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/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #MiniScript | MiniScript | print mouse.x, mouse.y |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Nanoquery | Nanoquery | import Nanoquery.Util.Windows
// a function to handle the mouse moved event
def mouse_moved($caller, $e)
println "(" + $e.getX() + ", " + $e.getY() + ")"
end
// create a window, set the handler for mouse moved, and show it
w = new("Window")
w.setSize(500, 500)
w.setHandler(w.mouseMoved, mouse_moved)
w.show() |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #IS-BASIC | IS-BASIC | 100 PROGRAM "NQueens.bas"
110 TEXT 80
120 DO
130 INPUT PROMPT "Size of board (2-12): ":N$
140 LET N=VAL(N$)
150 LOOP UNTIL N>1 AND N<13
160 NUMERIC A(1 TO N),X(1 TO N),B(2 TO 2*N),C(-N+1 TO N-1)
170 LET SOL=0
180 CALL INIT(A):CALL INIT(B):CALL INIT(C)
190 CALL TRY(1)
200 PRINT SOL;"solutions."
210 END
220 DEF WRITE
230 LET S$="":LET SOL=SOL+1
240 FOR K=1 TO N
250 LET S$=S$&CHR$(64+K)&STR$(X(K))&" "
260 NEXT
270 PRINT S$
280 END DEF
290 DEF TRY(I)
300 NUMERIC J
310 FOR J=1 TO N
320 IF A(J) AND B(I+J) AND C(I-J) THEN
330 LET X(I)=J:LET A(J),B(I+J),C(I-J)=0
340 IF I<N THEN
350 CALL TRY(I+1)
360 ELSE
370 CALL WRITE
380 END IF
390 LET A(J),B(I+J),C(I-J)=1
400 END IF
410 NEXT
420 END DEF
430 DEF INIT(REF T)
440 FOR I=LBOUND(T) TO UBOUND(T)
450 LET T(I)=1
460 NEXT
470 END DEF |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Phix | Phix | constant ordinals = {"th","st","nd","rd"}
function Nth(integer n, bool apostrophe=false)
integer mod10 = mod(n,10)+1
if mod10>4 or mod(n,100)=mod10+9 then mod10 = 1 end if
return sprintf("%d%s",{n,repeat('\'',apostrophe)&ordinals[mod10]})
end function
constant ranges = {{0,25},{250,265},{1000,1025}}
for i=1 to length(ranges) do
for j=ranges[i][1] to ranges[i][2] do
if mod(j,10)=0 then puts(1,"\n") end if
printf(1," %6s",{Nth(j,i=2)})
end for
puts(1,"\n")
end for
|
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Plain_English | Plain English | To run:
Start up.
Show the Munchausen numbers up to 5000.
Wait for the escape key.
Shut down.
To show the Munchausen numbers up to a number:
If a counter is past the number, exit.
If the counter is Munchausen, convert the counter to a string; write the string to the console.
Repeat.
To decide if a number is Munchausen:
Privatize the number.
Find the sum of the digit self powers of the number.
If the number is the original number, say yes.
Say no.
To find the sum of the digit self powers of a number:
If the number is 0, exit.
Put 0 into a sum number.
Loop.
Divide the number by 10 giving a quotient and a remainder.
Put the quotient into the number.
Raise the remainder to the remainder.
Add the remainder to the sum.
If the number is 0, break.
Repeat.
Put the sum into the number. |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Julia | Julia | F(n) = n < 1 ? one(n) : n - M(F(n - 1))
M(n) = n < 1 ? zero(n) : n - F(M(n - 1)) |
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Julia | Julia | struct maybe x::Union{Real, Missing}; end
Base.show(io::IO, m::maybe) = print(io, m.x)
unit(x) = maybe(x)
bind(f, x) = unit(f(x.x))
f1(x) = 5x
f2(x) = x + 4
a = unit(3)
b = unit(missing)
println(a, " -> ", bind(f2, bind(f1, a)))
println(b, " -> ", bind(f2, bind(f1, b)))
|
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Kotlin | Kotlin | // version 1.2.10
import java.util.Optional
/* doubles 'i' before wrapping it */
fun getOptionalInt(i: Int) = Optional.of(2 * i)
/* returns an 'A' repeated 'i' times wrapped in an Optional<String> */
fun getOptionalString(i: Int) = Optional.of("A".repeat(i))
/* does same as above if i > 0, otherwise returns an empty Optional<String> */
fun getOptionalString2(i: Int) =
Optional.ofNullable(if (i > 0) "A".repeat(i) else null)
fun main(args: Array<String>) {
/* prints 10 'A's */
println(getOptionalInt(5).flatMap(::getOptionalString).get())
/* prints 4 'A's */
println(getOptionalInt(2).flatMap(::getOptionalString2).get())
/* prints 'false' as there is no value present in the Optional<String> instance */
println(getOptionalInt(0).flatMap(::getOptionalString2).isPresent)
} |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <math.h>
double pi(double tolerance)
{
double x, y, val, error;
unsigned long sampled = 0, hit = 0, i;
do {
/* don't check error every turn, make loop tight */
for (i = 1000000; i; i--, sampled++) {
x = rand() / (RAND_MAX + 1.0);
y = rand() / (RAND_MAX + 1.0);
if (x * x + y * y < 1) hit ++;
}
val = (double) hit / sampled;
error = sqrt(val * (1 - val) / sampled) * 4;
val *= 4;
/* some feedback, or user gets bored */
fprintf(stderr, "Pi = %f +/- %5.3e at %ldM samples.\r",
val, error, sampled/1000000);
} while (!hit || error > tolerance);
/* !hit is for completeness's sake; if no hit after 1M samples,
your rand() is BROKEN */
return val;
}
int main()
{
printf("Pi is %f\n", pi(3e-4)); /* set to 1e-4 for some fun */
return 0;
} |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #MATLAB | MATLAB | function testMTF
symTable = 'abcdefghijklmnopqrstuvwxyz';
inStr = {'broood' 'bananaaa' 'hiphophiphop'};
for k = 1:length(inStr)
outArr = encodeMTF(inStr{k}, symTable);
outStr = decodeMTF(outArr, symTable);
fprintf('%s: [ %s]\n', inStr{k}, sprintf('%d ', outArr))
fprintf('%scorrectly decoded to %s\n', char('in'.*~strcmp(outStr, inStr{k})), outStr)
end
end
function arr = encodeMTF(str, symTable)
n = length(str);
arr = zeros(1, n);
for k = 1:n
arr(k) = find(str(k) == symTable, 1);
symTable = [symTable(arr(k)) symTable(1:arr(k)-1) symTable(arr(k)+1:end)];
end
arr = arr-1; % Change to zero-indexed array
end
function str = decodeMTF(arr, symTable)
arr = arr+1; % Change to one-indexed array
n = length(arr);
str = char(zeros(1, n));
for k = 1:n
str(k) = symTable(arr(k));
symTable = [symTable(arr(k)) symTable(1:arr(k)-1) symTable(arr(k)+1:end)];
end
end |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #AWK | AWK | # usage: awk -f morse.awk [inputfile]
BEGIN { FS="";
m="A.-B-...C-.-.D-..E.F..-.G--.H....I..J.---K-.-L.-..M--N-.";
m=m "O---P.--.Q--.-R.-.S...T-U..-V...-W.--X-..-Y-.--Z--.. ";
}
{ for(i=1; i<=NF; i++)
{
c=toupper($i); n=1; b=".";
while((c!=b)&&(b!=" ")) { b=substr(m,n,1); n++; }
b=substr(m,n,1);
while((b==".")||(b=="-")) { printf("%s",b); n++; b=substr(m,n,1); }
printf("|");
}
printf("\n");
}
|
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #AutoHotkey | AutoHotkey | #SingleInstance, Force
Iterations = 1000
Loop, %Iterations%
{
If Monty_Hall(1)
Correct_Change++
Else
Incorrect_Change++
If Monty_Hall(2)
Correct_Random++
Else
Incorrect_Random++
If Monty_Hall(3)
Correct_Stay++
Else
Incorrect_Stay++
}
Percent_Change := round(Correct_Change / Iterations * 100)
Percent_Stay := round(Correct_Stay / Iterations * 100)
Percent_Random := round(Correct_Random / Iterations * 100)
MsgBox,, Monty Hall Problem, These are the results:`r`n`r`nWhen I changed my guess, I got %Correct_Change% of %Iterations% (that's %Incorrect_Change% incorrect). That's %Percent_Change%`% correct.`r`n`r`nWhen I randomly changed my guess, I got %Correct_Random% of %Iterations% (that's %Incorrect_Random% incorrect). That's %Percent_Random%`% correct.`r`n`r`nWhen I stayed with my first guess, I got %Correct_Stay% of %Iterations% (that's %Incorrect_Stay% incorrect). That's %Percent_Stay%`% correct.
ExitApp
Monty_Hall(Mode) ;Mode is 1 for change, 2 for random, or 3 for stay
{
Random, guess, 1, 3
Random, actual, 1, 3
Random, rand, 1, 2
show := guess = actual ? guess = 3 ? guess - rand : guess = 1 ? guess+rand : guess + 2*rand - 3 : 6 - guess - actual
Mode := Mode = 2 ? 2*rand - 1: Mode
Return, Mode = 1 ? 6 - guess - show = actual : guess = actual
} |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #C.2B.2B | C++ | #include <iostream>
int mul_inv(int a, int b)
{
int b0 = b, t, q;
int x0 = 0, x1 = 1;
if (b == 1) return 1;
while (a > 1) {
q = a / b;
t = b, b = a % b, a = t;
t = x0, x0 = x1 - q * x0, x1 = t;
}
if (x1 < 0) x1 += b0;
return x1;
}
int main(void) {
std::cout << mul_inv(42, 2017) << std::endl;
return 0;
} |
http://rosettacode.org/wiki/Monads/Writer_monad | Monads/Writer monad | The Writer monad is a programming design pattern which makes it possible to compose functions which return their result values paired with a log string. The final result of a composed function yields both a value, and a concatenation of the logs from each component function application.
Demonstrate in your programming language the following:
Construct a Writer monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that monad (or just use what the language already provides)
Write three simple functions: root, addOne, and half
Derive Writer monad versions of each of these functions
Apply a composition of the Writer versions of root, addOne, and half to the integer 5, deriving both a value for the Golden Ratio φ, and a concatenated log of the function applications (starting with the initial value, and followed by the application of root, etc.) | #zkl | zkl | class Writer{
fcn init(x){ var X=x, logText=Data(Void," init \U2192; ",x.toString()) }
fcn unit(text) { logText.append(text); self }
fcn lift(f,name){ unit("\n %s \U2192; %s".fmt(name,X=f(X))) }
fcn bind(f,name){ lift.fp(f,name) }
fcn toString{ "Result = %s\n%s".fmt(X,logText.text) }
fcn root{ lift(fcn(x){ x.sqrt() },"root") }
fcn half{ lift('/(2),"half") }
fcn inc { lift('+(1),"inc") }
} |
http://rosettacode.org/wiki/Monads/List_monad | Monads/List monad | A Monad is a combination of a data-type with two helper functions written for that type.
The data-type can be of any kind which can contain values of some other type – common examples are lists, records, sum-types, even functions or IO streams. The two special functions, mathematically known as eta and mu, but usually given more expressive names like 'pure', 'return', or 'yield' and 'bind', abstract away some boilerplate needed for pipe-lining or enchaining sequences of computations on values held in the containing data-type.
The bind operator in the List monad enchains computations which return their values wrapped in lists. One application of this is the representation of indeterminacy, with returned lists representing a set of possible values. An empty list can be returned to express incomputability, or computational failure.
A sequence of two list monad computations (enchained with the use of bind) can be understood as the computation of a cartesian product.
The natural implementation of bind for the List monad is a composition of concat and map, which, used with a function which returns its value as a (possibly empty) list, provides for filtering in addition to transformation or mapping.
Demonstrate in your programming language the following:
Construct a List Monad by writing the 'bind' function and the 'pure' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> List Int and Int -> List String
Compose the two functions with bind | #uBasic.2F4tH | uBasic/4tH | s := "[" : Push 5, 4, 3
Do While Used ()
y = Set (x, Pop ()) + 1
s = Join (s, Str (Set (z, y * 2)), ", " )
Loop
Print Show (Set (s, Join (Clip (s, 2), "]")))
|
http://rosettacode.org/wiki/Monads/List_monad | Monads/List monad | A Monad is a combination of a data-type with two helper functions written for that type.
The data-type can be of any kind which can contain values of some other type – common examples are lists, records, sum-types, even functions or IO streams. The two special functions, mathematically known as eta and mu, but usually given more expressive names like 'pure', 'return', or 'yield' and 'bind', abstract away some boilerplate needed for pipe-lining or enchaining sequences of computations on values held in the containing data-type.
The bind operator in the List monad enchains computations which return their values wrapped in lists. One application of this is the representation of indeterminacy, with returned lists representing a set of possible values. An empty list can be returned to express incomputability, or computational failure.
A sequence of two list monad computations (enchained with the use of bind) can be understood as the computation of a cartesian product.
The natural implementation of bind for the List monad is a composition of concat and map, which, used with a function which returns its value as a (possibly empty) list, provides for filtering in addition to transformation or mapping.
Demonstrate in your programming language the following:
Construct a List Monad by writing the 'bind' function and the 'pure' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> List Int and Int -> List String
Compose the two functions with bind | #Wren | Wren | class Mlist {
construct new(value) { _value = value }
value { _value }
bind(f) { f.call(_value) }
static unit(lst) { Mlist.new(lst) }
}
var increment = Fn.new { |lst|
var lst2 = lst.map { |v| v + 1 }.toList
return Mlist.unit(lst2)
}
var double = Fn.new { |lst|
var lst2 = lst.map { |v| v * 2 }.toList
return Mlist.unit(lst2)
}
var ml1 = Mlist.unit([3, 4, 5])
var ml2 = ml1.bind(increment).bind(double)
System.print("%(ml1.value) -> %(ml2.value)") |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Tcl | Tcl | proc multilist {value args} {
set res $value
foreach dim [lreverse $args] {
set res [lrepeat $dim $res]
}
return $res
} |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Vlang | Vlang | smd := [][]string{len: 4, init: []string{len:2, init: 'Hello'}}
imd := [][]int{len: 3, init: []int{len:4, init: it}}
mut mmd := [][]f64{len: 5, init: []f64{len: 5}}
mmd[0][2] = 2.0
mmd[1][3] = 4.2
mmd[2][4] = 1.8
mmd[3][0] = 5.0
mut omd := [][]bool{} // initialize without defining size
omd << [true, false, true]
println(smd)
println(imd)
println(mmd)
println(omd) |
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #AutoHotkey | AutoHotkey | Gui, -MinimizeBox
Gui, Margin, 0, 0
Gui, Font, s9, Fixedsys
Gui, Add, Edit, h0 w0
Gui, Add, Edit, w432 r14 -VScroll
Gosub, Table
Gui, Show,, Multiplication Table
Return
GuiClose:
GuiEscape:
ExitApp
Return
Table:
; top row
Table := " x |"
Loop, 12
Table .= SubStr(" " A_Index, -3)
Table .= "`n"
; underlines
Table .= "----+"
Loop, 48
Table .= "-"
Table .= "`n"
; table
Loop, 12 { ; rows
Table .= SubStr(" " Row := A_Index, -2) " |"
Loop, 12 ; columns
Table .= SubStr(" " (A_Index >= Row ? A_Index * Row : ""), -3)
Table .= "`n"
}
GuiControl,, Edit2, %Table%
Return |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Raku | Raku | use Clifford;
my @height = <1.47 1.50 1.52 1.55 1.57 1.60 1.63 1.65 1.68 1.70 1.73 1.75 1.78 1.80 1.83>;
my @weight = <52.21 53.12 54.48 55.84 57.20 58.57 59.93 61.29 63.11 64.47 66.28 68.10 69.92 72.19 74.46>;
my $w = [+] @weight Z* @e;
my $h0 = [+] @e[^@weight];
my $h1 = [+] @height Z* @e;
my $h2 = [+] (@height X** 2) Z* @e;
my $I = $h0∧$h1∧$h2;
my $I2 = ($I·$I.reversion).Real;
say "α = ", ($w∧$h1∧$h2)·$I.reversion/$I2;
say "β = ", ($w∧$h2∧$h0)·$I.reversion/$I2;
say "γ = ", ($w∧$h0∧$h1)·$I.reversion/$I2; |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Haskell | Haskell | mulfac :: (Num a, Enum a) => a -> [a]
mulfac k = 1 : s
where
s = [1 .. k] <> zipWith (*) s [k + 1 ..]
-- For single n:
mulfac1 :: (Num a, Enum a) => a -> a -> a
mulfac1 k n = product [n, n - k .. 1]
main :: IO ()
main =
mapM_
(print . take 10 . tail . mulfac)
[1 .. 5]
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Nim | Nim | import gintro/[glib, gobject, gtk, gio]
import gintro/gdk except Window
#---------------------------------------------------------------------------------------------------
proc onButtonPress(window: ApplicationWindow; event: Event; data: pointer): bool =
echo event.getCoords()
result = true
#---------------------------------------------------------------------------------------------------
proc activate(app: Application) =
## Activate the application.
let window = app.newApplicationWindow()
window.setTitle("Mouse position")
window.setSizeRequest(640, 480)
discard window.connect("button-press-event", onButtonPress, pointer(nil))
window.showAll()
#———————————————————————————————————————————————————————————————————————————————————————————————————
let app = newApplication(Application, "Rosetta.MousePosition")
discard app.connect("activate", activate)
discard app.run() |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #OCaml | OCaml | open Xlib
let () =
let d = xOpenDisplay "" in
(* ask for active window (no error check);
the client must be freedesktop compliant *)
let _, _, _, _, props =
xGetWindowProperty_window d
(xDefaultRootWindow d)
(xInternAtom d "_NET_ACTIVE_WINDOW" true)
0 1 false AnyPropertyType
in
let _, _, _, child, _ = xQueryPointer d props in
begin match child with
| Some(_, childx, childy) ->
Printf.printf "relative to active window: %d,%d\n%!" childx childy;
| None ->
print_endline "the pointer is not on the same screen as the specified window"
end;
xCloseDisplay d;
;; |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #J | J | perm =: ! A.&i. ] NB. all permutations of integers 0 to y
comb2 =: (, #: I.@,@(</)&i.)~ NB. all size 2 combinations of integers 0 to y
mask =: [ */@:~:&(|@-/) {
queenst=: comb2 (] #"1~ mask)&.|: perm |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #PHP | PHP | function nth($num) {
$os = "th";
if ($num % 100 <= 10 or $num % 100 > 20) {
switch ($num % 10) {
case 1:
$os = "st";
break;
case 2:
$os = "nd";
break;
case 3:
$os = "rd";
break;
}
}
return $num . $os;
}
foreach ([[0,25], [250,265], [1000,1025]] as $i) {
while ($i[0] <= $i[1]) {
echo nth($i[0]) . " ";
$i[0]++;
}
echo "\n";
} |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #PowerBASIC | PowerBASIC | #COMPILE EXE
#DIM ALL
#COMPILER PBCC 6
DECLARE FUNCTION GetTickCount LIB "kernel32.dll" ALIAS "GetTickCount"() AS DWORD
FUNCTION PBMAIN () AS LONG
LOCAL i, j, n, sum, ten1, ten2, t AS DWORD
LOCAL n0, n1, n2, n3, n4, n5, n6, n7, n8, n9 AS DWORD
LOCAL s1, s2, s3, s4, s5, s6, s7, s8 AS DWORD
DIM pow(9) AS DWORD, num(9) AS DWORD
LOCAL pb AS BYTE PTR
LOCAL number AS STRING
t = GetTickCount()
ten2 = 10
FOR i = 1 TO 9
pow(i) = i
FOR j = 2 TO i
pow(i) *= i
NEXT j
NEXT i
FOR n = 1 TO 11
FOR n9 = 0 TO n
FOR n8 = 0 TO n - n9
s8 = n9 + n8
FOR n7 = 0 TO n - s8
s7 = s8 + n7
FOR n6 = 0 TO n - s7
s6 = s7 + n6
FOR n5 = 0 TO n - s6
s5 = s6 + n5
FOR n4 = 0 TO n - s5
s4 = s5 + n4
FOR n3 = 0 TO n - s4
s3 = s4 + n3
FOR n2 = 0 TO n - s3
s2 = s3 + n2
FOR n1 = 0 TO n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
SELECT CASE AS LONG sum
CASE ten1 TO ten2 - 1
number = LTRIM$(STR$(sum))
pb = STRPTR(number)
MAT num() = ZER
FOR i = 0 TO n -1
j = @pb[i] - 48
INCR num(j)
NEXT i
IF n0 = num(0) AND n1 = num(1) AND n2 = num(2) AND _
n3 = num(3) AND n4 = num(4) AND n5 = num(5) AND _
n6 = num(6) AND n7 = num(7) AND n8 = num(8) AND _
n9 = num(9) THEN CON.PRINT STR$(sum)
END SELECT
NEXT n1
NEXT n2
NEXT n3
NEXT n4
NEXT n5
NEXT n6
NEXT n7
NEXT n8
NEXT n9
ten1 = ten2
ten2 *= 10
NEXT n
t = GetTickCount() - t
CON.PRINT "execution time:" & STR$(t) & " ms; hit any key to end program"
CON.WAITKEY$
END FUNCTION |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Kotlin | Kotlin | // version 1.0.6
fun f(n: Int): Int =
when {
n == 0 -> 1
else -> n - m(f(n - 1))
}
fun m(n: Int): Int =
when {
n == 0 -> 0
else -> n - f(m(n - 1))
}
fun main(args: Array<String>) {
val n = 24
print("n :")
for (i in 0..n) print("%3d".format(i))
println()
println("-".repeat(78))
print("F :")
for (i in 0..24) print("%3d".format(f(i)))
println()
print("M :")
for (i in 0..24) print("%3d".format(m(i)))
println()
} |
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Lua | Lua |
-- None is represented by an empty table. Some is represented by any
-- array with one element. You SHOULD NOT compare maybe values with the
-- Lua operator == because it will give incorrect results. Use the
-- functions isMaybe(), isNone(), and isSome().
-- define once for efficiency, to avoid many different empty tables
local NONE = {}
local function unit(x) return { x } end
local Some = unit
local function isNone(mb) return #mb == 0 end
local function isSome(mb) return #mb == 1 end
local function isMaybe(mb) return isNone(mb) or isSome(mb) end
-- inverse of Some(), extract the value from the maybe; get(Some(x)) === x
local function get(mb) return mb[1] end
function maybeToStr(mb)
return isNone(mb) and "None" or ("Some " .. tostring(get(mb)))
end
local function bind(mb, ...) -- monadic bind for multiple functions
local acc = mb
for _, fun in ipairs({...}) do -- fun should be a monadic function
assert(type(fun) == "function")
if isNone(acc) then return NONE
else acc = fun(get(acc)) end
end
return acc
end
local function fmap(mb, ...) -- monadic fmap for multiple functions
local acc = mb
for _, fun in ipairs({...}) do -- fun should be a regular function
assert(type(fun) == "function")
if isNone(acc) then return NONE
else acc = Some(fun(get(acc))) end
end
return acc
end
-- ^^^ End of generic maybe monad functionality ^^^
--- vvv Start of example code vvv
local function time2(x) return x * 2 end
local function plus1(x) return x + 1 end
local answer
answer = fmap(Some(3), time2, plus1, time2)
assert(get(answer)==14)
answer = fmap(NONE, time2, plus1, time2)
assert(isNone(answer))
local function safeReciprocal(x)
if x ~= 0 then return Some(1/x) else return NONE end
end
local function safeRoot(x)
if x >= 0 then return Some(math.sqrt(x)) else return NONE end
end
local function safeLog(x)
if x > 0 then return Some(math.log(x)) else return NONE end
end
local function safeComputation(x)
return bind(safeReciprocal(x), safeRoot, safeLog)
end
local function map(func, table)
local result = {}
for key, val in pairs(table) do
result[key] = func(val)
end
return result
end
local inList = {-2, -1, -0.5, 0, math.exp (-1), 1, 2, math.exp (1), 3, 4, 5}
print("input:", table.concat(map(tostring, inList), ", "), "\n")
local outList = map(safeComputation, inList)
print("output:", table.concat(map(maybeToStr, outList), ", "), "\n")
|
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #C.23 | C# | using System;
class Program {
static double MonteCarloPi(int n) {
int inside = 0;
Random r = new Random();
for (int i = 0; i < n; i++) {
if (Math.Pow(r.NextDouble(), 2)+ Math.Pow(r.NextDouble(), 2) <= 1) {
inside++;
}
}
return 4.0 * inside / n;
}
static void Main(string[] args) {
int value = 1000;
for (int n = 0; n < 5; n++) {
value *= 10;
Console.WriteLine("{0}:{1}", value.ToString("#,###").PadLeft(11, ' '), MonteCarloPi(value));
}
}
} |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #Nim | Nim | import algorithm, sequtils, strformat
const SymbolTable = toSeq('a'..'z')
func encode(s: string): seq[int] =
var symtable: seq[char] = SymbolTable
for c in s:
let idx = symtable.find(c)
result.add idx
symtable.rotateLeft(0..idx, -1)
func decode(s: seq[int]): string =
var symtable = SymbolTable
for idx in s:
result.add symtable[idx]
symtable.rotateLeft(0..idx, -1)
for word in ["broood", "babanaaa", "hiphophiphop"]:
let encoded = word.encode()
let decoded = encoded.decode()
let status = if decoded == word: "correctly" else: "incorrectly"
echo &"'{word}' encodes to {encoded} which {status} decodes to '{decoded}'." |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #bash | bash |
#!/bin/bash
# michaeltd 2019-11-29 https://github.com/michaeltd/dots/blob/master/dot.files/.bashrc.d/.var/morse.sh
# https://en.wikipedia.org/wiki/Morse_code
# International Morse Code
# 1. Length of dot is 1 unit
# 2. Length of dash is 3 units
# 3. The space between parts of the same letter is 1 unit
# 4. The space between letters is 3 units.
# 5. The space between words is 7 units.
################################################################################
alpha2morse() {
local -A alpha_assoc=( [A]='.-' [B]='-...' [C]='-.-.' [D]='-..' [E]='.' \
[F]='..-.' [G]='--.' [H]='....' [I]='..' [J]='.---' \
[K]='-.-' [L]='.-..' [M]='--' [N]='-.' [O]='---' \
[P]='.--.' [Q]='--.-' [R]='.-.' [S]='...' [T]='-' \
[U]='..-' [V]='...-' [W]='.--' [X]='-..-' [Y]='-.--' [Z]='--..' \
[0]='-----' [1]='.----' [2]='..---' [3]='...--' [4]='....-' \
[5]='.....' [6]='-....' [7]='--...' [8]='----..' [9]='----.' )
if [[ "${#}" -lt "1" ]]; then
echo -ne "Usage: ${FUNCNAME[0]} arguments...\n \
${FUNCNAME[0]} is an IMC transmitter. \n \
It'll transmit your messages to International Morse Code.\n" >&2
return 1
fi
while [[ -n "${1}" ]]; do
for (( i = 0; i < ${#1}; i++ )); do
local letter="${1:${i}:1}"
for (( y = 0; y < ${#alpha_assoc[${letter^^}]}; y++ )); do
case "${alpha_assoc[${letter^^}]:${y}:1}" in
".") echo -n "dot "; play -q -n -c2 synth .1 2> /dev/null || sleep .1 ;;
"-") echo -n "dash "; play -q -n -c2 synth .3 2> /dev/null || sleep .3 ;;
esac
sleep .1
done
echo
sleep .3
done
echo
sleep .7
shift
done
}
if [[ "${BASH_SOURCE[0]}" == "${0}" ]]; then
alpha2morse "${@}"
fi
|
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #AWK | AWK | #!/bin/gawk -f
# Monty Hall problem
BEGIN {
srand()
doors = 3
iterations = 10000
# Behind a door:
EMPTY = "empty"; PRIZE = "prize"
# Algorithm used
KEEP = "keep"; SWITCH="switch"; RAND="random";
#
}
function monty_hall( choice, algorithm ) {
# Set up doors
for ( i=0; i<doors; i++ ) {
door[i] = EMPTY
}
# One door with prize
door[int(rand()*doors)] = PRIZE
chosen = door[choice]
del door[choice]
#if you didn't choose the prize first time around then
# that will be the alternative
alternative = (chosen == PRIZE) ? EMPTY : PRIZE
if( algorithm == KEEP) {
return chosen
}
if( algorithm == SWITCH) {
return alternative
}
return rand() <0.5 ? chosen : alternative
}
function simulate(algo){
prizecount = 0
for(j=0; j< iterations; j++){
if( monty_hall( int(rand()*doors), algo) == PRIZE) {
prizecount ++
}
}
printf " Algorithm %7s: prize count = %i, = %6.2f%%\n", \
algo, prizecount,prizecount*100/iterations
}
BEGIN {
print "\nMonty Hall problem simulation:"
print doors, "doors,", iterations, "iterations.\n"
simulate(KEEP)
simulate(SWITCH)
simulate(RAND)
} |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Clojure | Clojure | (ns test-p.core
(:require [clojure.math.numeric-tower :as math]))
(defn extended-gcd
"The extended Euclidean algorithm--using Clojure code from RosettaCode for Extended Eucliean
(see http://en.wikipedia.orwiki/Extended_Euclidean_algorithm)
Returns a list containing the GCD and the Bézout coefficients
corresponding to the inputs with the result: gcd followed by bezout coefficients "
[a b]
(cond (zero? a) [(math/abs b) 0 1]
(zero? b) [(math/abs a) 1 0]
:else (loop [s 0
s0 1
t 1
t0 0
r (math/abs b)
r0 (math/abs a)]
(if (zero? r)
[r0 s0 t0]
(let [q (quot r0 r)]
(recur (- s0 (* q s)) s
(- t0 (* q t)) t
(- r0 (* q r)) r))))))
(defn mul_inv
" Get inverse using extended gcd. Extended GCD returns
gcd followed by bezout coefficients. We want the 1st coefficients
(i.e. second of extend-gcd result). We compute mod base so result
is between 0..(base-1) "
[a b]
(let [b (if (neg? b) (- b) b)
a (if (neg? a) (- b (mod (- a) b)) a)
egcd (extended-gcd a b)]
(if (= (first egcd) 1)
(mod (second egcd) b)
(str "No inverse since gcd is: " (first egcd)))))
(println (mul_inv 42 2017))
(println (mul_inv 40 1))
(println (mul_inv 52 -217))
(println (mul_inv -486 217))
(println (mul_inv 40 2018))
|
http://rosettacode.org/wiki/Monads/List_monad | Monads/List monad | A Monad is a combination of a data-type with two helper functions written for that type.
The data-type can be of any kind which can contain values of some other type – common examples are lists, records, sum-types, even functions or IO streams. The two special functions, mathematically known as eta and mu, but usually given more expressive names like 'pure', 'return', or 'yield' and 'bind', abstract away some boilerplate needed for pipe-lining or enchaining sequences of computations on values held in the containing data-type.
The bind operator in the List monad enchains computations which return their values wrapped in lists. One application of this is the representation of indeterminacy, with returned lists representing a set of possible values. An empty list can be returned to express incomputability, or computational failure.
A sequence of two list monad computations (enchained with the use of bind) can be understood as the computation of a cartesian product.
The natural implementation of bind for the List monad is a composition of concat and map, which, used with a function which returns its value as a (possibly empty) list, provides for filtering in addition to transformation or mapping.
Demonstrate in your programming language the following:
Construct a List Monad by writing the 'bind' function and the 'pure' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> List Int and Int -> List String
Compose the two functions with bind | #zkl | zkl | class MList{
fcn init(xs){ var list=vm.arglist }
fcn bind(f) { list=list.apply(f); self }
fcn toString{ list.toString() }
} |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Wren | Wren | import "/fmt" for Fmt
// create a 4 dimensional list of the required size and initialize successive elements to the values 1 to 120
var m = 1
var a4 = List.filled(5, null)
for (i in 0..4) {
a4[i] = List.filled(4, null)
for (j in 0..3) {
a4[i][j] = List.filled(3, null)
for (k in 0..2) {
a4[i][j][k] = List.filled(2, 0)
for (l in 0..1) {
a4[i][j][k][l] = m
m = m + 1
}
}
}
}
System.print("First element = %(a4[0][0][0][0])") // access and print value of first element
a4[0][0][0][0] = 121 // change value of first element
System.print()
// access and print values of all elements
for (i in 0..4) {
for (j in 0..3) {
for (k in 0..2) {
for (l in 0..1) {
Fmt.write("$4d", a4[i][j][k][l])
}
}
}
} |
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #AutoIt | AutoIt | #AutoIt Version: 3.2.10.0
$tableupto=12
$table=""
for $i = 1 To $tableupto
for $j = $i to $tableupto
$prod=string($i*$j)
if StringLen($prod) == 1 then
$prod = " "& $prod
EndIf
if StringLen($prod) == 2 then
$prod = " "& $prod
EndIf
$table = $table&" "&$prod
Next
$table = $table&" - "&$i&@CRLF
for $k = 1 to $i
$table = $table&" "
Next
Next
msgbox(0,"Multiplication Tables",$table) |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Ruby | Ruby | require 'matrix'
def regression_coefficients y, x
y = Matrix.column_vector y.map { |i| i.to_f }
x = Matrix.columns x.map { |xi| xi.map { |i| i.to_f }}
(x.t * x).inverse * x.t * y
end |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Icon_and_Unicon | Icon and Unicon | procedure main(A)
l := integer(A[1]) | 10
every writeRow(n := !l, [: mf(!10,n) :])
end
procedure writeRow(n, r)
writes(right(n,3),": ")
every writes(right(!r,8)|"\n")
end
procedure mf(n, m)
if n <= 0 then return 1
return n*mf(n-m, m)
end |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Octave | Octave | [X, Y, BUTTONS] = ginput(N); |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Oz | Oz | declare
[QTk] = {Module.link ['x-oz://system/wp/QTk.ozf']}
WindowClosed = {NewCell false}
Label
Window = {QTk.build
td(action:proc {$} WindowClosed := true {Window close} end
label(text:"" handle:Label))}
in
{Window show}
for while:{Not @WindowClosed} do
TopmostWindow = {List.last {String.tokens {Tk.return wm(stackorder '.')} & }}
Winfo = {Record.mapInd winfo(rootx:_ rooty:_ pointerx:_ pointery:_)
fun {$ I _}
{Tk.returnInt winfo(I TopmostWindow)}
end}
in
{Label set(text:"x: "#(Winfo.pointerx - Winfo.rootx)
#", y: "#(Winfo.pointery - Winfo.rooty))}
{Delay 250}
end |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Java | Java | public class NQueens {
private static int[] b = new int[8];
private static int s = 0;
static boolean unsafe(int y) {
int x = b[y];
for (int i = 1; i <= y; i++) {
int t = b[y - i];
if (t == x ||
t == x - i ||
t == x + i) {
return true;
}
}
return false;
}
public static void putboard() {
System.out.println("\n\nSolution " + (++s));
for (int y = 0; y < 8; y++) {
for (int x = 0; x < 8; x++) {
System.out.print((b[y] == x) ? "|Q" : "|_");
}
System.out.println("|");
}
}
public static void main(String[] args) {
int y = 0;
b[0] = -1;
while (y >= 0) {
do {
b[y]++;
} while ((b[y] < 8) && unsafe(y));
if (b[y] < 8) {
if (y < 7) {
b[++y] = -1;
} else {
putboard();
}
} else {
y--;
}
}
}
} |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Picat | Picat | nth2(N) = N.to_string() ++ Th =>
( tween(N) -> Th = "th"
; 1 = N mod 10 -> Th = "st"
; 2 = N mod 10 -> Th = "nd"
; 3 = N mod 10 -> Th = "rd"
; Th = "th" ).
tween(N) => Tween = N mod 100, between(11, 13, Tween). |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Pure | Pure | // split numer into digits
digits n::number = loop n [] with
loop n l = loop (n div 10) ((n mod 10):l) if n > 0;
= l otherwise; end;
munchausen n::int = (filter isMunchausen list) when
list = 1..n; end with
isMunchausen n = n == foldl (+) 0
(map (\d -> d^d)
(digits n)); end;
munchausen 5000; |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Lambdatalk | Lambdatalk |
{def F {lambda {:n} {if {= :n 0} then 1 else {- :n {M {F {- :n 1}}}} }}}
{def M {lambda {:n} {if {= :n 0} then 0 else {- :n {F {M {- :n 1}}}} }}}
{map F {serie 0 19}}
-> 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12
{map M {serie 0 19}}
-> 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12
|
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Nim | Nim | import options,math,sugar,strformat
template checkAndWrap(x:float,body:typed): untyped =
if x.classify in {fcNormal,fcZero,fcNegZero}:
some(body)
else:
none(typeof(body))
func reciprocal(x:float):Option[float] =
let res = 1 / x
res.checkAndWrap(res)
func log(x:float):Option[float] =
let res = ln(x)
res.checkAndWrap(res)
func format(x:float):Option[string] =
x.checkAndWrap(&"{x:.2f}")
#our bind function:
func `-->`[T,U](input:Option[T], f: T->Option[U]):Option[U] =
if input.isSome:
f(input.get)
else:
none(U)
when isMainModule:
for i in [0.9,0.0,-0.9,3.0]:
echo some(i) --> reciprocal --> log --> format |
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Perl | Perl | # 20201101 added Perl programming solution
use strict;
use warnings;
use Data::Monad::Maybe;
sub safeReciprocal { ( $_[0] == 0 ) ? nothing : just( 1 / $_[0] ) }
sub safeRoot { ( $_[0] < 0 ) ? nothing : just( sqrt( $_[0] ) ) }
sub safeLog { ( $_[0] <= 0 ) ? nothing : just( log ( $_[0] ) ) }
print join(' ', map {
my $safeLogRootReciprocal = just($_)->flat_map( \&safeReciprocal )
->flat_map( \&safeRoot )
->flat_map( \&safeLog );
$safeLogRootReciprocal->is_nothing ? "NaN" : $safeLogRootReciprocal->value;
} (-2, -1, -0.5, 0, exp (-1), 1, 2, exp(1), 3, 4, 5) ), "\n"; |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #C.2B.2B | C++ |
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<time.h>
using namespace std;
int main(){
int jmax=1000; // maximum value of HIT number. (Length of output file)
int imax=1000; // maximum value of random numbers for producing HITs.
double x,y; // Coordinates
int hit; // storage variable of number of HITs
srand(time(0));
for (int j=0;j<jmax;j++){
hit=0;
x=0; y=0;
for(int i=0;i<imax;i++){
x=double(rand())/double(RAND_MAX);
y=double(rand())/double(RAND_MAX);
if(y<=sqrt(1-pow(x,2))) hit+=1; } //Choosing HITs according to analytic formula of circle
cout<<""<<4*double(hit)/double(imax)<<endl; } // Print out Pi number
}
|
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #Perl | Perl | use strict;
use warnings;
sub encode {
my ($str) = @_;
my $table = join '', 'a' .. 'z';
map {
$table =~ s/(.*?)$_/$_$1/ or die;
length($1);
} split //, $str;
}
sub decode {
my $table = join '', 'a' .. 'z';
join "", map {
$table =~ s/(.{$_})(.)/$2$1/ or die;
$2;
} @_;
}
for my $test ( qw(broood bananaaa hiphophiphop) ) {
my @encoded = encode($test);
print "$test: @encoded\n";
my $decoded = decode(@encoded);
print "in" x ( $decoded ne $test );
print "correctly decoded to $decoded\n";
}
|
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #BASIC | BASIC | DECLARE SUB player (what AS STRING)
'this determines the length of the notes
'lower number = longer duration
CONST noteLen = 16
DIM tones(62) AS STRING
FOR n% = 0 TO 62
READ tones(n%)
NEXT n%
'set up the playing system
PLAY "t255o4l" + LTRIM$(STR$(noteLen))
LINE INPUT "String to convert to Morse code: "; x$
FOR n% = 1 TO LEN(x$)
c$ = UCASE$(MID$(x$, n%, 1))
PLAY "p" + LTRIM$(STR$(noteLen / 2)) + "."
SELECT CASE UCASE$(c$)
CASE " "
'since each char is effectively wrapped with 6 p's, we only need to add 1:
PLAY "p" + LTRIM$(STR$(noteLen))
PRINT " ";
CASE "!" TO "_"
PRINT tones(ASC(c$) - 33); " ";
player tones(ASC(c$) - 33)
CASE ELSE
PRINT "# ";
player "#"
END SELECT
PLAY "p" + LTRIM$(STR$(noteLen / 2)) + "."
NEXT n%
PRINT
'all the Morse codes in ASCII order, from "!" to "_"
DATA "-.-.--", ".-..-.", "#", "...-..-", "#", ".-...", ".----.", "-.--."
DATA "-.--.-", "#", ".-.-.", "--..--", "-....-", ".-.-.-", "-..-.", "-----"
DATA ".----", "..---", "...--", "....-", ".....", "-....", "--...", "---.."
DATA "----.", "---...", "-.-.-.", "#", "-...-", "#", "..--..", ".--.-.", ".-"
DATA "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-"
DATA ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-"
DATA "...-", ".--", "-..-", "-.--", "--..", "#", "#", "#", "#", "..--.-"
SUB player (what AS STRING)
FOR i% = 1 TO LEN(what)
z$ = MID$(what, i%, 1)
SELECT CASE z$
CASE "."
o$ = "g"
CASE "-"
o$ = "g" + LTRIM$(STR$(noteLen / 2)) + "."
CASE ELSE
o$ = "<<<<c>>>>"
END SELECT
PLAY o$
PLAY "p" + LTRIM$(STR$(noteLen))
NEXT i%
END SUB |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #BASIC | BASIC | RANDOMIZE TIMER
DIM doors(3) '0 is a goat, 1 is a car
CLS
switchWins = 0
stayWins = 0
FOR plays = 0 TO 32767
winner = INT(RND * 3) + 1
doors(winner) = 1'put a winner in a random door
choice = INT(RND * 3) + 1'pick a door, any door
DO
shown = INT(RND * 3) + 1
'don't show the winner or the choice
LOOP WHILE doors(shown) = 1 OR shown = choice
stayWins = stayWins + doors(choice) 'if you won by staying, count it
switchWins = switchWins + doors(3 - choice - shown) 'could have switched to win
doors(winner) = 0 'clear the doors for the next test
NEXT plays
PRINT "Switching wins"; switchWins; "times."
PRINT "Staying wins"; stayWins; "times." |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #CLU | CLU | mul_inv = proc (a, b: int) returns (int) signals (no_inverse)
if b<0 then b := -b end
if a<0 then a := b - (-a // b) end
t: int := 0
nt: int := 1
r: int := b
nr: int := a // b
while nr ~= 0 do
q: int := r / nr
t, nt := nt, t - q*nt
r, nr := nr, r - q*nr
end
if r>1 then signal no_inverse end
if t<0 then t := t+b end
return(t)
end mul_inv
start_up = proc ()
pair = struct[a, b: int]
tests: sequence[pair] := sequence[pair]$
[pair${a: 42, b: 2017},
pair${a: 40, b: 1},
pair${a: 52, b: -217},
pair${a: -486, b: 217},
pair${a: 40, b: 2018}]
po: stream := stream$primary_output()
for test: pair in sequence[pair]$elements(tests) do
stream$puts(po, int$unparse(test.a) || ", "
|| int$unparse(test.b) || " -> ")
stream$putl(po, int$unparse(mul_inv(test.a, test.b)))
except when no_inverse:
stream$putl(po, "no modular inverse")
end
end
end start_up |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Comal | Comal | 0010 FUNC mulinv#(a#,b#) CLOSED
0020 IF b#<0 THEN b#:=-b#
0030 IF a#<0 THEN a#:=b#-(-a# MOD b#)
0040 t#:=0;nt#:=1;r#:=b#;nr#:=a# MOD b#
0050 WHILE nr#<>0 DO
0060 q#:=r# DIV nr#
0070 tmp#:=nt#;nt#:=t#-q#*nt#;t#:=tmp#
0080 tmp#:=nr#;nr#:=r#-q#*nr#;r#:=tmp#
0090 ENDWHILE
0100 IF r#>1 THEN RETURN -1
0110 IF t#<0 THEN t#:+b#
0120 RETURN t#
0130 ENDFUNC mulinv#
0140 //
0150 WHILE NOT EOD DO
0160 READ a#,b#
0170 PRINT a#,", ",b#," -> ",mulinv#(a#,b#)
0180 ENDWHILE
0190 END
0200 //
0210 DATA 42,2017,40,1,52,-217,-486,217,40,2018 |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #X86-64_Assembly | X86-64 Assembly |
option casemap:none
printf proto :qword, :vararg
exit proto :dword
ROW_LEN equ (4*4)
MEM_SIZE equ 4
.data
;; A 2d array - 2 rows, 4 columns
;; int twodimen[2][4] = { {0,1,2,3},
;; {4,5,6,7}};
twodimen db 48 dup (0)
tpl db "%d",13,10,0
.code
main proc
local cols:qword
lea rbx, twodimen
mov cols, 0
;; Forgive me for the multiple loops, I'm just to lazy to
;; do the conditional jumps required for 2 for loops. -.-
@1:
mov rcx, cols
mov dword ptr [rbx+0*ROW_LEN + rcx*MEM_SIZE], ecx ;; first row, rcx column
inc cols
cmp cols, 3
jle @1
mov cols, 0
mov rdx, 4
@2:
mov rcx, cols
mov dword ptr [rbx+1*ROW_LEN + rcx*MEM_SIZE], edx ;; second row, rcx column
inc cols
inc edx
cmp cols, 3
jle @2
invoke printf, CSTR("--> Printing columns in row 1",10)
mov cols, 0
@3:
mov rcx, cols
mov esi, dword ptr [rbx+0*ROW_LEN + rcx*MEM_SIZE]
lea rdi, tpl
call printf
inc cols
cmp cols, 3
jle @3
invoke printf, CSTR("--> Printing columns in row 2",10)
mov cols, 0
@4:
mov rcx, cols
mov esi, dword ptr [rbx+1*ROW_LEN + rcx*MEM_SIZE]
lea rdi, tpl
call printf
inc cols
cmp cols, 3
jle @4
mov rax, 0
xor edi, edi
call exit
leave
ret
main endp
|
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #AWK | AWK |
BEGIN {
for(i=1;i<=12;i++){
for(j=1;j<=12;j++){
if(j>=i||j==1){printf "%4d",i*j}
else {printf " "}
}
print
}
} |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #SPSS | SPSS | set rng=mc seed=17760704.
new file.
input program.
vector x(4).
loop #i=1 to 200.
loop #j=1 to 4.
compute x(#j)=rv.normal(0,1).
end loop.
end case.
end loop.
end file.
end input program.
compute y=1.5+0.8*x1-0.7*x2+1.1*x3-1.7*x4+rv.normal(0,1).
execute. |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Stata | Stata | clear
set seed 17760704
set obs 200
forv i=1/4 {
gen x`i'=rnormal()
}
gen y=1.5+0.8*x1-0.7*x2+1.1*x3-1.7*x4+rnormal() |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #IS-BASIC | IS-BASIC | 100 PROGRAM "Multifac.bas"
110 FOR I=1 TO 5
120 PRINT "Degree";I;":";
130 FOR N=1 TO 10
140 PRINT MFACT(N,I);
150 NEXT
160 PRINT
170 NEXT
180 DEF MFACT(N,D)
190 NUMERIC I,RES
200 IF N<2 THEN LET MFACT=1:EXIT DEF
210 LET RES=N
220 FOR I=N-D TO 2 STEP-D
230 LET RES=RES*I
240 NEXT
250 LET MFACT=RES
260 END DEF |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Perl | Perl | use SDL;
use SDL::Events;
use SDLx::App;
my $app = SDLx::App->new;
$app->add_event_handler( sub {
my $event = shift;
if( $event->type == SDL_MOUSEMOTION ) {
printf( "x=%d y=%d\n", $event->motion_x, $event->motion_y );
$app->stop
}
} );
$app->run;
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Phix | Phix | -- demo\rosetta\Mouse_position.exw
include pGUI.e
Ihandle global_lbl, canvas_lbl, timer_lbl
function globalmotion_cb(integer x, integer y, atom /*pStatus*/)
IupSetStrAttribute(global_lbl,"TITLE","globalmotion_cb %d, %d",{x,y})
return IUP_DEFAULT
end function
function canvas_motion_cb(Ihandle /*canvas*/, integer x, integer y, atom pStatus)
IupSetStrAttribute(canvas_lbl,"TITLE","canvasmotion_cb %d, %d",{x,y})
return IUP_DEFAULT;
end function
function OnTimer(Ihandle /*ih*/)
integer {x,y} = IupGetIntInt(NULL,"CURSORPOS")
IupSetStrAttribute(timer_lbl,"TITLE","timer %d, %d",{x,y})
return IUP_IGNORE
end function
procedure main()
Ihandle separator1, separator2,
canvas, frame_1, frame_2,
dialog
IupOpen()
global_lbl = IupLabel("Move the mouse anywhere on the window","EXPAND=HORIZONTAL")
separator1 = IupLabel(NULL,"SEPARATOR=HORIZONTAL")
canvas_lbl = IupLabel("Move the mouse anywhere on the canvas","EXPAND=HORIZONTAL")
separator2 = IupLabel(NULL,"SEPARATOR=HORIZONTAL")
timer_lbl = IupLabel("This one runs on a three second timer","EXPAND=HORIZONTAL")
frame_1 = IupFrame(IupVbox({global_lbl,
separator1,
canvas_lbl,
separator2,
timer_lbl}),
"TITLE=IupLabel, SIZE=200x")
canvas = IupCanvas("MOTION_CB", Icallback("canvas_motion_cb"),
"EXPAND=HORIZONTAL, RASTERSIZE=200x200")
frame_2 = IupFrame(canvas, "TITLE=IupCanvas")
dialog = IupDialog(IupHbox({frame_1,frame_2}, "MARGIN=5x5, GAP=5"))
IupSetAttribute(dialog,"TITLE","Mouse motion");
IupSetGlobal("INPUTCALLBACKS", "Yes");
IupSetGlobalFunction("GLOBALMOTION_CB", Icallback("globalmotion_cb"));
Ihandle hTimer = IupTimer(Icallback("OnTimer"), 3000)
IupShow(dialog)
IupMainLoop()
IupClose()
end procedure
main() |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #JavaScript | JavaScript | function queenPuzzle(rows, columns) {
if (rows <= 0) {
return [[]];
} else {
return addQueen(rows - 1, columns);
}
}
function addQueen(newRow, columns, prevSolution) {
var newSolutions = [];
var prev = queenPuzzle(newRow, columns);
for (var i = 0; i < prev.length; i++) {
var solution = prev[i];
for (var newColumn = 0; newColumn < columns; newColumn++) {
if (!hasConflict(newRow, newColumn, solution))
newSolutions.push(solution.concat([newColumn]))
}
}
return newSolutions;
}
function hasConflict(newRow, newColumn, solution) {
for (var i = 0; i < newRow; i++) {
if (solution[i] == newColumn ||
solution[i] + i == newColumn + newRow ||
solution[i] - i == newColumn - newRow) {
return true;
}
}
return false;
}
console.log(queenPuzzle(8,8)); |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #PicoLisp | PicoLisp | (de rangeth (A B)
(mapcar
'((I)
(pack I
(if (member (% I 100) (11 12 13))
'th
(case (% I 10)
(1 'st)
(2 'nd)
(3 'rd)
(T 'th) ) ) ) )
(range A B) ) )
(prinl (glue " " (rangeth 0 25)))
(prinl (glue " " (rangeth 250 265)))
(prinl (glue " " (rangeth 1000 1025)))
(bye) |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #PureBasic | PureBasic | EnableExplicit
Declare main()
If OpenConsole("Munchausen_numbers")
main() : Input() : End
EndIf
Procedure main()
Define i.i,
sum.i,
number.i,
digit.i
For i = 1 To 5000
sum = 0
number = i
While number > 0
digit = number % 10
sum + Pow(digit, digit)
number / 10
Wend
If sum = i
PrintN(Str(i))
EndIf
Next
EndProcedure |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Liberty_BASIC | Liberty BASIC |
print "F sequence."
for i = 0 to 20
print f(i);" ";
next
print
print "M sequence."
for i = 0 to 20
print m(i);" ";
next
end
function f(n)
if n = 0 then
f = 1
else
f = n - m(f(n - 1))
end if
end function
function m(n)
if n = 0 then
m = 0
else
m = n - f(m(n - 1))
end if
end function
|
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Phix | Phix | function bindf(object m, integer f)
return f(m)
end function
function unit(object m)
return m
end function
function times_five(object l)
return iff(integer(l)?l*5:l)
end function
function plus_four(object l)
return iff(integer(l)?l+4:l)
end function
procedure test(object l)
printf(1,"%v -> %v\n", {l, bindf(bindf(l,times_five),plus_four)})
end procedure
test(3)
test("none")
|
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Python | Python | """A Maybe Monad. Requires Python >= 3.7 for type hints."""
from __future__ import annotations
from typing import Any
from typing import Callable
from typing import Generic
from typing import Optional
from typing import TypeVar
from typing import Union
T = TypeVar("T")
class Maybe(Generic[T]):
def __init__(self, value: Union[Optional[T], Maybe[T]] = None):
if isinstance(value, Maybe):
self.value: Optional[T] = value.value
else:
self.value = value
def __rshift__(self, func: Callable[[Optional[T]], Maybe[Any]]):
return self.bind(func)
def bind(self, func: Callable[[Optional[T]], Maybe[Any]]) -> Maybe[Any]:
return func(self.value)
def __str__(self):
return f"{self.__class__.__name__}({self.value!r})"
def plus_one(value: Optional[int]) -> Maybe[int]:
if value is not None:
return Maybe[int](value + 1)
return Maybe[int](None)
def currency(value: Optional[int]) -> Maybe[str]:
if value is not None:
return Maybe[str](f"${value}.00")
return Maybe[str](None)
if __name__ == "__main__":
test_cases = [1, 99, None, 4]
for case in test_cases:
m_int = Maybe[int](case)
result = m_int >> plus_one >> currency
# or..
# result = m_int.bind(plus_one).bind(currency)
print(f"{str(case):<4} -> {result}")
|
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Clojure | Clojure | (defn calc-pi [iterations]
(loop [x (rand) y (rand) in 0 total 1]
(if (< total iterations)
(recur (rand) (rand) (if (<= (+ (* x x) (* y y)) 1) (inc in) in) (inc total))
(double (* (/ in total) 4)))))
(doseq [x (take 5 (iterate #(* 10 %) 10))] (println (str (format "% 8d" x) ": " (calc-pi x)))) |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Common_Lisp | Common Lisp | (defun approximate-pi (n)
(/ (loop repeat n count (<= (abs (complex (random 1.0) (random 1.0))) 1.0)) n 0.25))
(dolist (n (loop repeat 5 for n = 1000 then (* n 10) collect n))
(format t "~%~8d -> ~f" n (approximate-pi n))) |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #Phix | Phix | with javascript_semantics
function encode(string s)
string symtab = "abcdefghijklmnopqrstuvwxyz"
sequence res = {}
for i=1 to length(s) do
integer ch = s[i], k = find(ch,symtab)
res &= k-1
-- for j=k to 2 by -1 do
-- symtab[j] = symtab[j-1]
-- end for
-- symtab[1] = ch
symtab[1..k] = ch&symtab[1..k-1]
end for
return res
end function
function decode(sequence s)
string symtab = "abcdefghijklmnopqrstuvwxyz"
string res = ""
for i=1 to length(s) do
integer k = s[i]+1
integer ch = symtab[k]
res &= ch
-- for j=k to 2 by -1 do
-- symtab[j] = symtab[j-1]
-- end for
-- symtab[1] = ch
symtab[1..k] = ch&symtab[1..k-1]
end for
return res
end function
procedure test(string s)
sequence e = encode(s)
string d = decode(e)
?{s,e,d,{"**ERROR**","ok"}[(s=d)+1]}
end procedure
test("broood")
test("bananaaa")
test("hiphophiphop")
|
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Befunge | Befunge | >~>48*-:0\`#@_2*::"!"%\"!"/3+g75v
^v('_')v!:-*57g+3/"!"\%"!":+1\-*<
^$$,*84_\#!:#:2#-%#15#\9#/*#2+#,<
##X)P)##Z*##3(D)5);(##8(/)A)8)9(#
($(&(*(2(B(A(?(;(3([)M)##1(##V)L)
$%1'-')&$$.''&2'&%$'%&0'#%%%#&,''
'(&*&#$&&*'$&)'%'/'########6)##$%
1'-')&$$.''&2'&%$'%&0'#%%%#&,'''(
&*&#$&&*'$&)'%'/'################ |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #BBC_BASIC | BBC BASIC | total% = 10000
FOR trial% = 1 TO total%
prize_door% = RND(3) : REM. The prize is behind this door
guess_door% = RND(3) : REM. The contestant guesses this door
IF prize_door% = guess_door% THEN
REM. The contestant guessed right, reveal either of the others
reveal_door% = RND(2)
IF prize_door% = 1 reveal_door% += 1
IF prize_door% = 2 AND reveal_door% = 2 reveal_door% = 3
ELSE
REM. The contestant guessed wrong, so reveal the non-prize door
reveal_door% = prize_door% EOR guess_door%
ENDIF
stick_door% = guess_door% : REM. The sticker doesn't change his mind
swap_door% = guess_door% EOR reveal_door% : REM. but the swapper does
IF stick_door% = prize_door% sticker% += 1
IF swap_door% = prize_door% swapper% += 1
NEXT trial%
PRINT "After a total of ";total%;" trials,"
PRINT "The 'sticker' won ";sticker%;" times (";INT(sticker%/total%*100);"%)"
PRINT "The 'swapper' won ";swapper%;" times (";INT(swapper%/total%*100);"%)" |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Common_Lisp | Common Lisp |
;;
;; Calculates the GCD of a and b based on the Extended Euclidean Algorithm. The function also returns
;; the Bézout coefficients s and t, such that gcd(a, b) = as + bt.
;;
;; The algorithm is described on page http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Iterative_method_2
;;
(defun egcd (a b)
(do ((r (cons b a) (cons (- (cdr r) (* (car r) q)) (car r))) ; (r+1 r) i.e. the latest is first.
(s (cons 0 1) (cons (- (cdr s) (* (car s) q)) (car s))) ; (s+1 s)
(u (cons 1 0) (cons (- (cdr u) (* (car u) q)) (car u))) ; (t+1 t)
(q nil))
((zerop (car r)) (values (cdr r) (cdr s) (cdr u))) ; exit when r+1 = 0 and return r s t
(setq q (floor (/ (cdr r) (car r)))))) ; inside loop; calculate the q
;;
;; Calculates the inverse module for a = 1 (mod m).
;;
;; Note: The inverse is only defined when a and m are coprimes, i.e. gcd(a, m) = 1.”
;;
(defun invmod (a m)
(multiple-value-bind (r s k) (egcd a m)
(unless (= 1 r) (error "invmod: Values ~a and ~a are not coprimes." a m))
s))
|
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #XPL0 | XPL0 |
int A(5,4,3,2);
[A(3,1,0,1):= 3100;
A(3,1,0,1):= A(3,1,0,1)+1;
IntOut(0, A(3,1,0,1));
] |
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #Axe | Axe | Fix 5
ClrDraw
For(I,1,10)
Text(I-1*9,0,I▶Dec)
Text(91,I*7+1,I▶Dec)
End
For(J,1,8)
For(I,J,10)
Text(I-1*9,J*7+1,I*J▶Dec)
End
End
HLine(7)
VLine(89)
DispGraph
getKeyʳ
Fix 4 |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Tcl | Tcl | package require math::linearalgebra
namespace eval multipleRegression {
namespace export regressionCoefficients
namespace import ::math::linearalgebra::*
# Matrix inversion is defined in terms of Gaussian elimination
# Note that we assume (correctly) that we have a square matrix
proc invert {matrix} {
solveGauss $matrix [mkIdentity [lindex [shape $matrix] 0]]
}
# Implement the Ordinary Least Squares method
proc regressionCoefficients {y x} {
matmul [matmul [invert [matmul $x [transpose $x]]] $x] $y
}
}
namespace import multipleRegression::regressionCoefficients |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #TI-83_BASIC | TI-83 BASIC | {1.47,1.50,1.52,1.55,1.57,1.60,1.63,1.65,1.68,1.70,1.73,1.75,1.78,1.80,1.83}→L₁
{52.21,53.12,54.48,55.84,57.20,58.57,59.93,61.29,63.11,64.47,66.28,68.10,69.92,72.19,74.46}→L₂
QuadReg L₁,L₂ |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #J | J |
NB. tacit implementation of the recursive c function
NB. int multifact(int n,int deg){return n<=deg?n:n*multifact(n-deg,deg);}
multifact=: [`([ * - $: ])@.(<~)
(a:,<' degree'),multifact table >:i.10
┌─────────┬──────────────────────────────────────┐
│ │ degree │
├─────────┼──────────────────────────────────────┤
│multifact│ 1 2 3 4 5 6 7 8 9 10│
├─────────┼──────────────────────────────────────┤
│ 1 │ 1 1 1 1 1 1 1 1 1 1│
│ 2 │ 2 2 2 2 2 2 2 2 2 2│
│ 3 │ 6 3 3 3 3 3 3 3 3 3│
│ 4 │ 24 8 4 4 4 4 4 4 4 4│
│ 5 │ 120 15 10 5 5 5 5 5 5 5│
│ 6 │ 720 48 18 12 6 6 6 6 6 6│
│ 7 │ 5040 105 28 21 14 7 7 7 7 7│
│ 8 │ 40320 384 80 32 24 16 8 8 8 8│
│ 9 │ 362880 945 162 45 36 27 18 9 9 9│
│10 │3628800 3840 280 120 50 40 30 20 10 10│
└─────────┴──────────────────────────────────────┘
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #PicoLisp | PicoLisp | (de mousePosition ()
(prog2
(prin "^[[?9h") # Mouse reporting on
(and
(= "^[" (key))
(key 200)
(key 200)
(key)
(cons
(- (char (key)) 32)
(- (char (key)) 32) ) )
(prin "^[[?9l") ) ) # Mouse reporting off |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #PureBasic | PureBasic | x = WindowMouseX(#MyWindow)
y = WindowMouseY(#MyWindow) |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Processing | Processing | void setup(){
size(640, 480);
}
void draw(){
// mouseX and mouseY provide the current mouse position
ellipse(mouseX, mouseY, 5, 5); // graphic output example
println("x:" + mouseX + " y:" + mouseY);
} |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #jq | jq | def single_solution_queens(n):
def q: "♛";
def init(k): reduce range(0;k) as $i ([]; . + ["."]);
def matrix(k): init(k) as $row | reduce range(0;k) as $i ([]; . + [$row]);
def place(stream; i; j):
# jq indexing is based on offsets but we are using the 1-based formulae:
reduce stream as $s (.; setpath([-1+($s|i), -1+($s|j)]; q) );
def even(k):
if ((k-2) % 6) != 0 then
place( range(1; 1+(k/2)); .; 2*. )
| place( range(1; 1+(k/2)); (k/2) + .; 2*. -1 )
else place( range(1; 1+(k/2)); .; 1 + ((2*. + (k/2) - 3) % k))
| place( range(1; 1+(n/2)); n + 1 - .; n - ((2*. + (n/2) - 3) % n))
end;
matrix(n) # the chess board
| if (n % 2) == 0 then even(n)
else even(n-1) | .[n-1][n-1] = q
end;
# Example:
def pp: reduce .[] as $row
(""; reduce $row[] as $x (.; . + $x) + "\n");
single_solution_queens(8) | pp |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #PL.2FI | PL/I | Nth: procedure options (main); /* 1 June 2014 */
declare i fixed (10);
do i = 0 to 25, 250 to 265, 1000 to 1025;
if i = 250 | i = 1000 then put skip (2);
put edit (enth(i)) (x(1), a);
end;
enth: procedure (i) returns (character (25) varying);
declare i fixed (10);
declare suffix character (2);
select (mod(i, 10));
when (1) suffix = 'st';
when (2) suffix = 'nd';
when (3) suffix = 'rd';
otherwise suffix = 'th';
end;
select (mod(i, 100));
when (11, 12, 13) suffix = 'th';
otherwise ;
end;
return ( trim(i) || suffix );
end enth;
end Nth; |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #PowerShell | PowerShell | function nth($inp){
$suffix = "th"
switch($inp % 10){
1{$suffix="st"}
2{$suffix="nd"}
3{$suffix="rd"}
}
return "$inp$suffix "
}
0..25 | %{Write-host -nonewline (nth "$_")};""
250..265 | %{Write-host -nonewline (nth "$_")};""
1000..1025 | %{Write-host -nonewline (nth "$_")};"" |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Python | Python | for i in range(5000):
if i == sum(int(x) ** int(x) for x in str(i)):
print(i) |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #LibreOffice_Basic | LibreOffice Basic | '// LibreOffice Basic Implementation of Hofstadter Female-Male sequences
'// Utility functions
sub setfont(strfont)
ThisComponent.getCurrentController.getViewCursor.charFontName = strfont
end sub
sub newline
oVC = thisComponent.getCurrentController.getViewCursor
oText = oVC.text
oText.insertControlCharacter(oVC, com.sun.star.text.ControlCharacter.PARAGRAPH_BREAK, False)
end sub
sub out(sString)
oVC = ThisComponent.getCurrentController.getViewCursor
oText = oVC.text
oText.insertString(oVC, sString, false)
end sub
sub outln(optional sString)
if not ismissing (sString) then out(sString)
newline
end sub
function intformat(n as integer,nlen as integer) as string
dim nstr as string
nstr = CStr(n)
while len(nstr) < nlen
nstr = " " & nstr
wend
intformat = nstr
end function
'// Hofstadter Female-Male function definitions
function F(n as long) as long
if n = 0 Then
F = 1
elseif n > 0 Then
F = n - M(F(n - 1))
endif
end function
function M(n)
if n = 0 Then
M = 0
elseif n > 0 Then
M = n - F(M(n - 1))
endif
end function
'// Hofstadter Female Male sequence demo routine
sub Hofstadter_Female_Male_Demo
'// Introductory Text
setfont("LM Roman 10")
outln("Rosetta Code Hofstadter Female and Male Sequence Challenge")
outln
out("Two functions are said to be mutually recursive if the first calls the second,")
outln(" and in turn the second calls the first.")
out("Write two mutually recursive functions that compute members of the Hofstadter")
outln(" Female and Male sequences defined as:")
outln
setfont("LM Mono Slanted 10")
outln(chr(9)+"F(0) = 1 ; M(0)=0")
outln(chr(9)+"F(n) = n - M(F(n-1)), n > 0")
outln(chr(9)+"M(n) = n - F(M(n-1)), n > 0")
outln
'// Sequence Generation
const nmax as long = 20
dim n as long
setfont("LM Mono 10")
out("n = "
for n = 0 to nmax
out(" " + intformat(n, 2))
next n
outln
out("F(n) = "
for n = 0 to nmax
out(" " + intformat(F(n),2))
next n
outln
out("M(n) = "
for n = 0 to nmax
out(" " + intformat(M(n), 2))
next n
outln
end sub
------------------------------
Output
------------------------------
n = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
F(n) = 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 13
M(n) = 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12 12
|
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Racket | Racket | #lang racket
(require syntax/parse/define)
(define (bind x f) (and x (f x)))
(define return identity)
;; error when arg = 0
(define reciprocal (curry / 1))
;; error when arg < 0
(define (root x) (if (< x 0) (error 'bad) (sqrt x)))
;; error whe arg <= 0
(define (ln x) (if (<= x 0) (error 'bad) (log x)))
(define (lift f check) (λ (x) (and (check x) (f x))))
(define safe-reciprocal (lift reciprocal (negate (curry equal? 0))))
(define safe-root (lift root (curry <= 0)))
(define safe-ln (lift ln (curry < 0)))
(define (safe-log-root-reciprocal x)
(bind (bind (bind x safe-reciprocal) safe-root) safe-ln))
(define tests `(-2 -1 -0.5 0 1 ,(exp -1) 1 2 ,(exp 1) 3 4 5))
(map safe-log-root-reciprocal tests)
(define-syntax-parser do-macro
[(_ [x {~datum <-} y] . the-rest) #'(bind y (λ (x) (do-macro . the-rest)))]
[(_ e) #'e])
(define (safe-log-root-reciprocal* x)
(do-macro [x <- (safe-reciprocal x)]
[x <- (safe-root x)]
[x <- (safe-ln x)]
(return x)))
(map safe-log-root-reciprocal* tests) |
http://rosettacode.org/wiki/Monads/Maybe_monad | Monads/Maybe monad | Demonstrate in your programming language the following:
Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String
Compose the two functions with bind
A Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.
A Maybe Monad is a monad which specifically encapsulates the type of an undefined value.
| #Raku | Raku | my $monad = <42>;
say 'Is $monad an Int?: ', $monad ~~ Int;
say 'Is $monad a Str?: ', $monad ~~ Str;
say 'Wait, what? What exactly is $monad?: ', $monad.^name; |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Crystal | Crystal | def approx_pi(throws)
times_inside = throws.times.count {Math.hypot(rand, rand) <= 1.0}
4.0 * times_inside / throws
end
[1000, 10_000, 100_000, 1_000_000, 10_000_000].each do |n|
puts "%8d samples: PI = %s" % [n, approx_pi(n)]
end |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #D | D | import std.stdio, std.random, std.math;
double pi(in uint nthrows) /*nothrow*/ @safe /*@nogc*/ {
uint inside;
foreach (immutable i; 0 .. nthrows)
if (hypot(uniform01, uniform01) <= 1)
inside++;
return 4.0 * inside / nthrows;
}
void main() {
foreach (immutable p; 1 .. 8)
writefln("%10s: %07f", 10 ^^ p, pi(10 ^^ p));
} |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #PHP | PHP | <?php
function symbolTable() {
$symbol = array();
for ($c = ord('a') ; $c <= ord('z') ; $c++) {
$symbol[$c - ord('a')] = chr($c);
}
return $symbol;
}
function mtfEncode($original, $symbol) {
$encoded = array();
for ($i = 0 ; $i < strlen($original) ; $i++) {
$char = $original[$i];
$position = array_search($char, $symbol);
$encoded[] = $position;
$mtf = $symbol[$position];
unset($symbol[$position]);
array_unshift($symbol, $mtf);
}
return $encoded;
}
function mtfDecode($encoded, $symbol) {
$decoded = '';
foreach ($encoded AS $position) {
$char = $symbol[$position];
$decoded .= $char;
unset($symbol[$position]);
array_unshift($symbol, $char);
}
return $decoded;
}
foreach (array('broood', 'bananaaa', 'hiphophiphop') AS $original) {
$encoded = mtfEncode($original, symbolTable());
$decoded = mtfDecode($encoded, symbolTable());
echo
$original,
' -> [', implode(',', $encoded), ']',
' -> ', $decoded,
' : ', ($original === $decoded ? 'OK' : 'Error'),
PHP_EOL;
} |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #C | C |
/*
David Lambert, 2010-Dec-09
filter producing morse beep commands.
build:
make morse
use:
$ echo tie a. | ./morse
beep -n -f 440 -l 300 -D 100 -n -D 200 -n -f 440 -l 100 -D 100 -n -f 440 -l 100 -D 100 -n -D 200 -n -f 440 -l 100 -D 100 -n -D 200 -n -D 400 -n -f 440 -l 100 -D 100 -n -f 440 -l 300 -D 100 -n -D 200 -n -f 440 -l 100 -D 100 -n -f 440 -l 300 -D 100 -n -f 440 -l 100 -D 100 -n -f 440 -l 300 -D 100 -n -f 440 -l 100 -D 100 -n -f 440 -l 300 -D 100 -n -D 200 -n -D 400 -n -D 400
bugs:
What is the space between letter and punctuation?
Demo truncates input lines at 71 characters or so.
*/
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BIND(A,L,H) ((L)<(A)?(A)<(H)?(A):(H):(L))
/*
BIND(-1,0,9) is 0
BIND( 7,0,9) is 7
BIND(77,0,9) is 9
*/
char
/* beep args for */
/* dit dah extra space */
dih[50],dah[50],medium[30],word[30],
*dd[2] = {dih,dah};
const char
*ascii = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789.,?'!/()&:;=+-_\"$@",
*itu[] = {
"13","3111","3131","311","1","1131","331","1111","11","1333","313","1311","33","31","333","1331","3313","131","111","3","113","1113","133","3113","3133","3311","33333","13333","11333","11133","11113","11111","31111","33111","33311","33331","131313","331133","113311","133331","313133","31131","31331","313313","13111","333111","313131","31113","13131","311113","113313","131131","1113113","133131"
};
void append(char*s,const char*morse) {
for (; *morse; ++morse)
strcat(s,dd['3'==*morse]);
strcat(s,medium);
}
char*translate(const char*i,char*o) {
const char*pc;
sprintf(o,"beep");
for (; *i; ++i)
if (NULL == (pc = strchr(ascii,toupper(*i))))
strcat(o,word);
else
append(o,itu[pc-ascii]);
strcat(o,word);
return o;
}
int main(int ac,char*av[]) {
char
sin[73],sout[100000];
int
dit = 100;
if (1 < ac) {
if (strlen(av[1]) != strspn(av[1],"0123456789"))
return 0*fprintf(stderr,"use: %s [duration] dit in ms, default %d\n",*av,dit);
dit = BIND(atoi(av[1]),1,1000);
}
sprintf(dah," -n -f 440 -l %d -D %d",3*dit,dit);
sprintf(dih," -n -f 440 -l %d -D %d",dit,dit);
sprintf(medium," -n -D %d",(3-1)*dit);
sprintf(word," -n -D %d",(7-(3-1)-1)*dit);
while (NULL != fgets(sin,72,stdin))
puts(translate(sin,sout));
return 0;
}
|
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #C | C | //Evidence of the Monty Hall solution.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define GAMES 3000000
int main(void){
unsigned i, j, k, choice, winsbyswitch=0, door[3];
srand(time(NULL)); //initialize random seed.
for(i=0; i<GAMES; i++){
door[0] = (!(rand()%2)) ? 1: 0; //give door 1 either a car or a goat randomly.
if(door[0]) door[1]=door[2]=0; //if 1st door has car, give other doors goats.
else{ door[1] = (!(rand()%2)) ? 1: 0; door[2] = (!door[1]) ? 1: 0; } //else, give 2nd door car or goat, give 3rd door what's left.
choice = rand()%3; //choose a random door.
//if the next door has a goat, and the following door has a car, or vice versa, you'd win if you switch.
if(((!(door[((choice+1)%3)])) && (door[((choice+2)%3)])) || (!(door[((choice+2)%3)]) && (door[((choice+1)%3)]))) winsbyswitch++;
}
printf("\nAfter %u games, I won %u by switching. That is %f%%. ", GAMES, winsbyswitch, (float)winsbyswitch*100.0/(float)i);
}
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Cowgol | Cowgol | include "cowgol.coh";
sub mulinv(a: int32, b: int32): (t: int32) is
if b<0 then b := -b; end if;
if a<0 then a := b - (-a % b); end if;
t := 0;
var nt: int32 := 1;
var r := b;
var nr := a % b;
while nr != 0 loop
var q := r / nr;
var tmp := nt; nt := t - q*nt; t := tmp;
tmp := nr; nr := r - q*nr; r := tmp;
end loop;
if r>1 then t := -1;
elseif t<0 then t := t + b;
end if;
end sub;
record Pair is
a: int32;
b: int32;
end record;
var data: Pair[] := {
{42, 2017},
{40, 1},
{52, -217},
{-486, 217},
{40, 2018}
};
var i: @indexof data := 0;
while i < @sizeof data loop
print_i32(data[i].a as uint32);
print(", ");
print_i32(data[i].b as uint32);
print(" -> ");
var mi := mulinv(data[i].a, data[i].b);
if mi<0
then print("no inverse");
else print_i32(mi as uint32);
end if;
print_nl();
i := i + 1;
end loop; |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Crystal | Crystal | def modinv(a0, m0)
return 1 if m0 == 1
a, m = a0, m0
x0, inv = 0, 1
while a > 1
inv -= (a // m) * x0
a, m = m, a % m
x0, inv = inv, x0
end
inv += m0 if inv < 0
inv
end |
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #BASIC | BASIC | CLS
'header row
PRINT " ";
FOR n = 1 TO 12
'do it this way for alignment purposes
o$ = " "
MID$(o$, LEN(o$) - LEN(STR$(n)) + 1) = STR$(n)
PRINT o$;
NEXT
PRINT : PRINT " "; STRING$(49, "-");
FOR n = 1 TO 12
PRINT
IF n < 10 THEN PRINT " ";
PRINT n; "|"; 'row labels
FOR m = 1 TO n - 1
PRINT " ";
NEXT
FOR m = n TO 12
'alignment again
o$ = " "
MID$(o$, LEN(o$) - LEN(STR$(m * n)) + 1) = STR$(m * n)
PRINT o$;
NEXT
NEXT |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Ursala | Ursala | regression_coefficients = lapack..dgelsd |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Visual_Basic_.NET | Visual Basic .NET | Module Module1
Sub Swap(Of T)(ByRef x As T, ByRef y As T)
Dim temp = x
x = y
y = temp
End Sub
Sub Require(condition As Boolean, message As String)
If condition Then
Return
End If
Throw New ArgumentException(message)
End Sub
Class Matrix
Private data As Double(,)
Private rowCount As Integer
Private colCount As Integer
Public Sub New(rows As Integer, cols As Integer)
Require(rows > 0, "Need at least one row")
rowCount = rows
Require(cols > 0, "Need at least one column")
colCount = cols
data = New Double(rows - 1, cols - 1) {}
End Sub
Public Sub New(source As Double(,))
Dim rows = source.GetLength(0)
Require(rows > 0, "Need at least one row")
rowCount = rows
Dim cols = source.GetLength(1)
Require(cols > 0, "Need at least one column")
colCount = cols
data = New Double(rows - 1, cols - 1) {}
For i = 1 To rows
For j = 1 To cols
data(i - 1, j - 1) = source(i - 1, j - 1)
Next
Next
End Sub
Default Public Property Index(i As Integer, j As Integer) As Double
Get
Return data(i, j)
End Get
Set(value As Double)
data(i, j) = value
End Set
End Property
Public Property Slice(i As Integer) As Double()
Get
Dim m(colCount - 1) As Double
For j = 1 To colCount
m(j - 1) = Index(i, j - 1)
Next
Return m
End Get
Set(value As Double())
Require(colCount = value.Length, "Slice must match the number of columns")
For j = 1 To colCount
Index(i, j - 1) = value(j - 1)
Next
End Set
End Property
Public Shared Operator *(m1 As Matrix, m2 As Matrix) As Matrix
Dim rc1 = m1.rowCount
Dim cc1 = m1.colCount
Dim rc2 = m2.rowCount
Dim cc2 = m2.colCount
Require(cc1 = rc2, "Cannot multiply if the first columns does not equal the second rows")
Dim result As New Matrix(rc1, cc2)
For i = 1 To rc1
For j = 1 To cc2
For k = 1 To rc2
result(i - 1, j - 1) += m1(i - 1, k - 1) * m2(k - 1, j - 1)
Next
Next
Next
Return result
End Operator
Public Function Transpose() As Matrix
Dim rc = rowCount
Dim cc = colCount
Dim trans As New Matrix(cc, rc)
For i = 1 To cc
For j = 1 To rc
trans(i - 1, j - 1) = Index(j - 1, i - 1)
Next
Next
Return trans
End Function
Public Sub ToReducedRowEchelonForm()
Dim lead = 0
Dim rc = rowCount
Dim cc = colCount
For r = 1 To rc
If cc <= lead Then
Return
End If
Dim i = r
While Index(i - 1, lead) = 0.0
i += 1
If rc = i Then
i = r
lead += 1
If cc = lead Then
Return
End If
End If
End While
Dim temp = Slice(i - 1)
Slice(i - 1) = Slice(r - 1)
Slice(r - 1) = temp
If Index(r - 1, lead) <> 0.0 Then
Dim div = Index(r - 1, lead)
For j = 1 To cc
Index(r - 1, j - 1) /= div
Next
End If
For k = 1 To rc
If k <> r Then
Dim mult = Index(k - 1, lead)
For j = 1 To cc
Index(k - 1, j - 1) -= Index(r - 1, j - 1) * mult
Next
End If
Next
lead += 1
Next
End Sub
Public Function Inverse() As Matrix
Require(rowCount = colCount, "Not a square matrix")
Dim len = rowCount
Dim aug As New Matrix(len, 2 * len)
For i = 1 To len
For j = 1 To len
aug(i - 1, j - 1) = Index(i - 1, j - 1)
Next
REM augment identity matrix to right
aug(i - 1, i + len - 1) = 1.0
Next
aug.ToReducedRowEchelonForm()
Dim inv As New Matrix(len, len)
For i = 1 To len
For j = len + 1 To 2 * len
inv(i - 1, j - len - 1) = aug(i - 1, j - 1)
Next
Next
Return inv
End Function
End Class
Function ConvertArray(source As Double()) As Double(,)
Dim dest(0, source.Length - 1) As Double
For i = 1 To source.Length
dest(0, i - 1) = source(i - 1)
Next
Return dest
End Function
Function MultipleRegression(y As Double(), x As Matrix) As Double()
Dim tm As New Matrix(ConvertArray(y))
Dim cy = tm.Transpose
Dim cx = x.Transpose
Return ((x * cx).Inverse * x * cy).Transpose.Slice(0)
End Function
Sub Print(v As Double())
Dim it = v.GetEnumerator()
Console.Write("[")
If it.MoveNext() Then
Console.Write(it.Current)
End If
While it.MoveNext
Console.Write(", ")
Console.Write(it.Current)
End While
Console.Write("]")
End Sub
Sub Main()
Dim y() = {1.0, 2.0, 3.0, 4.0, 5.0}
Dim x As New Matrix({{2.0, 1.0, 3.0, 4.0, 5.0}})
Dim v = MultipleRegression(y, x)
Print(v)
Console.WriteLine()
y = {3.0, 4.0, 5.0}
x = New Matrix({
{1.0, 2.0, 1.0},
{1.0, 1.0, 2.0}
})
v = MultipleRegression(y, x)
Print(v)
Console.WriteLine()
y = {52.21, 53.12, 54.48, 55.84, 57.2, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.1, 69.92, 72.19, 74.46}
Dim a = {1.47, 1.5, 1.52, 1.55, 1.57, 1.6, 1.63, 1.65, 1.68, 1.7, 1.73, 1.75, 1.78, 1.8, 1.83}
Dim xs(2, a.Length - 1) As Double
For i = 1 To a.Length
xs(0, i - 1) = 1.0
xs(1, i - 1) = a(i - 1)
xs(2, i - 1) = a(i - 1) * a(i - 1)
Next
x = New Matrix(xs)
v = MultipleRegression(y, x)
Print(v)
Console.WriteLine()
End Sub
End Module |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Java | Java | public class MultiFact {
private static long multiFact(long n, int deg){
long ans = 1;
for(long i = n; i > 0; i -= deg){
ans *= i;
}
return ans;
}
public static void main(String[] args){
for(int deg = 1; deg <= 5; deg++){
System.out.print("degree " + deg + ":");
for(long n = 1; n <= 10; n++){
System.out.print(" " + multiFact(n, deg));
}
System.out.println();
}
}
} |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Python | Python |
import Tkinter as tk
def showxy(event):
xm, ym = event.x, event.y
str1 = "mouse at x=%d y=%d" % (xm, ym)
# show cordinates in title
root.title(str1)
# switch color to red if mouse enters a set location range
x,y, delta = 100, 100, 10
frame.config(bg='red'
if abs(xm - x) < delta and abs(ym - y) < delta
else 'yellow')
root = tk.Tk()
frame = tk.Frame(root, bg= 'yellow', width=300, height=200)
frame.bind("<Motion>", showxy)
frame.pack()
root.mainloop()
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Racket | Racket |
#lang racket/gui
(define-values [point _] (get-current-mouse-state))
(send point get-x)
(send point get-y)
|
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Julia | Julia | """
# EightQueensPuzzle
Ported to **Julia** from examples in several languages from
here: https://hbfs.wordpress.com/2009/11/10/is-python-slow
"""
module EightQueensPuzzle
export Board, solve!
mutable struct Board
cols::Int
nodes::Int
diag45::Int
diag135::Int
solutions::Int
Board() = new(0, 0, 0, 0, 0)
end
"Marks occupancy."
function mark!(b::Board, k::Int, j::Int)
b.cols ⊻= (1 << j)
b.diag135 ⊻= (1 << (j+k))
b.diag45 ⊻= (1 << (32+j-k))
end
"Tests if a square is menaced."
function test(b::Board, k::Int, j::Int)
b.cols & (1 << j) +
b.diag135 & (1 << (j+k)) +
b.diag45 & (1 << (32+j-k)) == 0
end
"Backtracking solver."
function solve!(b::Board, niv::Int, dx::Int)
if niv > 0
for i in 0:dx-1
if test(b, niv, i) == true
mark!(b, niv, i)
solve!(b, niv-1, dx)
mark!(b, niv, i)
end
end
else
for i in 0:dx-1
if test(b, 0, i) == true
b.solutions += 1
end
end
end
b.nodes += 1
b.solutions
end
end # module
using .EightQueensPuzzle
for n = 1:17
b = Board()
@show n
print("elapsed:")
solutions = @time solve!(b, n-1, n)
@show solutions
println()
end
|
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