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/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #C | C | #include<stdlib.h>
#include<stdio.h>
long totient(long n){
long tot = n,i;
for(i=2;i*i<=n;i+=2){
if(n%i==0){
while(n%i==0)
n/=i;
tot-=tot/i;
}
if(i==2)
i=1;
}
if(n>1)
tot-=tot/n;
return tot;
}
long* perfectTotients(long n){
long *ptList = (long*)malloc(n*sizeof(long)), m,count=0,sum,tot;
for(m=1;count<n;m++){
tot = m;
sum = 0;
while(tot != 1){
tot = totient(tot);
sum += tot;
}
if(sum == m)
ptList[count++] = m;
}
return ptList;
}
long main(long argC, char* argV[])
{
long *ptList,i,n;
if(argC!=2)
printf("Usage : %s <number of perfect Totient numbers required>",argV[0]);
else{
n = atoi(argV[1]);
ptList = perfectTotients(n);
printf("The first %d perfect Totient numbers are : \n[",n);
for(i=0;i<n;i++)
printf(" %d,",ptList[i]);
printf("\b]");
}
return 0;
}
|
http://rosettacode.org/wiki/Playing_cards | Playing cards | Task
Create a data structure and the associated methods to define and manipulate a deck of playing cards.
The deck should contain 52 unique cards.
The methods must include the ability to:
make a new deck
shuffle (randomize) the deck
deal from the deck
print the current contents of a deck
Each card must have a pip value and a suit value which constitute the unique value of the card.
Related tasks:
Card shuffles
Deal cards_for_FreeCell
War Card_Game
Poker hand_analyser
Go Fish
| #E | E | defmodule Card do
defstruct pip: nil, suit: nil
end
defmodule Playing_cards do
@pips ~w[2 3 4 5 6 7 8 9 10 Jack Queen King Ace]a
@suits ~w[Clubs Hearts Spades Diamonds]a
@pip_value Enum.with_index(@pips)
@suit_value Enum.with_index(@suits)
def deal( n_cards, deck ), do: Enum.split( deck, n_cards )
def deal( n_hands, n_cards, deck ) do
Enum.reduce(1..n_hands, {[], deck}, fn _,{acc,d} ->
{hand, new_d} = deal(n_cards, d)
{[hand | acc], new_d}
end)
end
def deck, do: (for x <- @suits, y <- @pips, do: %Card{suit: x, pip: y})
def print( cards ), do: IO.puts (for x <- cards, do: "\t#{inspect x}")
def shuffle( deck ), do: Enum.shuffle( deck )
def sort_pips( cards ), do: Enum.sort_by( cards, &@pip_value[&1.pip] )
def sort_suits( cards ), do: Enum.sort_by( cards, &(@suit_value[&1.suit]) )
def task do
shuffled = shuffle( deck )
{hand, new_deck} = deal( 3, shuffled )
{hands, _deck} = deal( 2, 3, new_deck )
IO.write "Hand:"
print( hand )
IO.puts "Hands:"
for x <- hands, do: print(x)
end
end
Playing_cards.task |
http://rosettacode.org/wiki/Pi | Pi |
Create a program to continually calculate and output the next decimal digit of
π
{\displaystyle \pi }
(pi).
The program should continue forever (until it is aborted by the user) calculating and outputting each decimal digit in succession.
The output should be a decimal sequence beginning 3.14159265 ...
Note: this task is about calculating pi. For information on built-in pi constants see Real constants and functions.
Related Task Arithmetic-geometric mean/Calculate Pi
| #D | D | import std.stdio, std.conv, std.string;
struct PiDigits {
immutable uint nDigits;
int opApply(int delegate(ref string /*chunk of pi digit*/) dg){
// Maximum width for correct output, for type ulong.
enum size_t width = 9;
enum ulong scale = 10UL ^^ width;
enum ulong initDigit = 2UL * 10UL ^^ (width - 1);
enum string formatString = "%0" ~ text(width) ~ "d";
immutable size_t len = 10 * nDigits / 3;
auto arr = new ulong[len];
arr[] = initDigit;
ulong carry;
foreach (i; 0 .. nDigits / width) {
ulong sum;
foreach_reverse (j; 0 .. len) {
auto quo = sum * (j + 1) + scale * arr[j];
arr[j] = quo % (j*2 + 1);
sum = quo / (j*2 + 1);
}
auto yield = format(formatString, carry + sum/scale);
if (dg(yield))
break;
carry = sum % scale;
}
return 0;
}
}
void main() {
foreach (d; PiDigits(100))
writeln(d);
} |
http://rosettacode.org/wiki/Periodic_table | Periodic table | Task
Display the row and column in the periodic table of the given atomic number.
The periodic table
Let us consider the following periodic table representation.
__________________________________________________________________________
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
| |
|1 H He |
| |
|2 Li Be B C N O F Ne |
| |
|3 Na Mg Al Si P S Cl Ar |
| |
|4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr |
| |
|5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe |
| |
|6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn |
| |
|7 Fr Ra ° Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og |
|__________________________________________________________________________|
| |
| |
|8 Lantanoidi* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu |
| |
|9 Aktinoidi° Ak Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr |
|__________________________________________________________________________|
Example test cases;
1 -> 1 1
2 -> 1 18
29 -> 4 11
42 -> 5 6
57 -> 8 4
58 -> 8 5
72 -> 6 4
89 -> 9 4
Details;
The representation of the periodic table may be represented in various way. The one presented in this challenge does have the following property : Lantanides and Aktinoides are all in a dedicated row, hence there is no element that is placed at 6, 3 nor 7, 3.
You may take a look at the atomic number repartitions here.
The atomic number is at least 1, at most 118.
See also
the periodic table
This task was an idea from CompSciFact
The periodic table in ascii that was used as template
| #Python | Python |
def perta(atomic) -> (int, int):
NOBLES = 2, 10, 18, 36, 54, 86, 118
INTERTWINED = 0, 0, 0, 0, 0, 57, 89
INTERTWINING_SIZE = 14
LINE_WIDTH = 18
prev_noble = 0
for row, noble in enumerate(NOBLES):
if atomic <= noble: # we are at the good row. We now need to determine the column
nb_elem = noble - prev_noble # number of elements on that row
rank = atomic - prev_noble # rank of the input element among elements
if INTERTWINED[row] and INTERTWINED[row] <= atomic <= INTERTWINED[row] + INTERTWINING_SIZE: # lantanides or actinides
row += 2
col = rank + 1
else: # not a lantanide nor actinide
# handle empty spaces between 1-2, 4-5 and 12-13.
nb_empty = LINE_WIDTH - nb_elem # spaces count as columns
inside_left_element_rank = 2 if noble > 2 else 1
col = rank + (nb_empty if rank > inside_left_element_rank else 0)
break
prev_noble = noble
return row+1, col
# small test suite
TESTS = {
1: (1, 1),
2: (1, 18),
29: (4,11),
42: (5, 6),
58: (8, 5),
59: (8, 6),
57: (8, 4),
71: (8, 18),
72: (6, 4),
89: (9, 4),
90: (9, 5),
103: (9, 18),
}
for input, out in TESTS.items():
found = perta(input)
print('TEST:{:3d} -> '.format(input) + str(found) + (f' ; ERROR: expected {out}' if found != out else ''))
|
http://rosettacode.org/wiki/Pig_the_dice_game | Pig the dice game | The game of Pig is a multiplayer game played with a single six-sided die. The
object of the game is to reach 100 points or more.
Play is taken in turns. On each person's turn that person has the option of either:
Rolling the dice: where a roll of two to six is added to their score for that turn and the player's turn continues as the player is given the same choice again; or a roll of 1 loses the player's total points for that turn and their turn finishes with play passing to the next player.
Holding: the player's score for that round is added to their total and becomes safe from the effects of throwing a 1 (one). The player's turn finishes with play passing to the next player.
Task
Create a program to score for, and simulate dice throws for, a two-person game.
Related task
Pig the dice game/Player
| #Julia | Julia |
type PigPlayer
name::String
score::Int
strat::Function
end
function PigPlayer(a::String)
PigPlayer(a, 0, pig_manual)
end
function scoreboard(pps::Array{PigPlayer,1})
join(map(x->@sprintf("%s has %d", x.name, x.score), pps), " | ")
end
function pig_manual(pps::Array{PigPlayer,1}, pdex::Integer, pot::Integer)
pname = pps[pdex].name
print(pname, " there is ", @sprintf("%3d", pot), " in the pot. ")
print("<ret> to continue rolling? ")
return chomp(readline()) == ""
end
function pig_round(pps::Array{PigPlayer,1}, pdex::Integer)
pot = 0
rcnt = 0
while pps[pdex].strat(pps, pdex, pot)
rcnt += 1
roll = rand(1:6)
if roll == 1
return (0, rcnt, false)
else
pot += roll
end
end
return (pot, rcnt, true)
end
function pig_game(pps::Array{PigPlayer,1}, winscore::Integer=100)
pnum = length(pps)
pdex = pnum
println("Playing a game of Pig the Dice.")
while(pps[pdex].score < winscore)
pdex = rem1(pdex+1, pnum)
println(scoreboard(pps))
println(pps[pdex].name, " is now playing.")
(pot, rcnt, ispotwon) = pig_round(pps, pdex)
print(pps[pdex].name, " played ", rcnt, " rolls ")
if ispotwon
println("and scored ", pot, " points.")
pps[pdex].score += pot
else
println("and butsted.")
end
end
println(pps[pdex].name, " won, scoring ", pps[pdex].score, " points.")
end
pig_game([PigPlayer("Alice"), PigPlayer("Bob")])
|
http://rosettacode.org/wiki/Pernicious_numbers | Pernicious numbers | A pernicious number is a positive integer whose population count is a prime.
The population count is the number of ones in the binary representation of a non-negative integer.
Example
22 (which is 10110 in binary) has a population count of 3, which is prime, and therefore
22 is a pernicious number.
Task
display the first 25 pernicious numbers (in decimal).
display all pernicious numbers between 888,888,877 and 888,888,888 (inclusive).
display each list of integers on one line (which may or may not include a title).
See also
Sequence A052294 pernicious numbers on The On-Line Encyclopedia of Integer Sequences.
Rosetta Code entry population count, evil numbers, odious numbers.
| #Eiffel | Eiffel |
class
APPLICATION
create
make
feature
make
-- Test of is_pernicious_number.
local
test: LINKED_LIST [INTEGER]
i: INTEGER
do
create test.make
from
i := 1
until
test.count = 25
loop
if is_pernicious_number (i) then
test.extend (i)
end
i := i + 1
end
across
test as t
loop
io.put_string (t.item.out + " ")
end
io.new_line
across
888888877 |..| 888888888 as c
loop
if is_pernicious_number (c.item) then
io.put_string (c.item.out + " ")
end
end
end
is_pernicious_number (n: INTEGER): BOOLEAN
-- Is 'n' a pernicious_number?
require
positiv_input: n > 0
do
Result := is_prime (count_population (n))
end
feature{NONE}
count_population (n: INTEGER): INTEGER
-- Population count of 'n'.
require
positiv_input: n > 0
local
j: INTEGER
math: DOUBLE_MATH
do
create math
j := math.log_2 (n).ceiling + 1
across
0 |..| j as c
loop
if n.bit_test (c.item) then
Result := Result + 1
end
end
end
is_prime (n: INTEGER): BOOLEAN
--Is 'n' a prime number?
require
positiv_input: n > 0
local
i: INTEGER
max: REAL_64
math: DOUBLE_MATH
do
create math
if n = 2 then
Result := True
elseif n <= 1 or n \\ 2 = 0 then
Result := False
else
Result := True
max := math.sqrt (n)
from
i := 3
until
i > max
loop
if n \\ i = 0 then
Result := False
end
i := i + 2
end
end
end
end
|
http://rosettacode.org/wiki/Permutations/Rank_of_a_permutation | Permutations/Rank of a permutation | A particular ranking of a permutation associates an integer with a particular ordering of all the permutations of a set of distinct items.
For our purposes the ranking will assign integers
0..
(
n
!
−
1
)
{\displaystyle 0..(n!-1)}
to an ordering of all the permutations of the integers
0..
(
n
−
1
)
{\displaystyle 0..(n-1)}
.
For example, the permutations of the digits zero to 3 arranged lexicographically have the following rank:
PERMUTATION RANK
(0, 1, 2, 3) -> 0
(0, 1, 3, 2) -> 1
(0, 2, 1, 3) -> 2
(0, 2, 3, 1) -> 3
(0, 3, 1, 2) -> 4
(0, 3, 2, 1) -> 5
(1, 0, 2, 3) -> 6
(1, 0, 3, 2) -> 7
(1, 2, 0, 3) -> 8
(1, 2, 3, 0) -> 9
(1, 3, 0, 2) -> 10
(1, 3, 2, 0) -> 11
(2, 0, 1, 3) -> 12
(2, 0, 3, 1) -> 13
(2, 1, 0, 3) -> 14
(2, 1, 3, 0) -> 15
(2, 3, 0, 1) -> 16
(2, 3, 1, 0) -> 17
(3, 0, 1, 2) -> 18
(3, 0, 2, 1) -> 19
(3, 1, 0, 2) -> 20
(3, 1, 2, 0) -> 21
(3, 2, 0, 1) -> 22
(3, 2, 1, 0) -> 23
Algorithms exist that can generate a rank from a permutation for some particular ordering of permutations, and that can generate the same rank from the given individual permutation (i.e. given a rank of 17 produce (2, 3, 1, 0) in the example above).
One use of such algorithms could be in generating a small, random, sample of permutations of
n
{\displaystyle n}
items without duplicates when the total number of permutations is large. Remember that the total number of permutations of
n
{\displaystyle n}
items is given by
n
!
{\displaystyle n!}
which grows large very quickly: A 32 bit integer can only hold
12
!
{\displaystyle 12!}
, a 64 bit integer only
20
!
{\displaystyle 20!}
. It becomes difficult to take the straight-forward approach of generating all permutations then taking a random sample of them.
A question on the Stack Overflow site asked how to generate one million random and indivudual permutations of 144 items.
Task
Create a function to generate a permutation from a rank.
Create the inverse function that given the permutation generates its rank.
Show that for
n
=
3
{\displaystyle n=3}
the two functions are indeed inverses of each other.
Compute and show here 4 random, individual, samples of permutations of 12 objects.
Stretch goal
State how reasonable it would be to use your program to address the limits of the Stack Overflow question.
References
Ranking and Unranking Permutations in Linear Time by Myrvold & Ruskey. (Also available via Google here).
Ranks on the DevData site.
Another answer on Stack Overflow to a different question that explains its algorithm in detail.
Related tasks
Factorial_base_numbers_indexing_permutations_of_a_collection
| #Tcl | Tcl | # A functional swap routine
proc swap {v idx1 idx2} {
lset v $idx2 [lindex $v $idx1][lset v $idx1 [lindex $v $idx2];subst ""]
}
# Fill the integer array <vec> with the permutation at rank <rank>
proc computePermutation {vecName rank} {
upvar 1 $vecName vec
if {![info exist vec] || ![llength $vec]} return
set N [llength $vec]
for {set n 0} {$n < $N} {incr n} {lset vec $n $n}
for {set n $N} {$n>=1} {incr n -1} {
set r [expr {$rank % $n}]
set rank [expr {$rank / $n}]
set vec [swap $vec $r [expr {$n-1}]]
}
}
# Return the rank of the current permutation.
proc computeRank {vec} {
if {![llength $vec]} return
set inv [lrepeat [llength $vec] 0]
set i -1
foreach v $vec {lset inv $v [incr i]}
# First argument is lambda term
set mrRank1 {{f n vec inv} {
if {$n < 2} {return 0}
set s [lindex $vec [set n1 [expr {$n - 1}]]]
set vec [swap $vec $n1 [lindex $inv $n1]]
set inv [swap $inv $s $n1]
return [expr {$s + $n * [apply $f $f $n1 $vec $inv]}]
}}
return [apply $mrRank1 $mrRank1 [llength $vec] $vec $inv]
} |
http://rosettacode.org/wiki/Pierpont_primes | Pierpont primes | A Pierpont prime is a prime number of the form: 2u3v + 1 for some non-negative integers u and v .
A Pierpont prime of the second kind is a prime number of the form: 2u3v - 1 for some non-negative integers u and v .
The term "Pierpont primes" is generally understood to mean the first definition, but will be called "Pierpont primes of the first kind" on this page to distinguish them.
Task
Write a routine (function, procedure, whatever) to find Pierpont primes of the first & second kinds.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the first kind.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the second kind
If your language supports large integers, find and display here, on this page, the 250th Pierpont prime of the first kind and the 250th Pierpont prime of the second kind.
See also
Wikipedia - Pierpont primes
OEIS:A005109 - Class 1 -, or Pierpont primes
OEIS:A005105 - Class 1 +, or Pierpont primes of the second kind
| #Python | Python | import random
# Copied from https://rosettacode.org/wiki/Miller-Rabin_primality_test#Python
def is_Prime(n):
"""
Miller-Rabin primality test.
A return value of False means n is certainly not prime. A return value of
True means n is very likely a prime.
"""
if n!=int(n):
return False
n=int(n)
#Miller-Rabin test for prime
if n==0 or n==1 or n==4 or n==6 or n==8 or n==9:
return False
if n==2 or n==3 or n==5 or n==7:
return True
s = 0
d = n-1
while d%2==0:
d>>=1
s+=1
assert(2**s * d == n-1)
def trial_composite(a):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n-1:
return False
return True
for i in range(8):#number of trials
a = random.randrange(2, n)
if trial_composite(a):
return False
return True
def pierpont(ulim, vlim, first):
p = 0
p2 = 1
p3 = 1
pp = []
for v in xrange(vlim):
for u in xrange(ulim):
p = p2 * p3
if first:
p = p + 1
else:
p = p - 1
if is_Prime(p):
pp.append(p)
p2 = p2 * 2
p3 = p3 * 3
p2 = 1
pp.sort()
return pp
def main():
print "First 50 Pierpont primes of the first kind:"
pp = pierpont(120, 80, True)
for i in xrange(50):
print "%8d " % pp[i],
if (i - 9) % 10 == 0:
print
print "First 50 Pierpont primes of the second kind:"
pp2 = pierpont(120, 80, False)
for i in xrange(50):
print "%8d " % pp2[i],
if (i - 9) % 10 == 0:
print
print "250th Pierpont prime of the first kind:", pp[249]
print "250th Pierpont prime of the second kind:", pp2[249]
main() |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #Gambas | Gambas | Public Sub Main()
Dim sList As String[] = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"]
Print sList[Rand(0, 11)]
End |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #GAP | GAP | a := [2, 9, 4, 7, 5, 3];
Random(a); |
http://rosettacode.org/wiki/Primality_by_trial_division | Primality by trial division | Task
Write a boolean function that tells whether a given integer is prime.
Remember that 1 and all non-positive numbers are not prime.
Use trial division.
Even numbers greater than 2 may be eliminated right away.
A loop from 3 to √ n will suffice, but other loops are allowed.
Related tasks
count in factors
prime decomposition
AKS test for primes
factors of an integer
Sieve of Eratosthenes
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
| #VBScript | VBScript | Function IsPrime(n)
If n = 2 Then
IsPrime = True
ElseIf n <= 1 Or n Mod 2 = 0 Then
IsPrime = False
Else
IsPrime = True
For i = 3 To Int(Sqr(n)) Step 2
If n Mod i = 0 Then
IsPrime = False
Exit For
End If
Next
End If
End Function
For n = 1 To 50
If IsPrime(n) Then
WScript.StdOut.Write n & " "
End If
Next |
http://rosettacode.org/wiki/Phrase_reversals | Phrase reversals | Task
Given a string of space separated words containing the following phrase:
rosetta code phrase reversal
Reverse the characters of the string.
Reverse the characters of each individual word in the string, maintaining original word order within the string.
Reverse the order of each word of the string, maintaining the order of characters in each word.
Show your output here.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Go | Go | package main
import (
"fmt"
"strings"
)
const phrase = "rosetta code phrase reversal"
func revStr(s string) string {
rs := make([]rune, len(s))
i := len(s)
for _, r := range s {
i--
rs[i] = r
}
return string(rs[i:])
}
func main() {
fmt.Println("Reversed: ", revStr(phrase))
ws := strings.Fields(phrase)
for i, w := range ws {
ws[i] = revStr(w)
}
fmt.Println("Words reversed: ", strings.Join(ws, " "))
ws = strings.Fields(phrase)
last := len(ws) - 1
for i, w := range ws[:len(ws)/2] {
ws[i], ws[last-i] = ws[last-i], w
}
fmt.Println("Word order reversed:", strings.Join(ws, " "))
} |
http://rosettacode.org/wiki/Permutations/Derangements | Permutations/Derangements | A derangement is a permutation of the order of distinct items in which no item appears in its original place.
For example, the only two derangements of the three items (0, 1, 2) are (1, 2, 0), and (2, 0, 1).
The number of derangements of n distinct items is known as the subfactorial of n, sometimes written as !n.
There are various ways to calculate !n.
Task
Create a named function/method/subroutine/... to generate derangements of the integers 0..n-1, (or 1..n if you prefer).
Generate and show all the derangements of 4 integers using the above routine.
Create a function that calculates the subfactorial of n, !n.
Print and show a table of the counted number of derangements of n vs. the calculated !n for n from 0..9 inclusive.
Optional stretch goal
Calculate !20
Related tasks
Anagrams/Deranged anagrams
Best shuffle
Left_factorials
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Elixir | Elixir | defmodule Permutation do
def derangements(n) do
list = Enum.to_list(1..n)
Enum.filter(permutation(list), fn perm ->
Enum.zip(list, perm) |> Enum.all?(fn {a,b} -> a != b end)
end)
end
def subfact(0), do: 1
def subfact(1), do: 0
def subfact(n), do: (n-1) * (subfact(n-1) + subfact(n-2))
def permutation([]), do: [[]]
def permutation(list) do
for x <- list, y <- permutation(list -- [x]), do: [x|y]
end
end
IO.puts "derangements for n = 4"
Enum.each(Permutation.derangements(4), &IO.inspect &1)
IO.puts "\nNumber of derangements"
IO.puts " n derange subfact"
Enum.each(0..9, fn n ->
:io.format "~2w :~9w,~9w~n", [n, length(Permutation.derangements(n)), Permutation.subfact(n)]
end)
Enum.each(10..20, fn n ->
:io.format "~2w :~19w~n", [n, Permutation.subfact(n)]
end) |
http://rosettacode.org/wiki/Permutations_by_swapping | Permutations by swapping | Task
Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.
Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.
Show the permutations and signs of three items, in order of generation here.
Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.
Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.
References
Steinhaus–Johnson–Trotter algorithm
Johnson-Trotter Algorithm Listing All Permutations
Heap's algorithm
[1] Tintinnalogia
Related tasks
Matrix arithmetic
Gray code
| #Forth | Forth | S" fsl-util.fs" REQUIRED
S" fsl/dynmem.seq" REQUIRED
cell darray p{
: sgn
DUP 0 > IF
DROP 1
ELSE 0 < IF
-1
ELSE
0
THEN THEN ;
: arr-swap {: addr1 addr2 | tmp -- :}
addr1 @ TO tmp
addr2 @ addr1 !
tmp addr2 ! ;
: perms {: n xt | my-i k s -- :}
& p{ n 1+ }malloc malloc-fail? ABORT" perms :: out of memory"
0 p{ 0 } !
n 1+ 1 DO
I NEGATE p{ I } !
LOOP
1 TO s
BEGIN
1 n 1+ DO
p{ I } @ ABS
-1 +LOOP
n 1+ s xt EXECUTE
0 TO k
n 1+ 2 DO
p{ I } @ 0 < ( flag )
p{ I } @ ABS p{ I 1- } @ ABS > ( flag flag )
p{ I } @ ABS p{ k } @ ABS > ( flag flag flag )
AND AND IF
I TO k
THEN
LOOP
n 1 DO
p{ I } @ 0 > ( flag )
p{ I } @ ABS p{ I 1+ } @ ABS > ( flag flag )
p{ I } @ ABS p{ k } @ ABS > ( flag flag flag )
AND AND IF
I TO k
THEN
LOOP
k IF
n 1+ 1 DO
p{ I } @ ABS p{ k } @ ABS > IF
p{ I } @ NEGATE p{ I } !
THEN
LOOP
p{ k } @ sgn k + TO my-i
p{ k } p{ my-i } arr-swap
s NEGATE TO s
THEN
k 0 = UNTIL ;
: .perm ( p0 p1 p2 ... pn n s )
>R
." Perm: [ "
1 DO
. SPACE
LOOP
R> ." ] Sign: " . CR ;
3 ' .perm perms CR
4 ' .perm perms |
http://rosettacode.org/wiki/Permutations_by_swapping | Permutations by swapping | Task
Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.
Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.
Show the permutations and signs of three items, in order of generation here.
Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.
Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.
References
Steinhaus–Johnson–Trotter algorithm
Johnson-Trotter Algorithm Listing All Permutations
Heap's algorithm
[1] Tintinnalogia
Related tasks
Matrix arithmetic
Gray code
| #FreeBASIC | FreeBASIC | ' version 31-03-2017
' compile with: fbc -s console
Sub perms(n As ULong)
Dim As Long p(n), i, k, s = 1
For i = 1 To n
p(i) = -i
Next
Do
Print "Perm: [ ";
For i = 1 To n
Print Abs(p(i)); " ";
Next
Print "] Sign: "; s
k = 0
For i = 2 To n
If p(i) < 0 Then
If Abs(p(i)) > Abs(p(i -1)) Then
If Abs(p(i)) > Abs(p(k)) Then k = i
End If
End If
Next
For i = 1 To n -1
If p(i) > 0 Then
If Abs(p(i)) > Abs(p(i +1)) Then
If Abs(p(i)) > Abs(p(k)) Then k = i
End If
End If
Next
If k Then
For i = 1 To n
If Abs(p(i)) > Abs(p(k)) Then p(i) = -p(i)
Next
i = k + Sgn(p(k))
Swap p(k), p(i)
s = -s
End If
Loop Until k = 0
End Sub
' ------=< MAIN >=------
perms(3)
print
perms(4)
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
|
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #J | J | require'stats'
trmt=: 0.85 0.88 0.75 0.66 0.25 0.29 0.83 0.39 0.97
ctrl=: 0.68 0.41 0.1 0.49 0.16 0.65 0.32 0.92 0.28 0.98
difm=: -&mean
result=: trmt difm ctrl
all=: trmt(#@[ ({. difm }.) |:@([ (comb ~.@,"1 i.@])&# ,) { ,) ctrl
smoutput 'under: ','%',~":100*mean all <: result
smoutput 'over: ','%',~":100*mean all > result |
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #Java | Java | public class PermutationTest {
private static final int[] data = new int[]{
85, 88, 75, 66, 25, 29, 83, 39, 97,
68, 41, 10, 49, 16, 65, 32, 92, 28, 98
};
private static int pick(int at, int remain, int accu, int treat) {
if (remain == 0) return (accu > treat) ? 1 : 0;
return pick(at - 1, remain - 1, accu + data[at - 1], treat)
+ ((at > remain) ? pick(at - 1, remain, accu, treat) : 0);
}
public static void main(String[] args) {
int treat = 0;
double total = 1.0;
for (int i = 0; i <= 8; ++i) {
treat += data[i];
}
for (int i = 19; i >= 11; --i) {
total *= i;
}
for (int i = 9; i >= 1; --i) {
total /= i;
}
int gt = pick(19, 9, 0, treat);
int le = (int) (total - gt);
System.out.printf("<= : %f%% %d\n", 100.0 * le / total, le);
System.out.printf(" > : %f%% %d\n", 100.0 * gt / total, gt);
}
} |
http://rosettacode.org/wiki/Percentage_difference_between_images | Percentage difference between images | basic bitmap storage
Useful for comparing two JPEG images saved with a different compression ratios.
You can use these pictures for testing (use the full-size version of each):
50% quality JPEG
100% quality JPEG
link to full size 50% image
link to full size 100% image
The expected difference for these two images is 1.62125%
| #Ada | Ada | type Count is mod 2**64; |
http://rosettacode.org/wiki/Percolation/Mean_cluster_density | Percolation/Mean cluster density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
c
{\displaystyle c}
be a 2D boolean square matrix of
n
×
n
{\displaystyle n\times n}
values of either 1 or 0 where the
probability of any value being 1 is
p
{\displaystyle p}
, (and of 0 is therefore
1
−
p
{\displaystyle 1-p}
).
We define a cluster of 1's as being a group of 1's connected vertically or
horizontally (i.e., using the Von Neumann neighborhood rule) and bounded by either
0
{\displaystyle 0}
or by the limits of the matrix.
Let the number of such clusters in such a randomly constructed matrix be
C
n
{\displaystyle C_{n}}
.
Percolation theory states that
K
(
p
)
{\displaystyle K(p)}
(the mean cluster density) will satisfy
K
(
p
)
=
C
n
/
n
2
{\displaystyle K(p)=C_{n}/n^{2}}
as
n
{\displaystyle n}
tends to infinity. For
p
=
0.5
{\displaystyle p=0.5}
,
K
(
p
)
{\displaystyle K(p)}
is found numerically to approximate
0.065770
{\displaystyle 0.065770}
...
Task
Show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
for
p
=
0.5
{\displaystyle p=0.5}
and
for values of
n
{\displaystyle n}
up to at least
1000
{\displaystyle 1000}
.
Any calculation of
C
n
{\displaystyle C_{n}}
for finite
n
{\displaystyle n}
is subject to randomness, so an approximation should be
computed as the average of
t
{\displaystyle t}
runs, where
t
{\displaystyle t}
≥
5
{\displaystyle 5}
.
For extra credit, graphically show clusters in a
15
×
15
{\displaystyle 15\times 15}
,
p
=
0.5
{\displaystyle p=0.5}
grid.
Show your output here.
See also
s-Cluster on Wolfram mathworld. | #D | D | import std.stdio, std.algorithm, std.random, std.math, std.array,
std.range, std.ascii;
alias Cell = ubyte;
alias Grid = Cell[][];
enum Cell notClustered = 1; // Filled cell, but not in a cluster.
Grid initialize(Grid grid, in double prob, ref Xorshift rng) nothrow {
foreach (row; grid)
foreach (ref cell; row)
cell = Cell(rng.uniform01 < prob);
return grid;
}
void show(in Grid grid) {
immutable static cell2char = " #" ~ letters;
writeln('+', "-".replicate(grid.length), '+');
foreach (row; grid) {
write('|');
row.map!(c => c < cell2char.length ? cell2char[c] : '@').write;
writeln('|');
}
writeln('+', "-".replicate(grid.length), '+');
}
size_t countClusters(bool justCount=false)(Grid grid)
pure nothrow @safe @nogc {
immutable side = grid.length;
static if (justCount)
enum Cell clusterID = 2;
else
Cell clusterID = 1;
void walk(in size_t r, in size_t c) nothrow @safe @nogc {
grid[r][c] = clusterID; // Fill grid.
if (r < side - 1 && grid[r + 1][c] == notClustered) // Down.
walk(r + 1, c);
if (c < side - 1 && grid[r][c + 1] == notClustered) // Right.
walk(r, c + 1);
if (c > 0 && grid[r][c - 1] == notClustered) // Left.
walk(r, c - 1);
if (r > 0 && grid[r - 1][c] == notClustered) // Up.
walk(r - 1, c);
}
size_t nClusters = 0;
foreach (immutable r; 0 .. side)
foreach (immutable c; 0 .. side)
if (grid[r][c] == notClustered) {
static if (!justCount)
clusterID++;
nClusters++;
walk(r, c);
}
return nClusters;
}
double clusterDensity(Grid grid, in double prob, ref Xorshift rng) {
return grid.initialize(prob, rng).countClusters!true /
double(grid.length ^^ 2);
}
void showDemo(in size_t side, in double prob, ref Xorshift rng) {
auto grid = new Grid(side, side);
grid.initialize(prob, rng);
writefln("Found %d clusters in this %d by %d grid:\n",
grid.countClusters, side, side);
grid.show;
}
void main() {
immutable prob = 0.5;
immutable nIters = 5;
auto rng = Xorshift(unpredictableSeed);
showDemo(15, prob, rng);
writeln;
foreach (immutable i; iota(4, 14, 2)) {
immutable side = 2 ^^ i;
auto grid = new Grid(side, side);
immutable density = nIters
.iota
.map!(_ => grid.clusterDensity(prob, rng))
.sum / nIters;
writefln("n_iters=%3d, p=%4.2f, n=%5d, sim=%7.8f",
nIters, prob, side, density);
}
} |
http://rosettacode.org/wiki/Perfect_numbers | Perfect numbers | Write a function which says whether a number is perfect.
A perfect number is a positive integer that is the sum of its proper positive divisors excluding the number itself.
Equivalently, a perfect number is a number that is half the sum of all of its positive divisors (including itself).
Note: The faster Lucas-Lehmer test is used to find primes of the form 2n-1, all known perfect numbers can be derived from these primes
using the formula (2n - 1) × 2n - 1.
It is not known if there are any odd perfect numbers (any that exist are larger than 102000).
The number of known perfect numbers is 51 (as of December, 2018), and the largest known perfect number contains 49,724,095 decimal digits.
See also
Rational Arithmetic
Perfect numbers on OEIS
Odd Perfect showing the current status of bounds on odd perfect numbers.
| #360_Assembly | 360 Assembly | * Perfect numbers 15/05/2016
PERFECTN CSECT
USING PERFECTN,R13 prolog
SAVEAREA B STM-SAVEAREA(R15) "
DC 17F'0' "
STM STM R14,R12,12(R13) "
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
LA R6,2 i=2
LOOPI C R6,NN do i=2 to nn
BH ELOOPI
LR R1,R6 i
BAL R14,PERFECT
LTR R0,R0 if perfect(i)
BZ NOTPERF
XDECO R6,PG edit i
XPRNT PG,L'PG print i
NOTPERF LA R6,1(R6) i=i+1
B LOOPI
ELOOPI L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
PERFECT SR R9,R9 function perfect(n); sum=0
LA R7,1 j
LR R8,R1 n
SRA R8,1 n/2
LOOPJ CR R7,R8 do j=1 to n/2
BH ELOOPJ
LR R4,R1 n
SRDA R4,32
DR R4,R7 n/j
LTR R4,R4 if mod(n,j)=0
BNZ NOTMOD
AR R9,R7 sum=sum+j
NOTMOD LA R7,1(R7) j=j+1
B LOOPJ
ELOOPJ SR R0,R0 r0=false
CR R9,R1 if sum=n
BNE NOTEQ
BCTR R0,0 r0=true
NOTEQ BR R14 return(r0); end perfect
NN DC F'10000'
PG DC CL12' ' buffer
YREGS
END PERFECTN |
http://rosettacode.org/wiki/Percolation/Bond_percolation | Percolation/Bond percolation |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Given an
M
×
N
{\displaystyle M\times N}
rectangular array of cells numbered
c
e
l
l
[
0..
M
−
1
,
0..
N
−
1
]
{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}
, assume
M
{\displaystyle M}
is horizontal and
N
{\displaystyle N}
is downwards. Each
c
e
l
l
[
m
,
n
]
{\displaystyle \mathrm {cell} [m,n]}
is bounded by (horizontal) walls
h
w
a
l
l
[
m
,
n
]
{\displaystyle \mathrm {hwall} [m,n]}
and
h
w
a
l
l
[
m
+
1
,
n
]
{\displaystyle \mathrm {hwall} [m+1,n]}
; (vertical) walls
v
w
a
l
l
[
m
,
n
]
{\displaystyle \mathrm {vwall} [m,n]}
and
v
w
a
l
l
[
m
,
n
+
1
]
{\displaystyle \mathrm {vwall} [m,n+1]}
Assume that the probability of any wall being present is a constant
p
{\displaystyle p}
where
0.0
≤
p
≤
1.0
{\displaystyle 0.0\leq p\leq 1.0}
Except for the outer horizontal walls at
m
=
0
{\displaystyle m=0}
and
m
=
M
{\displaystyle m=M}
which are always present.
The task
Simulate pouring a fluid onto the top surface (
n
=
0
{\displaystyle n=0}
) where the fluid will enter any empty cell it is adjacent to if there is no wall between where it currently is and the cell on the other side of the (missing) wall.
The fluid does not move beyond the horizontal constraints of the grid.
The fluid may move “up” within the confines of the grid of cells. If the fluid reaches a bottom cell that has a missing bottom wall then the fluid can be said to 'drip' out the bottom at that point.
Given
p
{\displaystyle p}
repeat the percolation
t
{\displaystyle t}
times to estimate the proportion of times that the fluid can percolate to the bottom for any given
p
{\displaystyle p}
.
Show how the probability of percolating through the random grid changes with
p
{\displaystyle p}
going from
0.0
{\displaystyle 0.0}
to
1.0
{\displaystyle 1.0}
in
0.1
{\displaystyle 0.1}
increments and with the number of repetitions to estimate the fraction at any given
p
{\displaystyle p}
as
t
=
100
{\displaystyle t=100}
.
Use an
M
=
10
,
N
=
10
{\displaystyle M=10,N=10}
grid of cells for all cases.
Optionally depict fluid successfully percolating through a grid graphically.
Show all output on this page.
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
// cell states
#define FILL 1
#define RWALL 2 // right wall
#define BWALL 4 // bottom wall
typedef unsigned int c_t;
c_t *cells, *start, *end;
int m, n;
void make_grid(double p, int x, int y)
{
int i, j, thresh = RAND_MAX * p;
m = x, n = y;
// Allocate two addition rows to avoid checking bounds.
// Bottom row is also required by drippage
start = realloc(start, m * (n + 2) * sizeof(c_t));
cells = start + m;
for (i = 0; i < m; i++)
start[i] = BWALL | RWALL;
for (i = 0, end = cells; i < y; i++) {
for (j = x; --j; )
*end++ = (rand() < thresh ? BWALL : 0)
|(rand() < thresh ? RWALL : 0);
*end++ = RWALL | (rand() < thresh ? BWALL: 0);
}
memset(end, 0, sizeof(c_t) * m);
}
void show_grid(void)
{
int i, j;
for (j = 0; j < m; j++) printf("+--");
puts("+");
for (i = 0; i <= n; i++) {
putchar(i == n ? ' ' : '|');
for (j = 0; j < m; j++) {
printf((cells[i*m + j] & FILL) ? "[]" : " ");
putchar((cells[i*m + j] & RWALL) ? '|' : ' ');
}
putchar('\n');
if (i == n) return;
for (j = 0; j < m; j++)
printf((cells[i*m + j] & BWALL) ? "+--" : "+ ");
puts("+");
}
}
int fill(c_t *p)
{
if ((*p & FILL)) return 0;
*p |= FILL;
if (p >= end) return 1; // success: reached bottom row
return ( !(p[ 0] & BWALL) && fill(p + m) ) ||
( !(p[ 0] & RWALL) && fill(p + 1) ) ||
( !(p[-1] & RWALL) && fill(p - 1) ) ||
( !(p[-m] & BWALL) && fill(p - m) );
}
int percolate(void)
{
int i;
for (i = 0; i < m && !fill(cells + i); i++);
return i < m;
}
int main(void)
{
make_grid(.5, 10, 10);
percolate();
show_grid();
int cnt, i, p;
puts("\nrunning 10x10 grids 10000 times for each p:");
for (p = 1; p < 10; p++) {
for (cnt = i = 0; i < 10000; i++) {
make_grid(p / 10., 10, 10);
cnt += percolate();
//show_grid(); // don't
}
printf("p = %3g: %.4f\n", p / 10., (double)cnt / i);
}
free(start);
return 0;
} |
http://rosettacode.org/wiki/Percolation/Mean_run_density | Percolation/Mean run density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
v
{\displaystyle v}
be a vector of
n
{\displaystyle n}
values of either 1 or 0 where the probability of any
value being 1 is
p
{\displaystyle p}
; the probability of a value being 0 is therefore
1
−
p
{\displaystyle 1-p}
.
Define a run of 1s as being a group of consecutive 1s in the vector bounded
either by the limits of the vector or by a 0. Let the number of such runs in a given
vector of length
n
{\displaystyle n}
be
R
n
{\displaystyle R_{n}}
.
For example, the following vector has
R
10
=
3
{\displaystyle R_{10}=3}
[1 1 0 0 0 1 0 1 1 1]
^^^ ^ ^^^^^
Percolation theory states that
K
(
p
)
=
lim
n
→
∞
R
n
/
n
=
p
(
1
−
p
)
{\displaystyle K(p)=\lim _{n\to \infty }R_{n}/n=p(1-p)}
Task
Any calculation of
R
n
/
n
{\displaystyle R_{n}/n}
for finite
n
{\displaystyle n}
is subject to randomness so should be
computed as the average of
t
{\displaystyle t}
runs, where
t
≥
100
{\displaystyle t\geq 100}
.
For values of
p
{\displaystyle p}
of 0.1, 0.3, 0.5, 0.7, and 0.9, show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
.
Show your output here.
See also
s-Run on Wolfram mathworld. | #EchoLisp | EchoLisp |
;; count 1-runs - The vector is not stored
(define (runs p n)
(define ct 0)
(define run-1 #t)
(for ([i n])
(if (< (random) p)
(set! run-1 #t) ;; 0 case
(begin ;; 1 case
(when run-1 (set! ct (1+ ct)))
(set! run-1 #f))))
(// ct n))
;; mean of t counts
(define (truns p (n 1000 ) (t 1000))
(// (for/sum ([i t]) (runs p n)) t))
(define (task)
(for ([p (in-range 0.1 1.0 0.2)])
(writeln)
(writeln '🔸 'p p 'Kp (* p (- 1 p)))
(for ([n '(10 100 1000)])
(printf "\t-- n %5d → %d" n (truns p n)))))
|
http://rosettacode.org/wiki/Percolation/Site_percolation | Percolation/Site percolation |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Given an
M
×
N
{\displaystyle M\times N}
rectangular array of cells numbered
c
e
l
l
[
0..
M
−
1
,
0..
N
−
1
]
{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}
assume
M
{\displaystyle M}
is horizontal and
N
{\displaystyle N}
is downwards.
Assume that the probability of any cell being filled is a constant
p
{\displaystyle p}
where
0.0
≤
p
≤
1.0
{\displaystyle 0.0\leq p\leq 1.0}
The task
Simulate creating the array of cells with probability
p
{\displaystyle p}
and then
testing if there is a route through adjacent filled cells from any on row
0
{\displaystyle 0}
to any on row
N
{\displaystyle N}
, i.e. testing for site percolation.
Given
p
{\displaystyle p}
repeat the percolation
t
{\displaystyle t}
times to estimate the proportion of times that the fluid can percolate to the bottom for any given
p
{\displaystyle p}
.
Show how the probability of percolating through the random grid changes with
p
{\displaystyle p}
going from
0.0
{\displaystyle 0.0}
to
1.0
{\displaystyle 1.0}
in
0.1
{\displaystyle 0.1}
increments and with the number of repetitions to estimate the fraction at any given
p
{\displaystyle p}
as
t
>=
100
{\displaystyle t>=100}
.
Use an
M
=
15
,
N
=
15
{\displaystyle M=15,N=15}
grid of cells for all cases.
Optionally depict a percolation through a cell grid graphically.
Show all output on this page.
| #FreeBASIC | FreeBASIC | #define SOLID "#"
#define EMPTY " "
#define WET "v"
Dim Shared As String grid()
Dim Shared As Integer last, lastrow, m, n
Sub make_grid(x As Integer, y As Integer, p As Double)
m = x
n = y
Redim Preserve grid(x*(y+1)+1)
last = Len(grid)
Dim As Integer lastrow = last-n
Dim As Integer i, j
For i = 0 To x-1
For j = 1 To y
grid(1+i*(y+1)+j) = Iif(Rnd < p, EMPTY, SOLID)
Next j
Next i
End Sub
Function ff(i As Integer) As Boolean
If i <= 0 Or i >= last Or grid(i) <> EMPTY Then Return 0
grid(i) = WET
Return i >= lastrow Or (ff(i+m+1) Or ff(i+1) Or ff(i-1) Or ff(i-m-1))
End Function
Function percolate() As Integer
For i As Integer = 2 To m+1
If ff(i) Then Return 1
Next i
Return 0
End Function
Dim As Double p
Dim As Integer ip, i, cont
make_grid(15, 15, 0.55)
Print "15x15 grid:"
For i = 1 To Ubound(grid)
Print grid(i);
If i Mod 15 = 0 Then Print
Next i
Print !"\nrunning 10,000 tests for each case:"
For ip As Ubyte = 0 To 10
p = ip / 10
cont = 0
For i = 1 To 10000
make_grid(15, 15, p)
cont += percolate()
Next i
Print Using "p=#.#: #.####"; p; cont/10000
Next ip
Sleep |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #ABAP | ABAP | data: lv_flag type c,
lv_number type i,
lt_numbers type table of i.
append 1 to lt_numbers.
append 2 to lt_numbers.
append 3 to lt_numbers.
do.
perform permute using lt_numbers changing lv_flag.
if lv_flag = 'X'.
exit.
endif.
loop at lt_numbers into lv_number.
write (1) lv_number no-gap left-justified.
if sy-tabix <> '3'.
write ', '.
endif.
endloop.
skip.
enddo.
" Permutation function - this is used to permute:
" Can be used for an unbounded size set.
form permute using iv_set like lt_numbers
changing ev_last type c.
data: lv_len type i,
lv_first type i,
lv_third type i,
lv_count type i,
lv_temp type i,
lv_temp_2 type i,
lv_second type i,
lv_changed type c,
lv_perm type i.
describe table iv_set lines lv_len.
lv_perm = lv_len - 1.
lv_changed = ' '.
" Loop backwards through the table, attempting to find elements which
" can be permuted. If we find one, break out of the table and set the
" flag indicating a switch.
do.
if lv_perm <= 0.
exit.
endif.
" Read the elements.
read table iv_set index lv_perm into lv_first.
add 1 to lv_perm.
read table iv_set index lv_perm into lv_second.
subtract 1 from lv_perm.
if lv_first < lv_second.
lv_changed = 'X'.
exit.
endif.
subtract 1 from lv_perm.
enddo.
" Last permutation.
if lv_changed <> 'X'.
ev_last = 'X'.
exit.
endif.
" Swap tail decresing to get a tail increasing.
lv_count = lv_perm + 1.
do.
lv_first = lv_len + lv_perm - lv_count + 1.
if lv_count >= lv_first.
exit.
endif.
read table iv_set index lv_count into lv_temp.
read table iv_set index lv_first into lv_temp_2.
modify iv_set index lv_count from lv_temp_2.
modify iv_set index lv_first from lv_temp.
add 1 to lv_count.
enddo.
lv_count = lv_len - 1.
do.
if lv_count <= lv_perm.
exit.
endif.
read table iv_set index lv_count into lv_first.
read table iv_set index lv_perm into lv_second.
read table iv_set index lv_len into lv_third.
if ( lv_first < lv_third ) and ( lv_first > lv_second ).
lv_len = lv_count.
endif.
subtract 1 from lv_count.
enddo.
read table iv_set index lv_perm into lv_temp.
read table iv_set index lv_len into lv_temp_2.
modify iv_set index lv_perm from lv_temp_2.
modify iv_set index lv_len from lv_temp.
endform. |
http://rosettacode.org/wiki/Perfect_shuffle | Perfect shuffle | A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:
7♠ 8♠ 9♠ J♠ Q♠ K♠→7♠ 8♠ 9♠
J♠ Q♠ K♠→7♠ J♠ 8♠ Q♠ 9♠ K♠
When you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:
original:
1
2
3
4
5
6
7
8
after 1st shuffle:
1
5
2
6
3
7
4
8
after 2nd shuffle:
1
3
5
7
2
4
6
8
after 3rd shuffle:
1
2
3
4
5
6
7
8
The Task
Write a function that can perform a perfect shuffle on an even-sized list of values.
Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under "Test Cases" below.
You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all "cards" are unique within each deck.
Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.
Test Cases
input (deck size)
output (number of shuffles required)
8
3
24
11
52
8
100
30
1020
1018
1024
10
10000
300
| #APL | APL | faro ← ∊∘⍉(2,2÷⍨≢)⍴⊢
count ← {⍺←⍵ ⋄ ⍺≡r←⍺⍺ ⍵:1 ⋄ 1+⍺∇r}
(⊢,[1.5] (faro count ⍳)¨) 8 24 52 100 1020 1024 10000 |
http://rosettacode.org/wiki/Perfect_shuffle | Perfect shuffle | A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:
7♠ 8♠ 9♠ J♠ Q♠ K♠→7♠ 8♠ 9♠
J♠ Q♠ K♠→7♠ J♠ 8♠ Q♠ 9♠ K♠
When you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:
original:
1
2
3
4
5
6
7
8
after 1st shuffle:
1
5
2
6
3
7
4
8
after 2nd shuffle:
1
3
5
7
2
4
6
8
after 3rd shuffle:
1
2
3
4
5
6
7
8
The Task
Write a function that can perform a perfect shuffle on an even-sized list of values.
Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under "Test Cases" below.
You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all "cards" are unique within each deck.
Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.
Test Cases
input (deck size)
output (number of shuffles required)
8
3
24
11
52
8
100
30
1020
1018
1024
10
10000
300
| #Arturo | Arturo | perfectShuffle: function [deckSize][
deck: 1..deckSize
original: new deck
halfDeck: deckSize/2
i: 1
while [true][
deck: flatten couple first.n: halfDeck deck last.n: halfDeck deck
if deck = original -> return i
i: i+1
]
]
loop [8 24 52 100 1020 1024 10000] 's ->
print [
pad.right join @["Perfect shuffles required for deck size " to :string s ":"] 48
perfectShuffle s
] |
http://rosettacode.org/wiki/Perlin_noise | Perlin noise | The Perlin noise is a kind of gradient noise invented by Ken Perlin around the end of the twentieth century and still currently heavily used in computer graphics, most notably to procedurally generate textures or heightmaps.
The Perlin noise is basically a pseudo-random mapping of
R
d
{\displaystyle \mathbb {R} ^{d}}
into
R
{\displaystyle \mathbb {R} }
with an integer
d
{\displaystyle d}
which can be arbitrarily large but which is usually 2, 3, or 4.
Either by using a dedicated library or by implementing the algorithm, show that the Perlin noise (as defined in 2002 in the Java implementation below) of the point in 3D-space with coordinates 3.14, 42, 7 is 0.13691995878400012.
Note: this result assumes 64 bit IEEE-754 floating point calculations. If your language uses a different floating point representation, make a note of it and calculate the value accurate to 15 decimal places, or your languages accuracy threshold if it is less. Trailing zeros need not be displayed.
| #Fortran | Fortran |
PROGRAM PERLIN
IMPLICIT NONE
INTEGER :: i
INTEGER, DIMENSION(0:511) :: p
INTEGER, DIMENSION(0:255), PARAMETER :: permutation = (/151,160,137,91,90,15, &
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23, &
190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33, &
88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166, &
77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244, &
102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196, &
135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123, &
5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42, &
223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9, &
129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228, &
251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107, &
49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254, &
138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180/)
DO i=0, 255
p(i) = permutation(i)
p(256+i) = permutation(i)
END DO
WRITE(*,"(F19.17)") NOISE(3.14d0, 42d0, 7d0)
CONTAINS
DOUBLE PRECISION FUNCTION NOISE(x_in, y_in, z_in)
DOUBLE PRECISION, INTENT(IN) :: x_in, y_in, z_in
DOUBLE PRECISION :: x, y, z
INTEGER :: xx, yy, zz, a, aa, ab, b, ba, bb
DOUBLE PRECISION :: u, v, w
x = x_in
y = y_in
z = z_in
xx = IAND(FLOOR(x), 255)
yy = IAND(FLOOR(y), 255)
zz = IAND(FLOOR(z), 255)
x = x - FLOOR(x)
y = y - FLOOR(y)
z = z - FLOOR(z)
u = FADE(x)
v = FADE(y)
w = FADE(z)
a = p(xx) + yy
aa = p(a) + zz
ab = p(a+1) + zz
b = p(xx+1) + yy
ba = p(b) + zz
bb = p(b+1) + zz
NOISE = LERP(w, LERP(v, LERP(u, GRAD(p(aa), x, y, z), &
GRAD(p(ba), x-1, y, z)), &
LERP(u, GRAD(p(ab), x, y-1, z), &
GRAD(p(bb), x-1, y-1, z))), &
LERP(v, LERP(u, GRAD(p(aa+1), x, y, z-1), &
GRAD(p(ba+1), x-1, y, z-1)), &
LERP(u, GRAD(p(ab+1), x, y-1, z-1), &
GRAD(p(bb+1), x-1, y-1, z-1))))
END FUNCTION
DOUBLE PRECISION FUNCTION FADE(t)
DOUBLE PRECISION, INTENT(IN) :: t
FADE = t ** 3 * (t * ( t * 6 - 15) + 10)
END FUNCTION
DOUBLE PRECISION FUNCTION LERP(t, a, b)
DOUBLE PRECISION, INTENT(IN) :: t, a, b
LERP = a + t * (b - a)
END FUNCTION
DOUBLE PRECISION FUNCTION GRAD(hash, x, y, z)
INTEGER, INTENT(IN) :: hash
DOUBLE PRECISION, INTENT(IN) :: x, y, z
INTEGER :: h
DOUBLE PRECISION :: u, v
h = IAND(hash, 15)
u = MERGE(x, y, h < 8)
v = MERGE(y, MERGE(x, z, h == 12 .OR. h == 14), h < 4)
GRAD = MERGE(u, -u, IAND(h, 1) == 0) + MERGE(v, -v, IAND(h, 2) == 0)
END FUNCTION
END PROGRAM
|
http://rosettacode.org/wiki/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #C.2B.2B | C++ | #include <cassert>
#include <iostream>
#include <vector>
class totient_calculator {
public:
explicit totient_calculator(int max) : totient_(max + 1) {
for (int i = 1; i <= max; ++i)
totient_[i] = i;
for (int i = 2; i <= max; ++i) {
if (totient_[i] < i)
continue;
for (int j = i; j <= max; j += i)
totient_[j] -= totient_[j] / i;
}
}
int totient(int n) const {
assert (n >= 1 && n < totient_.size());
return totient_[n];
}
bool is_prime(int n) const {
return totient(n) == n - 1;
}
private:
std::vector<int> totient_;
};
bool perfect_totient_number(const totient_calculator& tc, int n) {
int sum = 0;
for (int m = n; m > 1; ) {
int t = tc.totient(m);
sum += t;
m = t;
}
return sum == n;
}
int main() {
totient_calculator tc(10000);
int count = 0, n = 1;
std::cout << "First 20 perfect totient numbers:\n";
for (; count < 20; ++n) {
if (perfect_totient_number(tc, n)) {
if (count > 0)
std::cout << ' ';
++count;
std::cout << n;
}
}
std::cout << '\n';
return 0;
} |
http://rosettacode.org/wiki/Playing_cards | Playing cards | Task
Create a data structure and the associated methods to define and manipulate a deck of playing cards.
The deck should contain 52 unique cards.
The methods must include the ability to:
make a new deck
shuffle (randomize) the deck
deal from the deck
print the current contents of a deck
Each card must have a pip value and a suit value which constitute the unique value of the card.
Related tasks:
Card shuffles
Deal cards_for_FreeCell
War Card_Game
Poker hand_analyser
Go Fish
| #Elixir | Elixir | defmodule Card do
defstruct pip: nil, suit: nil
end
defmodule Playing_cards do
@pips ~w[2 3 4 5 6 7 8 9 10 Jack Queen King Ace]a
@suits ~w[Clubs Hearts Spades Diamonds]a
@pip_value Enum.with_index(@pips)
@suit_value Enum.with_index(@suits)
def deal( n_cards, deck ), do: Enum.split( deck, n_cards )
def deal( n_hands, n_cards, deck ) do
Enum.reduce(1..n_hands, {[], deck}, fn _,{acc,d} ->
{hand, new_d} = deal(n_cards, d)
{[hand | acc], new_d}
end)
end
def deck, do: (for x <- @suits, y <- @pips, do: %Card{suit: x, pip: y})
def print( cards ), do: IO.puts (for x <- cards, do: "\t#{inspect x}")
def shuffle( deck ), do: Enum.shuffle( deck )
def sort_pips( cards ), do: Enum.sort_by( cards, &@pip_value[&1.pip] )
def sort_suits( cards ), do: Enum.sort_by( cards, &(@suit_value[&1.suit]) )
def task do
shuffled = shuffle( deck )
{hand, new_deck} = deal( 3, shuffled )
{hands, _deck} = deal( 2, 3, new_deck )
IO.write "Hand:"
print( hand )
IO.puts "Hands:"
for x <- hands, do: print(x)
end
end
Playing_cards.task |
http://rosettacode.org/wiki/Pi | Pi |
Create a program to continually calculate and output the next decimal digit of
π
{\displaystyle \pi }
(pi).
The program should continue forever (until it is aborted by the user) calculating and outputting each decimal digit in succession.
The output should be a decimal sequence beginning 3.14159265 ...
Note: this task is about calculating pi. For information on built-in pi constants see Real constants and functions.
Related Task Arithmetic-geometric mean/Calculate Pi
| #Delphi | Delphi |
unit Pi_BBC_Main;
interface
uses
Classes, Controls, Forms, Dialogs, StdCtrls;
type
TForm1 = class(TForm)
btnRunSpigotAlgo: TButton;
memScreen: TMemo;
procedure btnRunSpigotAlgoClick(Sender: TObject);
procedure FormCreate(Sender: TObject);
private
fScreenWidth : integer;
fLineBuffer : string;
procedure ClearText();
procedure AddText( const s : string);
procedure FlushText();
end;
var
Form1: TForm1;
implementation
{$R *.dfm}
uses SysUtils;
// Button clicked to run algorithm
procedure TForm1.btnRunSpigotAlgoClick(Sender: TObject);
var
// BBC Basic variables. Delphi longint is 32 bits.
B : array of longint;
A, C, D, E, I, L, M, P : longint;
// Added for Delphi version
temp : string;
h, j, t : integer;
begin
fScreenWidth := 80;
ClearText();
M := 5368709; // floor( (2^31 - 1)/400 )
// DIM B%(M%) in BBC Basic declares an array [0..M%], i.e. M% + 1 elements
SetLength( B, M + 1);
for I := 0 to M do B[I] := 20;
E := 0;
L := 2;
// FOR C% = M% TO 14 STEP -7
// In Delphi (or at least Delphi 7) the step size in a for loop has to be 1.
// So the BBC Basic FOR loop has been replaced by a repeat loop.
C := M;
repeat
D := 0;
A := C*2 - 1;
for P := C downto 1 do begin
D := D*P + B[P]*$64; // hex notation copied from BBC version
B[P] := D mod A;
D := D div A;
dec( A, 2);
end;
// The BBC CASE statement here amounts to a series of if ... else
if (D = 99) then begin
E := E*100 + D;
inc( L, 2);
end
else if (C = M) then begin
AddText( SysUtils.Format( '%2.1f', [1.0*(D div 100) / 10.0] ));
E := D mod 100;
end
else begin
// PRINT RIGHT$(STRING$(L%,"0") + STR$(E% + D% DIV 100),L%);
// This can't be done so concisely in Delphi 7
SetLength( temp, L);
for j := 1 to L do temp[j] := '0';
temp := temp + SysUtils.IntToStr( E + D div 100);
t := Length( temp);
AddText( Copy( temp, t - L + 1, L));
E := D mod 100;
L := 2;
end;
dec( C, 7);
until (C < 14);
FlushText();
// Delphi addition: Write screen output to a file for checking
h := SysUtils.FileCreate( 'C:\Delphi\PiDigits.txt'); // h = file handle
for j := 0 to memScreen.Lines.Count - 1 do
SysUtils.FileWrite( h, memScreen.Lines[j][1], Length( memScreen.Lines[j]));
SysUtils.FileClose( h);
end;
{=========================== Auxiliary routines ===========================}
// Form created
procedure TForm1.FormCreate(Sender: TObject);
begin
fScreenWidth := 80; // in case not set by the algotithm
ClearText();
end;
// This Delphi version builds each screen line in a buffer and puts
// the line into the TMemo when the buffer is full.
// This is faster than writing to the TMemo a few characters at a time,
// but note that the buffer must be flushed at the end of the program.
procedure TForm1.ClearText();
begin
memScreen.Lines.Clear();
fLineBuffer := '';
end;
procedure TForm1.AddText( const s : string);
var
nrChars, nrLeft : integer;
begin
nrChars := Length( s);
nrLeft := fScreenWidth - Length( fLineBuffer); // nr chars left in line
if (nrChars <= nrLeft) then fLineBuffer := fLineBuffer + s
else begin
fLineBuffer := fLineBuffer + Copy( s, 1, nrLeft);
memScreen.Lines.Add( fLineBuffer);
fLineBuffer := Copy( s, nrLeft + 1, nrChars - nrLeft);
end;
end;
procedure TForm1.FlushText();
begin
if (Length(fLineBuffer) > 0) then begin
memScreen.Lines.Add( fLineBuffer);
fLineBuffer := '';
end;
end;
end.
|
http://rosettacode.org/wiki/Periodic_table | Periodic table | Task
Display the row and column in the periodic table of the given atomic number.
The periodic table
Let us consider the following periodic table representation.
__________________________________________________________________________
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
| |
|1 H He |
| |
|2 Li Be B C N O F Ne |
| |
|3 Na Mg Al Si P S Cl Ar |
| |
|4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr |
| |
|5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe |
| |
|6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn |
| |
|7 Fr Ra ° Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og |
|__________________________________________________________________________|
| |
| |
|8 Lantanoidi* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu |
| |
|9 Aktinoidi° Ak Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr |
|__________________________________________________________________________|
Example test cases;
1 -> 1 1
2 -> 1 18
29 -> 4 11
42 -> 5 6
57 -> 8 4
58 -> 8 5
72 -> 6 4
89 -> 9 4
Details;
The representation of the periodic table may be represented in various way. The one presented in this challenge does have the following property : Lantanides and Aktinoides are all in a dedicated row, hence there is no element that is placed at 6, 3 nor 7, 3.
You may take a look at the atomic number repartitions here.
The atomic number is at least 1, at most 118.
See also
the periodic table
This task was an idea from CompSciFact
The periodic table in ascii that was used as template
| #Raku | Raku | my $b = 18;
my @offset = 16, 10, 10, (2×$b)+1, (-2×$b)-15, (2×$b)+1, (-2×$b)-15;
my @span = flat ^8 Zxx <1 3 8 44 15 17 15 15>;
for <1 2 29 42 57 58 72 89 90 103> -> $n {
printf "%3d: %2d, %2d\n", $n, map {$_+1}, ($n-1 + [+] @offset.head(@span[$n-1])).polymod($b).reverse;
} |
http://rosettacode.org/wiki/Periodic_table | Periodic table | Task
Display the row and column in the periodic table of the given atomic number.
The periodic table
Let us consider the following periodic table representation.
__________________________________________________________________________
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
| |
|1 H He |
| |
|2 Li Be B C N O F Ne |
| |
|3 Na Mg Al Si P S Cl Ar |
| |
|4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr |
| |
|5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe |
| |
|6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn |
| |
|7 Fr Ra ° Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og |
|__________________________________________________________________________|
| |
| |
|8 Lantanoidi* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu |
| |
|9 Aktinoidi° Ak Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr |
|__________________________________________________________________________|
Example test cases;
1 -> 1 1
2 -> 1 18
29 -> 4 11
42 -> 5 6
57 -> 8 4
58 -> 8 5
72 -> 6 4
89 -> 9 4
Details;
The representation of the periodic table may be represented in various way. The one presented in this challenge does have the following property : Lantanides and Aktinoides are all in a dedicated row, hence there is no element that is placed at 6, 3 nor 7, 3.
You may take a look at the atomic number repartitions here.
The atomic number is at least 1, at most 118.
See also
the periodic table
This task was an idea from CompSciFact
The periodic table in ascii that was used as template
| #Wren | Wren | import "./fmt" for Fmt
var limits = [3..10, 11..18, 19..36, 37..54, 55..86, 87..118]
var periodicTable = Fn.new { |n|
if (n < 1 || n > 118) Fiber.abort("Atomic number is out of range.")
if (n == 1) return [1, 1]
if (n == 2) return [1, 18]
if (n >= 57 && n <= 71) return [8, n - 53]
if (n >= 89 && n <= 103) return [9, n - 85]
var row
var start
var end
for (i in 0...limits.count) {
var limit = limits[i]
if (n >= limit.from && n <= limit.to) {
row = i + 2
start = limit.from
end = limit.to
break
}
}
if (n < start + 2 || row == 4 || row == 5) return [row, n - start + 1]
return [row, n - end + 18]
}
for (n in [1, 2, 29, 42, 57, 58, 59, 71, 72, 89, 90, 103, 113]) {
var rc = periodicTable.call(n)
Fmt.print("Atomic number $3d -> $d, $-2d", n, rc[0], rc[1])
} |
http://rosettacode.org/wiki/Pig_the_dice_game | Pig the dice game | The game of Pig is a multiplayer game played with a single six-sided die. The
object of the game is to reach 100 points or more.
Play is taken in turns. On each person's turn that person has the option of either:
Rolling the dice: where a roll of two to six is added to their score for that turn and the player's turn continues as the player is given the same choice again; or a roll of 1 loses the player's total points for that turn and their turn finishes with play passing to the next player.
Holding: the player's score for that round is added to their total and becomes safe from the effects of throwing a 1 (one). The player's turn finishes with play passing to the next player.
Task
Create a program to score for, and simulate dice throws for, a two-person game.
Related task
Pig the dice game/Player
| #Kotlin | Kotlin | // version 1.1.2
fun main(Args: Array<String>) {
print("Player 1 - Enter your name : ")
val name1 = readLine()!!.trim().let { if (it == "") "PLAYER 1" else it.toUpperCase() }
print("Player 2 - Enter your name : ")
val name2 = readLine()!!.trim().let { if (it == "") "PLAYER 2" else it.toUpperCase() }
val names = listOf(name1, name2)
val r = java.util.Random()
val totals = intArrayOf(0, 0)
var player = 0
while (true) {
println("\n${names[player]}")
println(" Your total score is currently ${totals[player]}")
var score = 0
while (true) {
print(" Roll or Hold r/h : ")
val rh = readLine()!![0].toLowerCase()
if (rh == 'h') {
totals[player] += score
println(" Your total score is now ${totals[player]}")
if (totals[player] >= 100) {
println(" So, ${names[player]}, YOU'VE WON!")
return
}
player = if (player == 0) 1 else 0
break
}
if (rh != 'r') {
println(" Must be 'r'or 'h', try again")
continue
}
val dice = 1 + r.nextInt(6)
println(" You have thrown a $dice")
if (dice == 1) {
println(" Sorry, your score for this round is now 0")
println(" Your total score remains at ${totals[player]}")
player = if (player == 0) 1 else 0
break
}
score += dice
println(" Your score for the round is now $score")
}
}
} |
http://rosettacode.org/wiki/Pernicious_numbers | Pernicious numbers | A pernicious number is a positive integer whose population count is a prime.
The population count is the number of ones in the binary representation of a non-negative integer.
Example
22 (which is 10110 in binary) has a population count of 3, which is prime, and therefore
22 is a pernicious number.
Task
display the first 25 pernicious numbers (in decimal).
display all pernicious numbers between 888,888,877 and 888,888,888 (inclusive).
display each list of integers on one line (which may or may not include a title).
See also
Sequence A052294 pernicious numbers on The On-Line Encyclopedia of Integer Sequences.
Rosetta Code entry population count, evil numbers, odious numbers.
| #Elixir | Elixir |
defmodule SieveofEratosthenes do
def init(lim) do
find_primes(2,lim,(2..lim))
end
def find_primes(count,lim,nums) when (count * count) > lim do
nums
end
def find_primes(count,lim,nums) when (count * count) <= lim do
e = Enum.reject(nums,&(rem(&1,count) == 0 and &1 > count))
find_primes(count+1,lim,e)
end
end
defmodule PerniciousNumbers do
def take(n) do
primes = SieveofEratosthenes.init(100)
Stream.iterate(1,&(&1+1))
|> Stream.filter(&(pernicious?(&1,primes)))
|> Enum.take(n)
|> IO.inspect
end
def between(a..b) do
primes = SieveofEratosthenes.init(100)
a..b
|> Stream.filter(&(pernicious?(&1,primes)))
|> Enum.to_list
|> IO.inspect
end
def ones(num) do
num
|> Integer.to_string(2)
|> String.codepoints
|> Enum.count(fn n -> n == "1" end)
end
def pernicious?(n,primes), do: Enum.member?(primes,ones(n))
end
|
http://rosettacode.org/wiki/Permutations/Rank_of_a_permutation | Permutations/Rank of a permutation | A particular ranking of a permutation associates an integer with a particular ordering of all the permutations of a set of distinct items.
For our purposes the ranking will assign integers
0..
(
n
!
−
1
)
{\displaystyle 0..(n!-1)}
to an ordering of all the permutations of the integers
0..
(
n
−
1
)
{\displaystyle 0..(n-1)}
.
For example, the permutations of the digits zero to 3 arranged lexicographically have the following rank:
PERMUTATION RANK
(0, 1, 2, 3) -> 0
(0, 1, 3, 2) -> 1
(0, 2, 1, 3) -> 2
(0, 2, 3, 1) -> 3
(0, 3, 1, 2) -> 4
(0, 3, 2, 1) -> 5
(1, 0, 2, 3) -> 6
(1, 0, 3, 2) -> 7
(1, 2, 0, 3) -> 8
(1, 2, 3, 0) -> 9
(1, 3, 0, 2) -> 10
(1, 3, 2, 0) -> 11
(2, 0, 1, 3) -> 12
(2, 0, 3, 1) -> 13
(2, 1, 0, 3) -> 14
(2, 1, 3, 0) -> 15
(2, 3, 0, 1) -> 16
(2, 3, 1, 0) -> 17
(3, 0, 1, 2) -> 18
(3, 0, 2, 1) -> 19
(3, 1, 0, 2) -> 20
(3, 1, 2, 0) -> 21
(3, 2, 0, 1) -> 22
(3, 2, 1, 0) -> 23
Algorithms exist that can generate a rank from a permutation for some particular ordering of permutations, and that can generate the same rank from the given individual permutation (i.e. given a rank of 17 produce (2, 3, 1, 0) in the example above).
One use of such algorithms could be in generating a small, random, sample of permutations of
n
{\displaystyle n}
items without duplicates when the total number of permutations is large. Remember that the total number of permutations of
n
{\displaystyle n}
items is given by
n
!
{\displaystyle n!}
which grows large very quickly: A 32 bit integer can only hold
12
!
{\displaystyle 12!}
, a 64 bit integer only
20
!
{\displaystyle 20!}
. It becomes difficult to take the straight-forward approach of generating all permutations then taking a random sample of them.
A question on the Stack Overflow site asked how to generate one million random and indivudual permutations of 144 items.
Task
Create a function to generate a permutation from a rank.
Create the inverse function that given the permutation generates its rank.
Show that for
n
=
3
{\displaystyle n=3}
the two functions are indeed inverses of each other.
Compute and show here 4 random, individual, samples of permutations of 12 objects.
Stretch goal
State how reasonable it would be to use your program to address the limits of the Stack Overflow question.
References
Ranking and Unranking Permutations in Linear Time by Myrvold & Ruskey. (Also available via Google here).
Ranks on the DevData site.
Another answer on Stack Overflow to a different question that explains its algorithm in detail.
Related tasks
Factorial_base_numbers_indexing_permutations_of_a_collection
| #Wren | Wren | import "random" for Random
import "/fmt" for Fmt
var swap = Fn.new { |a, i, j|
var t = a[i]
a[i] = a[j]
a[j] = t
}
var mrUnrank1 // recursive
mrUnrank1 = Fn.new { |rank, n, vec|
if (n < 1) return
var q = (rank/n).floor
var r = rank % n
swap.call(vec, r, n - 1)
mrUnrank1.call(q, n - 1, vec)
}
var mrRank1 // recursive
mrRank1 = Fn.new { |n, vec, inv|
if (n < 2) return 0
var s = vec[n-1]
swap.call(vec, n - 1, inv[n-1])
swap.call(inv, s, n - 1)
return s + n * mrRank1.call(n - 1, vec, inv)
}
var getPermutation = Fn.new { |rank, n, vec|
for (i in 0...n) vec[i] = i
mrUnrank1.call(rank, n, vec)
}
var getRank = Fn.new { |n, vec|
var v = List.filled(n, 0)
var inv = List.filled(n, 0)
for (i in 0...n) {
v[i] = vec[i]
inv[vec[i]] = i
}
return mrRank1.call(n, v, inv)
}
var tv = List.filled(3, 0)
for (r in 0..5) {
getPermutation.call(r, 3, tv)
Fmt.print("$2d -> $n -> $d", r, tv, getRank.call(3, tv))
}
System.print()
tv = List.filled(4, 0)
for (r in 0..23) {
getPermutation.call(r, 4, tv)
Fmt.print("$2d -> $n -> $d", r, tv, getRank.call(4, tv))
}
System.print()
tv = List.filled(12, 0)
var a = List.filled(4, 0)
var rand = Random.new()
var fact12 = (2..12).reduce(1) { |acc, i| acc * i }
for (i in 0..3) a[i] = rand.int(fact12)
for (r in a) {
getPermutation.call(r, 12, tv)
Fmt.print("$9d -> $n -> $d", r, tv, getRank.call(12, tv))
} |
http://rosettacode.org/wiki/Pierpont_primes | Pierpont primes | A Pierpont prime is a prime number of the form: 2u3v + 1 for some non-negative integers u and v .
A Pierpont prime of the second kind is a prime number of the form: 2u3v - 1 for some non-negative integers u and v .
The term "Pierpont primes" is generally understood to mean the first definition, but will be called "Pierpont primes of the first kind" on this page to distinguish them.
Task
Write a routine (function, procedure, whatever) to find Pierpont primes of the first & second kinds.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the first kind.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the second kind
If your language supports large integers, find and display here, on this page, the 250th Pierpont prime of the first kind and the 250th Pierpont prime of the second kind.
See also
Wikipedia - Pierpont primes
OEIS:A005109 - Class 1 -, or Pierpont primes
OEIS:A005105 - Class 1 +, or Pierpont primes of the second kind
| #Quackery | Quackery | [ stack ] is pierponts
[ stack ] is kind
[ stack ] is quantity
[ 1 - -2 * 1+ kind put
1+ quantity put
' [ -1 ] pierponts put
' [ 2 3 ] smoothwith
[ -1 peek
kind share +
dup isprime not iff
[ drop false ] done
pierponts share -1 peek
over = iff
[ drop false ] done
pierponts take
swap join
dup size swap
pierponts put
quantity share = ]
drop
quantity release
kind release
pierponts take
behead drop ] is pierpontprimes
say "Pierpont primes of the first kind." cr
50 1 pierpontprimes echo
cr cr
say "Pierpont primes of the second kind." cr
50 2 pierpontprimes echo |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #Go | Go | package main
import (
"fmt"
"math/rand"
"time"
)
var list = []string{"bleen", "fuligin", "garrow", "grue", "hooloovoo"}
func main() {
rand.Seed(time.Now().UnixNano())
fmt.Println(list[rand.Intn(len(list))])
} |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #Groovy | Groovy | def list = [25, 30, 1, 450, 3, 78]
def random = new Random();
(0..3).each {
def i = random.nextInt(list.size())
println "list[${i}] == ${list[i]}"
} |
http://rosettacode.org/wiki/Primality_by_trial_division | Primality by trial division | Task
Write a boolean function that tells whether a given integer is prime.
Remember that 1 and all non-positive numbers are not prime.
Use trial division.
Even numbers greater than 2 may be eliminated right away.
A loop from 3 to √ n will suffice, but other loops are allowed.
Related tasks
count in factors
prime decomposition
AKS test for primes
factors of an integer
Sieve of Eratosthenes
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
| #Wren | Wren | import "/fmt" for Fmt
var isPrime = Fn.new { |n|
if (n < 2) return false
if (n%2 == 0) return n == 2
var p = 3
while (p * p <= n) {
if (n%p == 0) return false
p = p + 2
}
return true
}
var tests = [2, 5, 12, 19, 57, 61, 97]
System.print("Are the following prime?")
for (test in tests) {
System.print("%(Fmt.d(2, test)) -> %(isPrime.call(test) ? "yes" : "no")")
} |
http://rosettacode.org/wiki/Phrase_reversals | Phrase reversals | Task
Given a string of space separated words containing the following phrase:
rosetta code phrase reversal
Reverse the characters of the string.
Reverse the characters of each individual word in the string, maintaining original word order within the string.
Reverse the order of each word of the string, maintaining the order of characters in each word.
Show your output here.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Groovy | Groovy | def phaseReverse = { text, closure -> closure(text.split(/ /)).join(' ')}
def text = 'rosetta code phrase reversal'
println "Original: $text"
println "Reversed: ${phaseReverse(text) { it.reverse().collect { it.reverse() } } }"
println "Reversed Words: ${phaseReverse(text) { it.collect { it.reverse() } } }"
println "Reversed Order: ${phaseReverse(text) { it.reverse() } }" |
http://rosettacode.org/wiki/Permutations/Derangements | Permutations/Derangements | A derangement is a permutation of the order of distinct items in which no item appears in its original place.
For example, the only two derangements of the three items (0, 1, 2) are (1, 2, 0), and (2, 0, 1).
The number of derangements of n distinct items is known as the subfactorial of n, sometimes written as !n.
There are various ways to calculate !n.
Task
Create a named function/method/subroutine/... to generate derangements of the integers 0..n-1, (or 1..n if you prefer).
Generate and show all the derangements of 4 integers using the above routine.
Create a function that calculates the subfactorial of n, !n.
Print and show a table of the counted number of derangements of n vs. the calculated !n for n from 0..9 inclusive.
Optional stretch goal
Calculate !20
Related tasks
Anagrams/Deranged anagrams
Best shuffle
Left_factorials
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #F.23 | F# |
// Generate derangements. Nigel Galloway: July 9th., 2019
let derange n=
let fG n i g=let e=Array.copy n in e.[i]<-n.[g]; e.[g]<-n.[i]; e
let rec derange n g α=seq{
match (α>0,n&&&(1<<<α)=0) with
(true,true)->for i in [0..α-1] do if n&&&(1<<<i)=0 then let g=(fG g α i) in yield! derange (n+(1<<<i)) g (α-1); yield! derange n g (α-1)
|(true,false)->yield! derange n g (α-1)
|(false,false)->yield g
|_->()}
derange 0 [|1..n|] (n-1)
|
http://rosettacode.org/wiki/Permutations_by_swapping | Permutations by swapping | Task
Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.
Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.
Show the permutations and signs of three items, in order of generation here.
Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.
Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.
References
Steinhaus–Johnson–Trotter algorithm
Johnson-Trotter Algorithm Listing All Permutations
Heap's algorithm
[1] Tintinnalogia
Related tasks
Matrix arithmetic
Gray code
| #Go | Go | package permute
// Iter takes a slice p and returns an iterator function. The iterator
// permutes p in place and returns the sign. After all permutations have
// been generated, the iterator returns 0 and p is left in its initial order.
func Iter(p []int) func() int {
f := pf(len(p))
return func() int {
return f(p)
}
}
// Recursive function used by perm, returns a chain of closures that
// implement a loopless recursive SJT.
func pf(n int) func([]int) int {
sign := 1
switch n {
case 0, 1:
return func([]int) (s int) {
s = sign
sign = 0
return
}
default:
p0 := pf(n - 1)
i := n
var d int
return func(p []int) int {
switch {
case sign == 0:
case i == n:
i--
sign = p0(p[:i])
d = -1
case i == 0:
i++
sign *= p0(p[1:])
d = 1
if sign == 0 {
p[0], p[1] = p[1], p[0]
}
default:
p[i], p[i-1] = p[i-1], p[i]
sign = -sign
i += d
}
return sign
}
}
} |
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #jq | jq | # combination(r) generates a stream of combinations of r items from the input array.
def combination(r):
if r > length or r < 0 then empty
elif r == length then .
else ( [.[0]] + (.[1:]|combination(r-1))),
( .[1:]|combination(r))
end;
|
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #Julia | Julia | using Combinatorics
meandiff(a::Vector{T}, b::Vector{T}) where T <: Real = mean(a) - mean(b)
function bifurcate(a::AbstractVector, sel::Vector{T}) where T <: Integer
x = a[sel]
asel = trues(length(a))
asel[sel] = false
y = a[asel]
return x, y
end
function permutation_test(treated::Vector{T}, control::Vector{T}) where T <: Real
effect0 = meandiff(treated, control)
pool = vcat(treated, control)
tlen = length(treated)
plen = length(pool)
better = worse = 0
for subset in combinations(1:plen, tlen)
t, c = bifurcate(pool, subset)
if effect0 < meandiff(t, c)
better += 1
else
worse += 1
end
end
return better, worse
end |
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #Kotlin | Kotlin | // version 1.1.2
val data = intArrayOf(
85, 88, 75, 66, 25, 29, 83, 39, 97,
68, 41, 10, 49, 16, 65, 32, 92, 28, 98
)
fun pick(at: Int, remain: Int, accu: Int, treat: Int): Int {
if (remain == 0) return if (accu > treat) 1 else 0
return pick(at - 1, remain - 1, accu + data[at - 1], treat) +
if (at > remain) pick(at - 1, remain, accu, treat) else 0
}
fun main(args: Array<String>) {
var treat = 0
var total = 1.0
for (i in 0..8) treat += data[i]
for (i in 19 downTo 11) total *= i
for (i in 9 downTo 1) total /= i
val gt = pick(19, 9, 0, treat)
val le = (total - gt).toInt()
System.out.printf("<= : %f%% %d\n", 100.0 * le / total, le)
System.out.printf(" > : %f%% %d\n", 100.0 * gt / total, gt)
} |
http://rosettacode.org/wiki/Percentage_difference_between_images | Percentage difference between images | basic bitmap storage
Useful for comparing two JPEG images saved with a different compression ratios.
You can use these pictures for testing (use the full-size version of each):
50% quality JPEG
100% quality JPEG
link to full size 50% image
link to full size 100% image
The expected difference for these two images is 1.62125%
| #AutoHotkey | AutoHotkey | startup()
dibSection := getPixels("lenna100.jpg")
dibSection2 := getPixels("lenna50.jpg") ; ("File-Lenna100.jpg")
pixels := dibSection.pBits
pixels2 := dibSection2.pBits
z := 0
loop % dibSection.width * dibSection.height * 4
{
x := numget(pixels - 1, A_Index, "uchar")
y := numget(pixels2 - 1, A_Index, "uchar")
z += abs(y - x)
}
msgbox % z / (dibSection.width * dibSection.height * 3 * 255 / 100 ) ; 1.626
return
CreateDIBSection2(hDC, nW, nH, bpp = 32, ByRef pBits = "")
{
dib := object()
NumPut(VarSetCapacity(bi, 40, 0), bi)
NumPut(nW, bi, 4)
NumPut(nH, bi, 8)
NumPut(bpp, NumPut(1, bi, 12, "UShort"), 0, "Ushort")
NumPut(0, bi,16)
hbm := DllCall("gdi32\CreateDIBSection", "Uint", hDC, "Uint", &bi, "Uint", 0, "UintP", pBits, "Uint", 0, "Uint", 0)
dib.hbm := hbm
dib.pBits := pBits
dib.width := nW
dib.height := nH
dib.bpp := bpp
dib.header := header
Return dib
}
startup()
{
global disposables
disposables := object()
disposables.pBitmaps := object()
disposables.hBitmaps := object()
If !(disposables.pToken := Gdip_Startup())
{
MsgBox, 48, gdiplus error!, Gdiplus failed to start. Please ensure you have gdiplus on your system
ExitApp
}
OnExit, gdipExit
}
gdipExit:
loop % disposables.hBitmaps._maxindex()
DllCall("DeleteObject", "Uint", disposables.hBitmaps[A_Index])
Gdip_Shutdown(disposables.pToken)
ExitApp
getPixels(imageFile)
{
global disposables ; hBitmap will be disposed later
pBitmapFile1 := Gdip_CreateBitmapFromFile(imageFile)
hbmi := Gdip_CreateHBITMAPFromBitmap(pBitmapFile1)
width := Gdip_GetImageWidth(pBitmapFile1)
height := Gdip_GetImageHeight(pBitmapFile1)
mDCo := DllCall("CreateCompatibleDC", "Uint", 0)
mDCi := DllCall("CreateCompatibleDC", "Uint", 0)
dibSection := CreateDIBSection2(mDCo, width, height)
hBMo := dibSection.hbm
oBM := DllCall("SelectObject", "Uint", mDCo, "Uint", hBMo)
iBM := DllCall("SelectObject", "Uint", mDCi, "Uint", hbmi)
DllCall("BitBlt", "Uint", mDCo, "int", 0, "int", 0, "int", width, "int", height, "Uint", mDCi, "int", 0, "int", 0, "Uint", 0x40000000 | 0x00CC0020)
DllCall("SelectObject", "Uint", mDCo, "Uint", oBM)
DllCall("DeleteDC", "Uint", 0, "Uint", mDCi)
DllCall("DeleteDC", "Uint", 0, "Uint", mDCo)
Gdip_DisposeImage(pBitmapFile1)
DllCall("DeleteObject", "Uint", hBMi)
disposables.hBitmaps._insert(hBMo)
return dibSection
}
#Include Gdip.ahk ; Thanks to tic (Tariq Porter) for his GDI+ Library
|
http://rosettacode.org/wiki/Percolation/Mean_cluster_density | Percolation/Mean cluster density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
c
{\displaystyle c}
be a 2D boolean square matrix of
n
×
n
{\displaystyle n\times n}
values of either 1 or 0 where the
probability of any value being 1 is
p
{\displaystyle p}
, (and of 0 is therefore
1
−
p
{\displaystyle 1-p}
).
We define a cluster of 1's as being a group of 1's connected vertically or
horizontally (i.e., using the Von Neumann neighborhood rule) and bounded by either
0
{\displaystyle 0}
or by the limits of the matrix.
Let the number of such clusters in such a randomly constructed matrix be
C
n
{\displaystyle C_{n}}
.
Percolation theory states that
K
(
p
)
{\displaystyle K(p)}
(the mean cluster density) will satisfy
K
(
p
)
=
C
n
/
n
2
{\displaystyle K(p)=C_{n}/n^{2}}
as
n
{\displaystyle n}
tends to infinity. For
p
=
0.5
{\displaystyle p=0.5}
,
K
(
p
)
{\displaystyle K(p)}
is found numerically to approximate
0.065770
{\displaystyle 0.065770}
...
Task
Show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
for
p
=
0.5
{\displaystyle p=0.5}
and
for values of
n
{\displaystyle n}
up to at least
1000
{\displaystyle 1000}
.
Any calculation of
C
n
{\displaystyle C_{n}}
for finite
n
{\displaystyle n}
is subject to randomness, so an approximation should be
computed as the average of
t
{\displaystyle t}
runs, where
t
{\displaystyle t}
≥
5
{\displaystyle 5}
.
For extra credit, graphically show clusters in a
15
×
15
{\displaystyle 15\times 15}
,
p
=
0.5
{\displaystyle p=0.5}
grid.
Show your output here.
See also
s-Cluster on Wolfram mathworld. | #EchoLisp | EchoLisp |
(define-constant BLACK (rgb 0 0 0.6))
(define-constant WHITE -1)
;; sets pixels to clusterize to WHITE
;; returns bit-map vector
(define (init-C n p )
(plot-size n n)
(define C (pixels->int32-vector )) ;; get canvas bit-map
(pixels-map (lambda (x y) (if (< (random) p) WHITE BLACK )) C)
C )
;; random color for new cluster
(define (new-color)
(hsv->rgb (random) 0.9 0.9))
;; make-region predicate
(define (in-cluster C x y)
(= (pixel-ref C x y) WHITE))
;; paint all adjacents to (x0,y0) with new color
(define (make-cluster C x0 y0)
(pixel-set! C x0 y0 (new-color))
(make-region in-cluster C x0 y0))
;; task
(define (make-clusters (n 400) (p 0.5))
(define Cn 0)
(define C null)
(for ((t 5)) ;; 5 iterations
(plot-clear)
(set! C (init-C n p))
(for* ((x0 n) (y0 n))
#:when (= (pixel-ref C x0 y0) WHITE)
(set! Cn (1+ Cn))
(make-cluster C x0 y0)))
(writeln 'n n 'Cn Cn 'density (// Cn (* n n) 5) )
(vector->pixels C)) ;; to screen
|
http://rosettacode.org/wiki/Perfect_numbers | Perfect numbers | Write a function which says whether a number is perfect.
A perfect number is a positive integer that is the sum of its proper positive divisors excluding the number itself.
Equivalently, a perfect number is a number that is half the sum of all of its positive divisors (including itself).
Note: The faster Lucas-Lehmer test is used to find primes of the form 2n-1, all known perfect numbers can be derived from these primes
using the formula (2n - 1) × 2n - 1.
It is not known if there are any odd perfect numbers (any that exist are larger than 102000).
The number of known perfect numbers is 51 (as of December, 2018), and the largest known perfect number contains 49,724,095 decimal digits.
See also
Rational Arithmetic
Perfect numbers on OEIS
Odd Perfect showing the current status of bounds on odd perfect numbers.
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program perfectNumber64.s */
/* use Euclide Formula : if M=(2puis p)-1 is prime M * (M+1)/2 is perfect see Wikipedia */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeConstantesARM64.inc"
.equ MAXI, 63
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResult: .asciz "Perfect : @ \n"
szMessOverflow: .asciz "Overflow in function isPrime.\n"
szCarriageReturn: .asciz "\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: // entry of program
mov x4,2 // start 2
mov x3,1 // counter 2 power
1: // begin loop
lsl x4,x4,1 // 2 power
sub x0,x4,1 // - 1
bl isPrime // is prime ?
cbz x0,2f // no
sub x0,x4,1 // yes
mul x1,x0,x4 // multiply m by m-1
lsr x0,x1,1 // divide by 2
bl displayPerfect // and display
2:
add x3,x3,1 // next power of 2
cmp x3,MAXI
blt 1b
100: // standard end of the program
mov x0,0 // return code
mov x8,EXIT // request to exit program
svc 0 // perform the system call
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrsMessResult: .quad sMessResult
/******************************************************************/
/* Display perfect number */
/******************************************************************/
/* x0 contains the number */
displayPerfect:
stp x1,lr,[sp,-16]! // save registers
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrsMessResult
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
bl affichageMess // display message
100:
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
qAdrsZoneConv: .quad sZoneConv
/***************************************************/
/* is a number prime ? */
/***************************************************/
/* x0 contains the number */
/* x0 return 1 if prime else 0 */
//2147483647 OK
//4294967297 NOK
//131071 OK
//1000003 OK
//10001363 OK
isPrime:
stp x1,lr,[sp,-16]! // save registres
stp x2,x3,[sp,-16]! // save registres
mov x2,x0
sub x1,x0,#1
cmp x2,0
beq 99f // return zero
cmp x2,2 // for 1 and 2 return 1
ble 2f
mov x0,#2
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // no prime
cmp x2,3
beq 2f
mov x0,#3
bl moduloPuR64
blt 100f // error overflow
cmp x0,#1
bne 99f
cmp x2,5
beq 2f
mov x0,#5
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,7
beq 2f
mov x0,#7
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,11
beq 2f
mov x0,#11
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,13
beq 2f
mov x0,#13
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
2:
cmn x0,0 // carry à zero no error
mov x0,1 // prime
b 100f
99:
cmn x0,0 // carry à zero no error
mov x0,#0 // prime
100:
ldp x2,x3,[sp],16 // restaur des 2 registres
ldp x1,lr,[sp],16 // restaur des 2 registres
ret
/**************************************************************/
/********************************************************/
/* Compute modulo de b power e modulo m */
/* Exemple 4 puissance 13 modulo 497 = 445 */
/********************************************************/
/* x0 number */
/* x1 exposant */
/* x2 modulo */
moduloPuR64:
stp x1,lr,[sp,-16]! // save registres
stp x3,x4,[sp,-16]! // save registres
stp x5,x6,[sp,-16]! // save registres
stp x7,x8,[sp,-16]! // save registres
stp x9,x10,[sp,-16]! // save registres
cbz x0,100f
cbz x1,100f
mov x8,x0
mov x7,x1
mov x6,1 // result
udiv x4,x8,x2
msub x9,x4,x2,x8 // remainder
1:
tst x7,1 // if bit = 1
beq 2f
mul x4,x9,x6
umulh x5,x9,x6
mov x6,x4
mov x0,x6
mov x1,x5
bl divisionReg128U // division 128 bits
cbnz x1,99f // overflow
mov x6,x3 // remainder
2:
mul x8,x9,x9
umulh x5,x9,x9
mov x0,x8
mov x1,x5
bl divisionReg128U
cbnz x1,99f // overflow
mov x9,x3
lsr x7,x7,1
cbnz x7,1b
mov x0,x6 // result
cmn x0,0 // carry à zero no error
b 100f
99:
ldr x0,qAdrszMessOverflow
bl affichageMess // display error message
cmp x0,0 // carry set error
mov x0,-1 // code erreur
100:
ldp x9,x10,[sp],16 // restaur des 2 registres
ldp x7,x8,[sp],16 // restaur des 2 registres
ldp x5,x6,[sp],16 // restaur des 2 registres
ldp x3,x4,[sp],16 // restaur des 2 registres
ldp x1,lr,[sp],16 // restaur des 2 registres
ret // retour adresse lr x30
qAdrszMessOverflow: .quad szMessOverflow
/***************************************************/
/* division d un nombre de 128 bits par un nombre de 64 bits */
/***************************************************/
/* x0 contient partie basse dividende */
/* x1 contient partie haute dividente */
/* x2 contient le diviseur */
/* x0 retourne partie basse quotient */
/* x1 retourne partie haute quotient */
/* x3 retourne le reste */
divisionReg128U:
stp x6,lr,[sp,-16]! // save registres
stp x4,x5,[sp,-16]! // save registres
mov x5,#0 // raz du reste R
mov x3,#128 // compteur de boucle
mov x4,#0 // dernier bit
1:
lsl x5,x5,#1 // on decale le reste de 1
tst x1,1<<63 // test du bit le plus à gauche
lsl x1,x1,#1 // on decale la partie haute du quotient de 1
beq 2f
orr x5,x5,#1 // et on le pousse dans le reste R
2:
tst x0,1<<63
lsl x0,x0,#1 // puis on decale la partie basse
beq 3f
orr x1,x1,#1 // et on pousse le bit de gauche dans la partie haute
3:
orr x0,x0,x4 // position du dernier bit du quotient
mov x4,#0 // raz du bit
cmp x5,x2
blt 4f
sub x5,x5,x2 // on enleve le diviseur du reste
mov x4,#1 // dernier bit à 1
4:
// et boucle
subs x3,x3,#1
bgt 1b
lsl x1,x1,#1 // on decale le quotient de 1
tst x0,1<<63
lsl x0,x0,#1 // puis on decale la partie basse
beq 5f
orr x1,x1,#1
5:
orr x0,x0,x4 // position du dernier bit du quotient
mov x3,x5
100:
ldp x4,x5,[sp],16 // restaur des 2 registres
ldp x6,lr,[sp],16 // restaur des 2 registres
ret // retour adresse lr x30
/********************************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/Percolation/Bond_percolation | Percolation/Bond percolation |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Given an
M
×
N
{\displaystyle M\times N}
rectangular array of cells numbered
c
e
l
l
[
0..
M
−
1
,
0..
N
−
1
]
{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}
, assume
M
{\displaystyle M}
is horizontal and
N
{\displaystyle N}
is downwards. Each
c
e
l
l
[
m
,
n
]
{\displaystyle \mathrm {cell} [m,n]}
is bounded by (horizontal) walls
h
w
a
l
l
[
m
,
n
]
{\displaystyle \mathrm {hwall} [m,n]}
and
h
w
a
l
l
[
m
+
1
,
n
]
{\displaystyle \mathrm {hwall} [m+1,n]}
; (vertical) walls
v
w
a
l
l
[
m
,
n
]
{\displaystyle \mathrm {vwall} [m,n]}
and
v
w
a
l
l
[
m
,
n
+
1
]
{\displaystyle \mathrm {vwall} [m,n+1]}
Assume that the probability of any wall being present is a constant
p
{\displaystyle p}
where
0.0
≤
p
≤
1.0
{\displaystyle 0.0\leq p\leq 1.0}
Except for the outer horizontal walls at
m
=
0
{\displaystyle m=0}
and
m
=
M
{\displaystyle m=M}
which are always present.
The task
Simulate pouring a fluid onto the top surface (
n
=
0
{\displaystyle n=0}
) where the fluid will enter any empty cell it is adjacent to if there is no wall between where it currently is and the cell on the other side of the (missing) wall.
The fluid does not move beyond the horizontal constraints of the grid.
The fluid may move “up” within the confines of the grid of cells. If the fluid reaches a bottom cell that has a missing bottom wall then the fluid can be said to 'drip' out the bottom at that point.
Given
p
{\displaystyle p}
repeat the percolation
t
{\displaystyle t}
times to estimate the proportion of times that the fluid can percolate to the bottom for any given
p
{\displaystyle p}
.
Show how the probability of percolating through the random grid changes with
p
{\displaystyle p}
going from
0.0
{\displaystyle 0.0}
to
1.0
{\displaystyle 1.0}
in
0.1
{\displaystyle 0.1}
increments and with the number of repetitions to estimate the fraction at any given
p
{\displaystyle p}
as
t
=
100
{\displaystyle t=100}
.
Use an
M
=
10
,
N
=
10
{\displaystyle M=10,N=10}
grid of cells for all cases.
Optionally depict fluid successfully percolating through a grid graphically.
Show all output on this page.
| #C.2B.2B | C++ | #include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
using namespace std;
class Grid {
public:
Grid(const double p, const int x, const int y) : m(x), n(y) {
const int thresh = static_cast<int>(RAND_MAX * p);
// Allocate two addition rows to avoid checking bounds.
// Bottom row is also required by drippage
start = new cell[m * (n + 2)];
cells = start + m;
for (auto i = 0; i < m; i++) start[i] = RBWALL;
end = cells;
for (auto i = 0; i < y; i++) {
for (auto j = x; --j;)
*end++ = (rand() < thresh ? BWALL : 0) | (rand() < thresh ? RWALL : 0);
*end++ = RWALL | (rand() < thresh ? BWALL : 0);
}
memset(end, 0u, sizeof(cell) * m);
}
~Grid() {
delete[] start;
cells = 0;
start = 0;
end = 0;
}
int percolate() const {
auto i = 0;
for (; i < m && !fill(cells + i); i++);
return i < m;
}
void show() const {
for (auto j = 0; j < m; j++)
cout << ("+-");
cout << '+' << endl;
for (auto i = 0; i <= n; i++) {
cout << (i == n ? ' ' : '|');
for (auto j = 0; j < m; j++) {
cout << ((cells[i * m + j] & FILL) ? "#" : " ");
cout << ((cells[i * m + j] & RWALL) ? '|' : ' ');
}
cout << endl;
if (i == n) return;
for (auto j = 0; j < m; j++)
cout << ((cells[i * m + j] & BWALL) ? "+-" : "+ ");
cout << '+' << endl;
}
}
private:
enum cell_state {
FILL = 1 << 0,
RWALL = 1 << 1, // right wall
BWALL = 1 << 2, // bottom wall
RBWALL = RWALL | BWALL // right/bottom wall
};
typedef unsigned int cell;
bool fill(cell* p) const {
if ((*p & FILL)) return false;
*p |= FILL;
if (p >= end) return true; // success: reached bottom row
return (!(p[0] & BWALL) && fill(p + m)) || (!(p[0] & RWALL) && fill(p + 1))
||(!(p[-1] & RWALL) && fill(p - 1)) || (!(p[-m] & BWALL) && fill(p - m));
}
cell* cells;
cell* start;
cell* end;
const int m;
const int n;
};
int main() {
const auto M = 10, N = 10;
const Grid grid(.5, M, N);
grid.percolate();
grid.show();
const auto C = 10000;
cout << endl << "running " << M << "x" << N << " grids " << C << " times for each p:" << endl;
for (auto p = 1; p < M; p++) {
auto cnt = 0, i = 0;
for (; i < C; i++)
cnt += Grid(p / static_cast<double>(M), M, N).percolate();
cout << "p = " << p / static_cast<double>(M) << ": " << static_cast<double>(cnt) / i << endl;
}
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/Percolation/Mean_run_density | Percolation/Mean run density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
v
{\displaystyle v}
be a vector of
n
{\displaystyle n}
values of either 1 or 0 where the probability of any
value being 1 is
p
{\displaystyle p}
; the probability of a value being 0 is therefore
1
−
p
{\displaystyle 1-p}
.
Define a run of 1s as being a group of consecutive 1s in the vector bounded
either by the limits of the vector or by a 0. Let the number of such runs in a given
vector of length
n
{\displaystyle n}
be
R
n
{\displaystyle R_{n}}
.
For example, the following vector has
R
10
=
3
{\displaystyle R_{10}=3}
[1 1 0 0 0 1 0 1 1 1]
^^^ ^ ^^^^^
Percolation theory states that
K
(
p
)
=
lim
n
→
∞
R
n
/
n
=
p
(
1
−
p
)
{\displaystyle K(p)=\lim _{n\to \infty }R_{n}/n=p(1-p)}
Task
Any calculation of
R
n
/
n
{\displaystyle R_{n}/n}
for finite
n
{\displaystyle n}
is subject to randomness so should be
computed as the average of
t
{\displaystyle t}
runs, where
t
≥
100
{\displaystyle t\geq 100}
.
For values of
p
{\displaystyle p}
of 0.1, 0.3, 0.5, 0.7, and 0.9, show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
.
Show your output here.
See also
s-Run on Wolfram mathworld. | #Factor | Factor | USING: formatting fry io kernel math math.ranges math.statistics
random sequences ;
IN: rosetta-code.mean-run-density
: rising? ( ? ? -- ? ) [ f = ] [ t = ] bi* and ;
: count-run ( n ? ? -- m ? )
2dup rising? [ [ 1 + ] 2dip ] when nip ;
: runs ( n p -- n )
[ 0 f ] 2dip '[ random-unit _ < count-run ] times drop ;
: rn ( n p -- x ) over [ runs ] dip /f ;
: sim ( n p -- avg )
[ 1000 ] 2dip [ rn ] 2curry replicate mean ;
: theory ( p -- x ) 1 over - * ;
: result ( n p -- )
[ swap ] [ sim ] [ nip theory ] 2tri 2dup - abs
"%.1f %-5d %.4f %.4f %.4f\n" printf ;
: test ( p -- )
{ 100 1,000 10,000 } [ swap result ] with each nl ;
: header ( -- )
"1000 tests each:\np n K p(1-p) diff" print ;
: main ( -- ) header .1 .9 .2 <range> [ test ] each ;
MAIN: main |
http://rosettacode.org/wiki/Percolation/Mean_run_density | Percolation/Mean run density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
v
{\displaystyle v}
be a vector of
n
{\displaystyle n}
values of either 1 or 0 where the probability of any
value being 1 is
p
{\displaystyle p}
; the probability of a value being 0 is therefore
1
−
p
{\displaystyle 1-p}
.
Define a run of 1s as being a group of consecutive 1s in the vector bounded
either by the limits of the vector or by a 0. Let the number of such runs in a given
vector of length
n
{\displaystyle n}
be
R
n
{\displaystyle R_{n}}
.
For example, the following vector has
R
10
=
3
{\displaystyle R_{10}=3}
[1 1 0 0 0 1 0 1 1 1]
^^^ ^ ^^^^^
Percolation theory states that
K
(
p
)
=
lim
n
→
∞
R
n
/
n
=
p
(
1
−
p
)
{\displaystyle K(p)=\lim _{n\to \infty }R_{n}/n=p(1-p)}
Task
Any calculation of
R
n
/
n
{\displaystyle R_{n}/n}
for finite
n
{\displaystyle n}
is subject to randomness so should be
computed as the average of
t
{\displaystyle t}
runs, where
t
≥
100
{\displaystyle t\geq 100}
.
For values of
p
{\displaystyle p}
of 0.1, 0.3, 0.5, 0.7, and 0.9, show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
.
Show your output here.
See also
s-Run on Wolfram mathworld. | #Fortran | Fortran |
! loosely translated from python. We do not need to generate and store the entire vector at once.
! compilation: gfortran -Wall -std=f2008 -o thisfile thisfile.f08
program percolation_mean_run_density
implicit none
integer :: i, p10, n2, n, t
real :: p, limit, sim, delta
data n,p,t/100,0.5,500/
write(6,'(a3,a5,4a7)')'t','p','n','p(1-p)','sim','delta%'
do p10=1,10,2
p = p10/10.0
limit = p*(1-p)
write(6,'()')
do n2=10,15,2
n = 2**n2
sim = 0
do i=1,t
sim = sim + mean_run_density(n,p)
end do
sim = sim/t
if (limit /= 0) then
delta = abs(sim-limit)/limit
else
delta = sim
end if
delta = delta * 100
write(6,'(i3,f5.2,i7,2f7.3,f5.1)')t,p,n,limit,sim,delta
end do
end do
contains
integer function runs(n, p)
integer, intent(in) :: n
real, intent(in) :: p
real :: harvest
logical :: q
integer :: count, i
count = 0
q = .false.
do i=1,n
call random_number(harvest)
if (harvest < p) then
q = .true.
else
if (q) count = count+1
q = .false.
end if
end do
runs = count
end function runs
real function mean_run_density(n, p)
integer, intent(in) :: n
real, intent(in) :: p
mean_run_density = real(runs(n,p))/real(n)
end function mean_run_density
end program percolation_mean_run_density
|
http://rosettacode.org/wiki/Percolation/Site_percolation | Percolation/Site percolation |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Given an
M
×
N
{\displaystyle M\times N}
rectangular array of cells numbered
c
e
l
l
[
0..
M
−
1
,
0..
N
−
1
]
{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}
assume
M
{\displaystyle M}
is horizontal and
N
{\displaystyle N}
is downwards.
Assume that the probability of any cell being filled is a constant
p
{\displaystyle p}
where
0.0
≤
p
≤
1.0
{\displaystyle 0.0\leq p\leq 1.0}
The task
Simulate creating the array of cells with probability
p
{\displaystyle p}
and then
testing if there is a route through adjacent filled cells from any on row
0
{\displaystyle 0}
to any on row
N
{\displaystyle N}
, i.e. testing for site percolation.
Given
p
{\displaystyle p}
repeat the percolation
t
{\displaystyle t}
times to estimate the proportion of times that the fluid can percolate to the bottom for any given
p
{\displaystyle p}
.
Show how the probability of percolating through the random grid changes with
p
{\displaystyle p}
going from
0.0
{\displaystyle 0.0}
to
1.0
{\displaystyle 1.0}
in
0.1
{\displaystyle 0.1}
increments and with the number of repetitions to estimate the fraction at any given
p
{\displaystyle p}
as
t
>=
100
{\displaystyle t>=100}
.
Use an
M
=
15
,
N
=
15
{\displaystyle M=15,N=15}
grid of cells for all cases.
Optionally depict a percolation through a cell grid graphically.
Show all output on this page.
| #Go | Go | package main
import (
"bytes"
"fmt"
"math/rand"
"time"
)
func main() {
const (
m, n = 15, 15
t = 1e4
minp, maxp, Δp = 0, 1, 0.1
)
rand.Seed(2) // Fixed seed for repeatable example grid
g := NewGrid(.5, m, n)
g.Percolate()
fmt.Println(g)
rand.Seed(time.Now().UnixNano()) // could pick a better seed
for p := float64(minp); p < maxp; p += Δp {
count := 0
for i := 0; i < t; i++ {
g := NewGrid(p, m, n)
if g.Percolate() {
count++
}
}
fmt.Printf("p=%.2f, %.4f\n", p, float64(count)/t)
}
}
const (
full = '.'
used = '#'
empty = ' '
)
type grid struct {
cell [][]byte // row first, i.e. [y][x]
}
func NewGrid(p float64, xsize, ysize int) *grid {
g := &grid{cell: make([][]byte, ysize)}
for y := range g.cell {
g.cell[y] = make([]byte, xsize)
for x := range g.cell[y] {
if rand.Float64() < p {
g.cell[y][x] = full
} else {
g.cell[y][x] = empty
}
}
}
return g
}
func (g *grid) String() string {
var buf bytes.Buffer
// Don't really need to call Grow but it helps avoid multiple
// reallocations if the size is large.
buf.Grow((len(g.cell) + 2) * (len(g.cell[0]) + 3))
buf.WriteByte('+')
for _ = range g.cell[0] {
buf.WriteByte('-')
}
buf.WriteString("+\n")
for y := range g.cell {
buf.WriteByte('|')
buf.Write(g.cell[y])
buf.WriteString("|\n")
}
buf.WriteByte('+')
ly := len(g.cell) - 1
for x := range g.cell[ly] {
if g.cell[ly][x] == used {
buf.WriteByte(used)
} else {
buf.WriteByte('-')
}
}
buf.WriteByte('+')
return buf.String()
}
func (g *grid) Percolate() bool {
for x := range g.cell[0] {
if g.use(x, 0) {
return true
}
}
return false
}
func (g *grid) use(x, y int) bool {
if y < 0 || x < 0 || x >= len(g.cell[0]) || g.cell[y][x] != full {
return false // Off the edges, empty, or used
}
g.cell[y][x] = used
if y+1 == len(g.cell) {
return true // We're on the bottom
}
// Try down, right, left, up in that order.
return g.use(x, y+1) ||
g.use(x+1, y) ||
g.use(x-1, y) ||
g.use(x, y-1)
} |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #Action.21 | Action! | PROC PrintArray(BYTE ARRAY a BYTE len)
BYTE i
FOR i=0 TO len-1
DO
PrintB(a(i))
OD
Print(" ")
RETURN
BYTE FUNC NextPermutation(BYTE ARRAY a BYTE len)
BYTE i,j,k,tmp
i=len-1
WHILE i>0 AND a(i-1)>a(i)
DO
i==-1
OD
j=i
k=len-1
WHILE j<k
DO
tmp=a(j) a(j)=a(k) a(k)=tmp
j==+1 k==-1
OD
IF i=0 THEN
RETURN (0)
FI
j=i
WHILE a(j)<a(i-1)
DO
j==+1
OD
tmp=a(i-1) a(i-1)=a(j) a(j)=tmp
RETURN (1)
PROC Main()
DEFINE len="5"
BYTE ARRAY a(len)
BYTE RMARGIN=$53,oldRMARGIN
BYTE i
oldRMARGIN=RMARGIN
RMARGIN=37 ;change right margin on the screen
FOR i=0 TO len-1
DO
a(i)=i
OD
DO
PrintArray(a,len)
UNTIL NextPermutation(a,len)=0
OD
RMARGIN=oldRMARGIN ;restore right margin on the screen
RETURN |
http://rosettacode.org/wiki/Perfect_shuffle | Perfect shuffle | A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:
7♠ 8♠ 9♠ J♠ Q♠ K♠→7♠ 8♠ 9♠
J♠ Q♠ K♠→7♠ J♠ 8♠ Q♠ 9♠ K♠
When you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:
original:
1
2
3
4
5
6
7
8
after 1st shuffle:
1
5
2
6
3
7
4
8
after 2nd shuffle:
1
3
5
7
2
4
6
8
after 3rd shuffle:
1
2
3
4
5
6
7
8
The Task
Write a function that can perform a perfect shuffle on an even-sized list of values.
Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under "Test Cases" below.
You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all "cards" are unique within each deck.
Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.
Test Cases
input (deck size)
output (number of shuffles required)
8
3
24
11
52
8
100
30
1020
1018
1024
10
10000
300
| #AutoHotkey | AutoHotkey | Shuffle(cards){
n := cards.MaxIndex()/2, res := []
loop % n
res.push(cards[A_Index]), res.push(cards[round(A_Index + n)])
return res
} |
http://rosettacode.org/wiki/Perfect_shuffle | Perfect shuffle | A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:
7♠ 8♠ 9♠ J♠ Q♠ K♠→7♠ 8♠ 9♠
J♠ Q♠ K♠→7♠ J♠ 8♠ Q♠ 9♠ K♠
When you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:
original:
1
2
3
4
5
6
7
8
after 1st shuffle:
1
5
2
6
3
7
4
8
after 2nd shuffle:
1
3
5
7
2
4
6
8
after 3rd shuffle:
1
2
3
4
5
6
7
8
The Task
Write a function that can perform a perfect shuffle on an even-sized list of values.
Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under "Test Cases" below.
You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all "cards" are unique within each deck.
Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.
Test Cases
input (deck size)
output (number of shuffles required)
8
3
24
11
52
8
100
30
1020
1018
1024
10
10000
300
| #C | C | /* ===> INCLUDES <============================================================*/
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
/* ===> CONSTANTS <===========================================================*/
#define N_DECKS 7
const int kDecks[N_DECKS] = { 8, 24, 52, 100, 1020, 1024, 10000 };
/* ===> FUNCTION PROTOTYPES <=================================================*/
int CreateDeck( int **deck, int nCards );
void InitDeck( int *deck, int nCards );
int DuplicateDeck( int **dest, const int *orig, int nCards );
int InitedDeck( int *deck, int nCards );
int ShuffleDeck( int *deck, int nCards );
void FreeDeck( int **deck );
/* ===> FUNCTION DEFINITIONS <================================================*/
int main() {
int i, nCards, nShuffles;
int *deck = NULL;
for( i=0; i<N_DECKS; ++i ) {
nCards = kDecks[i];
if( !CreateDeck(&deck,nCards) ) {
fprintf( stderr, "Error: malloc() failed!\n" );
return 1;
}
InitDeck( deck, nCards );
nShuffles = 0;
do {
ShuffleDeck( deck, nCards );
++nShuffles;
} while( !InitedDeck(deck,nCards) );
printf( "Cards count: %d, shuffles required: %d.\n", nCards, nShuffles );
FreeDeck( &deck );
}
return 0;
}
int CreateDeck( int **deck, int nCards ) {
int *tmp = NULL;
if( deck != NULL )
tmp = malloc( nCards*sizeof(*tmp) );
return tmp!=NULL ? (*deck=tmp)!=NULL : 0; /* (?success) (:failure) */
}
void InitDeck( int *deck, int nCards ) {
if( deck != NULL ) {
int i;
for( i=0; i<nCards; ++i )
deck[i] = i;
}
}
int DuplicateDeck( int **dest, const int *orig, int nCards ) {
if( orig != NULL && CreateDeck(dest,nCards) ) {
memcpy( *dest, orig, nCards*sizeof(*orig) );
return 1; /* success */
}
else {
return 0; /* failure */
}
}
int InitedDeck( int *deck, int nCards ) {
int i;
for( i=0; i<nCards; ++i )
if( deck[i] != i )
return 0; /* not inited */
return 1; /* inited */
}
int ShuffleDeck( int *deck, int nCards ) {
int *copy = NULL;
if( DuplicateDeck(©,deck,nCards) ) {
int i, j;
for( i=j=0; i<nCards/2; ++i, j+=2 ) {
deck[j] = copy[i];
deck[j+1] = copy[i+nCards/2];
}
FreeDeck( © );
return 1; /* success */
}
else {
return 0; /* failure */
}
}
void FreeDeck( int **deck ) {
if( *deck != NULL ) {
free( *deck );
*deck = NULL;
}
}
|
http://rosettacode.org/wiki/Perlin_noise | Perlin noise | The Perlin noise is a kind of gradient noise invented by Ken Perlin around the end of the twentieth century and still currently heavily used in computer graphics, most notably to procedurally generate textures or heightmaps.
The Perlin noise is basically a pseudo-random mapping of
R
d
{\displaystyle \mathbb {R} ^{d}}
into
R
{\displaystyle \mathbb {R} }
with an integer
d
{\displaystyle d}
which can be arbitrarily large but which is usually 2, 3, or 4.
Either by using a dedicated library or by implementing the algorithm, show that the Perlin noise (as defined in 2002 in the Java implementation below) of the point in 3D-space with coordinates 3.14, 42, 7 is 0.13691995878400012.
Note: this result assumes 64 bit IEEE-754 floating point calculations. If your language uses a different floating point representation, make a note of it and calculate the value accurate to 15 decimal places, or your languages accuracy threshold if it is less. Trailing zeros need not be displayed.
| #FreeBASIC | FreeBASIC |
#define floor(x) ((x*2.0-0.5) Shr 1)
Dim Shared As Byte p(256) => { _
151, 160, 137, 91, 90, 15, 131, 13, 201, 95, 96, 53, 194, 233, 7, 225, _
140, 36, 103, 30, 69, 142, 8, 99, 37, 240, 21, 10, 23, 190, 6, 148, _
247, 120, 234, 75, 0, 26, 197, 62, 94, 252, 219, 203, 117, 35, 11, 32, _
57, 177, 33, 88, 237, 149, 56, 87, 174, 20, 125, 136, 171, 168, 68, 175, _
74, 165, 71, 134, 139, 48, 27, 166, 77, 146, 158, 231, 83, 111, 229, 122, _
60, 211, 133, 230, 220, 105, 92, 41, 55, 46, 245, 40, 244, 102, 143, 54, _
65, 25, 63, 161, 1, 216, 80, 73, 209, 76, 132, 187, 208, 89, 18, 169, _
200, 196, 135, 130, 116, 188, 159, 86, 164, 100, 109, 198, 173, 186, 3, 64, _
52, 217, 226, 250, 124, 123, 5, 202, 38, 147, 118, 126, 255, 82, 85, 212, _
207, 206, 59, 227, 47, 16, 58, 17, 182, 189, 28, 42, 223, 183, 170, 213, _
119, 248, 152, 2, 44, 154, 163, 70, 221, 153, 101, 155, 167, 43, 172, 9, _
129, 22, 39, 253, 19, 98, 108, 110, 79, 113, 224, 232, 178, 185, 112, 104, _
218, 246, 97, 228, 251, 34, 242, 193, 238, 210, 144, 12, 191, 179, 162, 241, _
81, 51, 145, 235, 249, 14, 239, 107, 49, 192, 214, 31, 181, 199, 106, 157, _
184, 84, 204, 176, 115, 121, 50, 45, 127, 4, 150, 254, 138, 236, 205, 93, _
222, 114, 67, 29, 24, 72, 243, 141, 128, 195, 78, 66, 215, 61, 156, 180}
Function fade(t As Double) As Double
Return t * t * t * (t * (t * 6 - 15) + 10)
End Function
Function lerp(t As Double, a As Double, b As Double) As Double
Return a + t * (b - a)
End Function
Function grad(hash As Integer, x As Double, y As Double, z As Double) As Double
Select Case (hash And 15)
Case 0 : Return x+y
Case 1 : Return y-x
Case 2 : Return x-y
Case 3 : Return -x-y
Case 4 : Return x+z
Case 5 : Return z-x
Case 6 : Return x-z
Case 7 : Return -x-y ' maybe -x-z?
Case 8 : Return y+z
Case 9 : Return z-y
Case 10 : Return y-z
Case 11 : Return -y-z
Case 12 : Return x+y
Case 13 : Return z-y
Case 14 : Return y-x
Case 15 : Return -y-z
End Select
End Function
Function noise(x As Double, y As Double, z As Double) As Double
Dim As Integer a = Int(x) And 255, b = Int(y) And 255, c = Int(z) And 255
x -= floor(x) : y -= floor(y) : z -= floor(z)
Dim As Double u = fade(x), v = fade(y), w = fade(z)
Dim As Integer A0 = p(a)+b, A1 = p(A0)+c, A2 = p(A0+1)+c
Dim As Integer B0 = p(a+1)+b, B1 = p(B0)+c, B2 = p(B0+1)+c
Return lerp(w, lerp(v, lerp(u, grad(p(A1), x, y, z), _
grad(p(B1), x-1, y, z)), _
lerp(u, grad(p(A2), x, y-1, z), _
grad(p(B2), x-1, y-1, z))), _
lerp(v, lerp(u, grad(p(A1+1), x, y, z-1), _
grad(p(B1+1), x-1, y, z-1)), _
lerp(u, grad(p(A2+1), x, y-1, z-1), _
grad(p(B2+1), x-1, y-1, z-1))))
End Function
Print Using "El Ruido Perlin para (3.14, 42, 7) es #.################"; noise(3.14, 42, 7)
Sleep
|
http://rosettacode.org/wiki/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #CLU | CLU | gcd = proc (a, b: int) returns (int)
while b ~= 0 do
a, b := b, a//b
end
return(a)
end gcd
totient = proc (n: int) returns (int)
tot: int := 0
for i: int in int$from_to(1,n-1) do
if gcd(n,i)=1 then tot := tot + 1 end
end
return(tot)
end totient
perfect = proc (n: int) returns (bool)
sum: int := 0
x: int := n
while true do
x := totient(x)
sum := sum + x
if x=1 then break end
end
return(sum = n)
end perfect
start_up = proc ()
po: stream := stream$primary_output()
seen: int := 0
n: int := 3
while seen < 20 do
if perfect(n) then
stream$puts(po, int$unparse(n) || " ")
seen := seen + 1
end
n := n + 2
end
end start_up |
http://rosettacode.org/wiki/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #Cowgol | Cowgol | include "cowgol.coh";
sub gcd(a: uint16, b: uint16): (r: uint16) is
while b != 0 loop
r := a;
a := b;
b := r % b;
end loop;
r := a;
end sub;
sub totient(n: uint16): (tot: uint16) is
var i: uint16 := 1;
tot := 0;
while i < n loop
if gcd(n,i) == 1 then
tot := tot + 1;
end if;
i := i + 1;
end loop;
end sub;
sub totientSum(n: uint16): (sum: uint16) is
var x := n;
sum := 0;
while x > 1 loop
x := totient(x);
sum := sum + x;
end loop;
end sub;
var seen: uint8 := 0;
var n: uint16 := 3;
while seen < 20 loop
if totientSum(n) == n then
print_i16(n);
print_char(' ');
seen := seen + 1;
end if;
n := n + 2;
end loop;
print_nl(); |
http://rosettacode.org/wiki/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #Delphi | Delphi |
program Perfect_totient_numbers;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
function totient(n: Integer): Integer;
begin
var tot := n;
var i := 2;
while i * i <= n do
begin
if (n mod i) = 0 then
begin
while (n mod i) = 0 do
n := n div i;
dec(tot, tot div i);
end;
if i = 2 then
i := 1;
inc(i, 2);
end;
if n > 1 then
dec(tot, tot div n);
Result := tot;
end;
begin
var perfect: TArray<Integer>;
var n := 1;
while length(perfect) < 20 do
begin
var tot := n;
var sum := 0;
while tot <> 1 do
begin
tot := totient(tot);
inc(sum, tot);
end;
if sum = n then
begin
SetLength(perfect, Length(perfect) + 1);
perfect[High(perfect)] := n;
end;
inc(n, 2);
end;
writeln('The first 20 perfect totient numbers are:');
write('[');
for var e in perfect do
write(e, ' ');
writeln(']');
readln;
end. |
http://rosettacode.org/wiki/Playing_cards | Playing cards | Task
Create a data structure and the associated methods to define and manipulate a deck of playing cards.
The deck should contain 52 unique cards.
The methods must include the ability to:
make a new deck
shuffle (randomize) the deck
deal from the deck
print the current contents of a deck
Each card must have a pip value and a suit value which constitute the unique value of the card.
Related tasks:
Card shuffles
Deal cards_for_FreeCell
War Card_Game
Poker hand_analyser
Go Fish
| #Erlang | Erlang |
-module( playing_cards ).
-export( [deal/2, deal/3, deck/0, print/1, shuffle/1, sort_pips/1, sort_suites/1, task/0] ).
-record( card, {pip, suite} ).
-spec( deal( N_cards::integer(), Deck::[#card{}]) -> {Hand::[#card{}], Deck::[#card{}]} ).
deal( N_cards, Deck ) -> lists:split( N_cards, Deck ).
-spec( deal( N_hands::integer(), N_cards::integer(), Deck::[#card{}]) -> {List_of_hands::[[#card{}]], Deck::[#card{}]} ).
deal( N_hands, N_cards, Deck ) -> lists:foldl( fun deal_hands/2, {lists:duplicate(N_hands, []), Deck}, lists:seq(1, N_cards * N_hands) ).
deck() -> [#card{suite=X, pip=Y} || X <- suites(), Y <- pips()].
print( Cards ) -> [io:fwrite( " ~p", [X]) || X <- Cards], io:nl().
shuffle( Deck ) -> knuth_shuffle:list( Deck ).
sort_pips( Cards ) -> lists:keysort( #card.pip, Cards ).
sort_suites( Cards ) -> lists:keysort( #card.suite, Cards ).
task() ->
Deck = deck(),
Shuffled = shuffle( Deck ),
{Hand, New_deck} = deal( 3, Shuffled ),
{Hands, _Deck} = deal( 2, 3, New_deck ),
io:fwrite( "Hand:" ),
print( Hand ),
io:fwrite( "Hands:~n" ),
[print(X) || X <- Hands].
deal_hands( _N, {[Hand | T], [Card | Deck]} ) -> {T ++ [[Card | Hand]], Deck}.
pips() -> ["2", "3", "4", "5", "6", "7", "8", "9", "10", "Jack", "Queen", "King", "Ace"].
suites() -> ["Clubs", "Hearts", "Spades", "Diamonds"].
|
http://rosettacode.org/wiki/Pi | Pi |
Create a program to continually calculate and output the next decimal digit of
π
{\displaystyle \pi }
(pi).
The program should continue forever (until it is aborted by the user) calculating and outputting each decimal digit in succession.
The output should be a decimal sequence beginning 3.14159265 ...
Note: this task is about calculating pi. For information on built-in pi constants see Real constants and functions.
Related Task Arithmetic-geometric mean/Calculate Pi
| #EDSAC_order_code | EDSAC order code |
[EDSAC program, Initial Orders 2.
Calculates digits of pi by spigot algorithm.
Easily edited to calculate digits of e (see "if pi" and "if e" in comments).
Based on http://pi314.net/eng/goutte.php
See also https://www.cut-the-knot.org/Curriculum/Algorithms/SpigotForPi.shtml
Uses 17-bit values throughout.
Array index and counters are stored in the address field,
i.e. with the least significant bit in bit 1.
For integer arithmetic, the least significant bit is bit 0.
Variables don't need to be initialized at load time, so they overwrite
locations 6..12 of initial orders to save space.]
[6] P F [index into remainder array]
[7] P F [carry in the spigot algorithm]
[8] P F [negative count of digits]
[9] P F [pending digit, always < 9]
[10] P F [negative count of pending 9's]
[11] P F [9 or 0 in top 5 bits, for printing]
[12] P F [negative count of characters in current line]
[Array corresponding to Remainder row on http://pi314.net/eng/goutte.php]
T 53 K [refer to array via B parameter]
P 13 F [start array immediately after variables]
[Subroutine for short (17-bit) integer division.
Input: dividend at 4F, divisor at 5F.
Output: remainder at 4F, quotient at 5F.
Working locations 0F, 1F. 37 locations.]
T 987 K
GKA3FT34@A5FUFT35@A4FRDS35@G13@T1FA35@LDE4@T1FT5FA4FS35@G22@
T4FA5FA36@T5FT1FAFS35@E34@T1FA35@RDT35@A5FLDT5FE15@EFPFPD
T 845 K
G K
[Constants]
[Enter the editable numbers as addresses, e.g. P 100 F for 100.
Reducing the maximum array index will make the program take less time.
A maximum index of 831 (i.e. using all available memory) will give
252 correct digits of pi, or 2070 correct digits of e.]
[0] P 831 F [maximum array index <-- EDIT HERE, don't exceed 831]
[1] P 252 F [number of digits <-- EDIT HERE]
[2] P 72 F [digits per line <-- EDIT HERE]
[3] P D [short-value 1]
[4] P 5 F [short-value 10]
[5] J F [10*(2^12)]
[6] X F [add to T order to make V order]
[7] M F [used to convert T order to A order]
[8] # F [figures shift (for printer)]
[9] @ F [carriage return]
[10] & F [line feed]
[Main routine. Enter with acc = 0.]
[11] O 8 @ [set printer to figures; also used to print '9']
S 2 @ [load negative characters per line]
T 12 F [initialize character count]
S 1 @ [load negated number of digits]
T 8 F [initialize digit count]
T 10 F [clear negative count of 9's]
S 2 F [load -2 (any value < -1 would do)]
T 9 F [initialize digit buffer]
[Start algorithm: fill the remainder array with 2's (or 1's for e)
The code is a bit neater if we work backwards.]
A @ [maximum index]
[20] A 81 @ [make T order for array entry]
T 23 @ [plant in code]
[if pi] A 2 F [acc := 2]
[if e] [A 3 @] [acc := 1]
[23] T F [store in array entry]
A 23 @ [dec address in array]
S 2 F
S 81 @ [finished array?]
E 20 @ [loop back if not
Outer loop. Here for next digit.]
[28] T F [clear acc]
[Multiply remainder array by 10.
NB To preserve integer scaling, we need product times 2^16.]
H 5 @ [mult reg := 10*(2^12)]
A @ [acc := maximum index]
[31] A 81 @ [make T order for array entry]
U 37 @ [plant in code]
A 6 @ [convert to V order, same address]
T 35 @ [plant in code]
[35] V F [acc := array entry * 10*(2^12)]
L 4 F [shift to complete mult by 10*(2^16)]
[37] T F [store result in array]
A 37 @ [load T order]
S 2 F [dec address]
S 81 @ [test for done]
E 31 @ [loop back if not]
T F [clear acc]
T 7 F [clear carry]
A @ [acc := maximum index]
T 6 F [initialize array index]
[Inner loop to get next digit.
Work backwards through remainder array.]
[46] T F [clear acc]
A 6 F [load index]
A 81 @ [make T order for array entry]
U 61 @ [plant in code]
A 7 @ [convert to A order]
T 52 @ [plant in code]
[52] A F [load array element]
A 7 F [add carry from last time round loop]
T 4 F [sum to 4F for division routine]
A 6 F [acc := index as address = 2*(index as integer)]
[if pi] A 3 @ [plus 1]
[if e] [R D] [shift right, address --> integer]
T 5 F [to 5F for division routine (for e, 5F = index/2)]
[58] A 58 @ [call routine to divide 4F by 5F]
G 987 F
A 4 F [load remainder]
[61] T F [update element of remainder array]
[if pi: 4 orders]
H 5 F [mult reg := quotient]
V 6 F [multiply by index
NB need to shift 15 left to preserve integer scaling]
L F [shift 13 left]
L 1 F [shift 2 more left (for e, just use quotient)
[if e: 1 order, plus 3 no-ops to keep addresses the same]
[A 5 F] [load quotient]
[XF XF XF]
T 7 F [update carry for next time round loop]
A 6 F [load index]
S 2 F [decrement]
U 6 F [store back]
[We want to terminate after doing index = 1]
S 2 F [dec again]
E 46 @ [jump back if index >= 1]
[Treatment of index = 0 is different]
T F [clear acc]
A B [load rem{0)]
A 7 F [add carry]
T 4 F [sum to 4F for division routine]
A 4 @ [load 10]
T 5 F [to 5F for division routine]
[78] A 78 @ [call division routine]
G 987 F
A 4 F [load remainder]
[81] T B [store in rem{0}; also used to manufacture orders]
[82] A 82 @ [call subroutine to deal with quotient (clears acc)]
G 93 @
A 8 F [load negative digit count]
A 2 F [increment]
U 8 F [store back]
G 28 @ [if not yet 0, loop for next digit]
[Fake a zero digit to flush the last genuine digit(s)]
T 5 F [store fake digit 0 in 5F]
[89] A 89 @ [call subroutine to deal with digit]
G 93 @
O 8 @ [set figures: dummy character to flush print buffer]
Z F [stop]
[Subroutine to handle buffering and printing of digits.
Here with quotient from spigot algorithm still in 5F.
The quotient at 5F is usually the new decimal digit.
But the quotient can be 10, in which case we must treat it as 0
and ripple a carry through the previously-computed digits.
Hence the need for buffering.]
[93] A 3 F [make and plant return link]
T 130 @
A 5 F [load quotient]
S 4 @ [subtract 10]
E 105 @ [jump if quotient >= 10]
A 3 @ [add 1]
G 109 @ [jump if quotient < 9]
[Here if quotient = 9. Update count of 9's,
don't do anything with the buffer.]
T F [clear acc]
A 10 F [load negative count of 9's]
S 2 F [subtract 1]
T 10 F [update count]
E 130 @ [exit with acc = 0]
[Here if quotient >= 10. Take digit = quotient - 10,
and ripple a carry through the buffered digits.]
[105] T 5 F [store (quotient - 10) formed above]
T 11 F [store 0 to print '0' not '9']
A 3 @ [add 1 to buffered digit]
E 112 @ [join common code]
[Here if quotient < 9. Flush the stored digits.]
[109] T F [clear acc]
A 11 @ [load any O order (code for O is 9)]
T 11 F [store to print '9']
[112] A 9 F [load buffered digit, plus 1 if quotient >= 10]
G 118 @ [skip printing if buffer is empty]
L 1024 F [shift digits to top 5 bits]
T 1 F [store in 1F for printing]
[116] A 116 @ [call print routine]
G 131 @
[118] T F [clear acc]
A 5 F [load quotient (possibly modified as above)]
T 9 F [store in buffer]
[121] A 10 F [load negative count of 9's]
E 130 @ [if none, exit with acc = 0]
A 2 F [inc count]
T 10 F
A 11 F [load 9 (or 0 if there's a carry)]
T 1 F [to 1F for printing]
[127] A 127 @ [call print routine (clears acc)]
G 131 @
E 121 @ [jump back (always)]
[130] E F [return to caller]
[Subroutine to print character at 1F.
Also prints CR LF if necessary.]
[131] A 3 F [make and plant link for return]
T 141 @
A 12 F [load negative character count]
G 138 @ [jump if not end of line]
S 2 @ [reset character count]
O 9 @ [print CR LF]
O 10 @
[138] O 1 F [print character]
A 2 F [add 1]
T 12 F
[141] E F
E 11 Z [define entry point]
P F [acc = 0 on entry]
|
http://rosettacode.org/wiki/Periodic_table | Periodic table | Task
Display the row and column in the periodic table of the given atomic number.
The periodic table
Let us consider the following periodic table representation.
__________________________________________________________________________
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
| |
|1 H He |
| |
|2 Li Be B C N O F Ne |
| |
|3 Na Mg Al Si P S Cl Ar |
| |
|4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr |
| |
|5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe |
| |
|6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn |
| |
|7 Fr Ra ° Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og |
|__________________________________________________________________________|
| |
| |
|8 Lantanoidi* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu |
| |
|9 Aktinoidi° Ak Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr |
|__________________________________________________________________________|
Example test cases;
1 -> 1 1
2 -> 1 18
29 -> 4 11
42 -> 5 6
57 -> 8 4
58 -> 8 5
72 -> 6 4
89 -> 9 4
Details;
The representation of the periodic table may be represented in various way. The one presented in this challenge does have the following property : Lantanides and Aktinoides are all in a dedicated row, hence there is no element that is placed at 6, 3 nor 7, 3.
You may take a look at the atomic number repartitions here.
The atomic number is at least 1, at most 118.
See also
the periodic table
This task was an idea from CompSciFact
The periodic table in ascii that was used as template
| #XPL0 | XPL0 | proc ShowPosn(N); \Show row and column for element
int N, M, A, B, I, R, C;
[A:= [ 1, 2, 5, 13, 57, 72, 89, 104]; \magic numbers
B:= [-1, 15, 25, 35, 72, 21, 58, 7];
I:= 7;
while A(I) > N do I:= I-1;
M:= N + B(I);
R:= M/18 +1;
C:= rem(0) +1;
IntOut(0, N); Text(0, " -> ");
IntOut(0, R); Text(0, ", ");
IntOut(0, C); CrLf(0);
];
int Element, I;
[Element:= [1, 2, 29, 42, 57, 58, 72, 89, 90, 103];
for I:= 0 to 10-1 do ShowPosn(Element(I));
] |
http://rosettacode.org/wiki/Pig_the_dice_game | Pig the dice game | The game of Pig is a multiplayer game played with a single six-sided die. The
object of the game is to reach 100 points or more.
Play is taken in turns. On each person's turn that person has the option of either:
Rolling the dice: where a roll of two to six is added to their score for that turn and the player's turn continues as the player is given the same choice again; or a roll of 1 loses the player's total points for that turn and their turn finishes with play passing to the next player.
Holding: the player's score for that round is added to their total and becomes safe from the effects of throwing a 1 (one). The player's turn finishes with play passing to the next player.
Task
Create a program to score for, and simulate dice throws for, a two-person game.
Related task
Pig the dice game/Player
| #Lua | Lua | local numPlayers = 2
local maxScore = 100
local scores = { }
for i = 1, numPlayers do
scores[i] = 0 -- total safe score for each player
end
math.randomseed(os.time())
print("Enter a letter: [h]old or [r]oll?")
local points = 0 -- points accumulated in current turn
local p = 1 -- start with first player
while true do
io.write("\nPlayer "..p..", your score is ".. scores[p]..", with ".. points.." temporary points. ")
local reply = string.sub(string.lower(io.read("*line")), 1, 1)
if reply == 'r' then
local roll = math.random(6)
io.write("You rolled a " .. roll)
if roll == 1 then
print(". Too bad. :(")
p = (p % numPlayers) + 1
points = 0
else
points = points + roll
end
elseif reply == 'h' then
scores[p] = scores[p] + points
if scores[p] >= maxScore then
print("Player "..p..", you win with a score of "..scores[p])
break
end
print("Player "..p..", your new score is " .. scores[p])
p = (p % numPlayers) + 1
points = 0
end
end
|
http://rosettacode.org/wiki/Pernicious_numbers | Pernicious numbers | A pernicious number is a positive integer whose population count is a prime.
The population count is the number of ones in the binary representation of a non-negative integer.
Example
22 (which is 10110 in binary) has a population count of 3, which is prime, and therefore
22 is a pernicious number.
Task
display the first 25 pernicious numbers (in decimal).
display all pernicious numbers between 888,888,877 and 888,888,888 (inclusive).
display each list of integers on one line (which may or may not include a title).
See also
Sequence A052294 pernicious numbers on The On-Line Encyclopedia of Integer Sequences.
Rosetta Code entry population count, evil numbers, odious numbers.
| #F.23 | F# | open System
//Taken from https://gist.github.com/rmunn/bc49d32a586cdfa5bcab1c3e7b45d7ac
let bitcount (n : int) =
let count2 = n - ((n >>> 1) &&& 0x55555555)
let count4 = (count2 &&& 0x33333333) + ((count2 >>> 2) &&& 0x33333333)
let count8 = (count4 + (count4 >>> 4)) &&& 0x0f0f0f0f
(count8 * 0x01010101) >>> 24
//Modified from other examples to actually state the 1 is not prime
let isPrime n =
if n < 2 then
false
else
let sqrtn n = int <| sqrt (float n)
seq { 2 .. sqrtn n } |> Seq.exists(fun i -> n % i = 0) |> not
[<EntryPoint>]
let main _ =
[1 .. 100] |> Seq.filter (bitcount >> isPrime) |> Seq.take 25 |> Seq.toList |> printfn "%A"
[888888877 .. 888888888] |> Seq.filter (bitcount >> isPrime) |> Seq.toList |> printfn "%A"
0 // return an integer exit code |
http://rosettacode.org/wiki/Permutations/Rank_of_a_permutation | Permutations/Rank of a permutation | A particular ranking of a permutation associates an integer with a particular ordering of all the permutations of a set of distinct items.
For our purposes the ranking will assign integers
0..
(
n
!
−
1
)
{\displaystyle 0..(n!-1)}
to an ordering of all the permutations of the integers
0..
(
n
−
1
)
{\displaystyle 0..(n-1)}
.
For example, the permutations of the digits zero to 3 arranged lexicographically have the following rank:
PERMUTATION RANK
(0, 1, 2, 3) -> 0
(0, 1, 3, 2) -> 1
(0, 2, 1, 3) -> 2
(0, 2, 3, 1) -> 3
(0, 3, 1, 2) -> 4
(0, 3, 2, 1) -> 5
(1, 0, 2, 3) -> 6
(1, 0, 3, 2) -> 7
(1, 2, 0, 3) -> 8
(1, 2, 3, 0) -> 9
(1, 3, 0, 2) -> 10
(1, 3, 2, 0) -> 11
(2, 0, 1, 3) -> 12
(2, 0, 3, 1) -> 13
(2, 1, 0, 3) -> 14
(2, 1, 3, 0) -> 15
(2, 3, 0, 1) -> 16
(2, 3, 1, 0) -> 17
(3, 0, 1, 2) -> 18
(3, 0, 2, 1) -> 19
(3, 1, 0, 2) -> 20
(3, 1, 2, 0) -> 21
(3, 2, 0, 1) -> 22
(3, 2, 1, 0) -> 23
Algorithms exist that can generate a rank from a permutation for some particular ordering of permutations, and that can generate the same rank from the given individual permutation (i.e. given a rank of 17 produce (2, 3, 1, 0) in the example above).
One use of such algorithms could be in generating a small, random, sample of permutations of
n
{\displaystyle n}
items without duplicates when the total number of permutations is large. Remember that the total number of permutations of
n
{\displaystyle n}
items is given by
n
!
{\displaystyle n!}
which grows large very quickly: A 32 bit integer can only hold
12
!
{\displaystyle 12!}
, a 64 bit integer only
20
!
{\displaystyle 20!}
. It becomes difficult to take the straight-forward approach of generating all permutations then taking a random sample of them.
A question on the Stack Overflow site asked how to generate one million random and indivudual permutations of 144 items.
Task
Create a function to generate a permutation from a rank.
Create the inverse function that given the permutation generates its rank.
Show that for
n
=
3
{\displaystyle n=3}
the two functions are indeed inverses of each other.
Compute and show here 4 random, individual, samples of permutations of 12 objects.
Stretch goal
State how reasonable it would be to use your program to address the limits of the Stack Overflow question.
References
Ranking and Unranking Permutations in Linear Time by Myrvold & Ruskey. (Also available via Google here).
Ranks on the DevData site.
Another answer on Stack Overflow to a different question that explains its algorithm in detail.
Related tasks
Factorial_base_numbers_indexing_permutations_of_a_collection
| #zkl | zkl | fcn computePermutation(items,rank){ // in place permutation of items
foreach n in ([items.len()..1,-1]){
r:=rank%n;
rank/=n;
items.swap(r,n-1);
}
items
} |
http://rosettacode.org/wiki/Pierpont_primes | Pierpont primes | A Pierpont prime is a prime number of the form: 2u3v + 1 for some non-negative integers u and v .
A Pierpont prime of the second kind is a prime number of the form: 2u3v - 1 for some non-negative integers u and v .
The term "Pierpont primes" is generally understood to mean the first definition, but will be called "Pierpont primes of the first kind" on this page to distinguish them.
Task
Write a routine (function, procedure, whatever) to find Pierpont primes of the first & second kinds.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the first kind.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the second kind
If your language supports large integers, find and display here, on this page, the 250th Pierpont prime of the first kind and the 250th Pierpont prime of the second kind.
See also
Wikipedia - Pierpont primes
OEIS:A005109 - Class 1 -, or Pierpont primes
OEIS:A005105 - Class 1 +, or Pierpont primes of the second kind
| #Raku | Raku | use ntheory:from<Perl5> <is_prime>;
sub pierpont ($kind is copy = 1) {
fail "Unknown type: $kind. Must be one of 1 (default) or 2" if $kind !== 1|2;
$kind = -1 if $kind == 2;
my $po3 = 3;
my $add-one = 3;
my @iterators = [1,2,4,8 … *].iterator, [3,9,27 … *].iterator;
my @head = @iterators».pull-one;
gather {
loop {
my $key = @head.pairs.min( *.value ).key;
my $min = @head[$key];
@head[$key] = @iterators[$key].pull-one;
take $min + $kind if "{$min + $kind}".&is_prime;
if $min >= $add-one {
@iterators.push: ([|((2,4,8).map: * * $po3) … *]).iterator;
$add-one = @head[+@iterators - 1] = @iterators[+@iterators - 1].pull-one;
$po3 *= 3;
}
}
}
}
say "First 50 Pierpont primes of the first kind:\n" ~ pierpont[^50].rotor(10)».fmt('%8d').join: "\n";
say "\nFirst 50 Pierpont primes of the second kind:\n" ~ pierpont(2)[^50].rotor(10)».fmt('%8d').join: "\n";
say "\n250th Pierpont prime of the first kind: " ~ pierpont[249];
say "\n250th Pierpont prime of the second kind: " ~ pierpont(2)[249]; |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #GW-BASIC | GW-BASIC | 10 RANDOMIZE TIMER : REM set random number seed to something arbitrary
20 DIM ARR(10) : REM initialise array
30 FOR I = 1 TO 10
40 ARR(I) = I*I : REM squares of the integers is OK as a demo
50 NEXT I
60 C = 1 + INT(RND*10) : REM get a random index from 1 to 10 inclusive
70 PRINT ARR(C) |
http://rosettacode.org/wiki/Primality_by_trial_division | Primality by trial division | Task
Write a boolean function that tells whether a given integer is prime.
Remember that 1 and all non-positive numbers are not prime.
Use trial division.
Even numbers greater than 2 may be eliminated right away.
A loop from 3 to √ n will suffice, but other loops are allowed.
Related tasks
count in factors
prime decomposition
AKS test for primes
factors of an integer
Sieve of Eratosthenes
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
| #XPL0 | XPL0 | func Prime(N); \Return 'true' if N is a prime number
int N;
int I;
[if N <= 1 then return false;
for I:= 2 to sqrt(N) do
if rem(N/I) = 0 then return false;
return true;
];
int Num;
repeat Num:= IntIn(0);
Text(0, if Prime(Num) then "is " else "not ");
Text(0, "prime^M^J");
until Num = 0 |
http://rosettacode.org/wiki/Phrase_reversals | Phrase reversals | Task
Given a string of space separated words containing the following phrase:
rosetta code phrase reversal
Reverse the characters of the string.
Reverse the characters of each individual word in the string, maintaining original word order within the string.
Reverse the order of each word of the string, maintaining the order of characters in each word.
Show your output here.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Haskell | Haskell | reverseString, reverseEachWord, reverseWordOrder :: String -> String
reverseString = reverse
reverseEachWord = wordLevel (fmap reverse)
reverseWordOrder = wordLevel reverse
wordLevel :: ([String] -> [String]) -> String -> String
wordLevel f = unwords . f . words
main :: IO ()
main =
(putStrLn . unlines) $
[reverseString, reverseEachWord, reverseWordOrder] <*>
["rosetta code phrase reversal"] |
http://rosettacode.org/wiki/Permutations/Derangements | Permutations/Derangements | A derangement is a permutation of the order of distinct items in which no item appears in its original place.
For example, the only two derangements of the three items (0, 1, 2) are (1, 2, 0), and (2, 0, 1).
The number of derangements of n distinct items is known as the subfactorial of n, sometimes written as !n.
There are various ways to calculate !n.
Task
Create a named function/method/subroutine/... to generate derangements of the integers 0..n-1, (or 1..n if you prefer).
Generate and show all the derangements of 4 integers using the above routine.
Create a function that calculates the subfactorial of n, !n.
Print and show a table of the counted number of derangements of n vs. the calculated !n for n from 0..9 inclusive.
Optional stretch goal
Calculate !20
Related tasks
Anagrams/Deranged anagrams
Best shuffle
Left_factorials
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Factor | Factor | USING: combinators formatting io kernel math math.combinatorics
prettyprint sequences ;
IN: rosetta-code.derangements
: !n ( n -- m )
{
{ 0 [ 1 ] }
{ 1 [ 0 ] }
[ [ 1 - !n ] [ 2 - !n + ] [ 1 - * ] tri ]
} case ;
: derangements ( n -- seq )
<iota> dup [ [ = ] 2map [ f = ] all? ] with
filter-permutations ;
"4 derangements" print 4 derangements . nl
"n count calc\n= ====== ======" print
10 <iota> [
dup [ derangements length ] [ !n ] bi
"%d%8d%8d\n" printf
] each nl
"!20 = " write 20 !n . |
http://rosettacode.org/wiki/Permutations_by_swapping | Permutations by swapping | Task
Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.
Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.
Show the permutations and signs of three items, in order of generation here.
Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.
Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.
References
Steinhaus–Johnson–Trotter algorithm
Johnson-Trotter Algorithm Listing All Permutations
Heap's algorithm
[1] Tintinnalogia
Related tasks
Matrix arithmetic
Gray code
| #Haskell | Haskell | sPermutations :: [a] -> [([a], Int)]
sPermutations = flip zip (cycle [-1, 1]) . foldr aux [[]]
where
aux x items = do
(f, item) <- zip (repeat id) items
f (insertEv x item)
insertEv x [] = [[x]]
insertEv x l@(y:ys) = (x : l) : ((y :) <$> insertEv x ys)
main :: IO ()
main = do
putStrLn "3 items:"
mapM_ print $ sPermutations [1 .. 3]
putStrLn "\n4 items:"
mapM_ print $ sPermutations [1 .. 4] |
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #M2000_Interpreter | M2000 Interpreter |
Module Checkit {
Global data(), treat=0
data()=(85, 88, 75, 66, 25, 29, 83, 39, 97,68, 41, 10, 49, 16, 65, 32, 92, 28, 98)
Function pick(at, remain, accu) {
If remain Else =If(accu>treat->1,0):Exit
=pick(at-1,remain-1,accu+data(at-1))+If(at>remain->pick(at-1, remain, accu),0)
}
total=1
For i=0 to 8 {treat+=data(i)}
For i=19 to 11 {total*=i}
For i=9 to 1 {total/=i}
gt=pick(19,9,0)
le=total-gt
Print Format$("<= : {0:1}% {1}", 100*le/total, le)
Print Format$(" > : {0:1}% {1}", 100*gt/total, gt)
}
Checkit
|
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | "<=: " <> ToString[#1] <> " " <> ToString[100. #1/#2] <> "%\n >: " <>
ToString[#2 - #1] <> " " <> ToString[100. (1 - #1/#2)] <> "%" &[
Count[Total /@ Subsets[Join[#1, #2], {Length@#1}],
n_ /; n <= Total@#1],
Binomial[Length@#1 + Length@#2, Length@#1]] &[{85, 88, 75, 66, 25,
29, 83, 39, 97}, {68, 41, 10, 49, 16, 65, 32, 92, 28, 98}] |
http://rosettacode.org/wiki/Percentage_difference_between_images | Percentage difference between images | basic bitmap storage
Useful for comparing two JPEG images saved with a different compression ratios.
You can use these pictures for testing (use the full-size version of each):
50% quality JPEG
100% quality JPEG
link to full size 50% image
link to full size 100% image
The expected difference for these two images is 1.62125%
| #BBC_BASIC | BBC BASIC | hbm1% = FNloadimage("C:lenna50.jpg")
hbm2% = FNloadimage("C:lenna100.jpg")
SYS "CreateCompatibleDC", @memhdc% TO hdc1%
SYS "CreateCompatibleDC", @memhdc% TO hdc2%
SYS "SelectObject", hdc1%, hbm1%
SYS "SelectObject", hdc2%, hbm2%
diff% = 0
FOR y% = 0 TO 511
FOR x% = 0 TO 511
SYS "GetPixel", hdc1%, x%, y% TO rgb1%
SYS "GetPixel", hdc2%, x%, y% TO rgb2%
diff% += ABS((rgb1% AND &FF) - (rgb2% AND &FF))
diff% += ABS((rgb1% >> 8 AND &FF) - (rgb2% >> 8 AND &FF))
diff% += ABS((rgb1% >> 16) - (rgb2% >> 16))
NEXT
NEXT y%
PRINT "Image difference = "; 100 * diff% / 512^2 / 3 / 255 " %"
SYS "DeleteDC", hdc1%
SYS "DeleteDC", hdc2%
SYS "DeleteObject", hbm1%
SYS "DeleteObject", hbm2%
END
DEF FNloadimage(file$)
LOCAL iid{}, hbm%, pic%, ole%, olpp%, text%
DIM iid{a%,b%,c%,d%}, text% LOCAL 513
iid.a% = &7BF80980 : REM. 128-bit iid
iid.b% = &101ABF32
iid.c% = &AA00BB8B
iid.d% = &AB0C3000
SYS "MultiByteToWideChar", 0, 0, file$, -1, text%, 256
SYS "LoadLibrary", "OLEAUT32.DLL" TO ole%
SYS "GetProcAddress", ole%, "OleLoadPicturePath" TO olpp%
IF olpp%=0 THEN = 0
SYS olpp%, text%, 0, 0, 0, iid{}, ^pic% : REM. OleLoadPicturePath
IF pic%=0 THEN = 0
SYS !(!pic%+12), pic%, ^hbm% : REM. IPicture::get_Handle
SYS "FreeLibrary", ole%
= hbm% |
http://rosettacode.org/wiki/Percentage_difference_between_images | Percentage difference between images | basic bitmap storage
Useful for comparing two JPEG images saved with a different compression ratios.
You can use these pictures for testing (use the full-size version of each):
50% quality JPEG
100% quality JPEG
link to full size 50% image
link to full size 100% image
The expected difference for these two images is 1.62125%
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <math.h>
/* #include "imglib.h" */
#define RED_C 0
#define GREEN_C 1
#define BLUE_C 2
#define GET_PIXEL(IMG, X, Y) ((IMG)->buf[ (Y) * (IMG)->width + (X) ])
int main(int argc, char **argv)
{
image im1, im2;
double totalDiff = 0.0;
unsigned int x, y;
if ( argc < 3 )
{
fprintf(stderr, "usage:\n%s FILE1 FILE2\n", argv[0]);
exit(1);
}
im1 = read_image(argv[1]);
if ( im1 == NULL ) exit(1);
im2 = read_image(argv[2]);
if ( im2 == NULL ) { free_img(im1); exit(1); }
if ( (im1->width != im2->width) || (im1->height != im2->height) )
{
fprintf(stderr, "width/height of the images must match!\n");
} else {
/* BODY: the "real" part! */
for(x=0; x < im1->width; x++)
{
for(y=0; y < im1->width; y++)
{
totalDiff += fabs( GET_PIXEL(im1, x, y)[RED_C] - GET_PIXEL(im2, x, y)[RED_C] ) / 255.0;
totalDiff += fabs( GET_PIXEL(im1, x, y)[GREEN_C] - GET_PIXEL(im2, x, y)[GREEN_C] ) / 255.0;
totalDiff += fabs( GET_PIXEL(im1, x, y)[BLUE_C] - GET_PIXEL(im2, x, y)[BLUE_C] ) / 255.0;
}
}
printf("%lf\n", 100.0 * totalDiff / (double)(im1->width * im1->height * 3) );
/* BODY ends */
}
free_img(im1);
free_img(im2);
} |
http://rosettacode.org/wiki/Percolation/Mean_cluster_density | Percolation/Mean cluster density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
c
{\displaystyle c}
be a 2D boolean square matrix of
n
×
n
{\displaystyle n\times n}
values of either 1 or 0 where the
probability of any value being 1 is
p
{\displaystyle p}
, (and of 0 is therefore
1
−
p
{\displaystyle 1-p}
).
We define a cluster of 1's as being a group of 1's connected vertically or
horizontally (i.e., using the Von Neumann neighborhood rule) and bounded by either
0
{\displaystyle 0}
or by the limits of the matrix.
Let the number of such clusters in such a randomly constructed matrix be
C
n
{\displaystyle C_{n}}
.
Percolation theory states that
K
(
p
)
{\displaystyle K(p)}
(the mean cluster density) will satisfy
K
(
p
)
=
C
n
/
n
2
{\displaystyle K(p)=C_{n}/n^{2}}
as
n
{\displaystyle n}
tends to infinity. For
p
=
0.5
{\displaystyle p=0.5}
,
K
(
p
)
{\displaystyle K(p)}
is found numerically to approximate
0.065770
{\displaystyle 0.065770}
...
Task
Show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
for
p
=
0.5
{\displaystyle p=0.5}
and
for values of
n
{\displaystyle n}
up to at least
1000
{\displaystyle 1000}
.
Any calculation of
C
n
{\displaystyle C_{n}}
for finite
n
{\displaystyle n}
is subject to randomness, so an approximation should be
computed as the average of
t
{\displaystyle t}
runs, where
t
{\displaystyle t}
≥
5
{\displaystyle 5}
.
For extra credit, graphically show clusters in a
15
×
15
{\displaystyle 15\times 15}
,
p
=
0.5
{\displaystyle p=0.5}
grid.
Show your output here.
See also
s-Cluster on Wolfram mathworld. | #Factor | Factor | USING: combinators formatting generalizations kernel math
math.matrices random sequences ;
IN: rosetta-code.mean-cluster-density
CONSTANT: p 0.5
CONSTANT: iterations 5
: rand-bit-matrix ( n probability -- matrix )
dupd [ random-unit > 1 0 ? ] curry make-matrix ;
: flood-fill ( x y matrix -- )
3dup ?nth ?nth 1 = [
[ [ -1 ] 3dip nth set-nth ] [
{
[ [ 1 + ] 2dip ]
[ [ 1 - ] 2dip ]
[ [ 1 + ] dip ]
[ [ 1 - ] dip ]
} [ flood-fill ] map-compose 3cleave
] 3bi
] [ 3drop ] if ;
: count-clusters ( matrix -- Cn )
0 swap dup dim matrix-coordinates flip concat [
first2 rot 3dup ?nth ?nth 1 = [ flood-fill 1 + ]
[ 3drop ] if
] with each ;
: mean-cluster-density ( matrix -- mcd )
[ count-clusters ] [ dim first sq / ] bi ;
: simulate ( n -- avg-mcd )
iterations swap [ p rand-bit-matrix mean-cluster-density ]
curry replicate sum iterations / ;
: main ( -- )
{ 4 64 256 1024 4096 } [
[ iterations p ] dip dup simulate
"iterations = %d p = %.1f n = %4d sim = %.5f\n" printf
] each ;
MAIN: main |
http://rosettacode.org/wiki/Percolation/Mean_cluster_density | Percolation/Mean cluster density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
c
{\displaystyle c}
be a 2D boolean square matrix of
n
×
n
{\displaystyle n\times n}
values of either 1 or 0 where the
probability of any value being 1 is
p
{\displaystyle p}
, (and of 0 is therefore
1
−
p
{\displaystyle 1-p}
).
We define a cluster of 1's as being a group of 1's connected vertically or
horizontally (i.e., using the Von Neumann neighborhood rule) and bounded by either
0
{\displaystyle 0}
or by the limits of the matrix.
Let the number of such clusters in such a randomly constructed matrix be
C
n
{\displaystyle C_{n}}
.
Percolation theory states that
K
(
p
)
{\displaystyle K(p)}
(the mean cluster density) will satisfy
K
(
p
)
=
C
n
/
n
2
{\displaystyle K(p)=C_{n}/n^{2}}
as
n
{\displaystyle n}
tends to infinity. For
p
=
0.5
{\displaystyle p=0.5}
,
K
(
p
)
{\displaystyle K(p)}
is found numerically to approximate
0.065770
{\displaystyle 0.065770}
...
Task
Show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
for
p
=
0.5
{\displaystyle p=0.5}
and
for values of
n
{\displaystyle n}
up to at least
1000
{\displaystyle 1000}
.
Any calculation of
C
n
{\displaystyle C_{n}}
for finite
n
{\displaystyle n}
is subject to randomness, so an approximation should be
computed as the average of
t
{\displaystyle t}
runs, where
t
{\displaystyle t}
≥
5
{\displaystyle 5}
.
For extra credit, graphically show clusters in a
15
×
15
{\displaystyle 15\times 15}
,
p
=
0.5
{\displaystyle p=0.5}
grid.
Show your output here.
See also
s-Cluster on Wolfram mathworld. | #Go | Go | package main
import (
"fmt"
"math/rand"
"time"
)
var (
n_range = []int{4, 64, 256, 1024, 4096}
M = 15
N = 15
)
const (
p = .5
t = 5
NOT_CLUSTERED = 1
cell2char = " #abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
)
func newgrid(n int, p float64) [][]int {
g := make([][]int, n)
for y := range g {
gy := make([]int, n)
for x := range gy {
if rand.Float64() < p {
gy[x] = 1
}
}
g[y] = gy
}
return g
}
func pgrid(cell [][]int) {
for n := 0; n < N; n++ {
fmt.Print(n%10, ") ")
for m := 0; m < M; m++ {
fmt.Printf(" %c", cell2char[cell[n][m]])
}
fmt.Println()
}
}
func cluster_density(n int, p float64) float64 {
cc := clustercount(newgrid(n, p))
return float64(cc) / float64(n) / float64(n)
}
func clustercount(cell [][]int) int {
walk_index := 1
for n := 0; n < N; n++ {
for m := 0; m < M; m++ {
if cell[n][m] == NOT_CLUSTERED {
walk_index++
walk_maze(m, n, cell, walk_index)
}
}
}
return walk_index - 1
}
func walk_maze(m, n int, cell [][]int, indx int) {
cell[n][m] = indx
if n < N-1 && cell[n+1][m] == NOT_CLUSTERED {
walk_maze(m, n+1, cell, indx)
}
if m < M-1 && cell[n][m+1] == NOT_CLUSTERED {
walk_maze(m+1, n, cell, indx)
}
if m > 0 && cell[n][m-1] == NOT_CLUSTERED {
walk_maze(m-1, n, cell, indx)
}
if n > 0 && cell[n-1][m] == NOT_CLUSTERED {
walk_maze(m, n-1, cell, indx)
}
}
func main() {
rand.Seed(time.Now().Unix())
cell := newgrid(N, .5)
fmt.Printf("Found %d clusters in this %d by %d grid\n\n",
clustercount(cell), N, N)
pgrid(cell)
fmt.Println()
for _, n := range n_range {
M = n
N = n
sum := 0.
for i := 0; i < t; i++ {
sum += cluster_density(n, p)
}
sim := sum / float64(t)
fmt.Printf("t=%3d p=%4.2f n=%5d sim=%7.5f\n", t, p, n, sim)
}
} |
http://rosettacode.org/wiki/Perfect_numbers | Perfect numbers | Write a function which says whether a number is perfect.
A perfect number is a positive integer that is the sum of its proper positive divisors excluding the number itself.
Equivalently, a perfect number is a number that is half the sum of all of its positive divisors (including itself).
Note: The faster Lucas-Lehmer test is used to find primes of the form 2n-1, all known perfect numbers can be derived from these primes
using the formula (2n - 1) × 2n - 1.
It is not known if there are any odd perfect numbers (any that exist are larger than 102000).
The number of known perfect numbers is 51 (as of December, 2018), and the largest known perfect number contains 49,724,095 decimal digits.
See also
Rational Arithmetic
Perfect numbers on OEIS
Odd Perfect showing the current status of bounds on odd perfect numbers.
| #Action.21 | Action! | PROC Main()
DEFINE MAXNUM="10000"
CARD ARRAY pds(MAXNUM+1)
CARD i,j
FOR i=2 TO MAXNUM
DO
pds(i)=1
OD
FOR i=2 TO MAXNUM
DO
FOR j=i+i TO MAXNUM STEP i
DO
pds(j)==+i
OD
OD
FOR i=2 TO MAXNUM
DO
IF pds(i)=i THEN
PrintCE(i)
FI
OD
RETURN |
http://rosettacode.org/wiki/Percolation/Bond_percolation | Percolation/Bond percolation |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Given an
M
×
N
{\displaystyle M\times N}
rectangular array of cells numbered
c
e
l
l
[
0..
M
−
1
,
0..
N
−
1
]
{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}
, assume
M
{\displaystyle M}
is horizontal and
N
{\displaystyle N}
is downwards. Each
c
e
l
l
[
m
,
n
]
{\displaystyle \mathrm {cell} [m,n]}
is bounded by (horizontal) walls
h
w
a
l
l
[
m
,
n
]
{\displaystyle \mathrm {hwall} [m,n]}
and
h
w
a
l
l
[
m
+
1
,
n
]
{\displaystyle \mathrm {hwall} [m+1,n]}
; (vertical) walls
v
w
a
l
l
[
m
,
n
]
{\displaystyle \mathrm {vwall} [m,n]}
and
v
w
a
l
l
[
m
,
n
+
1
]
{\displaystyle \mathrm {vwall} [m,n+1]}
Assume that the probability of any wall being present is a constant
p
{\displaystyle p}
where
0.0
≤
p
≤
1.0
{\displaystyle 0.0\leq p\leq 1.0}
Except for the outer horizontal walls at
m
=
0
{\displaystyle m=0}
and
m
=
M
{\displaystyle m=M}
which are always present.
The task
Simulate pouring a fluid onto the top surface (
n
=
0
{\displaystyle n=0}
) where the fluid will enter any empty cell it is adjacent to if there is no wall between where it currently is and the cell on the other side of the (missing) wall.
The fluid does not move beyond the horizontal constraints of the grid.
The fluid may move “up” within the confines of the grid of cells. If the fluid reaches a bottom cell that has a missing bottom wall then the fluid can be said to 'drip' out the bottom at that point.
Given
p
{\displaystyle p}
repeat the percolation
t
{\displaystyle t}
times to estimate the proportion of times that the fluid can percolate to the bottom for any given
p
{\displaystyle p}
.
Show how the probability of percolating through the random grid changes with
p
{\displaystyle p}
going from
0.0
{\displaystyle 0.0}
to
1.0
{\displaystyle 1.0}
in
0.1
{\displaystyle 0.1}
increments and with the number of repetitions to estimate the fraction at any given
p
{\displaystyle p}
as
t
=
100
{\displaystyle t=100}
.
Use an
M
=
10
,
N
=
10
{\displaystyle M=10,N=10}
grid of cells for all cases.
Optionally depict fluid successfully percolating through a grid graphically.
Show all output on this page.
| #D | D | import std.stdio, std.random, std.array, std.range, std.algorithm;
struct Grid {
// Not enforced by runtime and type system:
// a Cell must contain only the flags bits.
alias Cell = uint;
enum : Cell { // Cell states (bit flags).
empty = 0,
filled = 1,
rightWall = 2,
bottomWall = 4
}
const size_t nc, nr;
Cell[] cells;
this(in size_t nRows, in size_t nCols) pure nothrow {
nr = nRows;
nc = nCols;
// Allocate two addition rows to avoid checking bounds.
// Bottom row is also required by drippage.
cells = new Cell[nc * (nr + 2)];
}
void initialize(in double prob, ref Xorshift rng) {
cells[0 .. nc] = bottomWall | rightWall; // First row.
uint pos = nc;
foreach (immutable r; 1 .. nr + 1) {
foreach (immutable c; 1 .. nc)
cells[pos++] = (uniform01 < prob ?bottomWall : empty) |
(uniform01 < prob ? rightWall : empty);
cells[pos++] = rightWall |
(uniform01 < prob ? bottomWall : empty);
}
cells[$ - nc .. $] = empty; // Last row.
}
bool percolate() pure nothrow @nogc {
bool fill(in size_t i) pure nothrow @nogc {
if (cells[i] & filled)
return false;
cells[i] |= filled;
if (i >= cells.length - nc)
return true; // Success: reached bottom row.
return (!(cells[i] & bottomWall) && fill(i + nc)) ||
(!(cells[i] & rightWall) && fill(i + 1)) ||
(!(cells[i - 1] & rightWall) && fill(i - 1)) ||
(!(cells[i - nc] & bottomWall) && fill(i - nc));
}
return iota(nc, nc + nc).any!fill;
}
void show() const {
writeln("+-".replicate(nc), '+');
foreach (immutable r; 1 .. nr + 2) {
write(r == nr + 1 ? ' ' : '|');
foreach (immutable c; 0 .. nc) {
immutable cell = cells[r * nc + c];
write((cell & filled) ? (r <= nr ? '#' : 'X') : ' ');
write((cell & rightWall) ? '|' : ' ');
}
writeln;
if (r == nr + 1)
return;
foreach (immutable c; 0 .. nc)
write((cells[r * nc + c] & bottomWall) ? "+-" : "+ ");
'+'.writeln;
}
}
}
void main() {
enum uint nr = 10, nc = 10; // N. rows and columns of the grid.
enum uint nTries = 10_000; // N. simulations for each probability.
enum uint nStepsProb = 10; // N. steps of probability.
auto rng = Xorshift(2);
auto g = Grid(nr, nc);
g.initialize(0.5, rng);
g.percolate;
g.show;
writefln("\nRunning %dx%d grids %d times for each p:",
nr, nc, nTries);
foreach (immutable p; 0 .. nStepsProb) {
immutable probability = p / double(nStepsProb);
uint nPercolated = 0;
foreach (immutable i; 0 .. nTries) {
g.initialize(probability, rng);
nPercolated += g.percolate;
}
writefln("p = %0.2f: %.4f",
probability, nPercolated / double(nTries));
}
} |
http://rosettacode.org/wiki/Percolation/Mean_run_density | Percolation/Mean run density |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Let
v
{\displaystyle v}
be a vector of
n
{\displaystyle n}
values of either 1 or 0 where the probability of any
value being 1 is
p
{\displaystyle p}
; the probability of a value being 0 is therefore
1
−
p
{\displaystyle 1-p}
.
Define a run of 1s as being a group of consecutive 1s in the vector bounded
either by the limits of the vector or by a 0. Let the number of such runs in a given
vector of length
n
{\displaystyle n}
be
R
n
{\displaystyle R_{n}}
.
For example, the following vector has
R
10
=
3
{\displaystyle R_{10}=3}
[1 1 0 0 0 1 0 1 1 1]
^^^ ^ ^^^^^
Percolation theory states that
K
(
p
)
=
lim
n
→
∞
R
n
/
n
=
p
(
1
−
p
)
{\displaystyle K(p)=\lim _{n\to \infty }R_{n}/n=p(1-p)}
Task
Any calculation of
R
n
/
n
{\displaystyle R_{n}/n}
for finite
n
{\displaystyle n}
is subject to randomness so should be
computed as the average of
t
{\displaystyle t}
runs, where
t
≥
100
{\displaystyle t\geq 100}
.
For values of
p
{\displaystyle p}
of 0.1, 0.3, 0.5, 0.7, and 0.9, show the effect of varying
n
{\displaystyle n}
on the accuracy of simulated
K
(
p
)
{\displaystyle K(p)}
.
Show your output here.
See also
s-Run on Wolfram mathworld. | #FreeBASIC | FreeBASIC | Function run_test(p As Double, longitud As Integer, runs As Integer) As Double
Dim As Integer r, l, cont = 0
Dim As Integer v, pv
For r = 1 To runs
pv = 0
For l = 1 To longitud
v = Rnd < p
cont += Iif(pv < v, 1, 0)
pv = v
Next l
Next r
Return (cont/runs/longitud)
End Function
Print "Running 1000 tests each:"
Print " p n K p(1-p) delta"
Print String(46,"-")
Dim As Double K, p, p1p
Dim As Integer n, ip
For ip = 1 To 10 Step 2
p = ip / 10
p1p = p * (1-p)
n = 100
While n <= 100000
K = run_test(p, n, 1000)
Print Using !"#.# ###### #.#### #.#### +##.#### (##.## \b%)"; _
p; n; K; p1p; K-p1p; (K-p1p)/p1p*100
n *= 10
Wend
Print
Next ip
Sleep |
http://rosettacode.org/wiki/Percolation/Site_percolation | Percolation/Site percolation |
Percolation Simulation
This is a simulation of aspects of mathematical percolation theory.
For other percolation simulations, see Category:Percolation Simulations, or:
1D finite grid simulation
Mean run density
2D finite grid simulations
Site percolation | Bond percolation | Mean cluster density
Given an
M
×
N
{\displaystyle M\times N}
rectangular array of cells numbered
c
e
l
l
[
0..
M
−
1
,
0..
N
−
1
]
{\displaystyle \mathrm {cell} [0..M-1,0..N-1]}
assume
M
{\displaystyle M}
is horizontal and
N
{\displaystyle N}
is downwards.
Assume that the probability of any cell being filled is a constant
p
{\displaystyle p}
where
0.0
≤
p
≤
1.0
{\displaystyle 0.0\leq p\leq 1.0}
The task
Simulate creating the array of cells with probability
p
{\displaystyle p}
and then
testing if there is a route through adjacent filled cells from any on row
0
{\displaystyle 0}
to any on row
N
{\displaystyle N}
, i.e. testing for site percolation.
Given
p
{\displaystyle p}
repeat the percolation
t
{\displaystyle t}
times to estimate the proportion of times that the fluid can percolate to the bottom for any given
p
{\displaystyle p}
.
Show how the probability of percolating through the random grid changes with
p
{\displaystyle p}
going from
0.0
{\displaystyle 0.0}
to
1.0
{\displaystyle 1.0}
in
0.1
{\displaystyle 0.1}
increments and with the number of repetitions to estimate the fraction at any given
p
{\displaystyle p}
as
t
>=
100
{\displaystyle t>=100}
.
Use an
M
=
15
,
N
=
15
{\displaystyle M=15,N=15}
grid of cells for all cases.
Optionally depict a percolation through a cell grid graphically.
Show all output on this page.
| #Haskell | Haskell | {-# LANGUAGE OverloadedStrings #-}
import Control.Monad
import Control.Monad.Random
import Data.Array.Unboxed
import Data.List
import Formatting
type Field = UArray (Int, Int) Char
-- Start percolating some seepage through a field.
-- Recurse to continue percolation with new seepage.
percolateR :: [(Int, Int)] -> Field -> (Field, [(Int,Int)])
percolateR [] f = (f, [])
percolateR seep f =
let ((xLo,yLo),(xHi,yHi)) = bounds f
validSeep = filter (\p@(x,y) -> x >= xLo
&& x <= xHi
&& y >= yLo
&& y <= yHi
&& f!p == ' ') $ nub $ sort seep
neighbors (x,y) = [(x,y-1), (x,y+1), (x-1,y), (x+1,y)]
in percolateR
(concatMap neighbors validSeep)
(f // map (\p -> (p,'.')) validSeep)
-- Percolate a field. Return the percolated field.
percolate :: Field -> Field
percolate start =
let ((_,_),(xHi,_)) = bounds start
(final, _) = percolateR [(x,0) | x <- [0..xHi]] start
in final
-- Generate a random field.
initField :: Int -> Int -> Double -> Rand StdGen Field
initField w h threshold = do
frnd <- fmap (\rv -> if rv<threshold then ' ' else '#') <$> getRandoms
return $ listArray ((0,0), (w-1, h-1)) frnd
-- Get a list of "leaks" from the bottom of a field.
leaks :: Field -> [Bool]
leaks f =
let ((xLo,_),(xHi,yHi)) = bounds f
in [f!(x,yHi)=='.'| x <- [xLo..xHi]]
-- Run test once; Return bool indicating success or failure.
oneTest :: Int -> Int -> Double -> Rand StdGen Bool
oneTest w h threshold =
or.leaks.percolate <$> initField w h threshold
-- Run test multple times; Return the number of tests that pass.
multiTest :: Int -> Int -> Int -> Double -> Rand StdGen Double
multiTest testCount w h threshold = do
results <- replicateM testCount $ oneTest w h threshold
let leakyCount = length $ filter id results
return $ fromIntegral leakyCount / fromIntegral testCount
-- Display a field with walls and leaks.
showField :: Field -> IO ()
showField a = do
let ((xLo,yLo),(xHi,yHi)) = bounds a
mapM_ print [ [ a!(x,y) | x <- [xLo..xHi]] | y <- [yLo..yHi]]
main :: IO ()
main = do
g <- getStdGen
let w = 15
h = 15
threshold = 0.6
(startField, g2) = runRand (initField w h threshold) g
putStrLn ("Unpercolated field with " ++ show threshold ++ " threshold.")
putStrLn ""
showField startField
putStrLn ""
putStrLn "Same field after percolation."
putStrLn ""
showField $ percolate startField
let testCount = 10000
densityCount = 10
putStrLn ""
putStrLn ( "Results of running percolation test " ++ show testCount
++ " times with thresholds ranging from 0/" ++ show densityCount
++ " to " ++ show densityCount ++ "/" ++ show densityCount ++ " .")
let densities = [0..densityCount]
tests = sequence [multiTest testCount w h v
| density <- densities,
let v = fromIntegral density / fromIntegral densityCount ]
results = zip densities (evalRand tests g2)
mapM_ print [format ("p=" % int % "/" % int % " -> " % fixed 4) density densityCount x | (density,x) <- results] |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #Ada | Ada | generic
N: positive;
package Generic_Perm is
subtype Element is Positive range 1 .. N;
type Permutation is array(Element) of Element;
procedure Set_To_First(P: out Permutation; Is_Last: out Boolean);
procedure Go_To_Next(P: in out Permutation; Is_Last: out Boolean);
end Generic_Perm; |
http://rosettacode.org/wiki/Perfect_shuffle | Perfect shuffle | A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:
7♠ 8♠ 9♠ J♠ Q♠ K♠→7♠ 8♠ 9♠
J♠ Q♠ K♠→7♠ J♠ 8♠ Q♠ 9♠ K♠
When you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:
original:
1
2
3
4
5
6
7
8
after 1st shuffle:
1
5
2
6
3
7
4
8
after 2nd shuffle:
1
3
5
7
2
4
6
8
after 3rd shuffle:
1
2
3
4
5
6
7
8
The Task
Write a function that can perform a perfect shuffle on an even-sized list of values.
Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under "Test Cases" below.
You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all "cards" are unique within each deck.
Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.
Test Cases
input (deck size)
output (number of shuffles required)
8
3
24
11
52
8
100
30
1020
1018
1024
10
10000
300
| #C.23 | C# | using System;
using System.Collections.Generic;
using System.Linq;
public static class PerfectShuffle
{
static void Main()
{
foreach (int input in new [] {8, 24, 52, 100, 1020, 1024, 10000}) {
int[] numbers = Enumerable.Range(1, input).ToArray();
Console.WriteLine($"{input} cards: {ShuffleThrough(numbers).Count()}");
}
IEnumerable<T[]> ShuffleThrough<T>(T[] original) {
T[] copy = (T[])original.Clone();
do {
yield return copy = Shuffle(copy);
} while (!Enumerable.SequenceEqual(original, copy));
}
}
public static T[] Shuffle<T>(T[] array) {
if (array.Length % 2 != 0) throw new ArgumentException("Length must be even.");
int half = array.Length / 2;
T[] result = new T[array.Length];
for (int t = 0, l = 0, r = half; l < half; t+=2, l++, r++) {
result[t] = array[l];
result[t+1] = array[r];
}
return result;
}
} |
http://rosettacode.org/wiki/Perlin_noise | Perlin noise | The Perlin noise is a kind of gradient noise invented by Ken Perlin around the end of the twentieth century and still currently heavily used in computer graphics, most notably to procedurally generate textures or heightmaps.
The Perlin noise is basically a pseudo-random mapping of
R
d
{\displaystyle \mathbb {R} ^{d}}
into
R
{\displaystyle \mathbb {R} }
with an integer
d
{\displaystyle d}
which can be arbitrarily large but which is usually 2, 3, or 4.
Either by using a dedicated library or by implementing the algorithm, show that the Perlin noise (as defined in 2002 in the Java implementation below) of the point in 3D-space with coordinates 3.14, 42, 7 is 0.13691995878400012.
Note: this result assumes 64 bit IEEE-754 floating point calculations. If your language uses a different floating point representation, make a note of it and calculate the value accurate to 15 decimal places, or your languages accuracy threshold if it is less. Trailing zeros need not be displayed.
| #GLSL | GLSL |
float rand(vec2 c){
return fract(sin(dot(c.xy ,vec2(12.9898,78.233))) * 43758.5453);
}
float noise(vec2 p, float freq ){
float unit = screenWidth/freq;
vec2 ij = floor(p/unit);
vec2 xy = mod(p,unit)/unit;
//xy = 3.*xy*xy-2.*xy*xy*xy;
xy = .5*(1.-cos(PI*xy));
float a = rand((ij+vec2(0.,0.)));
float b = rand((ij+vec2(1.,0.)));
float c = rand((ij+vec2(0.,1.)));
float d = rand((ij+vec2(1.,1.)));
float x1 = mix(a, b, xy.x);
float x2 = mix(c, d, xy.x);
return mix(x1, x2, xy.y);
}
float pNoise(vec2 p, int res){
float persistance = .5;
float n = 0.;
float normK = 0.;
float f = 4.;
float amp = 1.;
int iCount = 0;
for (int i = 0; i<50; i++){
n+=amp*noise(p, f);
f*=2.;
normK+=amp;
amp*=persistance;
if (iCount == res) break;
iCount++;
}
float nf = n/normK;
return nf*nf*nf*nf;
}
|
http://rosettacode.org/wiki/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #Draco | Draco | proc nonrec gcd(word a, b) word:
word c;
while b ~= 0 do
c := a;
a := b;
b := c % b
od;
a
corp
proc nonrec totient(word n) word:
word r, i;
r := 0;
for i from 1 upto n-1 do
if gcd(n,i) = 1 then r := r+1 fi
od;
r
corp
proc nonrec perfect(word n) bool:
word sum, x;
sum := 0;
x := n;
while
x := totient(x);
sum := sum + x;
x ~= 1
do od;
sum = n
corp
proc nonrec main() void:
word seen, n;
seen := 0;
n := 3;
while seen < 20 do
if perfect(n) then
write(n, " ");
seen := seen + 1
fi;
n := n + 2
od
corp |
http://rosettacode.org/wiki/Perfect_totient_numbers | Perfect totient numbers | Generate and show here, the first twenty Perfect totient numbers.
Related task
Totient function
Also see
the OEIS entry for perfect totient numbers.
mrob list of the first 54
| #Factor | Factor | USING: formatting kernel lists lists.lazy math
math.primes.factors ;
: perfect? ( n -- ? )
[ 0 ] dip dup [ dup 2 < ] [ totient tuck [ + ] 2dip ] until
drop = ;
20 1 lfrom [ perfect? ] lfilter ltake list>array
"%[%d, %]\n" printf |
http://rosettacode.org/wiki/Playing_cards | Playing cards | Task
Create a data structure and the associated methods to define and manipulate a deck of playing cards.
The deck should contain 52 unique cards.
The methods must include the ability to:
make a new deck
shuffle (randomize) the deck
deal from the deck
print the current contents of a deck
Each card must have a pip value and a suit value which constitute the unique value of the card.
Related tasks:
Card shuffles
Deal cards_for_FreeCell
War Card_Game
Poker hand_analyser
Go Fish
| #Factor | Factor | USING: formatting grouping io kernel math qw random sequences
vectors ;
IN: rosetta-code.playing-cards
CONSTANT: pips qw{ A 2 3 4 5 6 7 8 9 10 J Q K }
CONSTANT: suits qw{ ♥ ♣ ♦ ♠ }
: <deck> ( -- vec ) 52 <iota> >vector ;
: card>str ( n -- str )
13 /mod [ suits nth ] [ pips nth ] bi* prepend ;
: print-deck ( seq -- )
13 group [ [ card>str "%-4s" printf ] each nl ] each ;
<deck> ! make new deck
randomize ! shuffle the deck
dup pop drop ! deal from the deck (and discard)
print-deck ! print the deck |
http://rosettacode.org/wiki/Pi | Pi |
Create a program to continually calculate and output the next decimal digit of
π
{\displaystyle \pi }
(pi).
The program should continue forever (until it is aborted by the user) calculating and outputting each decimal digit in succession.
The output should be a decimal sequence beginning 3.14159265 ...
Note: this task is about calculating pi. For information on built-in pi constants see Real constants and functions.
Related Task Arithmetic-geometric mean/Calculate Pi
| #Elixir | Elixir | defmodule Pi do
def calc, do: calc(1,0,1,1,3,3,0)
defp calc(q,r,t,k,n,l,c) when c==50 do
IO.write "\n"
calc(q,r,t,k,n,l,0)
end
defp calc(q,r,t,k,n,l,c) when (4*q + r - t) < n*t do
IO.write n
calc(q*10, 10*(r-n*t), t, k, div(10*(3*q+r), t) - 10*n, l, c+1)
end
defp calc(q,r,t,k,_n,l,c) do
calc(q*k, (2*q+r)*l, t*l, k+1, div(q*7*k+2+r*l, t*l), l+2, c)
end
end
Pi.calc |
http://rosettacode.org/wiki/Pig_the_dice_game | Pig the dice game | The game of Pig is a multiplayer game played with a single six-sided die. The
object of the game is to reach 100 points or more.
Play is taken in turns. On each person's turn that person has the option of either:
Rolling the dice: where a roll of two to six is added to their score for that turn and the player's turn continues as the player is given the same choice again; or a roll of 1 loses the player's total points for that turn and their turn finishes with play passing to the next player.
Holding: the player's score for that round is added to their total and becomes safe from the effects of throwing a 1 (one). The player's turn finishes with play passing to the next player.
Task
Create a program to score for, and simulate dice throws for, a two-person game.
Related task
Pig the dice game/Player
| #M2000_Interpreter | M2000 Interpreter |
Module GamePig {
Print "Game of Pig"
dice=-1
res$=""
Player1points=0
Player1sum=0
Player2points=0
Player2sum=0
HaveWin=False
score()
\\ for simulation, feed the keyboard buffer with R and H
simulate$=String$("R", 500)
For i=1 to 100
Insert Random(1,Len(simulate$)) simulate$="H"
Next i
Keyboard simulate$
\\ end simulation
while res$<>"Q" {
Print "Player 1 turn"
PlayerTurn(&Player1points, &player1sum)
if res$="Q" then exit
Player1sum+=Player1points
Score()
Print "Player 2 turn"
PlayerTurn(&Player2points,&player2sum)
if res$="Q" then exit
Player2sum+=Player2points
Score()
}
If HaveWin then {
Score()
If Player1Sum>Player2sum then {
Print "Player 1 Win"
} Else Print "Player 2 Win"
}
Sub Rolling()
dice=random(1,6)
Print "dice=";dice
End Sub
Sub PlayOrQuit()
Print "R -Roling Q -Quit"
Repeat {
res$=Ucase$(Key$)
} Until Instr("RQ", res$)>0
End Sub
Sub PlayAgain()
Print "R -Roling H -Hold Q -Quit"
Repeat {
res$=Ucase$(Key$)
} Until Instr("RHQ", res$)>0
End Sub
Sub PlayerTurn(&playerpoints, &sum)
PlayOrQuit()
If res$="Q" then Exit Sub
playerpoints=0
Rolling()
While dice<>1 and res$="R" {
playerpoints+=dice
if dice>1 and playerpoints+sum>100 then {
sum+=playerpoints
HaveWin=True
res$="Q"
} Else {
PlayAgain()
if res$="R" then Rolling()
}
}
if dice=1 then playerpoints=0
End Sub
Sub Score()
Print "Player1 points="; Player1sum
Print "Player2 points="; Player2sum
End Sub
}
GamePig
|
http://rosettacode.org/wiki/Pernicious_numbers | Pernicious numbers | A pernicious number is a positive integer whose population count is a prime.
The population count is the number of ones in the binary representation of a non-negative integer.
Example
22 (which is 10110 in binary) has a population count of 3, which is prime, and therefore
22 is a pernicious number.
Task
display the first 25 pernicious numbers (in decimal).
display all pernicious numbers between 888,888,877 and 888,888,888 (inclusive).
display each list of integers on one line (which may or may not include a title).
See also
Sequence A052294 pernicious numbers on The On-Line Encyclopedia of Integer Sequences.
Rosetta Code entry population count, evil numbers, odious numbers.
| #Factor | Factor | USING: lists lists.lazy math.bitwise math.primes math.ranges
prettyprint sequences ;
: pernicious? ( n -- ? ) bit-count prime? ;
25 0 lfrom [ pernicious? ] lfilter ltake list>array . ! print first 25 pernicious numbers
888,888,877 888,888,888 [a,b] [ pernicious? ] filter . ! print pernicious numbers in range |
http://rosettacode.org/wiki/Pierpont_primes | Pierpont primes | A Pierpont prime is a prime number of the form: 2u3v + 1 for some non-negative integers u and v .
A Pierpont prime of the second kind is a prime number of the form: 2u3v - 1 for some non-negative integers u and v .
The term "Pierpont primes" is generally understood to mean the first definition, but will be called "Pierpont primes of the first kind" on this page to distinguish them.
Task
Write a routine (function, procedure, whatever) to find Pierpont primes of the first & second kinds.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the first kind.
Use the routine to find and display here, on this page, the first 50 Pierpont primes of the second kind
If your language supports large integers, find and display here, on this page, the 250th Pierpont prime of the first kind and the 250th Pierpont prime of the second kind.
See also
Wikipedia - Pierpont primes
OEIS:A005109 - Class 1 -, or Pierpont primes
OEIS:A005105 - Class 1 +, or Pierpont primes of the second kind
| #REXX | REXX | /*REXX program finds and displays Pierpont primes of the first and second kinds. */
parse arg n . /*obtain optional argument from the CL.*/
if n=='' | n=="," then n= 50 /*Not specified? Then use the default.*/
numeric digits n /*ensure enough decimal digs (bit int).*/
big= copies(9, digits() ) /*BIG: used as a max number (a limit).*/
@.= '2nd'; @.1= '1st'
do t=1 to -1 by -2; usum= 0; vsum= 0; s= 0 /*T is 1, then -1.*/
#= 0 /*number of Pierpont primes (so far). */
$=; do j=0 until #>=n /*$: the list " " " " */
if usum<=s then usum= get(2, 3); if vsum<=s then vsum= get(3, 2)
s= min(vsum, usum); if \isPrime(s) then iterate /*get min; Not prime? */
#= # + 1; $= $ s /*bump counter; append.*/
end /*j*/
say
w= length(word($, #) ) /*biggest prime length.*/
say center(n " Pierpont primes of the " @.t ' kind', max(10 *(w+1), 80), "═")
do p=1 by 10 to #; _=; do k=p for 10; _= _ right( word($, k), w)
end /*k*/
if _\=='' then say substr( strip(_, "T"), 2)
end /*p*/
end /*t*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPrime: procedure; parse arg x '' -1 _; if x<17 then return wordpos(x,"2 3 5 7 11 13")>0
if _==5 then return 0; if x//2==0 then return 0 /*not prime. */
if x//3==0 then return 0; if x//7==0 then return 0 /* " " */
do j=11 by 6 until j*j>x /*skip ÷ 3's.*/
if x//j==0 then return 0; if x//(j+2)==0 then return 0 /*not prime. */
end /*j*/; return 1 /*it's prime.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
get: parse arg c1,c2; m=big; do ju=0; pu= c1**ju; if pu+t>s then return min(m, pu+t)
do jv=0; pv= c2**jv; if pv >s then iterate ju
_= pu*pv + t; if _ >s then m= min(_, m)
end /*jv*/
end /*ju*/ /*see the RETURN (above). */ |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #Hare | Hare | use fmt;
use math::random;
use datetime;
export fn main() void = {
const array = ["one", "two", "three", "four", "five"];
const seed = datetime::now();
const seed = datetime::nsec(&seed);
let r = math::random::init(seed: u32);
fmt::printfln("{}", array[math::random::u32n(&r, len(array): u32)])!;
}; |
http://rosettacode.org/wiki/Pick_random_element | Pick random element | Demonstrate how to pick a random element from a list.
| #Haskell | Haskell | import System.Random (randomRIO)
pick :: [a] -> IO a
pick xs = fmap (xs !!) $ randomRIO (0, length xs - 1)
x <- pick [1, 2, 3] |
http://rosettacode.org/wiki/Primality_by_trial_division | Primality by trial division | Task
Write a boolean function that tells whether a given integer is prime.
Remember that 1 and all non-positive numbers are not prime.
Use trial division.
Even numbers greater than 2 may be eliminated right away.
A loop from 3 to √ n will suffice, but other loops are allowed.
Related tasks
count in factors
prime decomposition
AKS test for primes
factors of an integer
Sieve of Eratosthenes
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
| #Yabasic | Yabasic | for i = 1 to 99
if isPrime(i) print str$(i), " ";
next i
print
end
sub isPrime(v)
if v < 2 return False
if mod(v, 2) = 0 return v = 2
if mod(v, 3) = 0 return v = 3
d = 5
while d * d <= v
if mod(v, d) = 0 then return False else d = d + 2 : fi
wend
return True
end sub |
http://rosettacode.org/wiki/Phrase_reversals | Phrase reversals | Task
Given a string of space separated words containing the following phrase:
rosetta code phrase reversal
Reverse the characters of the string.
Reverse the characters of each individual word in the string, maintaining original word order within the string.
Reverse the order of each word of the string, maintaining the order of characters in each word.
Show your output here.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #IS-BASIC | IS-BASIC | 100 PROGRAM "ReverseS.bas"
110 LET S$="Rosetta Code Pharse Reversal"
120 PRINT S$
130 PRINT REVERSE$(S$)
140 PRINT REVERSEW$(S$)
150 PRINT REVERSEC$(S$)
160 DEF REVERSE$(S$)
170 LET T$=""
180 FOR I=LEN(S$) TO 1 STEP-1
190 LET T$=T$&S$(I)
200 NEXT
210 LET REVERSE$=T$
220 END DEF
230 DEF REVERSEW$(S$)
240 LET T$="":LET PE=LEN(S$)
250 FOR PS=PE TO 1 STEP-1
260 IF PS=1 OR S$(PS)=" " THEN LET T$=T$&" ":LET T$=T$<RIM$(S$(PS:PE)):LET PE=PS-1
270 NEXT
280 LET REVERSEW$=LTRIM$(T$)
290 END DEF
300 DEF REVERSEC$(S$)
310 LET T$="":LET PS=1
320 FOR PE=1 TO LEN(S$)
330 IF PE=LEN(S$) OR S$(PE)=" " THEN LET T$=T$&" ":LET T$=T$&REVERSE$(RTRIM$(S$(PS:PE))):LET PS=PE+1
340 NEXT
350 LET REVERSEC$=LTRIM$(T$)
360 END DEF |
http://rosettacode.org/wiki/Phrase_reversals | Phrase reversals | Task
Given a string of space separated words containing the following phrase:
rosetta code phrase reversal
Reverse the characters of the string.
Reverse the characters of each individual word in the string, maintaining original word order within the string.
Reverse the order of each word of the string, maintaining the order of characters in each word.
Show your output here.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #J | J | getWords=: (' '&splitstring) :. (' '&joinstring)
reverseString=: |.
reverseWords=: |.&.>&.getWords
reverseWordOrder=: |.&.getWords |
http://rosettacode.org/wiki/Permutations/Derangements | Permutations/Derangements | A derangement is a permutation of the order of distinct items in which no item appears in its original place.
For example, the only two derangements of the three items (0, 1, 2) are (1, 2, 0), and (2, 0, 1).
The number of derangements of n distinct items is known as the subfactorial of n, sometimes written as !n.
There are various ways to calculate !n.
Task
Create a named function/method/subroutine/... to generate derangements of the integers 0..n-1, (or 1..n if you prefer).
Generate and show all the derangements of 4 integers using the above routine.
Create a function that calculates the subfactorial of n, !n.
Print and show a table of the counted number of derangements of n vs. the calculated !n for n from 0..9 inclusive.
Optional stretch goal
Calculate !20
Related tasks
Anagrams/Deranged anagrams
Best shuffle
Left_factorials
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #FreeBASIC | FreeBASIC | ' version 08-04-2017
' compile with: fbc -s console
Sub Subfactorial(a() As ULongInt)
Dim As ULong i
Dim As ULongInt num
For i = 0 To UBound(a)
num = num * i
If (i And 1) = 1 Then
num -= 1
Else
num += 1
End If
a(i) = num
Next
End Sub
' Heap's algorithm non-recursive
Function perms_derange(n As ULong, flag As Long = 0) As ULongInt
' fast upto n < 12
If n = 0 Then Return 1
Dim As ULong i, j, c1, count
Dim As ULong a(0 To n -1), c(0 To n -1)
For j = 0 To n -1
a(j) = j
Next
While i < n
If c(i) < i Then
If (i And 1) = 0 Then
Swap a(0), a(i)
Else
Swap a(c(i)), a(i)
End If
For j = 0 To n -1
If a(j) = j Then j = 99
Next
If j < 99 Then
count += 1
If flag = 0 Then
c1 += 1
For j = 0 To n -1
Print a(j);
Next
If c1 > 12 Then
Print : c1 = 0
Else
Print " ";
End If
End If
End If
c(i) += 1
i = 0
Else
c(i) = 0
i += 1
End If
Wend
If flag = 0 AndAlso c1 <> 0 Then Print
Return count
End Function
' ------=< MAIN >=------
Dim As ULong i, n = 4
Dim As ULongInt subfac(20)
Subfactorial(subfac())
Print "permutations derangements for n = "; n
i = perms_derange(n)
Print "count returned = "; i; " , !"; n; " calculated = "; subfac(n)
Print
Print "count counted subfactorial"
Print "---------------------------"
For i = 0 To 9
Print Using " ###: ######## ########"; i; perms_derange(i, 1); subfac(i)
Next
For i = 10 To 20
Print Using " ###: ###################"; i; subfac(i)
Next
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End |
http://rosettacode.org/wiki/Permutations_by_swapping | Permutations by swapping | Task
Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.
Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.
Show the permutations and signs of three items, in order of generation here.
Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.
Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.
References
Steinhaus–Johnson–Trotter algorithm
Johnson-Trotter Algorithm Listing All Permutations
Heap's algorithm
[1] Tintinnalogia
Related tasks
Matrix arithmetic
Gray code
| #Icon_and_Unicon | Icon and Unicon | procedure main(A)
every write("Permutations of length ",n := !A) do
every p := permute(n) do write("\t",showList(p[1])," -> ",right(p[2],2))
end
procedure permute(n)
items := [[]]
every (j := 1 to n, new_items := []) do {
every item := items[i := 1 to *items] do {
if *item = 0 then put(new_items, [j])
else if i%2 = 0 then
every k := 1 to *item+1 do {
new_item := item[1:k] ||| [j] ||| item[k:0]
put(new_items, new_item)
}
else
every k := *item+1 to 1 by -1 do {
new_item := item[1:k] ||| [j] ||| item[k:0]
put(new_items, new_item)
}
}
items := new_items
}
suspend (i := 0, [!items, if (i+:=1)%2 = 0 then 1 else -1])
end
procedure showList(A)
every (s := "[") ||:= image(!A)||", "
return s[1:-2]||"]"
end |
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #Nim | Nim | import strformat
const data = [85, 88, 75, 66, 25,
29, 83, 39, 97, 68,
41, 10, 49, 16, 65,
32, 92, 28, 98]
func pick(at, remain, accu, treat: int): int =
if remain == 0:
return if accu > treat: 1 else: 0
return pick(at - 1, remain - 1, accu + data[at - 1], treat) +
(if at > remain: pick(at - 1, remain, accu, treat) else: 0)
var treat = 0
var le, gt = 0
var total = 1.0
for i in countup(0, 8):
treat += data[i]
for i in countdown(19, 11):
total *= float(i)
for i in countdown(9, 1):
total /= float(i)
gt = pick(19, 9, 0, treat)
le = int(total - float(gt))
echo fmt"<= : {100.0 * float(le) / total:.6f}% {le}"
echo fmt" > : {100.0 * float(gt) / total:.6f}% {gt}" |
http://rosettacode.org/wiki/Permutation_test | Permutation test | Permutation test
You are encouraged to solve this task according to the task description, using any language you may know.
A new medical treatment was tested on a population of
n
+
m
{\displaystyle n+m}
volunteers, with each volunteer randomly assigned either to a group of
n
{\displaystyle n}
treatment subjects, or to a group of
m
{\displaystyle m}
control subjects.
Members of the treatment group were given the treatment,
and members of the control group were given a placebo.
The effect of the treatment or placebo on each volunteer
was measured and reported in this table.
Table of experimental results
Treatment group
Control group
85
68
88
41
75
10
66
49
25
16
29
65
83
32
39
92
97
28
98
Write a program that performs a
permutation test to judge
whether the treatment had a significantly stronger effect than the
placebo.
Do this by considering every possible alternative assignment from the same pool of volunteers to a treatment group of size
n
{\displaystyle n}
and a control group of size
m
{\displaystyle m}
(i.e., the same group sizes used in the actual experiment but with the group members chosen differently), while assuming that each volunteer's effect remains constant regardless.
Note that the number of alternatives will be the binomial coefficient
(
n
+
m
n
)
{\displaystyle {\tbinom {n+m}{n}}}
.
Compute the mean effect for each group and the difference in means between the groups in every case by subtracting the mean of the control group from the mean of the treatment group.
Report the percentage of alternative groupings for which the difference in means is less or equal to the actual experimentally observed difference in means, and the percentage for which it is greater.
Note that they should sum to 100%.
Extremely dissimilar values are evidence of an effect not entirely due
to chance, but your program need not draw any conclusions.
You may assume the experimental data are known at compile time if
that's easier than loading them at run time. Test your solution on the
data given above.
| #Perl | Perl | #!/usr/bin/perl
use warnings;
use strict;
use List::Util qw{ sum };
sub means {
my @groups = @_;
return map sum(@$_) / @$_, @groups;
}
sub following {
my $pattern = shift;
my $orig_count = grep $_, @$pattern;
my $count;
do {
my $i = $#{$pattern};
until (0 > $i) {
$pattern->[$i] = $pattern->[$i] ? 0 : 1;
last if $pattern->[$i];
--$i;
}
$count = grep $_, @$pattern;
} until $count == $orig_count or not $count;
undef @$pattern unless $count;
}
my @groups;
my $i = 0;
while (<DATA>) {
chomp;
$i++, next if /^$/;
push @{ $groups[$i] }, $_;
}
my @orig_means = means(@groups);
my $orig_cmp = $orig_means[0] - $orig_means[1];
my $pattern = [ (0) x @{ $groups[0] },
(1) x @{ $groups[1] }
];
my @cmp = (0) x 3;
while (@$pattern) {
my @perms = map { my $g = $_;
[ (@{ $groups[0] }, @{ $groups[1] } ) [ grep $pattern->[$_] == $g, 0 .. $#{$pattern} ] ];
} 0, 1;
my @means = means(@perms);
$cmp[ ($means[0] - $means[1]) <=> $orig_cmp ]++;
} continue {
following($pattern);
}
my $all = sum(@cmp);
my $length = length $all;
for (0, -1, 1) {
printf "%-7s %${length}d %6.3f%%\n",
(qw(equal greater less))[$_], $cmp[$_], 100 * $cmp[$_] / $all;
}
__DATA__
85
88
75
66
25
29
83
39
97
68
41
10
49
16
65
32
92
28
98 |
Subsets and Splits
Select Specific Languages Codes
Retrieves specific programming language names and codes from training data, providing basic filtering but limited analytical value beyond identifying these particular languages.