Permutations with repetitions
- 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
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>-- PERMUTATIONS WITH REPETITION ----------------------------------------------
-- permutationsWithRepetition :: Int -> [a] -> a on permutationsWithRepetition(n, xs)
if length of xs > 0 then foldl1(curry(my cartesianProduct)'s |λ|(xs), replicate(n, xs)) else {} end if
end permutationsWithRepetition
-- TEST ----------------------------------------------------------------------
on run
permutationsWithRepetition(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 ---------------------------------------------------------
-- cartesianProduct :: [a] -> [b] -> a, b on cartesianProduct(xs, ys)
script on |λ|(x) script on |λ|(y) {{x} & y} end |λ| end script concatMap(result, ys) end |λ| end script concatMap(result, xs)
end cartesianProduct
-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)
script append on |λ|(a, b) a & b end |λ| end script foldl(append, {}, map(f, xs))
end concatMap
-- curry :: (Script|Handler) -> Script on curry(f)
script on |λ|(a) script on |λ|(b) |λ|(a, b) of mReturn(f) end |λ| end script end |λ| end script
end curry
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs repeat with i from 1 to lng set v to |λ|(v, item i of xs, i, xs) end repeat return v end tell
end foldl
-- foldl1 :: (a -> a -> a) -> [a] -> a on foldl1(f, xs)
if length of xs > 0 then foldl(f, item 1 of xs, tail(xs)) else {} end if
end foldl1
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- 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
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then items 2 thru -1 of xs else {} end if
end tail</lang>
- Output:
<lang AppleScript>{{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}</lang>
Partial evaluation
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>-- Nth PERMUTATION WITH REPETITION -------------------------------------------
-- nthPermutationWithRepn :: [a] -> Int -> Int -> [a] on nthPermutationWithRepn(xs, groupSize, iIndex)
set intBase to length of xs set intSetSize to intBase ^ groupSize if intBase < 1 or iIndex > intSetSize then {} else set baseElems to inBaseElements(xs, iIndex) set intZeros to groupSize - (length of baseElems) if intZeros > 0 then replicate(intZeros, item 1 of xs) & baseElems else baseElems end if end if
end nthPermutationWithRepn
-- inBaseElements :: [a] -> Int -> [String] on inBaseElements(xs, n)
set intBase to length of xs script nextDigit on |λ|(residue) set {divided, remainder} to quotRem(residue, intBase) {valid:divided > 0, value:(item (remainder + 1) of xs), new:divided} end |λ| end script reverse of unfoldr(nextDigit, n)
end inBaseElements
-- TEST ----------------------------------------------------------------------
on run
script on |λ|(x) nthPermutationWithRepn({"X", "Y", "Z"}, 4, x) end |λ| end script -- 30th to 35th members of the series map(result, enumFromTo(30, 35))
end run
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)
if m > n then set d to -1 else set d to 1 end if set lst to {} repeat with i from m to n by d set end of lst to i end repeat return lst
end enumFromTo
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- quotRem :: Integral a => a -> a -> (a, a) on quotRem(m, n)
{m div n, m mod n}
end quotRem
-- 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
-- unfoldr :: (b -> Maybe (a, b)) -> b -> [a] on unfoldr(f, v)
set mf to mReturn(f) set lst to {} set recM to mf's |λ|(v) repeat while (valid of recM) is true set end of lst to value of recM set recM to mf's |λ|(new of recM) end repeat lst & value of recM
end unfoldr
-- until :: (a -> Bool) -> (a -> a) -> a -> a on |until|(p, f, x)
set mp to mReturn(p) set v to x tell mReturn(f) repeat until mp's |λ|(v) set v to |λ|(v) end repeat end tell return v
end |until|</lang>
- Output:
<lang AppleScript>{{"Y", "X", "Y", "X"}, {"Y", "X", "Y", "Y"}, {"Y", "X", "Y", "Z"}, {"Y", "X", "Z", "X"}, {"Y", "X", "Z", "Y"}, {"Y", "X", "Z", "Z"}}</lang>
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], upto = 4, temp2; for( temp2 = 1 ; temp2 <= numbers; temp2++){ a[temp2]=1; } a[numbers]=0; temp=numbers, temp2; 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)
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 = 1, 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 590 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}}
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
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
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) 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>