Permutations with repetitions: Difference between revisions
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int temp; |
int temp; |
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int numbers=3; |
int numbers=3; |
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int a[numbers], upto = 4, temp2; |
int a[numbers+1], upto = 4, temp2; |
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for( temp2 = 1 ; temp2 <= numbers; temp2++){ |
for( temp2 = 1 ; temp2 <= numbers; temp2++){ |
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a[temp2]=1; |
a[temp2]=1; |
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} |
} |
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a[numbers]=0; |
a[numbers]=0; |
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temp=numbers |
temp=numbers; |
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while(1){ |
while(1){ |
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if(a[temp]==upto){ |
if(a[temp]==upto){ |
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Revision as of 05:51, 7 May 2019
- Task
Generate a sequence of permutations of n elements drawn from choice of k values.
This sequence will have elements, unless the program decides to terminate early.
Do not store all the intermediate values of the sequence, rather generate them as required, and pass the intermediate result to a deciding routine for combinations selection and/or early generator termination.
For example: When "cracking" a "combination" lock a sequence is required, but the sequence is terminated once a successful "combination" is found. This case is a good example of where it is not required to store all the intermediate permutations.
See Also:
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant Order Important Without replacement Task: Combinations Task: Permutations With replacement Task: Combinations with repetitions Task: Permutations with repetitions
AppleScript
Strict evaluation of the whole set
Permutations with repetitions, using strict evaluation, generating the entire set (where system constraints permit) with some degree of efficiency. For lazy or interruptible evaluation, see the second example below.
<lang AppleScript>-- e.g. replicateM(3, {1, 2})) -> -- {{1, 1, 1}, {1, 1, 2}, {1, 2, 1}, {1, 2, 2}, {2, 1, 1}, -- {2, 1, 2}, {2, 2, 1}, {2, 2, 2}}
-- replicateM :: Int -> [a] -> a on replicateM(n, xs)
script go
script cons
on |λ|(a, bs)
{a} & bs
end |λ|
end script
on |λ|(x)
if x ≤ 0 then
{{}}
else
liftA2List(cons, xs, |λ|(x - 1))
end if
end |λ|
end script
go's |λ|(n)
end replicateM
-- TEST ------------------------------------------------------------
on run
replicateM(2, {1, 2, 3})
-- {{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}
end run
-- GENERIC FUNCTIONS -----------------------------------------------
-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)
set lng to length of xs
set acc to {}
tell mReturn(f)
repeat with i from 1 to lng
set acc to acc & |λ|(item i of xs, i, xs)
end repeat
end tell
return acc
end concatMap
-- liftA2List :: (a -> b -> c) -> [a] -> [b] -> [c] on liftA2List(f, xs, ys)
script
property g : mReturn(f)'s |λ|
on |λ|(x)
script
on |λ|(y)
{g(x, y)}
end |λ|
end script
concatMap(result, ys)
end |λ|
end script
concatMap(result, xs)
end liftA2List
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn</lang>
- Output:
<lang AppleScript>{{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}</lang>
Lazy evaluation with a generator
Permutations with repetition by treating the elements as an ordered set, and writing a function from a zero-based index to the nth permutation. This allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set: <lang AppleScript>use AppleScript version "2.4" use framework "Foundation" use scripting additions
-- permutesWithRepns :: [a] -> Int -> Generator a on permutesWithRepns(xs, n)
script
property f : curry3(my nthPermutationWithRepn)'s |λ|(xs)'s |λ|(n)
property limit : (length of xs) ^ n
property i : -1
on |λ|()
set i to 1 + i
if i < limit then
return f's |λ|(i)
else
missing value
end if
end |λ|
end script
end permutesWithRepns
-- nthPermutationWithRepn :: [a] -> Int -> Int -> [a]
on nthPermutationWithRepn(xs, intGroup, intIndex)
set intBase to length of xs
if intIndex < (intBase ^ intGroup) then
set ds to baseDigits(intBase, xs, intIndex)
-- With any 'leading zeros' required by length
replicate(intGroup - (length of ds), item 1 of xs) & ds
else
missing value
end if
end nthPermutationWithRepn
-- baseDigits :: Int -> [a] -> [a]
on baseDigits(intBase, digits, n)
script
on |λ|(v)
if 0 = v then
Nothing()
else
Just(Tuple(item (1 + (v mod intBase)) of digits, ¬
v div intBase))
end if
end |λ|
end script
unfoldr(result, n)
end baseDigits
-- TEST ------------------------------------------------------------------
on run
set cs to "ACKR"
set wordLength to 5
set gen to permutesWithRepns(cs, wordLength)
set i to 0
set v to gen's |λ|() -- First permutation drawn from series
set alpha to v
set psi to alpha
repeat while missing value is not v
set s to concat(v)
if "crack" = toLower(s) then
return ("Permutation " & (i as text) & " of " & ¬
(((length of cs) ^ wordLength) as integer) as text) & ¬
": " & s & linefeed & ¬
"Found after searching from " & alpha & " thru " & psi
else
set i to 1 + i
set psi to v
end if
set v to gen's |λ|()
end repeat
end run
-- GENERIC ----------------------------------------------------------
-- Just :: a -> Maybe a on Just(x)
{type:"Maybe", Nothing:false, Just:x}
end Just
-- Nothing :: Maybe a on Nothing()
{type:"Maybe", Nothing:true}
end Nothing
-- Tuple (,) :: a -> b -> (a, b) on Tuple(a, b)
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple
-- concat :: a -> [a] -- concat :: [String] -> String on concat(xs)
set lng to length of xs
if 0 < lng and string is class of (item 1 of xs) then
set acc to ""
else
set acc to {}
end if
repeat with i from 1 to lng
set acc to acc & item i of xs
end repeat
acc
end concat
-- curry3 :: ((a, b, c) -> d) -> a -> b -> c -> d on curry3(f)
script
on |λ|(a)
script
on |λ|(b)
script
on |λ|(c)
|λ|(a, b, c) of mReturn(f)
end |λ|
end script
end |λ|
end script
end |λ|
end script
end curry3
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- Egyptian multiplication - progressively doubling a list, appending -- stages of doubling to an accumulator where needed for binary -- assembly of a target length -- replicate :: Int -> a -> [a] on replicate(n, a)
set out to {}
if n < 1 then return out
set dbl to {a}
repeat while (n > 1)
if (n mod 2) > 0 then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate
-- toLower :: String -> String on toLower(str)
set ca to current application
((ca's NSString's stringWithString:(str))'s ¬
lowercaseStringWithLocale:(ca's NSLocale's currentLocale())) as text
end toLower
-- > unfoldr (\b -> if b == 0 then Nothing else Just (b, b-1)) 10 -- > [10,9,8,7,6,5,4,3,2,1] -- unfoldr :: (b -> Maybe (a, b)) -> b -> [a] on unfoldr(f, v)
set xr to Tuple(v, v) -- (value, remainder)
set xs to {}
tell mReturn(f)
repeat -- Function applied to remainder.
set mb to |λ|(|2| of xr)
if Nothing of mb then
exit repeat
else -- New (value, remainder) tuple,
set xr to Just of mb
-- and value appended to output list.
set end of xs to |1| of xr
end if
end repeat
end tell
return xs
end unfoldr</lang>
- Output:
Permutation 589 of 1024: CRACK Found after searching from AAAAA thru ARACK
ALGOL 68
File: prelude_permutations_with_repetitions.a68<lang algol68># -*- coding: utf-8 -*- #
MODE PERMELEMLIST = FLEX[0]PERMELEM; MODE PERMELEMLISTYIELD = PROC(PERMELEMLIST)VOID;
PROC perm gen elemlist = (FLEX[]PERMELEMLIST master, PERMELEMLISTYIELD yield)VOID:(
[LWB master:UPB master]INT counter;
[LWB master:UPB master]PERMELEM out;
FOR i FROM LWB counter TO UPB counter DO
INT c = counter[i] := LWB master[i];
out[i] := master[i][c]
OD;
yield(out);
WHILE TRUE DO
INT next i := LWB counter;
counter[next i] +:= 1;
FOR i FROM LWB counter TO UPB counter WHILE counter[i]>UPB master[i] DO
INT c = counter[i] := LWB master[i];
out[i] := master[i][c];
next i := i + 1;
IF next i > UPB counter THEN done FI;
counter[next i] +:= 1
OD;
INT c = counter[next i];
out[next i] := master[next i][c];
yield(out)
OD;
done: SKIP
);
SKIP</lang>File: test_permutations_with_repetitions.a68<lang algol68>#!/usr/bin/a68g --script #
- -*- coding: utf-8 -*- #
MODE PERMELEM = STRING; PR READ "prelude_permutations_with_repetitions.a68" PR;
INT lead actor = 1, co star = 2; PERMELEMLIST actors list = ("Chris Ciaffa", "Keith Urban","Tom Cruise",
"Katie Holmes","Mimi Rogers","Nicole Kidman");
FLEX[0]PERMELEMLIST combination := (actors list, actors list, actors list, actors list);
FORMAT partner fmt = $g"; "$; test:(
- FOR PERMELEMELEM candidate in # perm gen elemlist(combination #) DO (#,
- (PERMELEMLIST candidate)VOID: (
printf((partner fmt,candidate));
IF candidate[lead actor] = "Keith Urban" AND candidate[co star]="Nicole Kidman" OR
candidate[co star] = "Keith Urban" AND candidate[lead actor]="Nicole Kidman" THEN
print((" => Sunday + Faith as extras", new line)); # children #
done
FI;
print(new line)
- OD #));
done: SKIP
)</lang>Output:
Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Keith Urban; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Tom Cruise; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Katie Holmes; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Mimi Rogers; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Nicole Kidman; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Chris Ciaffa; Keith Urban; Chris Ciaffa; Chris Ciaffa; Keith Urban; Keith Urban; Chris Ciaffa; Chris Ciaffa; Tom Cruise; Keith Urban; Chris Ciaffa; Chris Ciaffa; Katie Holmes; Keith Urban; Chris Ciaffa; Chris Ciaffa; Mimi Rogers; Keith Urban; Chris Ciaffa; Chris Ciaffa; Nicole Kidman; Keith Urban; Chris Ciaffa; Chris Ciaffa; => Sunday + Faith as extras
AutoHotkey
Use the function from http://rosettacode.org/wiki/Permutations#Alternate_Version with opt=1 <lang ahk>P(n,k="",opt=0,delim="",str="") { ; generate all n choose k permutations lexicographically ;1..n = range, or delimited list, or string to parse ; to process with a different min index, pass a delimited list, e.g. "0`n1`n2" ;k = length of result ;opt 0 = no repetitions ;opt 1 = with repetitions ;opt 2 = run for 1..k ;opt 3 = run for 1..k with repetitions ;str = string to prepend (used internally) ;returns delimited string, error message, or (if k > n) a blank string i:=0 If !InStr(n,"`n") If n in 2,3,4,5,6,7,8,9 Loop, %n% n := A_Index = 1 ? A_Index : n "`n" A_Index Else Loop, Parse, n, %delim% n := A_Index = 1 ? A_LoopField : n "`n" A_LoopField If (k = "") RegExReplace(n,"`n","",k), k++ If k is not Digit Return "k must be a digit." If opt not in 0,1,2,3 Return "opt invalid." If k = 0 Return str Else Loop, Parse, n, `n If (!InStr(str,A_LoopField) || opt & 1) s .= (!i++ ? (opt & 2 ? str "`n" : "") : "`n" ) . P(n,k-1,opt,delim,str . A_LoopField . delim) Return s }</lang>
C
<lang d>#include <stdio.h>
- include <stdlib.h>
int main(){ int temp; int numbers=3; int a[numbers+1], upto = 4, temp2; for( temp2 = 1 ; temp2 <= numbers; temp2++){ a[temp2]=1; } a[numbers]=0; temp=numbers; while(1){ if(a[temp]==upto){ temp--; if(temp==0) break; } else{ a[temp]++; while(temp<numbers){ temp++; a[temp]=1; }
printf("("); for( temp2 = 1 ; temp2 <= numbers; temp2++){ printf("%d", a[temp2]); } printf(")"); } } return 0; }</lang>
- Output:
(111)(112)(113)(114)(121)(122)(123)(124)(131)(132)(133)(134)(141)(142)(143)(144)(211)(212)(213)(214)(221)(222)(223)(224)(231)(232)(233)(234)(241)(242)(243)(244)(311)(312)(313)(314)(321)(322)(323)(324)(331)(332)(333)(334)(341)(342)(343)(344)(411)(412)(413)(414)(421)(422)(423)(424)(431)(432)(433)(434)(441)(442)(443)(444)
C++
<lang d>
- include <stdio.h>
- include <stdlib.h>
struct Generator {
public:
Generator(int s, int v)
: cSlots(s)
, cValues(v)
{
a = new int[s];
for (int i = 0; i < cSlots - 1; i++) {
a[i] = 1;
}
a[cSlots - 1] = 0;
nextInd = cSlots;
}
~Generator()
{
delete a;
}
bool doNext()
{
for (;;)
{
if (a[nextInd - 1] == cValues) {
nextInd--;
if (nextInd == 0)
return false;
}
else {
a[nextInd - 1]++;
while (nextInd < cSlots) {
nextInd++;
a[nextInd - 1] = 1;
}
return true;
}
}
}
void doPrint()
{
printf("(");
for (int i = 0; i < cSlots; i++) {
printf("%d", a[i]);
}
printf(")");
}
private:
int *a;
int cSlots;
int cValues;
int nextInd;
};
int main()
{
Generator g(3, 4);
while (g.doNext()) {
g.doPrint();
}
return 0;
}
</lang>
- Output:
(111)(112)(113)(114)(121)(122)(123)(124)(131)(132)(133)(134)(141)(142)(143)(144)(211)(212)(213)(214)(221)(222)(223)(224)(231)(232)(233)(234)(241)(242)(243)(244)(311)(312)(313)(314)(321)(322)(323)(324)(331)(332)(333)(334)(341)(342)(343)(344)(411)(412)(413)(414)(421)(422)(423)(424)(431)(432)(433)(434)(441)(442)(443)(444)
D
opApply Version
<lang d>import std.array;
struct PermutationsWithRepetitions(T) {
const T[] data; const int n;
int opApply(int delegate(ref T[]) dg) {
int result;
T[] aux;
if (n == 1) {
foreach (el; data) {
aux = [el];
result = dg(aux);
if (result) goto END;
}
} else {
foreach (el; data) {
foreach (p; PermutationsWithRepetitions(data, n - 1)) {
aux = el ~ p;
result = dg(aux);
if (result) goto END;
}
}
}
END:
return result;
}
}
auto permutationsWithRepetitions(T)(T[] data, in int n) pure nothrow in {
assert(!data.empty && n > 0);
} body {
return PermutationsWithRepetitions!T(data, n);
}
void main() {
import std.stdio, std.array; [1, 2, 3].permutationsWithRepetitions(2).array.writeln;
}</lang>
- Output:
[[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]]
Generator Range Version
<lang d>import std.stdio, std.array, std.concurrency;
Generator!(T[]) permutationsWithRepetitions(T)(T[] data, in uint n) in {
assert(!data.empty && n > 0);
} body {
return new typeof(return)({
if (n == 1) {
foreach (el; data)
yield([el]);
} else {
foreach (el; data)
foreach (perm; permutationsWithRepetitions(data, n - 1))
yield(el ~ perm);
}
});
}
void main() {
[1, 2, 3].permutationsWithRepetitions(2).writeln;
}</lang> The output is the same.
EchoLisp
<lang scheme> (lib 'sequences) ;; (indices ..) (lib 'list) ;; (list-permute ..)
- (indices range_1 ..range_k) returns a procrastinator (lazy sequence)
- which gives all combinations of indices_i in range_i.
- If all k ranges are equal to (0 ...n-1)
- (indices (make-vector k n))
- will give the n^k permutations with repetitions of the integers (0 ... n-1).
(define perms (indices (make-vector 2 3)))
(take perms #:all)
→ (#(0 0) #(0 1) #(0 2) #(1 0) #(1 1) #(1 2) #(2 0) #(2 1) #(2 2))
(length perms) → 9
- 6-permute the numbers (0 ....9)
(define perms (indices (make-vector 6 10))) (length perms) → 1000000
- passing the procrastinator to a routine
- which stops when sum = 22
(for ((p perms))
#:break (= (apply + (vector->list p)) 22) => p )
→ #( 0 0 0 4 9 9)
- to permute any objects, use (list-permute list permutation-vector/list)
(list-permute '(a b c d e) '(1 0 1 0 3 2 1))
→ (b a b a d c b)
(list-permute '(a b c d e) #(1 0 1 0 3 2 1))
→ (b a b a d c b)
</lang>
Elixir
<lang elixir>defmodule RC do
def perm_rep(list), do: perm_rep(list, length(list)) def perm_rep([], _), do: [[]] def perm_rep(_, 0), do: [[]] def perm_rep(list, i) do for x <- list, y <- perm_rep(list, i-1), do: [x|y] end
end
list = [1, 2, 3] Enum.each(1..3, fn n ->
IO.inspect RC.perm_rep(list,n)
end)</lang>
- Output:
[[1], [2], [3]] [[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]] [[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 1], [1, 2, 2], [1, 2, 3], [1, 3, 1], [1, 3, 2], [1, 3, 3], [2, 1, 1], [2, 1, 2], [2, 1, 3], [2, 2, 1], [2, 2, 2], [2, 2, 3], [2, 3, 1], [2, 3, 2], [2, 3, 3], [3, 1, 1], [3, 1, 2], [3, 1, 3], [3, 2, 1], [3, 2, 2], [3, 2, 3], [3, 3, 1], [3, 3, 2], [3, 3, 3]]
Erlang
<lang Erlang>-module(permute). -export([permute/1]).
permute(L) -> permute(L,length(L)). permute([],_) -> [[]]; permute(_,0) -> [[]]; permute(L,I) -> [[X|Y] || X<-L, Y<-permute(L,I-1)].</lang>
Go
<lang go>package main
import "fmt"
var (
n = 3
values = []string{"A", "B", "C", "D"}
k = len(values)
decide = func(p []string) bool {
return p[0] == "B" && p[1] == "C"
}
)
func main() {
pn := make([]int, n)
p := make([]string, n)
for {
// generate permutaton
for i, x := range pn {
p[i] = values[x]
}
// show progress
fmt.Println(p)
// pass to deciding function
if decide(p) {
return // terminate early
}
// increment permutation number
for i := 0; ; {
pn[i]++
if pn[i] < k {
break
}
pn[i] = 0
i++
if i == n {
return // all permutations generated
}
}
}
}</lang>
- Output:
[A A A] [B A A] [C A A] [D A A] [A B A] [B B A] [C B A] [D B A] [A C A] [B C A]
Haskell
<lang haskell>import Control.Monad (replicateM)
main = mapM_ print (replicateM 2 [1,2,3])</lang>
- Output:
[1,1] [1,2] [1,3] [2,1] [2,2] [2,3] [3,1] [3,2] [3,3]
J
Position in the sequence is an integer from i.n^k, for example:
<lang j> i.3^2 0 1 2 3 4 5 6 7 8</lang>
The sequence itself is expressed using (k#n)#: position, for example:
<lang j> (2#3)#:i.3^2 0 0 0 1 0 2 1 0 1 1 1 2 2 0 2 1 2 2</lang>
Partial sequences belong in a context where they are relevant and the sheer number of such possibilities make it inadvisable to generalize outside of those contexts. But anything that can generate integers will do. For example:
<lang j> (2#3)#:3 4 5 1 0 1 1 1 2</lang>
We might express this as a verb
<lang j>perm=: # #: i.@^~</lang>
with example use:
<lang j> 2 perm 3 0 0 0 1 0 2 1 0 ...</lang>
but the structural requirements of this task (passing intermediate results "when needed") mean that we are not looking for a word that does it all, but are instead looking for components that we can assemble in other contexts. This means that the language primitives are what's needed here.
Java
<lang java>import java.util.function.Predicate;
public class PermutationsWithRepetitions {
public static void main(String[] args) {
char[] chars = {'a', 'b', 'c', 'd'};
// looking for bba
permute(chars, 3, i -> i[0] == 1 && i[1] == 1 && i[2] == 0);
}
static void permute(char[] a, int k, Predicate<int[]> decider) {
int n = a.length;
if (k < 1 || k > n)
throw new IllegalArgumentException("Illegal number of positions.");
int[] indexes = new int[n];
int total = (int) Math.pow(n, k);
while (total-- > 0) {
for (int i = 0; i < n - (n - k); i++)
System.out.print(a[indexes[i]]);
System.out.println();
if (decider.test(indexes))
break;
for (int i = 0; i < n; i++) {
if (indexes[i] >= n - 1) {
indexes[i] = 0;
} else {
indexes[i]++;
break;
}
}
}
}
}</lang>
Output:
aaa baa caa daa aba bba
JavaScript
ES5
Permutations with repetitions, using strict evaluation, generating the entire set (where system constraints permit) with some degree of efficiency. For lazy or interruptible evaluation, see the second example below.
<lang JavaScript>(function () {
'use strict';
// permutationsWithRepetition :: Int -> [a] -> a var permutationsWithRepetition = function (n, as) { return as.length > 0 ? ( foldl1(curry(cartesianProduct)(as), replicate(n, as)) ) : []; };
// GENERIC FUNCTIONS -----------------------------------------------------
// cartesianProduct :: [a] -> [b] -> a, b var cartesianProduct = function (xs, ys) { return [].concat.apply([], xs.map(function (x) { return [].concat.apply([], ys.map(function (y) { return [ [x].concat(y) ]; })); })); };
// foldl1 :: (a -> a -> a) -> [a] -> a
var foldl1 = function (f, xs) {
return xs.length > 0 ? xs.slice(1)
.reduce(f, xs[0]) : [];
};
// replicate :: Int -> a -> [a]
var replicate = function (n, a) {
var v = [a],
o = [];
if (n < 1) return o;
while (n > 1) {
if (n & 1) o = o.concat(v);
n >>= 1;
v = v.concat(v);
}
return o.concat(v);
};
// curry :: ((a, b) -> c) -> a -> b -> c
var curry = function (f) {
return function (a) {
return function (b) {
return f(a, b);
};
};
};
// TEST -----------------------------------------------------------------
// show :: a -> String
var show = function (x) {
return JSON.stringify(x);
}; //, null, 2);
return show(permutationsWithRepetition(2, [1, 2, 3]));
//--> [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
})();</lang>
- Output:
<lang JavaScript>[[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]</lang>
Permutations with repetition by treating the elements as an ordered set, and writing a function from a zero-based index to the nth permutation. This allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set:
<lang JavaScript>(function () {
'use strict';
// nthPermutationWithRepn :: [a] -> Int -> Int -> [a]
var nthPermutationWithRepn = function (xs, groupSize, index) {
var intBase = xs.length,
intSetSize = Math.pow(intBase, groupSize),
lastIndex = intSetSize - 1; // zero-based
if (intBase < 1 || index > lastIndex) return undefined;
var baseElements = unfoldr(function (m) {
var v = m.new,
d = Math.floor(v / intBase);
return {
valid: d > 0,
value: xs[v % intBase],
new: d
};
}, index),
intZeros = groupSize - baseElements.length;
return intZeros > 0 ? replicate(intZeros, xs[0])
.concat(baseElements) : baseElements;
};
// GENERIC FUNCTIONS
// unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
var unfoldr = function (mf, v) {
var xs = [];
return [until(function (m) {
return !m.valid;
}, function (m) {
var m2 = mf(m);
return m2.valid && (xs = [m2.value].concat(xs)), m2;
}, {
valid: true,
value: v,
new: v
})
.value
].concat(xs);
};
// until :: (a -> Bool) -> (a -> a) -> a -> a
var until = function (p, f, x) {
var v = x;
while (!p(v)) {
v = f(v);
}
return v;
};
// replicate :: Int -> a -> [a]
var replicate = function (n, a) {
var v = [a],
o = [];
if (n < 1) return o;
while (n > 1) {
if (n & 1) o = o.concat(v);
n >>= 1;
v = v.concat(v);
}
return o.concat(v);
};
// show :: a -> String
var show = function (x) {
return JSON.stringify(x);
}; //, null, 2);
// curry :: Function -> Function
var curry = function (f) {
for (var lng = arguments.length,
args = Array(lng > 1 ? lng - 1 : 0),
iArg = 1; iArg < lng; iArg++) {
args[iArg - 1] = arguments[iArg];
}
var intArgs = f.length,
go = function (xs) {
return xs.length >= intArgs ? f.apply(null, xs) : function () {
return go(xs.concat([].slice.apply(arguments)));
};
};
return go([].slice.call(args, 1));
};
// range :: Int -> Int -> [Int]
var range = function (m, n) {
return Array.from({
length: Math.floor(n - m) + 1
}, function (_, i) {
return m + i;
});
};
// TEST
// Just items 30 to 35 in the (zero-indexed) series:
return show(range(30, 35)
.map(curry(nthPermutationWithRepn)(['X', 'Y', 'Z'], 4)));
})();</lang>
- Output:
["Y","X","Y","X"], ["Y","X","Y","Y"], ["Y","X","Y","Z"], ["Y","X","Z","X"], ["Y","X","Z","Y"], ["Y","X","Z","Z"]
ES6
Strict evaluation of the whole set
Permutations with repetitions, using strict evaluation, generating the entire set. For partial or interruptible evaluation, see the second example below.
A (strict) analogue of the (lazy) replicateM in Haskell.
<lang JavaScript>(() => {
'use strict';
// GENERIC FUNCTIONS
// replicateM n act performs the action n times, gathering the results.
// replicateM :: (Applicative m) => Int -> m a -> m [a]
const replicateM = (n, f) => {
const loop = x => x <= 0 ? [
[]
] : liftA2(cons, f, loop(x - 1));
return loop(n);
};
// Lift a binary function to actions.
// liftA2 :: Applicative f => (a -> b -> c) -> f a -> f b -> f c
const liftA2 = (f, a, b) =>
listApply(a.map(curry(f)), b);
// <*>
// listApply :: [(a -> b)] -> [a] -> [b]
const listApply = (fs, xs) =>
[].concat.apply([], fs.map(f =>
[].concat.apply([], xs.map(x => [f(x)]))));
// curry :: ((a, b) -> c) -> a -> b -> c const curry = f => a => b => f(a, b);
// cons :: a -> [a] -> [a] const cons = (x, xs) => [x].concat(xs);
// show :: a -> String; const show = JSON.stringify;
// TEST
return show(
replicateM(2, [1, 2, 3])
);
// -> [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
})();</lang>
- Output:
<lang JavaScript>[[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]</lang>
Lazy evaluation with a generator
Permutations with repetition by treating the elements as an ordered set, and writing a function from a zero-based index to the nth permutation. Wrapping this function in a generator allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set:
<lang JavaScript>(() => {
'use strict';
const main = () => {
// Generator object
const gen = permsWithRepn('ACKR', 5);
// Search without needing to generate whole set:
let
nxt = gen.next(),
i = 0,
alpha = nxt.value,
psi = alpha;
while (!nxt.done && 'crack' !== toLower(concat(nxt.value))) {
psi = nxt.value;
console.log(psi)
nxt = gen.next()
i++
}
console.log(nxt.value)
return (
'Generated ' + i + ' of ' + Math.pow(4, 5) +
' possible permutations,\n' +
'searching from: ' + show(alpha) + ' thru: ' + show(psi) +
'\nbefore finding: ' + show(nxt.value)
);
};
// PERMUTATION GENERATOR ------------------------------
// permsWithRepn :: [a] -> Int -> Generator [a]
function* permsWithRepn(xs, intGroup) {
const
vs = Array.from(xs),
intBase = vs.length,
intSet = Math.pow(intBase, intGroup);
if (0 < intBase) {
let index = 0;
while (index < intSet) {
const
ds = unfoldr(
v => 0 < v ? (() => {
const rd = quotRem(v, intBase);
return Just(Tuple(vs[rd[1]], rd[0]))
})() : Nothing(),
index++
);
yield replicate(
intGroup - ds.length,
vs[0]
).concat(ds);
};
}
};
// GENERIC FUNCTIONS ----------------------------------
// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});
// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});
// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// concat :: a -> [a] // concat :: [String] -> String const concat = xs => 0 < xs.length ? (() => { const unit = 'string' !== typeof xs[0] ? ( [] ) : ; return unit.concat.apply(unit, xs); })() : [];
// index (!!) :: [a] -> Int -> a // index (!!) :: String -> Int -> Char const index = (xs, i) => xs[i];
// quotRem :: Int -> Int -> (Int, Int)
const quotRem = (m, n) =>
Tuple(Math.floor(m / n), m % n);
// replicate :: Int -> a -> [a]
const replicate = (n, x) =>
Array.from({
length: n
}, () => x);
// show :: a -> String const show = x => JSON.stringify(x);
// toLower :: String -> String const toLower = s => s.toLocaleLowerCase();
// unfoldr(x => 0 !== x ? Just([x, x - 1]) : Nothing(), 10); // --> [10,9,8,7,6,5,4,3,2,1]
// unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
const unfoldr = (f, v) => {
let
xr = [v, v],
xs = [];
while (true) {
const mb = f(xr[1]);
if (mb.Nothing) {
return xs
} else {
xr = mb.Just;
xs.push(xr[0])
}
}
};
// MAIN --- return main();
})();</lang>
- Output:
Generated 589 of 1024 possible permutations, searching from: ["A","A","A","A","A"] thru: ["A","R","A","C","K"] before finding: ["C","R","A","C","K"]
jq
We first present a definition of permutations_with_replacement(n) that is compatible with jq 1.4. To interrupt the stream that it produces, however, requires a version of jq with break, which was introduced after the release of jq 1.4.
Definitions
We shall define permutations_with_replacements(n) in terms of a more general filter, combinations/0, defined as follows:
<lang jq># Input: an array, $in, of 0 or more arrays
- Output: a stream of arrays, c, with c[i] drawn from $in[i].
def combinations:
if length == 0 then [] else .[0][] as $x | (.[1:] | combinations) as $y | [$x] + $y end ;
- Input: an array of the k values from which to choose.
- Output: a stream of arrays of length n with elements drawn from the input array.
def permutations_with_replacements(n):
. as $in | [range(0; n) | $in] | combinations;</lang>
Example 1: Enumeration:
Count the number of 4-combinations of [0,1,2] by enumerating them, i.e., without creating a data structure to store them all. <lang jq>def count(stream): reduce stream as $i (0; .+1);
count([0,1,2] | permutations_with_replacements(4))
- output: 81</lang>
Example 2: Early termination of the generator:
Counting from 1, and terminating the generator when the item is found, what is the sequence number of ["c", "a", "b"] in the stream of 3-combinations of ["a","b","c"]? <lang jq># Input: the item to be matched
- Output: the index of the item in the stream (counting from 1);
- emit null if the item is not found
def sequence_number(stream):
. as $in
| (label $top
| foreach stream as $i (0; .+1; if $in == $i then ., break $top else empty end))
// null; # NOTE: "//" here is an operator
["c", "a", "b"] | sequence_number( ["a","b","c"] | permutations_with_replacements(3))
- output: 20</lang>
Julia
Implements a simil-Combinatorics.jl API.
<lang julia>struct WithRepetitionsPermutations{T}
a::T t::Int
end
with_repetitions_permutations(elements::T, len::Integer) where T =
WithRepetitionsPermutations{T}(unique(elements), len)
Base.iteratorsize(::WithRepetitionsPermutations) = Base.HasLength() Base.length(p::WithRepetitionsPermutations) = length(p.a) ^ p.t Base.iteratoreltype(::WithRepetitionsPermutations) = Base.HasEltype() Base.eltype(::WithRepetitionsPermutations{T}) where T = T Base.start(p::WithRepetitionsPermutations) = ones(Int, p.t) Base.done(p::WithRepetitionsPermutations, s::Vector{Int}) = s[end] > endof(p.a) function Base.next(p::WithRepetitionsPermutations, s::Vector{Int})
cur = p.a[s]
s[1] += 1
local i = 1
while i < endof(s) && s[i] > length(p.a)
s[i] = 1
s[i+1] += 1
i += 1
end
return cur, s
end
println("Permutations of [4, 5, 6] in 3:") foreach(println, collect(with_repetitions_permutations([4, 5, 6], 3)))</lang>
- Output:
Permutations of [4, 5, 6] in 3: [4, 4, 4] [5, 4, 4] [6, 4, 4] [4, 5, 4] [5, 5, 4] [6, 5, 4] [4, 6, 4] [5, 6, 4] [6, 6, 4] [4, 4, 5] [5, 4, 5] [6, 4, 5] [4, 5, 5] [5, 5, 5] [6, 5, 5] [4, 6, 5] [5, 6, 5] [6, 6, 5] [4, 4, 6] [5, 4, 6] [6, 4, 6] [4, 5, 6] [5, 5, 6] [6, 5, 6] [4, 6, 6] [5, 6, 6] [6, 6, 6]
K
enlist each from x on the left and each from x on the right where x is range 10 <lang k> ,/x/:\:x:!10 </lang>
Kotlin
<lang scala>// version 1.1.2
fun main(args: Array<String>) {
val n = 3
val values = charArrayOf('A', 'B', 'C', 'D')
val k = values.size
// terminate when first two characters of the permutation are 'B' and 'C' respectively
val decide = fun(pc: CharArray) = pc[0] == 'B' && pc[1] == 'C'
val pn = IntArray(n)
val pc = CharArray(n)
while (true) {
// generate permutation
for ((i, x) in pn.withIndex()) pc[i] = values[x]
// show progress
println(pc.contentToString())
// pass to deciding function
if (decide(pc)) return // terminate early
// increment permutation number
var i = 0
while (true) {
pn[i]++
if (pn[i] < k) break
pn[i++] = 0
if (i == n) return // all permutations generated
}
}
}</lang>
- Output:
[A, A, A] [B, A, A] [C, A, A] [D, A, A] [A, B, A] [B, B, A] [C, B, A] [D, B, A] [A, C, A] [B, C, A]
M2000 Interpreter
<lang M2000 Interpreter> Module Checkit {
a=("A","B","C","D")
n=len(a)
c1=lambda a, n, c (&f) ->{
=(array(a, c),)
c++
if c=n then c=0: f=true
}
m=n-2
While m >0 {
c3=lambda c2=c1, a, n, c (&f) -> {
f=false
=Cons((array(a, c),), c2(&f))
if f then {
c++
f=false
if c=n then c=0: f=true
}
}
c1=c3
m--
}
k=false
While not k {
r=c3(&k)
rr=each(r end to start)
While rr {
Print array$(rr),
}
Print
if array$(r, 2)="B" and array$(r,1)="C" then exit
}
} Checkit </lang>
- Output:
A A A B A A C A A D A A A B A B B A C B A D B A A C A B C A
Mathematica
<lang mathematica>Tuples[{1, 2, 3}, 2]</lang>
- Output:
{{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}
Maxima
<lang maxima>apply(cartesian_product,makelist({1,2,3}, 2));</lang>
- Output:
{[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]}
Perl
<lang perl>use Algorithm::Combinatorics qw/tuples_with_repetition/; print join(" ", map { "[@$_]" } tuples_with_repetition([qw/A B C/],2)), "\n";</lang>
- Output:
[A A] [A B] [A C] [B A] [B B] [B C] [C A] [C B] [C C]
Solving the crack problem: <lang perl>use Algorithm::Combinatorics qw/tuples_with_repetition/; my $iter = tuples_with_repetition([qw/A C K R/], 5); my $tries = 0; while (my $p = $iter->next) {
$tries++;
die "Found the combination after $tries tries!\n" if join("",@$p) eq "CRACK";
}</lang>
- Output:
Found the combination after 455 tries!
Perl 6
We can use the X operator ("cartesian product") to cross the list with itself.
For :
<lang perl6>my @k = <a b c>;
.say for @k X @k;</lang>
For arbitrary :
<lang perl6>my @k = <a b c>; my $n = 2;
.say for [X] @k xx $n;</lang>
- Output:
a a a b a c b a b b b c c a c b c c
Here is an other approach, counting all possibilities in base :
<lang perl6>my @k = <a b c>; my $n = 2;
say @k[.polymod: +@k xx $n-1] for ^@k**$n</lang>
- Output:
a a b a c a a b b b c b a c b c c c
Pascal
Create a list of indices into what ever you want, one by one. Doing it by addig one to a number with k-positions to base n. <lang pascal>program PermuWithRep; //permutations with repetitions //http://rosettacode.org/wiki/Permutations_with_repetitions {$IFDEF FPC}
{$Mode Delphi}{$Optimization ON}{$Align 16}{$Codealign proc=16,loop=4}
{$ELSE}
{$APPTYPE CONSOLE}// for Delphi
{$ENDIF} uses
sysutils;
type
tPermData = record
mdTup_n, //number of positions
mdTup_k:NativeInt; //number of different elements
mdTup :array of integer;
end;
function InitTuple(k,n:nativeInt):tPermData; begin
with result do
Begin
IF k> 0 then
Begin
mdTup_k:= k;
setlength(mdTup,k);
IF (n<0) then
mdTup_n := 0
else
mdTup_n := n;
end
else
Begin
mdTup_k := 1;
mdTup_n := k;
end;
end;
end;
procedure PermOut(const p:tPermData); var
i : nativeInt;
Begin
with p do
Begin
For i := 0 to mdTup_k-1 do
write(mdTup[i]:4);
end;
writeln;
end;
function NextPermWithRep(var perm:tPermData): boolean; // create next permutation by adding 1 and correct "carry" // returns false if finished var
pDg :^Integer; dg,le :nativeInt;
begin
WIth perm do
Begin
pDg := @mdTup[0];
le := mdTup_k;
repeat
dg := pDg^+1;
IF (dg<mdTup_n) then
Begin
pDg^ := dg;
BREAK;
end
else
pDg^ := 0;
dec(le);
inc(pDg);
until le<=0;
result := (dg<mdTup_n);
end;
end;
var
p: tPermData; cnt,k,n: nativeInt;
Begin
cnt := 0;
//k := 2;n := 3;
k := 10;n := 8;
p:= InitTuple(k,n);
IF (n<= 6) then
repeat
inc(cnt);
PermOut(p);
until Not(NextPermWithRep(p))
else
repeat
inc(cnt);
until Not(NextPermWithRep(p));
writeln('k: ',k,' n: ',n,' count ',cnt);
end.</lang>
- Output:
0 0 1 0 2 0 0 1 1 1 2 1 0 2 1 2 2 2 k: 2 n: 3 count 9 .. //speedtest Compiler /fpc/3.1.1/ppc386 "%f" -al -Xs -XX -O3 // i4330 3.5 Ghz k: 10 n: 8 count 1073741824 => 8^10 real 0m2.556s // without inc(cnt); real 0m2.288s-> 7,5 cycles per call //"old" compiler-version //real 0m3.465s /fpc/2.6.4/ppc386 "%f" -al -Xs -XX -O3
Phix
The task is equivalent to simply counting in base=length(set), from 1 to power(base,n).
Asking for the 0th permutation just returns the total number of permutations (ie "").
Results can be generated in any order, hence early termination is quite simply a non-issue.
<lang Phix>function permrep(sequence set, integer n, idx=0)
integer base = length(set),
nperm = power(base,n)
if idx=0 then
-- return the number of permutations
return nperm
end if
-- return the idx'th [1-based] permutation
if idx<1 or idx>nperm then ?9/0 end if
idx -= 1 -- make it 0-based
sequence res = ""
for i=1 to n do
res = prepend(res,set[mod(idx,base)+1])
idx = floor(idx/base)
end for
if idx!=0 then ?9/0 end if -- sanity check
return res
end function</lang> Some slightly excessive testing: <lang Phix>procedure show_all(sequence set, integer n)
integer l = permrep(set,n)
sequence s = repeat(0,l)
for i=1 to l do
s[i] = permrep(set,n,i)
end for
?s
end procedure
show_all("123",1) show_all("123",2) show_all("123",3) show_all("456",3) show_all({1,2,3},3) show_all({"bat","fox","cow"},2)
sequence s = {} for i=31 to 36 do
s = append(s,permrep("XYZ",4,i))
end for ?s
integer l = permrep("ACKR",5) for i=1 to l do
if permrep("ACKR",5,i)="CRACK" then -- 455
printf(1,"Permutation %d of %d: CRACK\n",{i,l})
exit
end if
end for --The 590th (one-based) permrep is KCARC, ie reverse(CRACK), matching the 589 result of 0-based idx solutions printf(1,"reverse(permrep(\"ACKR\",5,589+1):%s\n",{reverse(permrep("ACKR",5,590))})</lang>
- Output:
{"1","2","3"}
{"11","12","13","21","22","23","31","32","33"}
{"111","112","113","121","122","123","131","132","133","211","212","213","221","222","223","231","232","233","311","312","313","321","322","323","331","332","333"}
{"444","445","446","454","455","456","464","465","466","544","545","546","554","555","556","564","565","566","644","645","646","654","655","656","664","665","666"}
{{1,1,1},{1,1,2},{1,1,3},{1,2,1},{1,2,2},{1,2,3},{1,3,1},{1,3,2},{1,3,3},{2,1,1},{2,1,2},{2,1,3},{2,2,1},{2,2,2},{2,2,3},{2,3,1},{2,3,2},{2,3,3},{3,1,1},{3,1,2},{3,1,3},{3,2,1},{3,2,2},{3,2,3},{3,3,1},{3,3,2},{3,3,3}}
{{"bat","bat"},{"bat","fox"},{"bat","cow"},{"fox","bat"},{"fox","fox"},{"fox","cow"},{"cow","bat"},{"cow","fox"},{"cow","cow"}}
{"YXYX","YXYY","YXYZ","YXZX","YXZY","YXZZ"}
Permutation 455 of 1024: CRACK
reverse(permrep("ACKR",5,589+1):CRACK
PHP
<lang PHP><?php function permutate($values, $size, $offset) {
$count = count($values);
$array = array();
for ($i = 0; $i < $size; $i++) {
$selector = ($offset / pow($count,$i)) % $count;
$array[$i] = $values[$selector];
}
return $array;
}
function permutations($values, $size) {
$a = array();
$c = pow(count($values), $size);
for ($i = 0; $i<$c; $i++) {
$a[$i] = permutate($values, $size, $i);
}
return $a;
}
$permutations = permutations(['bat','fox','cow'], 2); foreach ($permutations as $permutation) {
echo join(',', $permutation)."\n";
} </lang>
- Output:
bat,bat fox,bat cow,bat bat,fox fox,fox cow,fox bat,cow fox,cow cow,cow
PicoLisp
<lang PicoLisp>(de permrep (N Lst)
(if (=0 N)
(cons NIL)
(mapcan
'((X)
(mapcar '((Y) (cons Y X)) Lst) )
(permrep (dec N) Lst) ) ) )</lang>
Python
Strict evaluation of the whole set
To evaluate the whole set of permutations, without the option to make complete evaluation conditional, we can reach for a generic replicateM function for lists:
<lang python>from functools import (reduce)
- main :: IO ()
def main():
print(
replicateM(2)([1, 2, 3])
)
- GENERIC FUNCTIONS ---------------------------------------
- replicateM :: Int -> [a] -> a
def replicateM(n):
def loop(f):
def go(x):
return [[]] if 0 >= x else (
liftA2List(lambda a, b: [a] + b)(f)(go(x - 1))
)
return go(n)
return lambda f: loop(f)
- liftA2List :: (a -> b -> c) -> [a] -> [b] -> [c]
def liftA2List(f):
return lambda xs: lambda ys: concatMap(
lambda x: concatMap(lambda y: [f(x, y)])(ys)
)(xs)
- concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
return lambda xs: (
reduce(lambda a, b: a + b, map(f, xs), [])
)
- MAIN ---
main()</lang>
- Output:
[[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]]
Lazy evaluation with a generator
Applying itertools.product
<lang python>from itertools import product
- check permutations until we find the word 'crack'
for x in product('ACRK', repeat=5):
w = .join(x) print w if w.lower() == 'crack': break</lang>
Writing a generator
Or, composing our own generator, by wrapping a function from an index in the range 0 .. ((distinct items to the power of groupSize) - 1) to a unique permutation. (Each permutation is equivalent to a 'number' in the base of the size of the set of distinct items, in which each distinct item functions as a 'digit'): <lang Python>from functools import (reduce) from itertools import (repeat)
- main :: IO ()
def main():
cs = 'ACKR'
wordLength = 5
gen = permutesWithRepns(cs)(wordLength)
for idx, xs in enumerate(gen):
s = .join(xs)
if 'crack' == s.lower():
break
print (
'Permutation ' + str(idx) + ' of ' +
str(len(cs)**wordLength) + ':', s
)
- permutesWithRepns :: [a] -> Int -> Generator a
def permutesWithRepns(xs):
def groupsOfSize(n):
f = nthPermWithRepn(xs)(n)
limit = len(xs)**n
i = 0
while (i < limit):
yield f(i)
i = 1 + i
return lambda n: groupsOfSize(n)
- Index as a 'number' in the base of the
- size of the set (of distinct values to be permuted),
- using each value as a 'digit'
- (leftmost value used as the 'zero')
- nthPermWithRepn :: [a] -> Int -> Int -> [a]
def nthPermWithRepn(xs):
def go(intGroup, index):
vs = list(xs)
intBase = len(vs)
intSet = intBase ** intGroup
return (
lambda ds=unfoldr(lambda v: (
(lambda qr=divmod(v, intBase):
Just((vs[qr[1]], qr[0])))()
) if 0 < v else Nothing()
)(index): (
list(repeat(vs[0], intGroup - len(ds))) + ds
)
)() if 0 < intBase and index < intSet else None
return lambda intGroup: lambda index: go(
intGroup, index
)
- GENERIC FUNCTIONS -------------------------------------
- Just :: a -> Maybe a
def Just(x):
return {type: 'Maybe', 'Nothing': False, 'Just': x}
- Nothing :: Maybe a
def Nothing():
return {type: 'Maybe', 'Nothing': True}
- concat :: a -> [a]
def concat(xs):
return (
reduce(
lambda a, b: a + b, xs,
if type(xs[0]) is str else []
) if xs else []
)
- unfoldr(lambda x: Just((x, x - 1)) if 0 != x else Nothing(), 10)]
- -> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
- unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
def unfoldr(f):
def go(v):
xr = (v, v)
xs = []
while True:
mb = f(xr[1])
if mb.get('Nothing'):
return xs
else:
xr = mb.get('Just')
xs.append(xr[0])
return xs
return lambda v: go(v)
main()</lang>
- Output:
Permutation 589 of 1024: CRACK
Racket
As a sequence
First we define a procedure that defines the sequence of the permutations. <lang Racket>#lang racket (define (permutations-with-repetitions/proc size items)
(define items-vector (list->vector items))
(define num (length items))
(define (pos->element pos)
(reverse
(for/list ([p (in-vector pos)])
(vector-ref items-vector p))))
(define (next-pos pos)
(let ([ret (make-vector size #f)])
(for/fold ([carry 1]) ((i (in-range size)))
(let ([tmp (+ (vector-ref pos i) carry)])
(if (= tmp num)
(begin
(vector-set! ret i 0)
#;carry 1)
(begin
(vector-set! ret i tmp)
#;carry 0))))
ret))
(define initial-pos (vector->immutable-vector (make-vector size 0)))
(define last-pos (vector->immutable-vector (make-vector size (sub1 num))))
(define (continue-after-pos+val? pos val)
(not (equal? pos last-pos)))
(make-do-sequence (lambda ()
(values pos->element
next-pos
initial-pos
#f
#f
continue-after-pos+val?))))
(sequence->list (permutations-with-repetitions/proc 2 '(1 2 3)))</lang>
- Output:
'((1 1) (1 2) (1 3) (2 1) (2 2) (2 3) (3 1) (3 2) (3 3))
As a sequence with for clause support
Now we define a more general version that can be used efficiently in as a for clause. In other uses it falls back to the sequence implementation. <lang Racket>(require (for-syntax racket))
(define-sequence-syntax in-permutations-with-repetitions
(lambda () #'permutations-with-repetitions/proc)
(lambda (stx)
(syntax-case stx ()
[[(element) (_ size/ex items/ex)]
#'[(element)
(:do-in ([(size) size/ex]
[(items) items/ex]
[(items-vector) (list->vector items/ex)]
[(num) (length items/ex)]
[(last-pos) (make-vector size/ex (sub1 (length items/ex)))])
(void)
([pos (make-vector size 0)])
#t
([(element) (reverse
(for/list ([p (in-vector pos)])
(vector-ref items-vector p)))])
#t
(not (equal? pos last-pos))
[(let ([ret (make-vector size #f)])
(for/fold ([carry 1]) ((i (in-range size)))
(let ([tmp (+ (vector-ref pos i) carry)])
(if (= tmp num)
(begin
(vector-set! ret i 0)
#;carry 1)
(begin
(vector-set! ret i tmp)
#;carry 0))))
ret)])]])))
(for/list ([element (in-permutations-with-repetitions 2 '(1 2 3))])
element)
(sequence->list (in-permutations-with-repetitions 2 '(1 2 3)))</lang>
- Output:
'((1 1) (1 2) (1 3) (2 1) (2 2) (2 3) (3 1) (3 2) (3 3)) '((1 1) (1 2) (1 3) (2 1) (2 2) (2 3) (3 1) (3 2) (3 3))
REXX
version 1
<lang rexx>/*REXX pgm generates/displays all permutations of N different objects taken M at a time.*/ parse arg things bunch inbetweenChars names
/* ╔════════════════════════════════════════════════════════════════╗ */
/* ║ inBetweenChars (optional) defaults to a [null]. ║ */
/* ║ names (optional) defaults to digits (and letters).║ */
/* ╚════════════════════════════════════════════════════════════════╝ */
call permSets things, bunch, inBetweenChars, names exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ p: return word( arg(1), 1) /*P function (Pick first arg of many).*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ permSets: procedure; parse arg x,y,between,uSyms /*X things taken Y at a time. */
@.=; sep= /*X can't be > length(@0abcs). */
@abc = 'abcdefghijklmnopqrstuvwxyz'; @abcU= @abc; upper @abcU
@abcS = @abcU || @abc; @0abcS= 123456789 || @abcS
do k=1 for x /*build a list of permutation symbols. */
_= p( word(uSyms, k) p( substr(@0abcS, k, 1) k) ) /*get/generate a symbol.*/
if length(_)\==1 then sep= '_' /*if not 1st character, then use sep. */
$.k= _ /*append the character to symbol list. */
end /*k*/
if between== then between= sep /*use the appropriate separator chars. */
call .permSet 1 /*start with the first permutation. */
return /* [↓] this is a recursive subroutine.*/
.permSet: procedure expose $. @. between x y; parse arg ?
if ?>y then do; _=@.1; do j=2 for y-1; _=_ || between || @.j; end; say _
end
else do q=1 for x /*build the permutation recursively. */
@.?= $.q; call .permSet ?+1
end /*q*/
return /*this is meant to be an anonymous sub.*/</lang>
- output when using the default inputs of: 3 2
11 12 13 21 22 23 31 32 33
- output when using the default inputs of : 3 2 , bat fox cow
bat,bat bat,fox bat,cow fox,bat fox,fox fox,cow cow,bat cow,fox cow,cow
version 2 (using Interpret)
Note: this REXX version will cause Regina REXX to fail (crash) if the expression to be INTERPRETed is too large (byte-wise).
PC/REXX and Personal REXX also fail, but for a smaller expression.
Please specify limitations. One could add:
If length(a)>implementation_dependent_limit Then
Say 'too large for this Rexx version'
Also note that the output isn't the same as REXX version 1 when the 1st argument is two digits or more, i.e.: 11 2
<lang rexx>/* REXX ***************************************************************
- Arguments and output as in REXX version 1 (for the samples shown there)
- For other elements (such as 11 2), please specify a separator
- Translating 10, 11, etc. to A, B etc. is left to the reader
- 12.05.2013 Walter Pachl
- 12-05-2013 Walter Pachl take care of bunch<=0 and other oddities
- /
Parse Arg things bunch sep names If datatype(things,'W') & datatype(bunch,'W') Then
Nop
Else
Call exit 'First two arguments must be integers >0'
If things= Then n=3; Else n=things If bunch= Then m=2; Else m=bunch If things<=0 Then Call exit 'specify a positive number of things' If bunch<=0 Then Call exit 'no permutations with' bunch 'elements!'
Select
When sep= Then ss='
When datatype(sep)='NUM' Then ss='copies(' ',sep)'
Otherwise ss='sep'
End
Do i=1 To n
If names<> Then Parse Var names e.i names Else e.i=i End
a='p=0;'; Do i=1 To m; a=a||'Do p'i'=1 To n;'; End a=a||'ol=e.p1'
Do i=2 To m; a=a||'||'ss'||e.p'i; End
a=a||'; say ol; p=p+1;'
Do i=1 To m; a=a||'end;'; End
a=a||'Say' p 'permutations' /* Say a */ Interpret a</lang>
version 3
This is a very simplistic version that is limited to nine things (N).
It essentially just executes a do loop and ignores any permutation out of range,
this is very wasteful of CPU processing time when using a larger N.
This version could easily be extended to N up to 15 (using hexadecimal arithmetic). <lang rexx>/*REXX pgm gens all permutations with repeats of N objects (<10) taken M at a time. */ parse arg N M . z= N**M $= left(1234567890, N) t= 0
do j=copies(1, M) until t==z
if verify(j, $)\==0 then iterate
t= t+1
say j
end /*j*/ /*stick a fork in it, we're all done. */</lang>
- output when using the following inputs: 3 2
11 12 13 21 22 23 31 32 33
Ring
<lang ring>
- Project : Permutations with repetitions
list1 = [["a", "b", "c"], ["a", "b", "c"]] list2 = [["1", "2", "3"], ["1", "2", "3"]] permutation(list1) permutation(list2)
func permutation(list1)
for n = 1 to len(list1[1])
for m = 1 to len(list1[2])
see list1[1][n] + " " + list1[2][m] + nl
next
next
see nl
</lang> Output:
a a a b a c b a b b b c c a c b c c 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3
Ruby
This is built in (Array#repeated_permutation): <lang ruby>rp = [1,2,3].repeated_permutation(2) # an enumerator (generator) p rp.to_a #=>[[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]]
- yield permutations until their sum happens to exceed 4, then quit:
p rp.take_while{|(a, b)| a + b < 5} #=>[[1, 1], [1, 2], [1, 3], [2, 1], [2, 2]]</lang>
Scala
<lang scala>package permutationsRep
object PermutationsRepTest extends Application {
/**
* Calculates all permutations taking n elements of the input List,
* with repetitions.
* Precondition: input.length > 0 && n > 0
*/
def permutationsWithRepetitions[T](input : List[T], n : Int) : List[List[T]] = {
require(input.length > 0 && n > 0)
n match {
case 1 => for (el <- input) yield List(el)
case _ => for (el <- input; perm <- permutationsWithRepetitions(input, n - 1)) yield el :: perm
}
}
println(permutationsWithRepetitions(List(1, 2, 3), 2))
}</lang>
- Output:
List(List(1, 1), List(1, 2), List(1, 3), List(2, 1), List(2, 2), List(2, 3), List(3, 1), List(3, 2), List(3, 3))
Sidef
<lang ruby>var k = %w(a b c) var n = 2
cartesian([k] * n, {|*a| say a.join(' ') })</lang>
- Output:
a a a b a c b a b b b c c a c b c c
Tcl
Iterative version
<lang tcl> proc permutate {values size offset} {
set count [llength $values]
set arr [list]
for {set i 0} {$i < $size} {incr i} {
set selector [expr [round [expr $offset / [pow $count $i]]] % $count];
lappend arr [lindex $values $selector]
}
return $arr
}
proc permutations {values size} {
set a [list]
set c [pow [llength $values] $size]
for {set i 0} {$i < $c} {incr i} {
set permutation [permutate $values $size $i]
lappend a $permutation
}
return $a
}
- Usage
permutations [list 1 2 3 4] 3 </lang>
Version without additional libraries
<lang tcl>package require Tcl 8.6
- Utility function to make procedures that define generators
proc generator {name arguments body} {
set body [list try $body on ok {} {return -code break}]
set lambda [list $arguments "yield \[info coroutine\];$body"]
proc $name args "tailcall \
coroutine gen_\[incr ::generate_ctr\] apply [list $lambda] {*}\$args" }
- How to generate permutations with repetitions
generator permutationsWithRepetitions {input n} {
if {[llength $input] == 0 || $n < 1} {error "bad arguments"}
if {![incr n -1]} {
foreach el $input { yield [list $el] }
} else {
foreach el $input { set g [permutationsWithRepetitions $input $n] while 1 { yield [list $el {*}[$g]] } }
}
}
- Demonstrate usage
set g [permutationsWithRepetitions {1 2 3} 2] while 1 {puts [$g]}</lang>
Alternate version with extra library package
<lang tcl>package require Tcl 8.6 package require generator
- How to generate permutations with repetitions
generator define permutationsWithRepetitions {input n} {
if {[llength $input] == 0 || $n < 1} {error "bad arguments"}
if {![incr n -1]} {
foreach el $input { generator yield [list $el] }
} else {
foreach el $input { set g [permutationsWithRepetitions $input $n] while 1 { generator yield [list $el {*}[$g]] } }
}
}
- Demonstrate usage
generator foreach val [permutationsWithRepetitions {1 2 3} 2] {
puts $val
}</lang>