)
)
row
)
leftPadded :: [[String]] -> [[String]]
leftPadded xs =
let w = maximum (length <$> xs)
in ((<>) =<< flip replicate [] . (-) w . length) <$> xs |
http://rosettacode.org/wiki/Maze_generation | Maze generation |
This page uses content from Wikipedia. The original article was at Maze generation algorithm. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Generate and show a maze, using the simple Depth-first search algorithm.
Start at a random cell.
Mark the current cell as visited, and get a list of its neighbors. For each neighbor, starting with a randomly selected neighbor:
If that neighbor hasn't been visited, remove the wall between this cell and that neighbor, and then recurse with that neighbor as the current cell.
Related tasks
Maze solving.
| #Action.21 | Action! | DEFINE TOP="0"
DEFINE RIGHT="1"
DEFINE BOTTOM="2"
DEFINE LEFT="3"
DEFINE WIDTH="160"
DEFINE HEIGHT="96"
DEFINE STACK_SIZE="5000"
BYTE ARRAY stack(STACK_SIZE)
INT stackSize
PROC InitStack()
stackSize=0
RETURN
BYTE FUNC IsEmpty()
IF stackSize=0 THEN
RETURN (1)
FI
RETURN (0)
BYTE FUNC IsFull()
IF stackSize>=STACK_SIZE THEN
RETURN (1)
FI
RETURN (0)
PROC Push(BYTE x,y)
IF IsFull() THEN Break() RETURN FI
stack(stackSize)=x stackSize==+1
stack(stackSize)=y stackSize==+1
RETURN
PROC Pop(BYTE POINTER x,y)
IF IsEmpty() THEN Break() RETURN FI
stackSize==-1 y^=stack(stackSize)
stackSize==-1 x^=stack(stackSize)
RETURN
PROC FillScreen()
BYTE POINTER ptr ;pointer to the screen memory
INT screenSize=[3840]
ptr=PeekC(88)
SetBlock(ptr,screenSize,$55)
Color=0
Plot(0,HEIGHT-1) DrawTo(WIDTH-1,HEIGHT-1) DrawTo(WIDTH-1,0)
RETURN
PROC GetNeighbors(BYTE x,y BYTE ARRAY n BYTE POINTER count)
DEFINE WALL="1"
count^=0
IF y>2 AND Locate(x,y-2)=WALL THEN
n(count^)=TOP count^==+1
FI
IF x<WIDTH-3 AND Locate(x+2,y)=WALL THEN
n(count^)=RIGHT count^==+1
FI
IF y<HEIGHT-3 AND Locate(x,y+2)=WALL THEN
n(count^)=BOTTOM count^==+1
FI
IF x>2 AND Locate(x-2,y)=WALL THEN
n(count^)=LEFT count^==+1
FI
RETURN
PROC Maze(BYTE x,y)
BYTE ARRAY stack,neighbors
BYTE dir,nCount
FillScreen()
Color=2
InitStack()
Push(x,y)
WHILE IsEmpty()=0
DO
Pop(@x,@y)
GetNeighbors(x,y,neighbors,@nCount)
IF nCount>0 THEN
Push(x,y)
Plot(x,y)
dir=neighbors(Rand(nCount))
IF dir=TOP THEN
y==-2
ELSEIF dir=RIGHT THEN
x==+2
ELSEIF dir=BOTTOM THEN
y==+2
ELSE
x==-2
FI
DrawTo(x,y)
Push(x,y)
FI
OD
RETURN
PROC Main()
BYTE CH=$02FC,COLOR0=$02C4,COLOR1=$02C5
BYTE x,y
Graphics(7+16)
COLOR0=$0A
COLOR1=$04
x=Rand((WIDTH RSH 1)-1) LSH 1+1
y=Rand((HEIGHT RSH 1)-1) LSH 1+1
Maze(x,y)
DO UNTIL CH#$FF OD
CH=$FF
RETURN |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #BBC_BASIC | BBC BASIC | DIM matrix(1,1), output(1,1)
matrix() = 3, 2, 2, 1
FOR power% = 0 TO 9
PROCmatrixpower(matrix(), output(), power%)
PRINT "matrix()^" ; power% " = "
FOR row% = 0 TO DIM(output(), 1)
FOR col% = 0 TO DIM(output(), 2)
PRINT output(row%,col%);
NEXT
PRINT
NEXT row%
NEXT power%
END
DEF PROCmatrixpower(src(), dst(), pow%)
LOCAL i%
dst() = 0
FOR i% = 0 TO DIM(dst(), 1) : dst(i%,i%) = 1 : NEXT
IF pow% THEN
FOR i% = 1 TO pow%
dst() = dst() . src()
NEXT
ENDIF
ENDPROC |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #BQN | BQN | MatMul ← +˝∘×⎉1‿∞
MatEx ← {𝕨 MatMul⍟(𝕩-1) 𝕨}
(>⟨3‿2
2‿1⟩) MatEx 1‿2‿3‿4‿10 |
http://rosettacode.org/wiki/Matrix_chain_multiplication | Matrix chain multiplication | Problem
Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. The number of operations required to compute the product of matrices A1, A2... An depends on the order of matrix multiplications, hence on where parens are put. Remember that the matrix product is associative, but not commutative, hence only the parens can be moved.
For instance, with four matrices, one can compute A(B(CD)), A((BC)D), (AB)(CD), (A(BC))D, (AB)C)D. The number of different ways to put the parens is a Catalan number, and grows exponentially with the number of factors.
Here is an example of computation of the total cost, for matrices A(5,6), B(6,3), C(3,1):
AB costs 5*6*3=90 and produces a matrix of dimensions (5,3), then (AB)C costs 5*3*1=15. The total cost is 105.
BC costs 6*3*1=18 and produces a matrix of dimensions (6,1), then A(BC) costs 5*6*1=30. The total cost is 48.
In this case, computing (AB)C requires more than twice as many operations as A(BC). The difference can be much more dramatic in real cases.
Task
Write a function which, given a list of the successive dimensions of matrices A1, A2... An, of arbitrary length, returns the optimal way to compute the matrix product, and the total cost. Any sensible way to describe the optimal solution is accepted. The input list does not duplicate shared dimensions: for the previous example of matrices A,B,C, one will only pass the list [5,6,3,1] (and not [5,6,6,3,3,1]) to mean the matrix dimensions are respectively (5,6), (6,3) and (3,1). Hence, a product of n matrices is represented by a list of n+1 dimensions.
Try this function on the following two lists:
[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
To solve the task, it's possible, but not required, to write a function that enumerates all possible ways to parenthesize the product. This is not optimal because of the many duplicated computations, and this task is a classic application of dynamic programming.
See also Matrix chain multiplication on Wikipedia.
| #11l | 11l | T Optimizer
[Int] dims
[[Int]] m, s
F (dims)
.dims = dims
F findMatrixChainOrder()
V n = .dims.len - 1
.m = [[0] * n] * n
.s = [[0] * n] * n
L(lg) 1 .< n
L(i) 0 .< n - lg
V j = i + lg
.m[i][j] = 7FFF'FFFF
L(k) i .< j
V cost = .m[i][k] + .m[k + 1][j] + .dims[i] * .dims[k + 1] * .dims[j + 1]
I cost < .m[i][j]
.m[i][j] = cost
.s[i][j] = k
F optimalChainOrder(i, j)
I i == j
R String(Char(code' i + ‘A’.code))
E
R ‘(’(.optimalChainOrder(i, .s[i][j]))‘’
‘’(.optimalChainOrder(.s[i][j] + 1, j))‘)’
V Dims1 = [5, 6, 3, 1]
V Dims2 = [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
V Dims3 = [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
L(dims) [Dims1, Dims2, Dims3]
V opt = Optimizer(dims)
opt.findMatrixChainOrder()
print(‘Dims: ’dims)
print(‘Order: ’opt.optimalChainOrder(0, dims.len - 2))
print(‘Cost: ’opt.m[0][dims.len - 2])
print(‘’) |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #Frege | Frege | module MazeSolver where
import frege.IO
import Data.Maybe
-- given two points, returns the average of them
average :: (Int, Int) -> (Int, Int) -> (Int, Int)
average (x, y) (x', y') = ((x + x') `div` 2, (y + y') `div` 2)
-- given a maze and a tuple of position and wall position, returns
-- true if the wall position is not blocked (first position is unused)
notBlocked :: [String] -> ((Int, Int), (Int, Int)) -> Bool
notBlocked maze (_, (x, y)) = (' ' == String.charAt (maze !! y) x)
-- given a list, a position, and an element, returns a new list
-- with the new element substituted at the position
substitute :: [a] -> Int -> a -> [a]
substitute orig pos el =
let (before, after) = splitAt pos orig
in before ++ [el] ++ tail after
-- like above, but for strings, since Frege strings are not
-- lists of characters
substituteString :: String -> Int -> String -> String
substituteString orig pos el =
let before = substr orig 0 pos
after = strtail orig (pos + 1)
in before ++ el ++ after
-- given a maze and a position, draw a '*' at that position in the maze
draw :: [String] -> (Int, Int) -> [String]
draw maze (x,y) = substitute maze y $ substituteString row x "*"
where row = maze !! y
-- given a maze, a previous position, and a list of tuples of potential
-- new positions and their wall positions, returns the solved maze, or
-- None if it cannot be solved
tryMoves :: [String] -> (Int, Int) -> [((Int, Int), (Int, Int))] -> Maybe [String]
tryMoves _ _ [] = Nothing
tryMoves maze prevPos ((newPos,wallPos):more) =
case solve' maze newPos prevPos
of Nothing -> tryMoves maze prevPos more
Just maze' -> Just $ foldl draw maze' [newPos, wallPos]
-- given a maze, a new position, and a previous position, returns
-- the solved maze, or None if it cannot be solved
-- (assumes goal is upper-left corner of maze)
solve' :: [String] -> (Int, Int) -> (Int, Int) -> Maybe [String]
solve' maze (2, 1) _ = Just maze
solve' maze (x, y) prevPos =
let newPositions = [(x, y - 2), (x + 4, y), (x, y + 2), (x - 4, y)]
notPrev pos' = pos' /= prevPos
newPositions' = filter notPrev newPositions
wallPositions = map (average (x,y)) newPositions'
zipped = zip newPositions' wallPositions
legalMoves = filter (notBlocked maze) zipped
in tryMoves maze (x,y) legalMoves
-- given a maze, returns a solved maze, or None if it cannot be solved
-- (starts at lower right corner and goes to upper left corner)
solve :: [String] -> Maybe [String]
solve maze = solve' (draw maze start) start (-1, -1)
where startx = (length $ head maze) - 3
starty = (length maze) - 2
start = (startx, starty)
-- takes unsolved maze on standard input, prints solved maze on standard output
main _ = do
isin <- stdin
isrin <- InputStreamReader.new isin
brin <- BufferedReader.fromISR isrin
lns <- BufferedReader.getlines brin
printStr $ unlines $ fromMaybe ["can't solve"] $ solve lns |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #Elixir | Elixir | defmodule Maximum do
def triangle_path(text) do
text
|> String.split("\n", trim: true)
|> Enum.map(fn line ->
line
|> String.split()
|> Enum.map(&String.to_integer(&1))
end)
|> Enum.reduce([], fn x,total ->
[0]++total++[0]
|> Enum.chunk_every( 2, 1)
|> Enum.map(&Enum.max(&1))
|> Enum.zip(x)
|> Enum.map(fn{a,b} -> a+b end)
end)
|> Enum.max()
end
end
text = """
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
"""
IO.puts Maximum.triangle_path(text)
|
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | Hash["Rosetta Code", "MD4", "HexString"] |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #Seed7 | Seed7 | $include "seed7_05.s7i";
const func string: middle3 (in integer: number) is func
result
var string: middle3 is "";
begin
middle3 := str(abs(number));
if not odd(length(middle3)) or length(middle3) < 3 then
middle3 := "error";
else
middle3 := middle3[succ((length(middle3) - 3) div 2) len 3];
end if;
end func;
const proc: main is func
local
var integer: number is 0;
var string: mid3 is "";
begin
for number range [] (123, 12345, 1234567, 987654321, 10001, -10001, -123,
-100, 100, -12345, 1, 2, -1, -10, 2002, -2002, 0) do
writeln(number <& ": " <& middle3(number));
end for;
end func; |
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #Clojure | Clojure | (apply str
(map (partial format "%02x")
(.digest (doto (java.security.MessageDigest/getInstance "MD5")
.reset
(.update (.getBytes "The quick brown fox jumps over the lazy dog")))))) |
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Modula-2 | Modula-2 | MODULE McNuggets;
FROM InOut IMPORT WriteCard, WriteString, WriteLn;
CONST Max = 100;
VAR a, b, c: CARDINAL;
nugget: ARRAY [0..Max] OF BOOLEAN;
BEGIN
FOR a := 0 TO Max DO
nugget[a] := FALSE;
END;
FOR a := 0 TO Max BY 6 DO
FOR b := a TO Max BY 9 DO
FOR c := b TO Max BY 20 DO
nugget[c] := TRUE;
END;
END;
END;
a := 100;
REPEAT DEC(a); UNTIL NOT nugget[a];
WriteString("Maximum non-McNuggets number: ");
WriteCard(a, 2);
WriteLn();
END McNuggets. |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #J | J | elems =: 6 4$' . .. ... ....----'
digit =: (elems{~5:<.0:>.15 10 5 0-~])`((4 4$_14{.'@')&[)@.(0&=)
mayan =: ":@(digit each@(20&#.^:_1)) |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #Java | Java |
import java.math.BigInteger;
public class MayanNumerals {
public static void main(String[] args) {
for ( long base10 : new long[] {4005, 8017, 326205, 886205, 1000000000, 1081439556L, 26960840421L, 503491211079L }) {
displayMyan(BigInteger.valueOf(base10));
System.out.printf("%n");
}
}
private static char[] digits = "0123456789ABCDEFGHJK".toCharArray();
private static BigInteger TWENTY = BigInteger.valueOf(20);
private static void displayMyan(BigInteger numBase10) {
System.out.printf("As base 10: %s%n", numBase10);
String numBase20 = "";
while ( numBase10.compareTo(BigInteger.ZERO) > 0 ) {
numBase20 = digits[numBase10.mod(TWENTY).intValue()] + numBase20;
numBase10 = numBase10.divide(TWENTY);
}
System.out.printf("As base 20: %s%nAs Mayan:%n", numBase20);
displayMyanLine1(numBase20);
displayMyanLine2(numBase20);
displayMyanLine3(numBase20);
displayMyanLine4(numBase20);
displayMyanLine5(numBase20);
displayMyanLine6(numBase20);
}
private static char boxUL = Character.toChars(9556)[0];
private static char boxTeeUp = Character.toChars(9574)[0];
private static char boxUR = Character.toChars(9559)[0];
private static char boxHorz = Character.toChars(9552)[0];
private static char boxVert = Character.toChars(9553)[0];
private static char theta = Character.toChars(952)[0];
private static char boxLL = Character.toChars(9562)[0];
private static char boxLR = Character.toChars(9565)[0];
private static char boxTeeLow = Character.toChars(9577)[0];
private static char bullet = Character.toChars(8729)[0];
private static char dash = Character.toChars(9472)[0];
private static void displayMyanLine1(String base20) {
char[] chars = base20.toCharArray();
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < chars.length ; i++ ) {
if ( i == 0 ) {
sb.append(boxUL);
}
for ( int j = 0 ; j < 4 ; j++ ) {
sb.append(boxHorz);
}
sb.append(i < chars.length-1 ? boxTeeUp : boxUR);
}
System.out.println(sb.toString());
}
private static String getBullet(int count) {
StringBuilder sb = new StringBuilder();
switch ( count ) {
case 1: sb.append(" " + bullet + " "); break;
case 2: sb.append(" " + bullet + bullet + " "); break;
case 3: sb.append("" + bullet + bullet + bullet + " "); break;
case 4: sb.append("" + bullet + bullet + bullet + bullet); break;
default: throw new IllegalArgumentException("Must be 1-4: " + count);
}
return sb.toString();
}
private static void displayMyanLine2(String base20) {
char[] chars = base20.toCharArray();
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < chars.length ; i++ ) {
if ( i == 0 ) {
sb.append(boxVert);
}
switch ( chars[i] ) {
case 'G': sb.append(getBullet(1)); break;
case 'H': sb.append(getBullet(2)); break;
case 'J': sb.append(getBullet(3)); break;
case 'K': sb.append(getBullet(4)); break;
default : sb.append(" ");
}
sb.append(boxVert);
}
System.out.println(sb.toString());
}
private static String DASH = getDash();
private static String getDash() {
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < 4 ; i++ ) {
sb.append(dash);
}
return sb.toString();
}
private static void displayMyanLine3(String base20) {
char[] chars = base20.toCharArray();
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < chars.length ; i++ ) {
if ( i == 0 ) {
sb.append(boxVert);
}
switch ( chars[i] ) {
case 'B': sb.append(getBullet(1)); break;
case 'C': sb.append(getBullet(2)); break;
case 'D': sb.append(getBullet(3)); break;
case 'E': sb.append(getBullet(4)); break;
case 'F': case 'G': case 'H': case 'J': case 'K':
sb.append(DASH); break;
default : sb.append(" ");
}
sb.append(boxVert);
}
System.out.println(sb.toString());
}
private static void displayMyanLine4(String base20) {
char[] chars = base20.toCharArray();
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < chars.length ; i++ ) {
if ( i == 0 ) {
sb.append(boxVert);
}
switch ( chars[i] ) {
case '6': sb.append(getBullet(1)); break;
case '7': sb.append(getBullet(2)); break;
case '8': sb.append(getBullet(3)); break;
case '9': sb.append(getBullet(4)); break;
case 'A': case 'B': case 'C': case 'D': case 'E':
case 'F': case 'G': case 'H': case 'J': case 'K':
sb.append(DASH); break;
default : sb.append(" ");
}
sb.append(boxVert);
}
System.out.println(sb.toString());
}
private static void displayMyanLine5(String base20) {
char[] chars = base20.toCharArray();
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < chars.length ; i++ ) {
if ( i == 0 ) {
sb.append(boxVert);
}
switch ( chars[i] ) {
case '0': sb.append(" " + theta + " "); break;
case '1': sb.append(getBullet(1)); break;
case '2': sb.append(getBullet(2)); break;
case '3': sb.append(getBullet(3)); break;
case '4': sb.append(getBullet(4)); break;
case '5': case '6': case '7': case '8': case '9':
case 'A': case 'B': case 'C': case 'D': case 'E':
case 'F': case 'G': case 'H': case 'J': case 'K':
sb.append(DASH); break;
default : sb.append(" ");
}
sb.append(boxVert);
}
System.out.println(sb.toString());
}
private static void displayMyanLine6(String base20) {
char[] chars = base20.toCharArray();
StringBuilder sb = new StringBuilder();
for ( int i = 0 ; i < chars.length ; i++ ) {
if ( i == 0 ) {
sb.append(boxLL);
}
for ( int j = 0 ; j < 4 ; j++ ) {
sb.append(boxHorz);
}
sb.append(i < chars.length-1 ? boxTeeLow : boxLR);
}
System.out.println(sb.toString());
}
}
|
http://rosettacode.org/wiki/Maze_generation | Maze generation |
This page uses content from Wikipedia. The original article was at Maze generation algorithm. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Generate and show a maze, using the simple Depth-first search algorithm.
Start at a random cell.
Mark the current cell as visited, and get a list of its neighbors. For each neighbor, starting with a randomly selected neighbor:
If that neighbor hasn't been visited, remove the wall between this cell and that neighbor, and then recurse with that neighbor as the current cell.
Related tasks
Maze solving.
| #Ada | Ada | generic
Height : Positive;
Width : Positive;
package Mazes is
type Maze_Grid is private;
procedure Initialize (Maze : in out Maze_Grid);
procedure Put (Item : Maze_Grid);
private
type Directions is (North, South, West, East);
type Cell_Walls is array (Directions) of Boolean;
type Cells is record
Walls : Cell_Walls := (others => True);
Visited : Boolean := False;
end record;
subtype Height_Type is Positive range 1 .. Height;
subtype Width_Type is Positive range 1 .. Width;
type Maze_Grid is array (Height_Type, Width_Type) of Cells;
end Mazes; |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #Burlesque | Burlesque | blsq ) {{1 1} {1 0}} 10 .*{mm}r[
{{89 55} {55 34}} |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #C | C | #include <math.h>
#include <stdio.h>
#include <stdlib.h>
typedef struct squareMtxStruct {
int dim;
double *cells;
double **m;
} *SquareMtx;
/* function for initializing row r of a new matrix */
typedef void (*FillFunc)( double *cells, int r, int dim, void *ff_data);
SquareMtx NewSquareMtx( int dim, FillFunc fillFunc, void *ff_data )
{
SquareMtx sm = malloc(sizeof(struct squareMtxStruct));
if (sm) {
int rw;
sm->dim = dim;
sm->cells = malloc(dim*dim * sizeof(double));
sm->m = malloc( dim * sizeof(double *));
if ((sm->cells != NULL) && (sm->m != NULL)) {
for (rw=0; rw<dim; rw++) {
sm->m[rw] = sm->cells + dim*rw;
fillFunc( sm->m[rw], rw, dim, ff_data );
}
}
else {
free(sm->m);
free(sm->cells);
free(sm);
printf("Square Matrix allocation failure\n");
return NULL;
}
}
else {
printf("Malloc failed for square matrix\n");
}
return sm;
}
void ffMatxSquare( double *cells, int rw, int dim, SquareMtx m0 )
{
int col, ix;
double sum;
double *m0rw = m0->m[rw];
for (col = 0; col < dim; col++) {
sum = 0.0;
for (ix=0; ix<dim; ix++)
sum += m0rw[ix] * m0->m[ix][col];
cells[col] = sum;
}
}
void ffMatxMulply( double *cells, int rw, int dim, SquareMtx mplcnds[] )
{
SquareMtx mleft = mplcnds[0];
SquareMtx mrigt = mplcnds[1];
double sum;
double *m0rw = mleft->m[rw];
int col, ix;
for (col = 0; col < dim; col++) {
sum = 0.0;
for (ix=0; ix<dim; ix++)
sum += m0rw[ix] * mrigt->m[ix][col];
cells[col] = sum;
}
}
void MatxMul( SquareMtx mr, SquareMtx left, SquareMtx rigt)
{
int rw;
SquareMtx mplcnds[2];
mplcnds[0] = left; mplcnds[1] = rigt;
for (rw = 0; rw < left->dim; rw++)
ffMatxMulply( mr->m[rw], rw, left->dim, mplcnds);
}
void ffIdentity( double *cells, int rw, int dim, void *v )
{
int col;
for (col=0; col<dim; col++) cells[col] = 0.0;
cells[rw] = 1.0;
}
void ffCopy(double *cells, int rw, int dim, SquareMtx m1)
{
int col;
for (col=0; col<dim; col++) cells[col] = m1->m[rw][col];
}
void FreeSquareMtx( SquareMtx m )
{
free(m->m);
free(m->cells);
free(m);
}
SquareMtx SquareMtxPow( SquareMtx m0, int exp )
{
SquareMtx v0 = NewSquareMtx(m0->dim, ffIdentity, NULL);
SquareMtx v1 = NULL;
SquareMtx base0 = NewSquareMtx( m0->dim, ffCopy, m0);
SquareMtx base1 = NULL;
SquareMtx mplcnds[2], t;
while (exp) {
if (exp % 2) {
if (v1)
MatxMul( v1, v0, base0);
else {
mplcnds[0] = v0; mplcnds[1] = base0;
v1 = NewSquareMtx(m0->dim, ffMatxMulply, mplcnds);
}
{t = v0; v0=v1; v1 = t;}
}
if (base1)
MatxMul( base1, base0, base0);
else
base1 = NewSquareMtx( m0->dim, ffMatxSquare, base0);
t = base0; base0 = base1; base1 = t;
exp = exp/2;
}
if (base0) FreeSquareMtx(base0);
if (base1) FreeSquareMtx(base1);
if (v1) FreeSquareMtx(v1);
return v0;
}
FILE *fout;
void SquareMtxPrint( SquareMtx mtx, const char *mn )
{
int rw, col;
int d = mtx->dim;
fprintf(fout, "%s dim:%d =\n", mn, mtx->dim);
for (rw=0; rw<d; rw++) {
fprintf(fout, " |");
for(col=0; col<d; col++)
fprintf(fout, "%8.5f ",mtx->m[rw][col] );
fprintf(fout, " |\n");
}
fprintf(fout, "\n");
}
void fillInit( double *cells, int rw, int dim, void *data)
{
double theta = 3.1415926536/6.0;
double c1 = cos( theta);
double s1 = sin( theta);
switch(rw) {
case 0:
cells[0]=c1; cells[1]=s1; cells[2]=0.0;
break;
case 1:
cells[0]=-s1; cells[1]=c1; cells[2]=0;
break;
case 2:
cells[0]=0.0; cells[1]=0.0; cells[2]=1.0;
break;
}
}
int main()
{
SquareMtx m0 = NewSquareMtx( 3, fillInit, NULL);
SquareMtx m1 = SquareMtxPow( m0, 5);
SquareMtx m2 = SquareMtxPow( m0, 9);
SquareMtx m3 = SquareMtxPow( m0, 2);
// fout = stdout;
fout = fopen("matrx_exp.txt", "w");
SquareMtxPrint(m0, "m0"); FreeSquareMtx(m0);
SquareMtxPrint(m1, "m0^5"); FreeSquareMtx(m1);
SquareMtxPrint(m2, "m0^9"); FreeSquareMtx(m2);
SquareMtxPrint(m3, "m0^2"); FreeSquareMtx(m3);
fclose(fout);
return 0;
} |
http://rosettacode.org/wiki/Matrix_chain_multiplication | Matrix chain multiplication | Problem
Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. The number of operations required to compute the product of matrices A1, A2... An depends on the order of matrix multiplications, hence on where parens are put. Remember that the matrix product is associative, but not commutative, hence only the parens can be moved.
For instance, with four matrices, one can compute A(B(CD)), A((BC)D), (AB)(CD), (A(BC))D, (AB)C)D. The number of different ways to put the parens is a Catalan number, and grows exponentially with the number of factors.
Here is an example of computation of the total cost, for matrices A(5,6), B(6,3), C(3,1):
AB costs 5*6*3=90 and produces a matrix of dimensions (5,3), then (AB)C costs 5*3*1=15. The total cost is 105.
BC costs 6*3*1=18 and produces a matrix of dimensions (6,1), then A(BC) costs 5*6*1=30. The total cost is 48.
In this case, computing (AB)C requires more than twice as many operations as A(BC). The difference can be much more dramatic in real cases.
Task
Write a function which, given a list of the successive dimensions of matrices A1, A2... An, of arbitrary length, returns the optimal way to compute the matrix product, and the total cost. Any sensible way to describe the optimal solution is accepted. The input list does not duplicate shared dimensions: for the previous example of matrices A,B,C, one will only pass the list [5,6,3,1] (and not [5,6,6,3,3,1]) to mean the matrix dimensions are respectively (5,6), (6,3) and (3,1). Hence, a product of n matrices is represented by a list of n+1 dimensions.
Try this function on the following two lists:
[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
To solve the task, it's possible, but not required, to write a function that enumerates all possible ways to parenthesize the product. This is not optimal because of the many duplicated computations, and this task is a classic application of dynamic programming.
See also Matrix chain multiplication on Wikipedia.
| #Ada | Ada |
package mat_chain is
type Vector is array (Natural range <>) of Integer;
procedure Chain_Multiplication (Dims : Vector);
end mat_chain;
|
http://rosettacode.org/wiki/Matrix_chain_multiplication | Matrix chain multiplication | Problem
Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. The number of operations required to compute the product of matrices A1, A2... An depends on the order of matrix multiplications, hence on where parens are put. Remember that the matrix product is associative, but not commutative, hence only the parens can be moved.
For instance, with four matrices, one can compute A(B(CD)), A((BC)D), (AB)(CD), (A(BC))D, (AB)C)D. The number of different ways to put the parens is a Catalan number, and grows exponentially with the number of factors.
Here is an example of computation of the total cost, for matrices A(5,6), B(6,3), C(3,1):
AB costs 5*6*3=90 and produces a matrix of dimensions (5,3), then (AB)C costs 5*3*1=15. The total cost is 105.
BC costs 6*3*1=18 and produces a matrix of dimensions (6,1), then A(BC) costs 5*6*1=30. The total cost is 48.
In this case, computing (AB)C requires more than twice as many operations as A(BC). The difference can be much more dramatic in real cases.
Task
Write a function which, given a list of the successive dimensions of matrices A1, A2... An, of arbitrary length, returns the optimal way to compute the matrix product, and the total cost. Any sensible way to describe the optimal solution is accepted. The input list does not duplicate shared dimensions: for the previous example of matrices A,B,C, one will only pass the list [5,6,3,1] (and not [5,6,6,3,3,1]) to mean the matrix dimensions are respectively (5,6), (6,3) and (3,1). Hence, a product of n matrices is represented by a list of n+1 dimensions.
Try this function on the following two lists:
[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
To solve the task, it's possible, but not required, to write a function that enumerates all possible ways to parenthesize the product. This is not optimal because of the many duplicated computations, and this task is a classic application of dynamic programming.
See also Matrix chain multiplication on Wikipedia.
| #C | C | #include <stdio.h>
#include <limits.h>
#include <stdlib.h>
int **m;
int **s;
void optimal_matrix_chain_order(int *dims, int n) {
int len, i, j, k, temp, cost;
n--;
m = (int **)malloc(n * sizeof(int *));
for (i = 0; i < n; ++i) {
m[i] = (int *)calloc(n, sizeof(int));
}
s = (int **)malloc(n * sizeof(int *));
for (i = 0; i < n; ++i) {
s[i] = (int *)calloc(n, sizeof(int));
}
for (len = 1; len < n; ++len) {
for (i = 0; i < n - len; ++i) {
j = i + len;
m[i][j] = INT_MAX;
for (k = i; k < j; ++k) {
temp = dims[i] * dims[k + 1] * dims[j + 1];
cost = m[i][k] + m[k + 1][j] + temp;
if (cost < m[i][j]) {
m[i][j] = cost;
s[i][j] = k;
}
}
}
}
}
void print_optimal_chain_order(int i, int j) {
if (i == j)
printf("%c", i + 65);
else {
printf("(");
print_optimal_chain_order(i, s[i][j]);
print_optimal_chain_order(s[i][j] + 1, j);
printf(")");
}
}
int main() {
int i, j, n;
int a1[4] = {5, 6, 3, 1};
int a2[13] = {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2};
int a3[12] = {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10};
int *dims_list[3] = {a1, a2, a3};
int sizes[3] = {4, 13, 12};
for (i = 0; i < 3; ++i) {
printf("Dims : [");
n = sizes[i];
for (j = 0; j < n; ++j) {
printf("%d", dims_list[i][j]);
if (j < n - 1) printf(", "); else printf("]\n");
}
optimal_matrix_chain_order(dims_list[i], n);
printf("Order : ");
print_optimal_chain_order(0, n - 2);
printf("\nCost : %d\n\n", m[0][n - 2]);
for (j = 0; j <= n - 2; ++j) free(m[j]);
free(m);
for (j = 0; j <= n - 2; ++j) free(s[j]);
free(s);
}
return 0;
} |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #Go | Go | package main
import (
"bytes"
"fmt"
"math/rand"
"time"
)
type maze struct {
c2 [][]byte // cells by row
h2 [][]byte // horizontal walls by row (ignore first row)
v2 [][]byte // vertical walls by row (ignore first of each column)
}
func newMaze(rows, cols int) *maze {
c := make([]byte, rows*cols) // all cells
h := bytes.Repeat([]byte{'-'}, rows*cols) // all horizontal walls
v := bytes.Repeat([]byte{'|'}, rows*cols) // all vertical walls
c2 := make([][]byte, rows) // cells by row
h2 := make([][]byte, rows) // horizontal walls by row
v2 := make([][]byte, rows) // vertical walls by row
for i := range h2 {
c2[i] = c[i*cols : (i+1)*cols]
h2[i] = h[i*cols : (i+1)*cols]
v2[i] = v[i*cols : (i+1)*cols]
}
return &maze{c2, h2, v2}
}
func (m *maze) String() string {
hWall := []byte("+---")
hOpen := []byte("+ ")
vWall := []byte("| ")
vOpen := []byte(" ")
rightCorner := []byte("+\n")
rightWall := []byte("|\n")
var b []byte
for r, hw := range m.h2 {
for _, h := range hw {
if h == '-' || r == 0 {
b = append(b, hWall...)
} else {
b = append(b, hOpen...)
if h != '-' && h != 0 {
b[len(b)-2] = h
}
}
}
b = append(b, rightCorner...)
for c, vw := range m.v2[r] {
if vw == '|' || c == 0 {
b = append(b, vWall...)
} else {
b = append(b, vOpen...)
if vw != '|' && vw != 0 {
b[len(b)-4] = vw
}
}
if m.c2[r][c] != 0 {
b[len(b)-2] = m.c2[r][c]
}
}
b = append(b, rightWall...)
}
for _ = range m.h2[0] {
b = append(b, hWall...)
}
b = append(b, rightCorner...)
return string(b)
}
func (m *maze) gen() {
m.g2(rand.Intn(len(m.c2)), rand.Intn(len(m.c2[0])))
}
const (
up = iota
dn
rt
lf
)
func (m *maze) g2(r, c int) {
m.c2[r][c] = ' '
for _, dir := range rand.Perm(4) {
switch dir {
case up:
if r > 0 && m.c2[r-1][c] == 0 {
m.h2[r][c] = 0
m.g2(r-1, c)
}
case lf:
if c > 0 && m.c2[r][c-1] == 0 {
m.v2[r][c] = 0
m.g2(r, c-1)
}
case dn:
if r < len(m.c2)-1 && m.c2[r+1][c] == 0 {
m.h2[r+1][c] = 0
m.g2(r+1, c)
}
case rt:
if c < len(m.c2[0])-1 && m.c2[r][c+1] == 0 {
m.v2[r][c+1] = 0
m.g2(r, c+1)
}
}
}
}
func main() {
rand.Seed(time.Now().UnixNano())
const height = 4
const width = 7
m := newMaze(height, width)
m.gen()
m.solve(
rand.Intn(height), rand.Intn(width),
rand.Intn(height), rand.Intn(width))
fmt.Print(m)
}
func (m *maze) solve(ra, ca, rz, cz int) {
var rSolve func(ra, ca, dir int) bool
rSolve = func(r, c, dir int) bool {
if r == rz && c == cz {
m.c2[r][c] = 'F'
return true
}
if dir != dn && m.h2[r][c] == 0 {
if rSolve(r-1, c, up) {
m.c2[r][c] = '^'
m.h2[r][c] = '^'
return true
}
}
if dir != up && r+1 < len(m.h2) && m.h2[r+1][c] == 0 {
if rSolve(r+1, c, dn) {
m.c2[r][c] = 'v'
m.h2[r+1][c] = 'v'
return true
}
}
if dir != lf && c+1 < len(m.v2[0]) && m.v2[r][c+1] == 0 {
if rSolve(r, c+1, rt) {
m.c2[r][c] = '>'
m.v2[r][c+1] = '>'
return true
}
}
if dir != rt && m.v2[r][c] == 0 {
if rSolve(r, c-1, lf) {
m.c2[r][c] = '<'
m.v2[r][c] = '<'
return true
}
}
return false
}
rSolve(ra, ca, -1)
m.c2[ra][ca] = 'S'
} |
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #Erlang | Erlang |
-mode(compile).
-import(lists, [foldl/3]).
main(_) ->
{ok, Tmat} = file:open("triangle.txt", [read, raw, {read_ahead, 16384}]),
Max = max_sum(Tmat, []),
io:format("The maximum total is ~b~n", [Max]).
max_sum(FD, Last) ->
case file:read_line(FD) of
eof -> foldl(fun erlang:max/2, 0, Last);
{ok, Line} ->
Current = [binary_to_integer(B) || B <- re:split(Line, "[ \n]"), byte_size(B) > 0],
max_sum(FD, fold_row(Last, Current))
end.
% The first argument has one more element than the second, so compute
% the initial sum so that both lists have identical length for fold_rest().
fold_row([], L) -> L;
fold_row([A|_] = Last, [B|Bs]) ->
[A+B | fold_rest(Last, Bs)].
% Both lists must have same length
fold_rest([A], [B]) -> [A+B];
fold_rest([A1 | [A2|_] = As], [B|Bs]) -> [B + max(A1,A2) | fold_rest(As, Bs)].
|
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #Nim | Nim | import strutils
const MD4Len = 16
proc MD4(d: cstring, n: culong, md: cstring = nil): cstring {.cdecl, dynlib: "libssl.so", importc.}
proc MD4(s: string): string =
result = ""
var s = MD4(s.cstring, s.len.culong)
for i in 0 ..< MD4Len:
result.add s[i].BiggestInt.toHex(2).toLower
echo MD4("Rosetta Code") |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #SenseTalk | SenseTalk | set inputs to [123, 12345, 1234567, 987654321,
10001, -10001, -123, -100, 100, -12345,
1, 2, -1, -10, 2002, -2002, 0]
repeat with each num in inputs
put num into output
put middle3digits of num into character 15 of output -- will autofill with "."s
put output
end repeat
to handle middle3digits of number
put every occurrence of <digit> in number into digits
if the number of items in digits is less than 3 then return "Error: too few digits"
if the number of items in digits is an even number then return "Error: even number of digits"
repeat while the number of items in digits is greater than 3
delete the first item of digits
delete the last item of digits
end repeat
return digits joined by empty
end middle3digits
|
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #Sidef | Sidef | func middle_three(n) {
var l = n.len;
if (l < 3) {
"#{n} is too short"
} elsif (l.is_even) {
"#{n} has an even number of digits"
} else {
"The three middle digits of #{n} are: " + n.digits.ft(l-3 / 2, l/2 + 1).join
}
}
var nums = %n(
123 12345 1234567 987654321 10001 -10001 -123 -100 100 -12345
1 2 -1 -10 2002 -2002 0
);
nums.each { say middle_three(_) }; |
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #COBOL | COBOL |
IDENTIFICATION DIVISION.
PROGRAM-ID. MD5.
AUTHOR. Bill Gunshannon
INSTALLATION. Home.
DATE-WRITTEN. 16 December 2021.
************************************************************
** Program Abstract:
** Use the md5sum utility and pass the HASH back using
** a temp file. Not elegant, but it works.
************************************************************
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
SELECT Tmp-MD5 ASSIGN TO "/tmp/MD5"
ORGANIZATION IS LINE SEQUENTIAL.
DATA DIVISION.
FILE SECTION.
FD Tmp-MD5
DATA RECORD IS MD5-Rec.
01 MD5-Rec PIC X(32).
WORKING-STORAGE SECTION.
01 Eof PIC X VALUE 'F'.
01 Str1.
05 Pre-cmd PIC X(8)
VALUE 'echo -n '.
05 Str1-complete.
10 Str1-Part1 PIC X(26)
VALUE 'The quick brown fox jumps'.
10 Str1-Part2 PIC X(19)
VALUE ' over the lazy dog'.
05 Post-cmd PIC X(20)
VALUE ' | md5sum > /tmp/MD5'.
01 Str1-MD5 PIC X(32).
PROCEDURE DIVISION.
Main-Program.
DISPLAY Str1-complete.
PERFORM Get-MD5.
DISPLAY Str1-MD5.
STOP RUN.
Get-MD5.
CALL "SYSTEM" USING Str1.
OPEN INPUT Tmp-MD5.
READ Tmp-MD5 INTO Str1-MD5.
CLOSE Tmp-MD5.
CALL "CBL_DELETE_FILE" USING '/tmp/MD5'.
|
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #MiniZinc | MiniZinc |
%McNuggets. Nigel Galloway, August 27th., 2019
var 0..99: n;
constraint forall(x in 0..16,y in 0..11,z in 0..5)(6*x + 9*y + 20*z!=n);
solve maximize n;
output [show(n)]
|
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #JavaScript | JavaScript | (() => {
'use strict';
const main = () =>
unlines(
map(mayanFramed,
[4005, 8017, 326205, 886205, 1081439556, 1000000, 1000000000]
)
);
// MAYAN NUMBERS --------------------------------------
// mayanFramed :: Int -> String
const mayanFramed = n =>
'\nMayan ' + n.toString() + ':\n\n' +
wikiTable({
style: 'text-align:center; background-color:#F0EDDE; ' +
'color:#605B4B; border:2px solid silver',
colwidth: '3em'
})(
mayanGlyph(n)
);
// mayanGlyph :: Int -> [[String]]
const mayanGlyph = n =>
filter(any(compose(not, isNull)),
transpose(leftPadded(
showIntAtBase(20, mayanDigit, n, [])
))
);
// mayanDigit :: Int -> [String]
const mayanDigit = n =>
0 !== n ? cons(
replicateString(rem(n, 5), '●'),
replicate(quot(n, 5), '━━')
) : ['Θ'];
// FORMATTING -----------------------------------------
// wikiTable :: Dict -> [[a]] -> String
const wikiTable = opts => rows => {
const colWidth = () =>
'colwidth' in opts ? (
'|style="width:' + opts.colwidth + ';"'
) : '';
return 0 < rows.length ? (
'{| ' + ['class', 'style'].reduce(
(a, k) => k in opts ? (
a + k + '="' + opts[k] + '" '
) : a, ''
) + '\n' + rows.map(
(row, i) => row.map(
x => (0 === i ? (
colWidth() + '| '
) : '|') + (x.toString() || ' ')
).join('\n')
).join('\n|-\n') + '\n|}\n\n'
) : '';
};
// leftPadded :: [[String]] -> [[String]]
const leftPadded = xs => {
const w = maximum(map(length, xs));
return map(
x => replicate(w - x.length, '').concat(x),
xs
);
};
// GENERIC FUNCTIONS ----------------------------------
// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// any :: (a -> Bool) -> [a] -> Bool
const any = p => xs => xs.some(p);
// comparing :: (a -> b) -> (a -> a -> Ordering)
const comparing = f =>
(x, y) => {
const
a = f(x),
b = f(y);
return a < b ? -1 : (a > b ? 1 : 0);
};
// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (f, g) => x => f(g(x));
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) =>
xs.reduce((a, x) => a.concat(f(x)), []);
// cons :: a -> [a] -> [a]
const cons = (x, xs) =>
Array.isArray(xs) ? (
[x].concat(xs)
) : 'GeneratorFunction' !== xs.constructor.constructor.name ? (
x + xs
) : ( // Existing generator wrapped with one additional element
function*() {
yield x;
let nxt = xs.next()
while (!nxt.done) {
yield nxt.value;
nxt = xs.next();
}
}
)();
// filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => xs.filter(f);
// foldl1 :: (a -> a -> a) -> [a] -> a
const foldl1 = (f, xs) =>
1 < xs.length ? xs.slice(1)
.reduce(f, xs[0]) : xs[0];
// isNull :: [a] -> Bool
// isNull :: String -> Bool
const isNull = xs =>
Array.isArray(xs) || ('string' === typeof xs) ? (
1 > xs.length
) : undefined;
// Returns Infinity over objects without finite length.
// This enables zip and zipWith to choose the shorter
// argument when one is non-finite, like cycle, repeat etc
// length :: [a] -> Int
const length = xs =>
(Array.isArray(xs) || 'string' === typeof xs) ? (
xs.length
) : Infinity;
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) =>
(Array.isArray(xs) ? (
xs
) : xs.split('')).map(f);
// maximum :: Ord a => [a] -> a
const maximum = xs =>
0 < xs.length ? (
foldl1((a, x) => x > a ? x : a, xs)
) : undefined;
// Ordering: (LT|EQ|GT):
// GT: 1 (or other positive n)
// EQ: 0
// LT: -1 (or other negative n)
// maximumBy :: (a -> a -> Ordering) -> [a] -> a
const maximumBy = (f, xs) =>
0 < xs.length ? (
xs.slice(1)
.reduce((a, x) => 0 < f(x, a) ? x : a, xs[0])
) : undefined;
// not :: Bool -> Bool
const not = b => !b;
// quot :: Int -> Int -> Int
const quot = (n, m) => Math.floor(n / m);
// quotRem :: Int -> Int -> (Int, Int)
const quotRem = (m, n) =>
Tuple(Math.floor(m / n), m % n);
// rem :: Int -> Int -> Int
const rem = (n, m) => n % m;
// replicate :: Int -> a -> [a]
const replicate = (n, x) =>
Array.from({
length: n
}, () => x);
// replicateString :: Int -> String -> String
const replicateString = (n, s) => s.repeat(n);
// showIntAtBase :: Int -> (Int -> [String])
// -> Int -> [[String]] -> [[String]]
const showIntAtBase = (base, toStr, n, rs) => {
const go = ([n, d], r) => {
const r_ = cons(toStr(d), r);
return 0 !== n ? (
go(Array.from(quotRem(n, base)), r_)
) : r_;
};
return go(Array.from(quotRem(n, base)), rs);
};
// transpose :: [[a]] -> [[a]]
const transpose = tbl => {
const
gaps = replicate(
length(maximumBy(comparing(length), tbl)), []
),
rows = map(xs => xs.concat(gaps.slice(xs.length)), tbl);
return map(
(_, col) => concatMap(row => [row[col]], rows),
rows[0]
);
};
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// MAIN ---
return main();
})(); |
http://rosettacode.org/wiki/Maze_generation | Maze generation |
This page uses content from Wikipedia. The original article was at Maze generation algorithm. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Generate and show a maze, using the simple Depth-first search algorithm.
Start at a random cell.
Mark the current cell as visited, and get a list of its neighbors. For each neighbor, starting with a randomly selected neighbor:
If that neighbor hasn't been visited, remove the wall between this cell and that neighbor, and then recurse with that neighbor as the current cell.
Related tasks
Maze solving.
| #Aime | Aime | grid_maze(data b, integer N)
{
data d;
N.times(bb_cast, d, "+---");
bb_cast(d, "+\n");
N.times(bb_cast, d, "| * ");
bb_cast(d, "|\n");
N.times(bb_copy, b, d);
b_size(d, N * 4 + 2);
bb_copy(b, d);
}
void
walk_cell(data b, integer N, line_size, x, y, list x_offsets, y_offsets)
{
integer i, p, q, r;
b_replace(b, y + x, ' ');
r = drand(3);
i = 0;
while (i < 4) {
p = x + x_offsets[q = (r + i) & 3];
q = y + y_offsets[q];
if (-1 < p && p < line_size
&& -1 < q && q < line_size * (N * 2 + 1)) {
if (b[q + p] == '*') {
walk_cell(b, N, line_size, p, q, x_offsets, y_offsets);
b[(q + y) / 2 + (p + x) / 2] = ' ';
if (p == x) {
b[(q + y) / 2 + p - 1] = ' ';
b[(q + y) / 2 + p + 1] = ' ';
}
}
}
i += 1;
}
}
walk_maze(data b, integer N)
{
integer line_size, x, y;
list x_offsets, y_offsets;
line_size = N * 4 + 1 + 1;
l_bill(x_offsets, 0, 4, 0, -4, 0);
l_bill(y_offsets, 0, 0, line_size * 2, 0, line_size * -2);
x = drand(N - 1) * 4 + 2;
y = line_size * (drand(N - 1) * 2 + 1);
walk_cell(b, N, line_size, x, y, x_offsets, y_offsets);
}
main(void)
{
data b;
grid_maze(b, 10);
walk_maze(b, 10);
o_(b);
0;
} |
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #C.23 | C# | using System;
using System.Collections;
using System.Collections.Generic;
using static System.Linq.Enumerable;
public static class MatrixExponentation
{
public static double[,] Identity(int size) {
double[,] matrix = new double[size, size];
for (int i = 0; i < size; i++) matrix[i, i] = 1;
return matrix;
}
public static double[,] Multiply(this double[,] left, double[,] right) {
if (left.ColumnCount() != right.RowCount()) throw new ArgumentException();
double[,] m = new double[left.RowCount(), right.ColumnCount()];
foreach (var (row, column) in from r in Range(0, m.RowCount()) from c in Range(0, m.ColumnCount()) select (r, c)) {
m[row, column] = Range(0, m.RowCount()).Sum(i => left[row, i] * right[i, column]);
}
return m;
}
public static double[,] Pow(this double[,] matrix, int exp) {
if (matrix.RowCount() != matrix.ColumnCount()) throw new ArgumentException("Matrix must be square.");
double[,] accumulator = Identity(matrix.RowCount());
for (int i = 0; i < exp; i++) {
accumulator = accumulator.Multiply(matrix);
}
return accumulator;
}
private static int RowCount(this double[,] matrix) => matrix.GetLength(0);
private static int ColumnCount(this double[,] matrix) => matrix.GetLength(1);
private static void Print(this double[,] m) {
foreach (var row in Rows()) {
Console.WriteLine("[ " + string.Join(" ", row) + " ]");
}
Console.WriteLine();
IEnumerable<IEnumerable<double>> Rows() =>
Range(0, m.RowCount()).Select(row => Range(0, m.ColumnCount()).Select(column => m[row, column]));
}
public static void Main() {
var matrix = new double[,] {
{ 3, 2 },
{ 2, 1 }
};
matrix.Pow(0).Print();
matrix.Pow(1).Print();
matrix.Pow(2).Print();
matrix.Pow(3).Print();
matrix.Pow(4).Print();
matrix.Pow(50).Print();
}
} |
http://rosettacode.org/wiki/Matrix_digital_rain | Matrix digital rain | Implement the Matrix Digital Rain visual effect from the movie "The Matrix" as described in Wikipedia.
Provided is a reference implementation in Common Lisp to be run in a terminal.
| #Ada | Ada | with Ada.Text_IO;
with Ada.Numerics.Discrete_Random;
with Ada.Strings.Fixed;
procedure Digital_Rain is
type Column_Type is range 0 .. 150;
type Row_Type is range 0 .. 25;
type Character_Range is new Character range 'A' .. 'Z';
subtype Delay_Range is Integer range 2 .. 6;
package Column_Random is new Ada.Numerics.Discrete_Random (Column_Type);
package Row_Random is new Ada.Numerics.Discrete_Random (Row_Type);
package Character_Random is new Ada.Numerics.Discrete_Random (Character_Range);
package Delay_Random is new Ada.Numerics.Discrete_Random (Delay_Range);
Column_Generator : Column_Random.Generator;
Row_Generator : Row_Random.Generator;
Character_Generator : Character_Random.Generator;
Delay_Generator : Delay_Random.Generator;
Esc : constant Character := Character'Val (8#33#);
Running : Boolean := True;
Char : Character;
use Ada.Text_IO;
protected Text_Out is
procedure Put_At (Row : Row_Type; Col : Column_Type;
Color : Integer; Item : Character);
end Text_Out;
protected body Text_Out is
procedure Put_At (Row : Row_Type; Col : Column_Type;
Color : Integer; Item : Character)
is
use Ada.Strings;
begin
Put (Esc & "[" &
Fixed.Trim (Row'Image, Left) & ";" &
Fixed.Trim (Col'Image, Left) & "H"); -- Place Cursor
Put (Esc & "[" & Fixed.Trim (Color'Image, Left) & "m"); -- Foreground color
Put (Item);
end Put_At;
end Text_Out;
task type Rainer;
task body Rainer is
Row : Row_Type;
Col : Column_Type;
Del : Delay_Range;
begin
Outer_Loop:
loop
Col := Column_Random.Random (Column_Generator);
Row := Row_Random.Random (Row_Generator);
for Rain in reverse Boolean loop
Del := Delay_Random.Random (Delay_Generator);
for R in 0 .. Row loop
if Rain then
if R in 1 .. Row then
Text_Out.Put_At (R - 1, Col, Color => 32, Item => Char);
end if;
Char := Character (Character_Random.Random (Character_Generator));
if R in 0 .. Row - 1 then
Text_Out.Put_At (R, Col, Color => 97, Item => Char);
end if;
else
Text_Out.Put_At (R, Col, Color => 30, Item => ' ');
end if;
delay 0.020 * Del;
exit Outer_Loop when not Running;
end loop;
end loop;
end loop Outer_Loop;
end Rainer;
Dummy : Character;
Rainers : array (1 .. 50) of Rainer;
begin
Put (ESC & "[?25l"); -- Hide the cursor
Put (ESC & "[40m"); -- Black background
Put (ESC & "[2J"); -- Clear Terminal
Get_Immediate (Dummy);
Running := False;
delay 0.200;
Put (ESC & "[?25h"); -- Restore the cursor
Put (ESC & "[0;0H"); -- Place cursor
Put (ESC & "[39m"); -- Default foreground
Put (ESC & "[49m"); -- Default backgound
Put (ESC & "[2J"); -- Clear Terminal
end Digital_Rain; |
http://rosettacode.org/wiki/Mastermind | Mastermind | Create a simple version of the board game: Mastermind.
It must be possible to:
choose the number of colors will be used in the game (2 - 20)
choose the color code length (4 - 10)
choose the maximum number of guesses the player has (7 - 20)
choose whether or not colors may be repeated in the code
The (computer program) game should display all the player guesses and the results of that guess.
Display (just an idea):
Feature
Graphic Version
Text Version
Player guess
Colored circles
Alphabet letters
Correct color & position
Black circle
X
Correct color
White circle
O
None
Gray circle
-
A text version example: 1. ADEF - XXO-
Translates to:
the first guess;
the four colors (ADEF);
result:
two correct colors and spot,
one correct color/wrong spot, one color isn't in the code.
Happy coding!
Related tasks
Bulls and cows
Bulls and cows/Player
Guess the number
Guess the number/With Feedback
| #11l | 11l | F encode(correct, guess)
[String] output_arr
L(i) 0 .< correct.len
output_arr [+]= I guess[i] == correct[i] {‘X’} E I guess[i] C correct {‘O’} E ‘-’
R output_arr
F safe_int_input(prompt, min_val, max_val)
L
V user_input_str = input(prompt)
X.try
V user_input = Int(user_input_str)
I user_input C min_val .. max_val
R user_input
X.catch ValueError
L.continue
F play_game()
print(‘Welcome to Mastermind.’)
print(‘You will need to guess a random code.’)
print(‘For each guess, you will receive a hint.’)
print(‘In this hint, X denotes a correct letter, and O a letter in the original string but in a different position.’)
print()
V number_of_letters = safe_int_input(‘Select a number of possible letters for the code (2-20): ’, 2, 20)
V code_length = safe_int_input(‘Select a length for the code (4-10): ’, 4, 10)
V letters = ‘ABCDEFGHIJKLMNOPQRST’[0 .< number_of_letters]
V code = ‘’
L 0 .< code_length
code ‘’= random:choice(letters)
[String] guesses
L
print()
V guess = input(‘Enter a guess of length #. (#.): ’.format(code_length, letters)).uppercase()
I guess.len != code_length | any(guess.map(char -> char !C @letters))
L.continue
E I guess == code
print("\nYour guess "guess‘ was correct!’)
L.break
E
guesses.append(‘#.: #. => #.’.format(guesses.len + 1, Array(guess).join(‘ ’), encode(code, guess).join(‘ ’)))
L(i_guess) guesses
print(‘------------------------------------’)
print(i_guess)
print(‘------------------------------------’)
play_game() |
http://rosettacode.org/wiki/Matrix_chain_multiplication | Matrix chain multiplication | Problem
Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. The number of operations required to compute the product of matrices A1, A2... An depends on the order of matrix multiplications, hence on where parens are put. Remember that the matrix product is associative, but not commutative, hence only the parens can be moved.
For instance, with four matrices, one can compute A(B(CD)), A((BC)D), (AB)(CD), (A(BC))D, (AB)C)D. The number of different ways to put the parens is a Catalan number, and grows exponentially with the number of factors.
Here is an example of computation of the total cost, for matrices A(5,6), B(6,3), C(3,1):
AB costs 5*6*3=90 and produces a matrix of dimensions (5,3), then (AB)C costs 5*3*1=15. The total cost is 105.
BC costs 6*3*1=18 and produces a matrix of dimensions (6,1), then A(BC) costs 5*6*1=30. The total cost is 48.
In this case, computing (AB)C requires more than twice as many operations as A(BC). The difference can be much more dramatic in real cases.
Task
Write a function which, given a list of the successive dimensions of matrices A1, A2... An, of arbitrary length, returns the optimal way to compute the matrix product, and the total cost. Any sensible way to describe the optimal solution is accepted. The input list does not duplicate shared dimensions: for the previous example of matrices A,B,C, one will only pass the list [5,6,3,1] (and not [5,6,6,3,3,1]) to mean the matrix dimensions are respectively (5,6), (6,3) and (3,1). Hence, a product of n matrices is represented by a list of n+1 dimensions.
Try this function on the following two lists:
[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
To solve the task, it's possible, but not required, to write a function that enumerates all possible ways to parenthesize the product. This is not optimal because of the many duplicated computations, and this task is a classic application of dynamic programming.
See also Matrix chain multiplication on Wikipedia.
| #C.23 | C# | using System;
class MatrixChainOrderOptimizer {
private int[,] m;
private int[,] s;
void OptimalMatrixChainOrder(int[] dims) {
int n = dims.Length - 1;
m = new int[n, n];
s = new int[n, n];
for (int len = 1; len < n; ++len) {
for (int i = 0; i < n - len; ++i) {
int j = i + len;
m[i, j] = Int32.MaxValue;
for (int k = i; k < j; ++k) {
int temp = dims[i] * dims[k + 1] * dims[j + 1];
int cost = m[i, k] + m[k + 1, j] + temp;
if (cost < m[i, j]) {
m[i, j] = cost;
s[i, j] = k;
}
}
}
}
}
void PrintOptimalChainOrder(int i, int j) {
if (i == j)
Console.Write((char)(i + 65));
else {
Console.Write("(");
PrintOptimalChainOrder(i, s[i, j]);
PrintOptimalChainOrder(s[i, j] + 1, j);
Console.Write(")");
}
}
static void Main() {
var mcoo = new MatrixChainOrderOptimizer();
var dimsList = new int[3][];
dimsList[0] = new int[4] {5, 6, 3, 1};
dimsList[1] = new int[13] {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2};
dimsList[2] = new int[12] {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10};
for (int i = 0; i < dimsList.Length; ++i) {
Console.Write("Dims : [");
int n = dimsList[i].Length;
for (int j = 0; j < n; ++j) {
Console.Write(dimsList[i][j]);
if (j < n - 1)
Console.Write(", ");
else
Console.WriteLine("]");
}
mcoo.OptimalMatrixChainOrder(dimsList[i]);
Console.Write("Order : ");
mcoo.PrintOptimalChainOrder(0, n - 2);
Console.WriteLine("\nCost : {0}\n", mcoo.m[0, n - 2]);
}
}
} |
http://rosettacode.org/wiki/Maze_solving | Maze solving | Task
For a maze generated by this task, write a function
that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths,
and a simple depth-first tree search can be used.
| #Haskell | Haskell | #!/usr/bin/runhaskell
import Data.Maybe (fromMaybe)
-- given two points, returns the average of them
average :: (Int, Int) -> (Int, Int) -> (Int, Int)
average (x, y) (x_, y_) = ((x + x_) `div` 2, (y + y_) `div` 2)
-- given a maze and a tuple of position and wall position, returns
-- true if the wall position is not blocked (first position is unused)
notBlocked :: [String] -> ((Int, Int), (Int, Int)) -> Bool
notBlocked maze (_, (x, y)) = ' ' == (maze !! y) !! x
-- given a list, a position, and an element, returns a new list
-- with the new element substituted at the position
-- (it seems such a function should exist in the standard library;
-- I must be missing it)
substitute :: [a] -> Int -> a -> [a]
substitute orig pos el =
let (before, after) = splitAt pos orig
in before ++ [el] ++ tail after
-- given a maze and a position, draw a '*' at that position in the maze
draw :: [String] -> (Int, Int) -> [String]
draw maze (x, y) =
let row = maze !! y
in substitute maze y $ substitute row x '*'
-- given a maze, a previous position, and a list of tuples of potential
-- new positions and their wall positions, returns the solved maze, or
-- None if it cannot be solved
tryMoves :: [String]
-> (Int, Int)
-> [((Int, Int), (Int, Int))]
-> Maybe [String]
tryMoves _ _ [] = Nothing
tryMoves maze prevPos ((newPos, wallPos):more) =
case solve_ maze newPos prevPos of
Nothing -> tryMoves maze prevPos more
Just maze_ -> Just $ foldl draw maze_ [newPos, wallPos]
-- given a maze, a new position, and a previous position, returns
-- the solved maze, or None if it cannot be solved
-- (assumes goal is upper-left corner of maze)
solve_ :: [String] -> (Int, Int) -> (Int, Int) -> Maybe [String]
solve_ maze (2, 1) _ = Just maze
solve_ maze pos@(x, y) prevPos =
let newPositions = [(x, y - 2), (x + 4, y), (x, y + 2), (x - 4, y)]
notPrev pos_ = pos_ /= prevPos
newPositions_ = filter notPrev newPositions
wallPositions = map (average pos) newPositions_
zipped = zip newPositions_ wallPositions
legalMoves = filter (notBlocked maze) zipped
in tryMoves maze pos legalMoves
-- given a maze, returns a solved maze, or None if it cannot be solved
-- (starts at lower right corner and goes to upper left corner)
solve :: [String] -> Maybe [String]
solve maze = solve_ (draw maze start) start (-1, -1)
where
startx = length (head maze) - 3
starty = length maze - 2
start = (startx, starty)
-- takes unsolved maze on standard input, prints solved maze on standard output
main =
let main_ = unlines . fromMaybe ["can_t solve"] . solve . lines
in interact main_
|
http://rosettacode.org/wiki/Maximum_triangle_path_sum | Maximum triangle path sum | Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:
55
94 48
95 30 96
77 71 26 67
One of such walks is 55 - 94 - 30 - 26.
You can compute the total of the numbers you have seen in such walk,
in this case it's 205.
Your problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.
Task
Find the maximum total in the triangle below:
55
94 48
95 30 96
77 71 26 67
97 13 76 38 45
07 36 79 16 37 68
48 07 09 18 70 26 06
18 72 79 46 59 79 29 90
20 76 87 11 32 07 07 49 18
27 83 58 35 71 11 25 57 29 85
14 64 36 96 27 11 58 56 92 18 55
02 90 03 60 48 49 41 46 33 36 47 23
92 50 48 02 36 59 42 79 72 20 82 77 42
56 78 38 80 39 75 02 71 66 66 01 03 55 72
44 25 67 84 71 67 11 61 40 57 58 89 40 56 36
85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52
06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15
27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93
Such numbers can be included in the solution code, or read from a "triangle.txt" file.
This task is derived from the Euler Problem #18.
| #ERRE | ERRE |
PROGRAM TRIANGLE_PATH
CONST ROW=18
DIM TRI[200]
!
! for rosettacode,org
!
FUNCTION MAX(X,Y)
MAX=-X*(X>=Y)-Y*(X<Y)
END FUNCTION
BEGIN
DATA(55)
DATA(94,48)
DATA(95,30,96)
DATA(77,71,26,67)
DATA(97,13,76,38,45)
DATA(7,36,79,16,37,68)
DATA(48,7,9,18,70,26,6)
DATA(18,72,79,46,59,79,29,90)
DATA(20,76,87,11,32,7,7,49,18)
DATA(27,83,58,35,71,11,25,57,29,85)
DATA(14,64,36,96,27,11,58,56,92,18,55)
DATA(2,90,3,60,48,49,41,46,33,36,47,23)
DATA(92,50,48,2,36,59,42,79,72,20,82,77,42)
DATA(56,78,38,80,39,75,2,71,66,66,1,3,55,72)
DATA(44,25,67,84,71,67,11,61,40,57,58,89,40,56,36)
DATA(85,32,25,85,57,48,84,35,47,62,17,1,1,99,89,52)
DATA(6,71,28,75,94,48,37,10,23,51,6,48,53,18,74,98,15)
DATA(27,2,92,23,8,71,76,84,15,52,92,63,81,10,44,10,69,93)
PRINT(CHR$(12);) !CLS
LUNG=ROW*(ROW+1)/2
FOR I%=0 TO LUNG-1 DO
READ(TRI[I%])
END FOR
BSE=(SQR(8*LUNG+1)-1)/2
STP=BSE-1
STEPC=0
FOR I%=LUNG-BSE-1 TO 0 STEP -1 DO
TRI[I%]=TRI[I%]+MAX(TRI[I%+STP],TRI[I%+STP+1])
STEPC=STEPC+1
IF STEPC=STP THEN
STP=STP-1
STEPC=0
END IF
END FOR
PRINT(TRI[0])
END PROGRAM
|
http://rosettacode.org/wiki/MD4 | MD4 | Find the MD4 message digest of a string of octets.
Use the ASCII encoded string “Rosetta Code” (without quotes).
You may either call an MD4 library, or implement MD4 in your language.
MD4 is an obsolete hash function that computes a 128-bit message digest that sometimes appears in obsolete protocols.
RFC 1320 specifies the MD4 algorithm. RFC 6150 declares that MD4 is obsolete.
| #PARI.2FGP | PARI/GP | #include <pari/pari.h>
#include <openssl/md4.h>
#define HEX(x) (((x) < 10)? (x)+'0': (x)-10+'a')
/*
* PARI/GP func: MD4 hash
*
* gp code: install("plug_md4", "s", "MD4", "<library path>");
*/
GEN plug_md4(char *text)
{
char md[MD4_DIGEST_LENGTH];
char hash[sizeof(md) * 2 + 1];
int i;
MD4((unsigned char*)text, strlen(text), (unsigned char*)md);
for (i = 0; i < sizeof(md); i++) {
hash[i+i] = HEX((md[i] >> 4) & 0x0f);
hash[i+i+1] = HEX(md[i] & 0x0f);
}
hash[sizeof(md) * 2] = 0;
return strtoGENstr(hash);
} |
http://rosettacode.org/wiki/Middle_three_digits | Middle three digits | Task
Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.
Note: The order of the middle digits should be preserved.
Your function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:
123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345
1, 2, -1, -10, 2002, -2002, 0
Show your output on this page.
| #SQL | SQL | ;WITH DATA
AS (SELECT CAST(ABS(NUMBER) AS NVARCHAR(MAX)) charNum,
NUMBER,
LEN(CAST(ABS(NUMBER) AS NVARCHAR(MAX))) LcharNum
FROM TABLE1)
SELECT CASE
WHEN ( LCHARNUM >= 3
AND LCHARNUM % 2 = 1 ) THEN SUBSTRING(CHARNUM, LCHARNUM / 2, 3)
ELSE 'Error!'
END Output,
NUMBER INPUT
FROM DATA |
http://rosettacode.org/wiki/MD5 | MD5 | Task
Encode a string using an MD5 algorithm. The algorithm can be found on Wikipedia.
Optionally, validate your implementation by running all of the test values in IETF RFC (1321) for MD5.
Additionally, RFC 1321 provides more precise information on the algorithm than the Wikipedia article.
Warning: MD5 has known weaknesses, including collisions and forged signatures. Users may consider a stronger alternative when doing production-grade cryptography, such as SHA-256 (from the SHA-2 family), or the upcoming SHA-3.
If the solution on this page is a library solution, see MD5/Implementation for an implementation from scratch.
| #Common_Lisp | Common Lisp | (ql:quickload 'ironclad)
(defun md5 (str)
(ironclad:byte-array-to-hex-string
(ironclad:digest-sequence :md5
(ironclad:ascii-string-to-byte-array str))))
(defvar *tests* '(""
"a"
"abc"
"message digest"
"abcdefghijklmnopqrstuvwxyz"
"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
"12345678901234567890123456789012345678901234567890123456789012345678901234567890"))
(dolist (msg *tests*)
(format T "~s: ~a~%" msg (md5 msg)))
|
http://rosettacode.org/wiki/McNuggets_problem | McNuggets problem | Wikipedia
The McNuggets version of the coin problem was introduced by Henri Picciotto,
who included it in his algebra textbook co-authored with Anita Wah. Picciotto
thought of the application in the 1980s while dining with his son at
McDonald's, working the problem out on a napkin. A McNugget number is
the total number of McDonald's Chicken McNuggets in any number of boxes.
In the United Kingdom, the original boxes (prior to the introduction of
the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.
Task
Calculate (from 0 up to a limit of 100) the largest non-McNuggets
number (a number n which cannot be expressed with 6x + 9y + 20z = n
where x, y and z are natural numbers).
| #Nim | Nim | const Limit = 100
var mcnuggets: array[0..Limit, bool]
for a in countup(0, Limit, 6):
for b in countup(a, Limit, 9):
for c in countup(b, Limit, 20):
mcnuggets[c] = true
for n in countdown(Limit, 0):
if not mcnuggets[n]:
echo "The largest non-McNuggets number is: ", n
break |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #jq | jq | def m:
{ ul: "╔",
uc: "╦",
ur: "╗",
ll: "╚",
lc: "╩",
lr: "╝",
hb: "═",
vb: "║",
m0: " Θ ",
m5: "────"
};
def mayan: [
" ",
" ∙ ",
" ∙∙ ",
"∙∙∙ ",
"∙∙∙∙"
];
# decimal number to base-20 array, most significant digit first
def dec2vig:
[recurse(if . > 0 then ./20|floor else empty end) | . % 20]
| if length > 1 then .[:-1] else . end
| reverse;
def vig2quin:
if . >= 20 then "Cannot convert a number >= 20." | error
else . as $n
| [mayan[0], mayan[0], mayan[0], mayan[0]]
| if $n == 0
then .[3] = m.m0
else (($n/5)|floor) as $fives
| ($n % 5) as $rem
| .[3-$fives] = mayan[$rem]
| reduce range(3; 3-$fives; -1) as $i (.; .[$i] = m.m5)
end
end;
def draw($mayans):
($mayans|length) as $lm
| (reduce range(0; $lm) as $i (m.ul;
. + m.hb * 4 + if $i < $lm - 1 then m.uc else m.ur + "\n" end ) )
+ (reduce range(1; 5) as $i ("";
. + m.vb
+ (reduce range(0; $lm) as $j ("";
. + $mayans[$j][$i-1] + m.vb)) + "\n" ) )
+ (reduce range(0; $lm) as $i (m.ll;
. + (m.hb * 4)
+ if ($i < $lm - 1) then m.lc else m.lr + "\n" end ) );
def decimal2mayan:
"Converting \(.) to Mayan:",
draw(dec2vig | map( vig2quin ) ),
"" ;
4005, 8017, 326205, 886205, 1081439556
| decimal2mayan |
http://rosettacode.org/wiki/Mayan_numerals | Mayan numerals | Task
Present numbers using the Mayan numbering system (displaying the Mayan numerals in a cartouche).
Mayan numbers
Normally, Mayan numbers are written vertically (top─to─bottom) with the most significant
numeral at the top (in the sense that decimal numbers are written left─to─right with the most significant
digit at the left). This task will be using a left─to─right (horizontal) format, mostly for familiarity and
readability, and to conserve screen space (when showing the output) on this task page.
Mayan numerals
Mayan numerals (a base─20 "digit" or glyph) are written in two orientations, this
task will be using the "vertical" format (as displayed below). Using the vertical format makes
it much easier to draw/construct the Mayan numerals (glyphs) with simple dots (.)
and hyphen (-); (however, round bullets (•) and long dashes (─)
make a better presentation on Rosetta Code).
Furthermore, each Mayan numeral (for this task) is to be displayed as a
cartouche (enclosed in a box) to make it easier to parse (read); the box may be
drawn with any suitable (ASCII or Unicode) characters that are presentable/visible in all web browsers.
Mayan numerals added to Unicode
Mayan numerals (glyphs) were added to the Unicode Standard in June of 2018 (this corresponds with
version 11.0). But since most web browsers don't support them at this time, this Rosetta Code
task will be constructing the glyphs with "simple" characters and/or ASCII art.
The "zero" glyph
The Mayan numbering system has the concept of zero, and should be shown by a glyph that represents
an upside─down (sea) shell, or an egg. The Greek letter theta (Θ) can be
used (which more─or─less, looks like an
egg). A commercial at symbol (@) could make a poor substitute.
Mayan glyphs (constructed)
The Mayan numbering system is
a [vigesimal (base 20)] positional numeral system.
The Mayan numerals (and some random numbers) shown in the vertical format would be shown as
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙ ║ ║ ║ ║
1──► ║ ║ 11──► ║────║ 21──► ║ ║ ║
║ ∙ ║ ║────║ ║ ∙ ║ ∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║ ∙∙ ║ ║ ║ ║
2──► ║ ║ 12──► ║────║ 22──► ║ ║ ║
║ ∙∙ ║ ║────║ ║ ∙ ║ ∙∙ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙ ║ ║ ║ ║
3──► ║ ║ 13──► ║────║ 40──► ║ ║ ║
║∙∙∙ ║ ║────║ ║ ∙∙ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║∙∙∙∙║ ║ ║ ║
4──► ║ ║ 14──► ║────║ 80──► ║ ║ ║
║∙∙∙∙║ ║────║ ║∙∙∙∙║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
5──► ║ ║ 15──► ║────║ 90──► ║ ║────║
║────║ ║────║ ║∙∙∙∙║────║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
6──► ║ ∙ ║ 16──► ║────║ 100──► ║ ║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║ ∙∙ ║ ║ ║ ║
║ ║ ║────║ ║ ║ ║
7──► ║ ∙∙ ║ 17──► ║────║ 200──► ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╗
║ ║ ║∙∙∙ ║ ║ ║ ║
║ ║ ║────║ 300──► ║────║ ║
8──► ║∙∙∙ ║ 18──► ║────║ ║────║ ║
║────║ ║────║ ║────║ Θ ║
╚════╝ ╚════╝ ╚════╩════╝
╔════╗ ╔════╗ ╔════╦════╦════╗
║ ║ ║∙∙∙∙║ ║ ║ ║ ║
║ ║ ║────║ 400──► ║ ║ ║ ║
9──► ║∙∙∙∙║ 19──► ║────║ ║ ║ ║ ║
║────║ ║────║ ║ ∙ ║ Θ ║ Θ ║
╚════╝ ╚════╝ ╚════╩════╩════╝
╔════╗ ╔════╦════╗ ╔════╦════╦════╦════╗
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
║ ║ ║ ║ ║ ║ ║ ║ ║ ║
10──► ║────║ 20──► ║ ║ ║ 16,000──► ║ ║ ║ ║ ║
║────║ ║ ∙ ║ Θ ║ ║ ∙∙ ║ Θ ║ Θ ║ Θ ║
╚════╝ ╚════╩════╝ ╚════╩════╩════╩════╝
Note that the Mayan numeral 13 in horizontal format would be shown as:
╔════╗
║ ││║
║ ∙││║
13──► ║ ∙││║ ◄─── this glyph form won't be used in this Rosetta Code task.
║ ∙││║
╚════╝
Other forms of cartouches (boxes) can be used for this task.
Task requirements
convert the following decimal numbers to Mayan numbers:
4,005
8,017
326,205
886,205
show a unique interesting/pretty/unusual/intriguing/odd/amusing/weird Mayan number
show all output here
Related tasks
Roman numerals/Encode ─── convert numeric values into Roman numerals
Roman numerals/Decode ─── convert Roman numerals into Arabic numbers
See also
The Wikipedia entry: [Mayan numerals]
| #Julia | Julia | using Gumbo
mayan_glyphs(x, y) = (x == 0 && y == 0) ? "\n<td>Θ</td>\n" : "<td>\n" * "●" ^ x * "<br />\n───" ^ y * "</td>\n"
inttomayan(n) = (s = string(n, base=20); map(ch -> reverse(divrem(parse(Int, ch, base=20), 5)), split(s, "")))
function testmayan()
startstring = """\n
<style>
table.roundedcorners {
border: 1px solid DarkOrange;
border-radius: 13px;
border-spacing: 1;
}
table.roundedcorners td,
table.roundedcorners th {
border: 2px solid DarkOrange;
border-radius: 13px;
border-bottom: 3px solid DarkOrange;
vertical-align: bottom;
text-align: center;
padding: 10px;
}
</style>
\n"""
txt = startstring
for n in [4005, 8017, 326205, 886205, 70913241, 2147483647]
txt *= "<h3>The Mayan representation for the integer $n is: </h3><table class=\"roundedcorners\"><tr>" *
join(map(x -> mayan_glyphs(x[1], x[2]), inttomayan(n))) * "</tr></table>\n\n"
end
println(parsehtml(txt))
end
testmayan()
|
http://rosettacode.org/wiki/Maze_generation | Maze generation |
This page uses content from Wikipedia. The original article was at Maze generation algorithm. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Generate and show a maze, using the simple Depth-first search algorithm.
Start at a random cell.
Mark the current cell as visited, and get a list of its neighbors. For each neighbor, starting with a randomly selected neighbor:
If that neighbor hasn't been visited, remove the wall between this cell and that neighbor, and then recurse with that neighbor as the current cell.
Related tasks
Maze solving.
| #APL | APL |
This example shows how to use GNU APL scripting.
#!/usr/local/bin/apl --script --
⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝
⍝ ⍝
⍝ mazeGen.apl 2022-01-07 19:47:35 (GMT-8) ⍝
⍝ ⍝
⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝
∇initPRNG
⍝⍝ Seed the internal PRNG used by APL ? operator
⎕RL ← +/ ⎕TS ⍝⍝ Not great... but good enough
∇
∇offs ← cellTo dir
⍝⍝ Return the offset (row col) to cell which lies in compass (dir)
offs ← ∊((¯1 0)(0 1)(1 0)(0 ¯1))[('nesw'⍳dir)]
∇
∇doMaze rc
⍝⍝ Main function
0 0 mazeGen rc ⍝⍝ Do the maze gen
m ⍝⍝ output result
∇
∇b ← m isVisited coord;mr;mc
→( ∨/ (coord[1] < 1) (coord[2] < 1) )/yes
→( ∨/ (coord > ⌊(⍴m)÷2) )/yes
b ← ' ' ∊ m[2×coord[1];2×coord[2]]
→0
yes:
b←1
∇
∇c mazeGen sz ;dirs;c;dir;cell;next
→(c≠(0 0))/gen
init:
c ← ?sz[1],?sz[2]
m ← mazeInit sz
gen:
cell ← c
dirs ← 'nesw'[4?4]
m[2×c[1];2×c[2]] ← ' ' ⍝ mark cell as visited
dir1:
dir ← dirs[1]
next ← cell + cellTo dir
→(m isVisited next)/dir2
m ← m openWall cell dir
next mazeGen sz
dir2:
dir ← dirs[2]
next ← cell + cellTo dir
→(m isVisited next)/dir3
m ← m openWall cell dir
next mazeGen sz
dir3:
dir ← dirs[3]
next ← cell + cellTo dir
→(m isVisited next)/dir4
m ← m openWall cell dir
next mazeGen sz
dir4:
dir ← dirs[4]
next ← cell + cellTo dir
→(m isVisited next)/done
m ← m openWall cell dir
next mazeGen sz
done:
∇
∇m ← mazeInit sz;rows;cols;r
⍝⍝ Init an ASCII grid which
⍝⍝ has all closed and unvisited cells:
⍝⍝
⍝⍝ +-+
⍝⍝ |.|
⍝⍝ +-+
⍝⍝
⍝⍝ @param sz - tuple (rows cols)
⍝⍝ @return m - ASCII representation of (rows × cols) closed maze cells
⍝⍝⍝⍝
initPRNG
(rows cols) ← sz
r ← ∊ (cols ⍴ ⊂"+-" ),"+"
r ← r,∊ (cols ⍴ ⊂"|." ),"|"
r ← (rows,(⍴r))⍴r
r ← ((2×rows),(1+2×cols))⍴r
r ← r⍪ (∊ (cols ⍴ ⊂"+-" ),"+")
m ← r
∇
∇r ← m openWall cellAndDir ;ri;ci;rw;cw;row;col;dir
(row col dir) ← ∊cellAndDir
ri ← 2×row
ci ← 2×col
(rw cw) ← (ri ci) + cellTo dir
m[rw;cw] ← ' ' ⍝ open wall in (dir)
r ← m
∇
⎕IO←1
doMaze 9 9
)OFF
|
http://rosettacode.org/wiki/Matrix-exponentiation_operator | Matrix-exponentiation operator | Most programming languages have a built-in implementation of exponentiation for integers and reals only.
Task
Demonstrate how to implement matrix exponentiation as an operator.
| #C.2B.2B | C++ | #include <complex>
#include <cmath>
#include <iostream>
using namespace std;
template<int MSize = 3, class T = complex<double> >
class SqMx {
typedef T Ax[MSize][MSize];
typedef SqMx<MSize, T> Mx;
private:
Ax a;
SqMx() { }
public:
SqMx(const Ax &_a) { // constructor with pre-defined array
for (int r = 0; r < MSize; r++)
for (int c = 0; c < MSize; c++)
a[r][c] = _a[r][c];
}
static Mx identity() {
Mx m;
for (int r = 0; r < MSize; r++)
for (int c = 0; c < MSize; c++)
m.a[r][c] = (r == c ? 1 : 0);
return m;
}
friend ostream &operator<<(ostream& os, const Mx &p)
{ // ugly print
for (int i = 0; i < MSize; i++) {
for (int j = 0; j < MSize; j++)
os << p.a[i][j] << ',';
os << endl;
}
return os;
}
Mx operator*(const Mx &b) {
Mx d;
for (int r = 0; r < MSize; r++)
for (int c = 0; c < MSize; c++) {
d.a[r][c] = 0;
for (int k = 0; k < MSize; k++)
d.a[r][c] += a[r][k] * b.a[k][c];
}
return d;
} |
http://rosettacode.org/wiki/Matrix_digital_rain | Matrix digital rain | Implement the Matrix Digital Rain visual effect from the movie "The Matrix" as described in Wikipedia.
Provided is a reference implementation in Common Lisp to be run in a terminal.
| #Batch_File | Batch File | :: Matrix Digital Rain Task from RosettaCode
:: Batch File Implementation
@echo off
setlocal enabledelayedexpansion
rem escape character (for Windows 10 VT100 escape sequences)
rem info: https://docs.microsoft.com/en-us/windows/console/console-virtual-terminal-sequences
for /f %%e in ('echo prompt $e^| cmd') do @set "esc=%%e"
rem set window size
set "col=120" %== please don't make this too large ==%
set "row=30" %== please don't make this too large ==%
mode con cols=%col% lines=%row%
rem set up the variables for display
set "rain_length=12"
for /l %%y in (1,1,%col%) do set "disp_col[%%y]= " %== what to display ==%
for /l %%y in (1,1,%col%) do set "ctr_col[%%y]=0" %== counter for rain length ==%
rem hide the cursor, and clear the screen
<nul set /p "=%esc%[?25l"
cls
:matrix_loop
for /l %%y in (1,1,%col%) do (
if !ctr_col[%%y]! equ 0 (
set "disp_col[%%y]= "
) else (
set /a "rnd_digit=!random! %% 10"
if !ctr_col[%%y]! equ 1 (
set "disp_col[%%y]=%esc%[97m!rnd_digit!%esc%[32m"
) else if !ctr_col[%%y]! equ 2 (
set "disp_col[%%y]=%esc%[92m!rnd_digit!%esc%[32m"
) else (
set "disp_col[%%y]=!rnd_digit!"
)
set /a "ctr_col[%%y]=(!ctr_col[%%y]! + 1) %% (%rain_length% + 1)"
)
rem drop rain randomly
set /a "rnd_drop=!random! %% 20"
if !rnd_drop! equ 0 set "ctr_col[%%y]=1"
)
set "disp_line=%esc%[32m"
for /l %%y in (1,1,%col%) do set "disp_line=!disp_line!!disp_col[%%y]!"
<nul set /p "=%esc%[1T%esc%[1;1H" %== scroll down and set cursor position to home ==%
echo(%disp_line%
goto matrix_loop |
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