Permutations with some identical elements: Difference between revisions
→{{header|Wren}}: Now uses Wren-perm module. |
|||
Line 404:
["AABBBC", "AABBCB", "AABCBB", "AACBBB", "ABABBC", "ABABCB", "ABACBB", "ABBABC", "ABBACB", "ABBBAC", "ABBBCA", "ABBCAB", "ABBCBA", "ABCABB", "ABCBAB", "ABCBBA", "ACABBB", "ACBABB", "ACBBAB", "ACBBBA", "BAABBC", "BAABCB", "BAACBB", "BABABC", "BABACB", "BABBAC", "BABBCA", "BABCAB", "BABCBA", "BACABB", "BACBAB", "BACBBA", "BBAABC", "BBAACB", "BBABAC", "BBABCA", "BBACAB", "BBACBA", "BBBAAC", "BBBACA", "BBBCAA", "BBCAAB", "BBCABA", "BBCBAA", "BCAABB", "BCABAB", "BCABBA", "BCBAAB", "BCBABA", "BCBBAA", "CAABBB", "CABABB", "CABBAB", "CABBBA", "CBAABB", "CBABAB", "CBABBA", "CBBAAB", "CBBABA", "CBBBAA"]
</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<lang Mathematica>ClearAll[PermsFromFrequencies]
PermsFromFrequencies[l_List] := Module[{digs},
digs = Flatten[MapIndexed[ReverseApplied[ConstantArray], l]];
Permutations[digs]
]
PermsFromFrequencies[{2, 3, 1}] // Column</lang>
{{out}}
<pre>{1,1,2,2,2,3}
{1,1,2,2,3,2}
{1,1,2,3,2,2}
{1,1,3,2,2,2}
{1,2,1,2,2,3}
{1,2,1,2,3,2}
{1,2,1,3,2,2}
{1,2,2,1,2,3}
{1,2,2,1,3,2}
{1,2,2,2,1,3}
{1,2,2,2,3,1}
{1,2,2,3,1,2}
{1,2,2,3,2,1}
{1,2,3,1,2,2}
{1,2,3,2,1,2}
{1,2,3,2,2,1}
{1,3,1,2,2,2}
{1,3,2,1,2,2}
{1,3,2,2,1,2}
{1,3,2,2,2,1}
{2,1,1,2,2,3}
{2,1,1,2,3,2}
{2,1,1,3,2,2}
{2,1,2,1,2,3}
{2,1,2,1,3,2}
{2,1,2,2,1,3}
{2,1,2,2,3,1}
{2,1,2,3,1,2}
{2,1,2,3,2,1}
{2,1,3,1,2,2}
{2,1,3,2,1,2}
{2,1,3,2,2,1}
{2,2,1,1,2,3}
{2,2,1,1,3,2}
{2,2,1,2,1,3}
{2,2,1,2,3,1}
{2,2,1,3,1,2}
{2,2,1,3,2,1}
{2,2,2,1,1,3}
{2,2,2,1,3,1}
{2,2,2,3,1,1}
{2,2,3,1,1,2}
{2,2,3,1,2,1}
{2,2,3,2,1,1}
{2,3,1,1,2,2}
{2,3,1,2,1,2}
{2,3,1,2,2,1}
{2,3,2,1,1,2}
{2,3,2,1,2,1}
{2,3,2,2,1,1}
{3,1,1,2,2,2}
{3,1,2,1,2,2}
{3,1,2,2,1,2}
{3,1,2,2,2,1}
{3,2,1,1,2,2}
{3,2,1,2,1,2}
{3,2,1,2,2,1}
{3,2,2,1,1,2}
{3,2,2,1,2,1}
{3,2,2,2,1,1}</pre>
=={{header|MiniZinc}}==
|
Revision as of 21:49, 2 August 2022
Sometimes you want to find all permutations of elements where some elements are repeated, e.g. you have 3 red balls, 2 blue balls and one black ball.
If you just do all permutations of the 6 elements, each permutation will be duplicated 12 times where you can't tell that the identical elements have switched places.
Given an input of the form [a1, a2, ···, ak] where ak denotes how many duplicates of element k you should have,
each ak > 0 and the sum of all ak is n.
You should get n! / (a1! × a2! ... × ak!) permutations as a result.
(You may, of course, denote the elements 0..k-1 if that works better.)
For example, the input [2,1] should give results (1,1,2), (1,2,1) and (2,1,1).
Alternatively, if zero-based: (0,0,1), (0,1,0) and (1,0,0).
- Task
List the permutations you get from the input [2, 3, 1].
Optionally output the permutations as strings where the first element is represented by A, the second by B and the third by C
(the example result would then be AAB, ABA and BAA).
- Related tasks
11l
<lang 11l>F shouldSwap(s, start, curr)
L(i) start .< curr I s[i] == s[curr] R 0B R 1B
F findPerms(ss, index, n, &res)
I index >= n res.append(ss) R
V s = copy(ss) L(i) index .< n I shouldSwap(s, index, i) swap(&s[index], &s[i]) findPerms(s, index + 1, n, &res) swap(&s[index], &s[i])
F createSlice(nums, charSet)
V result = ‘’ L(n) nums V i = L.index L 0 .< n result ‘’= charSet[i] R result
[String] res1, res2, res3 V nums = [2, 1]
V s = createSlice(nums, ‘12’) findPerms(s, 0, s.len, &res1) print(res1.join(‘ ’)) print()
nums = [2, 3, 1] s = createSlice(nums, ‘123’) findPerms(s, 0, s.len, &res2) L(val) res2
print(val, end' I (L.index + 1) % 10 == 0 {"\n"} E ‘ ’)
print()
s = createSlice(nums, ‘ABC’) findPerms(s, 0, s.len, &res3) L(val) res3
print(val, end' I (L.index + 1) % 10 == 0 {"\n"} E ‘ ’)</lang>
- Output:
112 121 211 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122321 122312 123221 123212 123122 132221 132212 132122 131222 211223 211232 211322 212123 212132 212213 212231 212321 212312 213221 213212 213122 221123 221132 221213 221231 221321 221312 222113 222131 222311 223121 223112 223211 231221 231212 231122 232121 232112 232211 312221 312212 312122 311222 321221 321212 321122 322121 322112 322211 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA
Dart
<lang dart> import 'dart:io';
void main() {
stdout.writeln(distinctPerms([2,3,1]).map((p) => alpha("ABC", p)).toList());
}
String alpha(String alphabet, List<int> perm) {
return perm.map((i) => alphabet[i]).join("");
}
Iterable<List<int>> distinctPerms(List<int> reps) sync* {
Iterable<List<int>> perms(List<List<int>> elements) sync* { if (elements.length == 1) { yield List.of(elements[0]); } else { for (int k = 0; k < elements.length; k++) { List<List<int>> tail = []; for (int i = 0; i < elements.length; i++) { if (i == k) { if (elements[i].length > 1) { tail.add(List.of(elements[i].skip(1))); } } else { tail.add(elements[i]); } } yield* perms(tail).map((t) { t.insert(0, elements[k][0]); return t; }); } } } List<List<int>> elements = []; for (int i = 0; i < reps.length; i++) { elements.add(List.filled(reps[i], i)); } yield* perms(elements);
} </lang>
- Output:
[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCAB, ABBCBA, ABCABB, ABCBAB, ABCBBA, ACABBB, ACBABB, ACBBAB, ACBBBA, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCAB, BABCBA, BACABB, BACBAB, BACBBA, BBAABC, BBAACB, BBABAC, BBABCA, BBACAB, BBACBA, BBBAAC, BBBACA, BBBCAA, BBCAAB, BBCABA, BBCBAA, BCAABB, BCABAB, BCABBA, BCBAAB, BCBABA, BCBBAA, CAABBB, CABABB, CABBAB, CABBBA, CBAABB, CBABAB, CBABBA, CBBAAB, CBBABA, CBBBAA]
Factor
Removing duplicates from the list of all permutations: <lang factor>USING: arrays grouping math math.combinatorics prettyprint sequences sets ;
- distinct-permutations ( seq -- seq )
[ CHAR: A + <array> ] map-index "" concat-as <permutations> members ;
{ 2 3 1 } distinct-permutations 10 group simple-table.</lang>
- Output:
AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA
Generating distinct permutations directly (more efficient in time and space):
<lang factor>USING: arrays io kernel locals math math.ranges sequences ; IN: rosetta-code.distinct-permutations
- should-swap? ( start curr seq -- ? )
[ nipd nth ] [ <slice> member? not ] 3bi ;
- .find-permutations ( seq index n -- )
index n >= [ seq write bl ] [ index n [a,b) [ :> i index i seq should-swap? [ index i seq exchange seq index 1 + n .find-permutations index i seq exchange ] when ] each ] if ;
- first-permutation ( nums charset -- seq )
[ <array> ] 2map "" concat-as ;
- .distinct-permutations ( nums charset -- )
first-permutation 0 over length .find-permutations nl ;
- main ( -- )
{ 2 1 } "12" { 2 3 1 } "123" { 2 3 1 } "ABC" [ .distinct-permutations ] 2tri@ ;
MAIN: main</lang>
- Output:
112 121 211 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122321 122312 123221 123212 123122 132221 132212 132122 131222 211223 211232 211322 212123 212132 212213 212231 212321 212312 213221 213212 213122 221123 221132 221213 221231 221321 221312 222113 222131 222311 223121 223112 223211 231221 231212 231122 232121 232112 232211 312221 312212 312122 311222 321221 321212 321122 322121 322112 322211 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA
Go
This is based on the C++ code here. <lang go>package main
import "fmt"
func shouldSwap(s []byte, start, curr int) bool {
for i := start; i < curr; i++ { if s[i] == s[curr] { return false } } return true
}
func findPerms(s []byte, index, n int, res *[]string) {
if index >= n { *res = append(*res, string(s)) return } for i := index; i < n; i++ { check := shouldSwap(s, index, i) if check { s[index], s[i] = s[i], s[index] findPerms(s, index+1, n, res) s[index], s[i] = s[i], s[index] } }
}
func createSlice(nums []int, charSet string) []byte {
var chars []byte for i := 0; i < len(nums); i++ { for j := 0; j < nums[i]; j++ { chars = append(chars, charSet[i]) } } return chars
}
func main() {
var res, res2, res3 []string nums := []int{2, 1} s := createSlice(nums, "12") findPerms(s, 0, len(s), &res) fmt.Println(res) fmt.Println()
nums = []int{2, 3, 1} s = createSlice(nums, "123") findPerms(s, 0, len(s), &res2) fmt.Println(res2) fmt.Println()
s = createSlice(nums, "ABC") findPerms(s, 0, len(s), &res3) fmt.Println(res3)
}</lang>
- Output:
[112 121 211] [112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122321 122312 123221 123212 123122 132221 132212 132122 131222 211223 211232 211322 212123 212132 212213 212231 212321 212312 213221 213212 213122 221123 221132 221213 221231 221321 221312 222113 222131 222311 223121 223112 223211 231221 231212 231122 232121 232112 232211 312221 312212 312122 311222 321221 321212 321122 322121 322112 322211] [AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA]
Haskell
<lang haskell>permutationsSomeIdentical :: [(a, Int)] -> a permutationsSomeIdentical [] = [[]] permutationsSomeIdentical xs =
[ x : ys | (x, xs_) <- select xs , ys <- permutationsSomeIdentical xs_ ] where select [] = [] select ((x, n):xs) = (x, xs_) : [ (y, (x, n) : cs) | (y, cs) <- select xs ] where xs_ | 1 == n = xs | otherwise = (x, n - 1) : xs
main :: IO () main = do
print $ permutationsSomeIdentical [(1, 2), (2, 1)] print $ permutationsSomeIdentical [(1, 2), (2, 3), (3, 1)] print $ permutationsSomeIdentical [('A', 2), ('B', 3), ('C', 1)]</lang>
- Output:
[[1,1,2],[1,2,1],[2,1,1]] [[1,1,2,2,2,3],[1,1,2,2,3,2],[1,1,2,3,2,2],[1,1,3,2,2,2],[1,2,1,2,2,3],[1,2,1,2,3,2],[1,2,1,3,2,2],[1,2,2,1,2,3],[1,2,2,1,3,2],[1,2,2,2,1,3],[1,2,2,2,3,1],[1,2,2,3,1,2],[1,2,2,3,2,1],[1,2,3,1,2,2],[1,2,3,2,1,2],[1,2,3,2,2,1],[1,3,1,2,2,2],[1,3,2,1,2,2],[1,3,2,2,1,2],[1,3,2,2,2,1],[2,1,1,2,2,3],[2,1,1,2,3,2],[2,1,1,3,2,2],[2,1,2,1,2,3],[2,1,2,1,3,2],[2,1,2,2,1,3],[2,1,2,2,3,1],[2,1,2,3,1,2],[2,1,2,3,2,1],[2,1,3,1,2,2],[2,1,3,2,1,2],[2,1,3,2,2,1],[2,2,1,1,2,3],[2,2,1,1,3,2],[2,2,1,2,1,3],[2,2,1,2,3,1],[2,2,1,3,1,2],[2,2,1,3,2,1],[2,2,2,1,1,3],[2,2,2,1,3,1],[2,2,2,3,1,1],[2,2,3,1,1,2],[2,2,3,1,2,1],[2,2,3,2,1,1],[2,3,1,1,2,2],[2,3,1,2,1,2],[2,3,1,2,2,1],[2,3,2,1,1,2],[2,3,2,1,2,1],[2,3,2,2,1,1],[3,1,1,2,2,2],[3,1,2,1,2,2],[3,1,2,2,1,2],[3,1,2,2,2,1],[3,2,1,1,2,2],[3,2,1,2,1,2],[3,2,1,2,2,1],[3,2,2,1,1,2],[3,2,2,1,2,1],[3,2,2,2,1,1]] ["AABBBC","AABBCB","AABCBB","AACBBB","ABABBC","ABABCB","ABACBB","ABBABC","ABBACB","ABBBAC","ABBBCA","ABBCAB","ABBCBA","ABCABB","ABCBAB","ABCBBA","ACABBB","ACBABB","ACBBAB","ACBBBA","BAABBC","BAABCB","BAACBB","BABABC","BABACB","BABBAC","BABBCA","BABCAB","BABCBA","BACABB","BACBAB","BACBBA","BBAABC","BBAACB","BBABAC","BBABCA","BBACAB","BBACBA","BBBAAC","BBBACA","BBBCAA","BBCAAB","BBCABA","BBCBAA","BCAABB","BCABAB","BCABBA","BCBAAB","BCBABA","BCBBAA","CAABBB","CABABB","CABBAB","CABBBA","CBAABB","CBABAB","CBABBA","CBBAAB","CBBABA","CBBBAA"]
jq
Works with gojq, the Go implementation of jq
First, we present the filters for generating the distinct permutations, and then some functions for pretty printing a solution for the task.
<lang jq># Given an array of counts of the nonnegative integers, produce an array reflecting the multiplicities:
- e.g. [3,1] => [0,0,0,1]
def items:
. as $in | reduce range(0;length) as $i ([]; . + [range(0;$in[$i])|$i]);
- distinct permutations of the input array, via insertion
def distinct_permutations:
# Given an array as input, generate a stream by inserting $x at different positions to the left def insert($x): ((index([$x]) // length) + 1) as $ix | range(0; $ix) as $pos | .[0:$pos] + [$x] + .[$pos:];
if length <= 1 then . else .[0] as $first | .[1:] | distinct_permutations | insert($first) end;</lang>
For pretty-printing the results: <lang jq># Input: an array
- Output: a stream of arrays
def nwise($n):
def w: if length <= $n then . else .[:$n], (.[$n:]|w) end; w;
def to_table:
nwise(10) | join(" ");</lang>
The task: <lang jq>[2,3,1] | items | [distinct_permutations | join("")] | to_table</lang>
- Output:
001112 010112 100112 011012 101012 110012 011102 101102 110102 111002 011120 101120 110120 111020 111200 001121 010121 100121 011021 101021 110021 011201 101201 110201 112001 011210 101210 110210 112010 112100 001211 010211 100211 012011 102011 120011 012101 102101 120101 121001 012110 102110 120110 121010 121100 002111 020111 200111 021011 201011 210011 021101 201101 210101 211001 021110 201110 210110 211010 211100
Julia
With the Combinatorics package, create all permutations and filter out the duplicates
<lang julia>using Combinatorics
catlist(spec) = mapreduce(i -> repeat([i], spec[i]), vcat, 1:length(spec))
alphastringfromintvector(v) = String([Char(Int('A') + i - 1) for i in v])
function testpermwithident(spec)
println("\nTesting $spec yielding:") for (i, p) in enumerate(unique(collect(permutations(catlist(spec))))) print(alphastringfromintvector(p), " ", i % 10 == 0 ? "\n" : "") end
end
testpermwithident([2, 3, 1])
</lang>
- Output:
Testing [2, 3, 1] yielding: AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA
Generate directly
<lang julia> alpha(s, v) = map(i -> s[i], v)
function distinctPerms(spec)
function perm(elements) if length(elements) == 1 deepcopy(elements) else [pushfirst!(p, elements[k][1]) for k in 1:length(elements) for p in perm(filter(dups -> length(dups) != 0, [ if i == k dups[2:end] else dups end for (i, dups) in enumerate(elements)])) ] end end elements = [fill(x...) for x in enumerate(spec)] perm(elements)
end
println(map(p -> join(alpha("ABC", p), ""), distinctPerms([2, 3, 1]))) </lang>
- Output:
["AABBBC", "AABBCB", "AABCBB", "AACBBB", "ABABBC", "ABABCB", "ABACBB", "ABBABC", "ABBACB", "ABBBAC", "ABBBCA", "ABBCAB", "ABBCBA", "ABCABB", "ABCBAB", "ABCBBA", "ACABBB", "ACBABB", "ACBBAB", "ACBBBA", "BAABBC", "BAABCB", "BAACBB", "BABABC", "BABACB", "BABBAC", "BABBCA", "BABCAB", "BABCBA", "BACABB", "BACBAB", "BACBBA", "BBAABC", "BBAACB", "BBABAC", "BBABCA", "BBACAB", "BBACBA", "BBBAAC", "BBBACA", "BBBCAA", "BBCAAB", "BBCABA", "BBCBAA", "BCAABB", "BCABAB", "BCABBA", "BCBAAB", "BCBABA", "BCBBAA", "CAABBB", "CABABB", "CABBAB", "CABBBA", "CBAABB", "CBABAB", "CBABBA", "CBBAAB", "CBBABA", "CBBBAA"]
Mathematica /Wolfram Language
<lang Mathematica>ClearAll[PermsFromFrequencies] PermsFromFrequencies[l_List] := Module[{digs},
digs = Flatten[MapIndexed[ReverseApplied[ConstantArray], l]]; Permutations[digs] ]
PermsFromFrequencies[{2, 3, 1}] // Column</lang>
- Output:
{1,1,2,2,2,3} {1,1,2,2,3,2} {1,1,2,3,2,2} {1,1,3,2,2,2} {1,2,1,2,2,3} {1,2,1,2,3,2} {1,2,1,3,2,2} {1,2,2,1,2,3} {1,2,2,1,3,2} {1,2,2,2,1,3} {1,2,2,2,3,1} {1,2,2,3,1,2} {1,2,2,3,2,1} {1,2,3,1,2,2} {1,2,3,2,1,2} {1,2,3,2,2,1} {1,3,1,2,2,2} {1,3,2,1,2,2} {1,3,2,2,1,2} {1,3,2,2,2,1} {2,1,1,2,2,3} {2,1,1,2,3,2} {2,1,1,3,2,2} {2,1,2,1,2,3} {2,1,2,1,3,2} {2,1,2,2,1,3} {2,1,2,2,3,1} {2,1,2,3,1,2} {2,1,2,3,2,1} {2,1,3,1,2,2} {2,1,3,2,1,2} {2,1,3,2,2,1} {2,2,1,1,2,3} {2,2,1,1,3,2} {2,2,1,2,1,3} {2,2,1,2,3,1} {2,2,1,3,1,2} {2,2,1,3,2,1} {2,2,2,1,1,3} {2,2,2,1,3,1} {2,2,2,3,1,1} {2,2,3,1,1,2} {2,2,3,1,2,1} {2,2,3,2,1,1} {2,3,1,1,2,2} {2,3,1,2,1,2} {2,3,1,2,2,1} {2,3,2,1,1,2} {2,3,2,1,2,1} {2,3,2,2,1,1} {3,1,1,2,2,2} {3,1,2,1,2,2} {3,1,2,2,1,2} {3,1,2,2,2,1} {3,2,1,1,2,2} {3,2,1,2,1,2} {3,2,1,2,2,1} {3,2,2,1,1,2} {3,2,2,1,2,1} {3,2,2,2,1,1}
MiniZinc
<lang MiniZinc> %Permutations with some identical elements. Nigel Galloway, September 9th., 2019 include "count.mzn"; enum N={A,B,C}; array [1..6] of var N: G; constraint count(G,A,2) /\ count(G,C,1); output [show(G)] </lang>
- Output:
minizinc --all-solutions produces:
[C, B, B, B, A, A] ---------- [B, C, B, B, A, A] ---------- [B, B, C, B, A, A] ---------- [B, B, B, C, A, A] ---------- [C, B, B, A, A, B] ---------- [B, C, B, A, A, B] ---------- [B, B, C, A, A, B] ---------- [B, B, B, A, A, C] ---------- [C, B, A, B, A, B] ---------- [B, C, A, B, A, B] ---------- [B, B, A, C, A, B] ---------- [B, B, A, B, A, C] ---------- [C, A, B, B, A, B] ---------- [B, A, C, B, A, B] ---------- [B, A, B, C, A, B] ---------- [B, A, B, B, A, C] ---------- [A, C, B, B, A, B] ---------- [A, B, C, B, A, B] ---------- [A, B, B, C, A, B] ---------- [A, B, B, B, A, C] ---------- [C, B, B, A, B, A] ---------- [B, C, B, A, B, A] ---------- [B, B, C, A, B, A] ---------- [B, B, B, A, C, A] ---------- [C, B, A, A, B, B] ---------- [B, C, A, A, B, B] ---------- [B, B, A, A, C, B] ---------- [B, B, A, A, B, C] ---------- [C, A, B, A, B, B] ---------- [B, A, C, A, B, B] ---------- [B, A, B, A, C, B] ---------- [B, A, B, A, B, C] ---------- [A, C, B, A, B, B] ---------- [A, B, C, A, B, B] ---------- [A, B, B, A, C, B] ---------- [A, B, B, A, B, C] ---------- [C, B, A, B, B, A] ---------- [B, C, A, B, B, A] ---------- [B, B, A, C, B, A] ---------- [B, B, A, B, C, A] ---------- [C, A, A, B, B, B] ---------- [B, A, A, C, B, B] ---------- [B, A, A, B, C, B] ---------- [B, A, A, B, B, C] ---------- [A, C, A, B, B, B] ---------- [A, B, A, C, B, B] ---------- [A, B, A, B, C, B] ---------- [A, B, A, B, B, C] ---------- [C, A, B, B, B, A] ---------- [B, A, C, B, B, A] ---------- [B, A, B, C, B, A] ---------- [B, A, B, B, C, A] ---------- [A, C, B, B, B, A] ---------- [A, B, C, B, B, A] ---------- [A, B, B, C, B, A] ---------- [A, B, B, B, C, A] ---------- [A, A, C, B, B, B] ---------- [A, A, B, C, B, B] ---------- [A, A, B, B, C, B] ---------- [A, A, B, B, B, C] ---------- ==========
Nim
<lang Nim>import strutils
func shouldSwap(s: string; start, curr: int): bool =
for i in start..<curr: if s[i] == s[curr]: return false return true
func findPerms(s: string; index, n: int; res: var seq[string]) =
if index >= n: res.add s return var s = s for i in index..<n: if s.shouldSwap(index, i): swap s[index], s[i] findPerms(s, index+1, n, res) swap s[index], s[i]
func createSlice(nums: openArray[int]; charSet: string): string =
for i, n in nums: for _ in 1..n: result.add charSet[i]
when isMainModule:
var res1, res2, res3: seq[string] var nums = @[2, 1]
var s = createSlice(nums, "12") s.findPerms(0, s.len, res1) echo res1.join(" ") echo()
nums = @[2, 3, 1] s = createSlice(nums, "123") findPerms(s, 0, s.len, res2) for i, val in res2: stdout.write val, if (i + 1) mod 10 == 0: '\n' else: ' ' echo()
s = createSlice(nums, "ABC") findPerms(s, 0, s.len, res3) for i, val in res3: stdout.write val, if (i + 1) mod 10 == 0: '\n' else: ' '</lang>
- Output:
112 121 211 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122321 122312 123221 123212 123122 132221 132212 132122 131222 211223 211232 211322 212123 212132 212213 212231 212321 212312 213221 213212 213122 221123 221132 221213 221231 221321 221312 222113 222131 222311 223121 223112 223211 231221 231212 231122 232121 232112 232211 312221 312212 312122 311222 321221 321212 321122 322121 322112 322211 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA
Pascal
modified version to get permutations of different lenght.
Most time consuming is appending to the stringlist.So I limited that to 10000
One can use the string directly in EvaluatePerm.
<lang pascal>program PermWithRep;//of different length
{$IFDEF FPC}
{$mode Delphi} {$Optimization ON,All}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF} uses
sysutils,classes {for stringlist};
const
cTotalSum = 16; cMaxCardsOnDeck = cTotalSum; CMaxCardsUsed = cTotalSum;
type
tDeckIndex = 0..cMaxCardsOnDeck-1; tSequenceIndex = 0..CMaxCardsUsed; tDiffCardCount = Byte;//A..Z
tSetElem = packed record Elemcount : tDeckIndex; Elem : tDiffCardCount; end;
tRemSet = array [low(tDeckIndex)..High(tDeckIndex)] of tSetElem; tpRemSet = ^tRemSet; tRemainSet = array [low(tSequenceIndex)..High(tSequenceIndex)] of tRemSet; tCardSequence = array [low(tSequenceIndex)..High(tSequenceIndex)] of tDiffCardCount;
var {$ALIGN 32}
RemainSets : tRemainSet; CardString : AnsiString; CS : pchar; sl :TStringList; gblMaxCardsUsed, gblMaxUsedIdx, gblPermCount : NativeInt;
//***************************************************************************** procedure Out_SL(const sl:TStringlIst;colCnt:NativeInt); var
j,i : NativeInt;
begin
j := colCnt; For i := 0 to sl.count-1 do Begin write(sl[i],' '); dec(j); if j= 0 then Begin writeln; j := colCnt; end; end; if j <> colCnt then writeln;
end;
procedure SetClear(var ioSet:tRemSet); begin
fillChar(ioSet[0],SizeOf(ioSet),#0);
end;
procedure SetInit(var ioSet:tRemSet;const inSet:tRemSet); var
i,j,k,sum : integer;
begin
ioSet := inSet; sum := 0; k := 0; write('Initial set : '); For i := Low(ioSet) to High(ioSet) do Begin j := inSet[i].ElemCount; if j <> 0 then inc(k); sum += j; For j := j downto 1 do write(chr(inSet[i].Elem)); end; gblMaxCardsUsed := sum; gblMaxUsedIdx := k; writeln(' lenght: ',sum,' different elements : ',k);
end;
procedure EvaluatePerm; Begin
//append maximal 10000 strings if gblPermCount < 10000 then sl.append(CS);
end;
procedure Permute(depth,MaxCardsUsed:NativeInt); var
pSetElem : tpRemSet;//^tSetElem; i : NativeInt;
begin
i := 0; pSetElem := @RemainSets[depth]; repeat if pSetElem^[i].Elemcount > 0 then begin //take one of the same elements of the stack //insert in result here string CS[depth] := chr(pSetElem^[i].Elem); //done one permutation IF depth = MaxCardsUsed then begin inc(gblpermCount); EvaluatePerm; end else begin RemainSets[depth+1]:=RemainSets[depth]; //remove one element dec(RemainSets[depth+1][i].ElemCount); Permute(depth+1,MaxCardsUsed); end; end; //move on to the next Elem inc(i); until i >= gblMaxUsedIdx;
end;
procedure Permutate(MaxCardsUsed:NativeInt); Begin
gblpermCount := 0; if MaxCardsUsed > gblMaxCardsUsed then MaxCardsUsed := gblMaxCardsUsed;
if MaxCardsUsed>0 then Begin setlength(CardString,MaxCardsUsed); CS := @CardString[1]; permute(0,MaxCardsUsed-1) end;
end;
var
Manifolds : tRemSet; j :nativeInt;
Begin
SetClear(Manifolds); with Manifolds[0] do begin Elemcount := 2; Elem := Ord('A'); end; with Manifolds[1] do begin Elemcount := 3; Elem := Ord('B'); end; with Manifolds[2] do begin Elemcount := 1; Elem := Ord('C'); end;
try sl := TStringList.create;
SetInit(RemainSets[0], Manifolds); j := gblMaxCardsUsed; writeln('Count of elements: ',j); while j > 1 do begin sl.clear; Permutate(j); writeln('Length ',j:2,' Permutations ',gblpermCount:7); Out_SL(sl,80 DIV (Length(CS)+1)); writeln; dec(j); end; //change to 1,2,3 Manifolds[0].Elem := Ord('1'); Manifolds[1].Elem := Ord('2'); Manifolds[2].Elem := Ord('3');
SetInit(RemainSets[0], Manifolds); j := gblMaxCardsUsed; writeln('Count of elements: ',j); while j > 1 do begin sl.clear; Permutate(j); writeln('Length ',j:2,' Permutations ',gblpermCount:7); Out_SL(sl,80 DIV (Length(CS)+1)); writeln; dec(j); end;
//extend by 3 more elements with Manifolds[3] do begin Elemcount := 2; Elem := Ord('4'); end; with Manifolds[4] do begin Elemcount := 3; Elem := Ord('5'); end; with Manifolds[5] do begin Elemcount := 1; Elem := Ord('6'); end; SetInit(RemainSets[0], Manifolds); j := gblMaxCardsUsed; writeln('Count of elements: ',j); sl.clear; Permutate(j); writeln('Length ',j:2,' Permutations ',gblpermCount:7); //Out_SL(sl,80 DIV (Length(CS)+1)); writeln;
except writeln(' Stringlist Error '); end; sl.free;
end.</lang>
- Output:
Initial set : AABBBC lenght: 6 Count of elements: 6 Length 6 Permutations 60 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA Length 5 Permutations 60 AABBB AABBC AABCB AACBB ABABB ABABC ABACB ABBAB ABBAC ABBBA ABBBC ABBCA ABBCB ABCAB ABCBA ABCBB ACABB ACBAB ACBBA ACBBB BAABB BAABC BAACB BABAB BABAC BABBA BABBC BABCA BABCB BACAB BACBA BACBB BBAAB BBAAC BBABA BBABC BBACA BBACB BBBAA BBBAC BBBCA BBCAA BBCAB BBCBA BCAAB BCABA BCABB BCBAA BCBAB BCBBA CAABB CABAB CABBA CABBB CBAAB CBABA CBABB CBBAA CBBAB CBBBA Length 4 Permutations 38 AABB AABC AACB ABAB ABAC ABBA ABBB ABBC ABCA ABCB ACAB ACBA ACBB BAAB BAAC BABA BABB BABC BACA BACB BBAA BBAB BBAC BBBA BBBC BBCA BBCB BCAA BCAB BCBA BCBB CAAB CABA CABB CBAA CBAB CBBA CBBB Length 3 Permutations 19 AAB AAC ABA ABB ABC ACA ACB BAA BAB BAC BBA BBB BBC BCA BCB CAA CAB CBA CBB Length 2 Permutations 8 AA AB AC BA BB BC CA CB Initial set : 112223 lenght: 6 Count of elements: 6 Length 6 Permutations 60 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211 Length 5 Permutations 60 11222 11223 11232 11322 12122 12123 12132 12212 12213 12221 12223 12231 12232 12312 12321 12322 13122 13212 13221 13222 21122 21123 21132 21212 21213 21221 21223 21231 21232 21312 21321 21322 22112 22113 22121 22123 22131 22132 22211 22213 22231 22311 22312 22321 23112 23121 23122 23211 23212 23221 31122 31212 31221 31222 32112 32121 32122 32211 32212 32221 Length 4 Permutations 38 1122 1123 1132 1212 1213 1221 1222 1223 1231 1232 1312 1321 1322 2112 2113 2121 2122 2123 2131 2132 2211 2212 2213 2221 2223 2231 2232 2311 2312 2321 2322 3112 3121 3122 3211 3212 3221 3222 Length 3 Permutations 19 112 113 121 122 123 131 132 211 212 213 221 222 223 231 232 311 312 321 322 Length 2 Permutations 8 11 12 13 21 22 23 31 32 Initial set : 112223445556 lenght: 12 Count of elements: 12 Length 12 Permutations 3326400
Perl
<lang perl>use ntheory qw<formultiperm>;
formultiperm { print join(,@_) . ' ' } [<1 1 2>]; print "\n\n"; formultiperm { print join(,@_) . ' ' } [<1 1 2 2 2 3>]; print "\n\n"; formultiperm { print join(,@_) . ' ' } [split //,'AABBBC']; print "\n";</lang>
- Output:
112 121 211 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA
Phix
with javascript_semantics function shouldSwap(string s, integer start, curr) for i=start to curr-1 do if s[i] == s[curr] then return false end if end for return true end function function findPerms(string s, integer i=1, sequence res={}) if i>length(s) then res = append(res, s) else for j=i to length(s) do if shouldSwap(s, i, j) then integer {si, sj} = {s[i], s[j]} s[i] = sj s[j] = si res = findPerms(s, i+1, res) s[i] = si s[j] = sj end if end for end if return res end function function createSlice(sequence nums, string charSet) string chars = "" for i=1 to length(nums) do chars &= repeat(charSet[i],nums[i]) end for return chars end function pp(findPerms(createSlice({2,1}, "12"))) -- (=== findPerms("112")) pp(findPerms(createSlice({2,3,1}, "123"))) -- (=== findPerms("112223")) pp(findPerms(createSlice({2,3,1}, "ABC"))) -- (=== findPerms("AABBBC"))
- Output:
{`112`, `121`, `211`} {`112223`, `112232`, `112322`, `113222`, `121223`, `121232`, `121322`, `122123`, `122132`, `122213`, `122231`, `122321`, `122312`, `123221`, `123212`, `123122`, `132221`, `132212`, `132122`, `131222`, `211223`, `211232`, `211322`, `212123`, `212132`, `212213`, `212231`, `212321`, `212312`, `213221`, `213212`, `213122`, `221123`, `221132`, `221213`, `221231`, `221321`, `221312`, `222113`, `222131`, `222311`, `223121`, `223112`, `223211`, `231221`, `231212`, `231122`, `232121`, `232112`, `232211`, `312221`, `312212`, `312122`, `311222`, `321221`, `321212`, `321122`, `322121`, `322112`, `322211`} {`AABBBC`, `AABBCB`, `AABCBB`, `AACBBB`, `ABABBC`, `ABABCB`, `ABACBB`, `ABBABC`, `ABBACB`, `ABBBAC`, `ABBBCA`, `ABBCBA`, `ABBCAB`, `ABCBBA`, `ABCBAB`, `ABCABB`, `ACBBBA`, `ACBBAB`, `ACBABB`, `ACABBB`, `BAABBC`, `BAABCB`, `BAACBB`, `BABABC`, `BABACB`, `BABBAC`, `BABBCA`, `BABCBA`, `BABCAB`, `BACBBA`, `BACBAB`, `BACABB`, `BBAABC`, `BBAACB`, `BBABAC`, `BBABCA`, `BBACBA`, `BBACAB`, `BBBAAC`, `BBBACA`, `BBBCAA`, `BBCABA`, `BBCAAB`, `BBCBAA`, `BCABBA`, `BCABAB`, `BCAABB`, `BCBABA`, `BCBAAB`, `BCBBAA`, `CABBBA`, `CABBAB`, `CABABB`, `CAABBB`, `CBABBA`, `CBABAB`, `CBAABB`, `CBBABA`, `CBBAAB`, `CBBBAA`}
You can also use a builtin function to produce exactly the same output (createSlice omitted for clarity)
with javascript_semantics requires("1.0.2") pp(permutes("112")) pp(permutes("112223")) pp(permutes("AABBBC"))
Python
Set filters out unique permutations <lang python>
- Aamrun, 5th October 2021
from itertools import permutations
numList = [2,3,1]
baseList = []
for i in numList:
for j in range(0,i): baseList.append(i)
stringDict = {'A':2,'B':3,'C':1}
baseString=""
for i in stringDict:
for j in range(0,stringDict[i]): baseString+=i
print("Permutations for " + str(baseList) + " : ") [print(i) for i in set(permutations(baseList))]
print("Permutations for " + baseString + " : ") [print(i) for i in set(permutations(baseString))] </lang>
- Output:
Permutations for [2, 2, 3, 3, 3, 1] : (2, 1, 3, 2, 3, 3) (3, 2, 3, 3, 1, 2) (3, 2, 3, 1, 2, 3) (3, 3, 1, 2, 2, 3) (3, 3, 2, 2, 3, 1) (3, 1, 3, 3, 2, 2) (3, 1, 3, 2, 3, 2) (2, 3, 3, 3, 1, 2) (3, 3, 1, 3, 2, 2) (3, 1, 2, 3, 2, 3) (3, 2, 2, 1, 3, 3) (3, 2, 3, 2, 1, 3) (3, 1, 3, 2, 2, 3) (2, 3, 1, 3, 2, 3) (3, 2, 1, 2, 3, 3) (2, 3, 3, 2, 1, 3) (2, 3, 1, 3, 3, 2) (3, 3, 2, 3, 2, 1) (3, 3, 2, 2, 1, 3) (2, 2, 3, 3, 3, 1) (2, 3, 1, 2, 3, 3) (3, 3, 2, 3, 1, 2) (3, 3, 3, 2, 2, 1) (2, 1, 2, 3, 3, 3) (2, 3, 2, 3, 3, 1) (2, 1, 3, 3, 3, 2) (2, 2, 3, 3, 1, 3) (3, 1, 2, 2, 3, 3) (2, 3, 2, 1, 3, 3) (3, 2, 1, 3, 3, 2) (1, 3, 3, 3, 2, 2) (3, 3, 3, 2, 1, 2) (2, 3, 2, 3, 1, 3) (3, 2, 2, 3, 3, 1) (1, 3, 2, 2, 3, 3) (2, 1, 3, 3, 2, 3) (3, 2, 1, 3, 2, 3) (1, 3, 3, 2, 2, 3) (1, 3, 3, 2, 3, 2) (1, 2, 3, 3, 3, 2) (2, 3, 3, 1, 3, 2) (3, 3, 2, 1, 3, 2) (1, 2, 3, 2, 3, 3) (3, 3, 2, 1, 2, 3) (3, 2, 3, 3, 2, 1) (1, 3, 2, 3, 3, 2) (1, 2, 3, 3, 2, 3) (2, 3, 3, 1, 2, 3) (3, 2, 2, 3, 1, 3) (1, 3, 2, 3, 2, 3) (3, 1, 2, 3, 3, 2) (2, 3, 3, 2, 3, 1) (3, 3, 1, 2, 3, 2) (2, 2, 3, 1, 3, 3) (3, 3, 3, 1, 2, 2) (1, 2, 2, 3, 3, 3) (3, 2, 3, 1, 3, 2) (2, 2, 1, 3, 3, 3) (2, 3, 3, 3, 2, 1) (3, 2, 3, 2, 3, 1) Permutations for AABBBC : ('B', 'B', 'A', 'C', 'B', 'A') ('B', 'C', 'B', 'A', 'B', 'A') ('A', 'B', 'A', 'B', 'B', 'C') ('B', 'A', 'B', 'B', 'C', 'A') ('C', 'B', 'A', 'B', 'B', 'A') ('A', 'A', 'B', 'C', 'B', 'B') ('A', 'B', 'B', 'C', 'A', 'B') ('B', 'C', 'B', 'B', 'A', 'A') ('B', 'B', 'A', 'A', 'C', 'B') ('A', 'B', 'C', 'B', 'B', 'A') ('A', 'C', 'A', 'B', 'B', 'B') ('B', 'B', 'B', 'A', 'A', 'C') ('C', 'B', 'B', 'A', 'B', 'A') ('A', 'A', 'C', 'B', 'B', 'B') ('C', 'A', 'A', 'B', 'B', 'B') ('B', 'C', 'A', 'A', 'B', 'B') ('A', 'B', 'B', 'A', 'C', 'B') ('B', 'B', 'C', 'A', 'B', 'A') ('A', 'A', 'B', 'B', 'B', 'C') ('C', 'B', 'B', 'B', 'A', 'A') ('B', 'C', 'A', 'B', 'B', 'A') ('C', 'A', 'B', 'A', 'B', 'B') ('B', 'A', 'A', 'B', 'C', 'B') ('B', 'B', 'B', 'C', 'A', 'A') ('B', 'B', 'C', 'B', 'A', 'A') ('B', 'A', 'B', 'C', 'A', 'B') ('A', 'C', 'B', 'B', 'B', 'A') ('B', 'B', 'A', 'A', 'B', 'C') ('C', 'B', 'A', 'A', 'B', 'B') ('C', 'A', 'B', 'B', 'A', 'B') ('B', 'A', 'A', 'C', 'B', 'B') ('B', 'C', 'B', 'A', 'A', 'B') ('B', 'B', 'A', 'B', 'A', 'C') ('A', 'B', 'B', 'A', 'B', 'C') ('B', 'A', 'A', 'B', 'B', 'C') ('A', 'B', 'B', 'C', 'B', 'A') ('B', 'A', 'C', 'B', 'B', 'A') ('C', 'B', 'A', 'B', 'A', 'B') ('B', 'A', 'B', 'A', 'C', 'B') ('A', 'B', 'A', 'B', 'C', 'B') ('A', 'B', 'C', 'A', 'B', 'B') ('B', 'B', 'B', 'A', 'C', 'A') ('A', 'B', 'B', 'B', 'A', 'C') ('B', 'B', 'A', 'C', 'A', 'B') ('A', 'B', 'C', 'B', 'A', 'B') ('C', 'A', 'B', 'B', 'B', 'A') ('A', 'A', 'B', 'B', 'C', 'B') ('B', 'A', 'B', 'B', 'A', 'C') ('A', 'B', 'A', 'C', 'B', 'B') ('A', 'C', 'B', 'A', 'B', 'B') ('B', 'A', 'C', 'A', 'B', 'B') ('B', 'B', 'A', 'B', 'C', 'A') ('B', 'C', 'A', 'B', 'A', 'B') ('C', 'B', 'B', 'A', 'A', 'B') ('B', 'A', 'B', 'A', 'B', 'C') ('B', 'A', 'B', 'C', 'B', 'A') ('B', 'A', 'C', 'B', 'A', 'B') ('A', 'B', 'B', 'B', 'C', 'A') ('B', 'B', 'C', 'A', 'A', 'B') ('A', 'C', 'B', 'B', 'A', 'B')
Raku
(formerly Perl 6)
<lang perl6>sub permutations-with-some-identical-elements ( @elements, @reps = () ) {
with @elements { (@reps ?? flat $_ Zxx @reps !! flat .keys.map(*+1) Zxx .values).permutations».join.unique } }
for (<2 1>,), (<2 3 1>,), (<A B C>, <2 3 1>), (<🦋 ⚽ 🙄>, <2 2 1>) {
put permutations-with-some-identical-elements |$_; say ;
}</lang>
- Output:
112 121 211 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA 🦋🦋⚽⚽🙄 🦋🦋⚽🙄⚽ 🦋🦋🙄⚽⚽ 🦋⚽🦋⚽🙄 🦋⚽🦋🙄⚽ 🦋⚽⚽🦋🙄 🦋⚽⚽🙄🦋 🦋⚽🙄🦋⚽ 🦋⚽🙄⚽🦋 🦋🙄🦋⚽⚽ 🦋🙄⚽🦋⚽ 🦋🙄⚽⚽🦋 ⚽🦋🦋⚽🙄 ⚽🦋🦋🙄⚽ ⚽🦋⚽🦋🙄 ⚽🦋⚽🙄🦋 ⚽🦋🙄🦋⚽ ⚽🦋🙄⚽🦋 ⚽⚽🦋🦋🙄 ⚽⚽🦋🙄🦋 ⚽⚽🙄🦋🦋 ⚽🙄🦋🦋⚽ ⚽🙄🦋⚽🦋 ⚽🙄⚽🦋🦋 🙄🦋🦋⚽⚽ 🙄🦋⚽🦋⚽ 🙄🦋⚽⚽🦋 🙄⚽🦋🦋⚽ 🙄⚽🦋⚽🦋 🙄⚽⚽🦋🦋
REXX
shows permutation list
<lang rexx>/*REXX program computes and displays the permutations with some identical elements. */ parse arg g /*obtain optional arguments from the CL*/ if g= | g="," then g= 2 3 1 /*Not specified? Then use the defaults*/
- = words(g) /*obtain the number of source items. */
@= left('ABCDEFGHIJKLMNOPQRSTUVWXYZ', #) /*@: the (output) letters to be used.*/ LO= /*LO: the start of the sequence. */ HI= /*HI: " end " " " */
do i=1 for #; @.i= word(g, i) /*get number of characters for an arg. */ LO= LO || copies(i, @.i) /*build the LO number for the range. */ HI= copies(i, @.i) || HI /* " " HI " " " " */ end /*i*/
$= /*initialize the output string to null.*/
do j=LO to HI /*generate the enumerated output string*/ if verify(j, LO)\==0 then iterate /*An invalid digital string? Then skip*/ do k=1 for # /*parse string for correct # of digits.*/ if countstr(k, j)\==@.k then iterate j /*Incorrect number of digits? Skip. */ end /*k*/ $= $ j /*append digital string to the list. */ end /*j*/ /*stick a fork in it, we're all done. */
say 'number of permutations: ' words($) say say strip(translate($, @, left(123456789, #) ) ) /*display the translated string to term*/</lang>
- output when using the inputs of: 2 1
number of permutations: 3 AAB ABA BAA
- output when using the default inputs: 2 3 1
number of permutations: 60 AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA
only shows permutation count
If any of the arguments is negative, the list of the permutations is suppressed, only the permutation count is shown. <lang rexx>/*REXX program computes and displays the permutations with some identical elements. */ parse arg g /*obtain optional arguments from the CL*/ if g= | g="," then g= 2 3 1 /*Not specified? Then use the defaults*/
- = words(g) /*obtain the number of source items. */
@= left('ABCDEFGHIJKLMNOPQRSTUVWXYZ', #) /*@: the (output) letters to be used.*/ show= 1 /*if = 1, will show permutation list. */ sum= 0 LO= /*LO: the start of the sequence. */ HI= /*HI: " end " " " */
do i=1 for #; y= word(g, i) /*get number of characters for an arg. */ show= show & y>=0; a= abs(y) /*Is it negative? Then don't show list*/ sum= sum + a; @.i= a /*use the absolute value of an argument*/ LO= LO || copies(i, @.i) /*build the LO number for the range. */ HI= copies(i, @.i) || HI /* " " HI " " " " */ end /*i*/
$= /*initialize the output string to null.*/ numeric digits max(9, sum) /*ensure enough numeric decimal digits.*/
do j=LO to HI /*generate the enumerated output string*/ if verify(j, LO)\==0 then iterate /*An invalid digital string? Then skip*/ do k=1 for # /*parse string for correct # of digits.*/ if countstr(k, j)\==@.k then iterate j /*Incorrect number of digits? Skip. */ end /*k*/ $= $ j /*append digital string to the list. */ end /*j*/ /*stick a fork in it, we're all done. */
say 'number of permutations: ' words($) /*# perms with some identical elements.*/ say if show then say strip(translate($, @, left(123456789, #) ) ) /*display translated str.*/</lang>
- output when using the inputs of: -2 3 1 1 1
number of permutations: 3360
- output when using the inputs of: -1 2 1 2 1
number of permutations: 1260
- output when using the inputs of: -1 2 3 2 1
number of permutations: 15120
- output when using the inputs of: -1 1 5 2
number of permutations: 1512
- output when using the inputs of: -1 1 1 1 6
number of permutations: 5040
Sidef
Simple implementation, by filtering out the duplicated permutations. <lang ruby>func permutations_with_some_identical_elements (reps) {
reps.map_kv {|k,v| v.of(k+1)... }.permutations.uniq
}
say permutations_with_some_identical_elements([2,1]).map{.join}.join(' ') say permutations_with_some_identical_elements([2,3,1]).map{.join}.join(' ')</lang>
- Output:
112 121 211 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211
More efficient approach, by generating the permutations without duplicates: <lang ruby>func next_uniq_perm (array) {
var k = array.end return ([], false) if (k < 0) var i = k-1
while ((i >= 0) && (array[i] >= array[i+1])) { --i }
return (array.flip, false) if (i == -1)
if (array[i+1] > array[k]) { array = [array.slice(0, i)..., array.slice(i+1, k).flip...] }
var j = i+1 while (array[i] >= array[j]) { j++ }
array.clone! array.swap(i,j)
return (array, true)
}
func unique_permutations(array, reps=[]) {
var perm = (reps ? reps : array).map_kv {|k,v| v.of { reps ? array[k] : (k+1) }... } var perms = [perm] loop { (perm, var more) = next_uniq_perm(perm) break if !more perms << perm } return perms
}
for a,b in ([[[2,1]], 2,3,1, [%w(A B C), [2,3,1]]]) {
say "\nPermutations with array = #{a}#{b ? \" and reps = #{b}\" : }:" say unique_permutations(a,b).map{.join}.join(' ')
}</lang>
- Output:
Permutations with array = [2, 1]: 112 121 211 Permutations with array = [2, 3, 1]: 112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211 Permutations with array = ["A", "B", "C"] and reps = [2, 3, 1]: AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA
Tailspin
Creates lots of new arrays, which might be wasteful. <lang tailspin> templates distinctPerms
templates perms when <[](1)> do $(1) ! otherwise def elements: $; 1..$::length -> \( def k: $; def tail: $elements -> \[i]( when <?($i <=$k>)> do $ -> \(<[](2..)> $(2..last)!\) ! otherwise $! \); $tail -> perms -> [$elements($k;1), $...] ! \) ! end perms $ -> \[i]([1..$ -> $i] !\) -> perms !
end distinctPerms
def alpha: ['ABC'...]; [[2,3,1] -> distinctPerms -> '$alpha($)...;' ] -> !OUT::write </lang>
- Output:
[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCAB, ABBCBA, ABCABB, ABCBAB, ABCBBA, ACABBB, ACBABB, ACBBAB, ACBBBA, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCAB, BABCBA, BACABB, BACBAB, BACBBA, BBAABC, BBAACB, BBABAC, BBABCA, BBACAB, BBACBA, BBBAAC, BBBACA, BBBCAA, BBCAAB, BBCABA, BBCBAA, BCAABB, BCABAB, BCABBA, BCBAAB, BCBABA, BCBBAA, CAABBB, CABABB, CABBAB, CABBBA, CBAABB, CBABAB, CBABBA, CBBAAB, CBBABA, CBBBAA]
Work in place (slightly modified from the Go solution to preserve lexical order)
<lang tailspin> templates distinctPerms
templates shouldSwap&{start:} when <?($@distinctPerms($start..~$) <~[<=$@distinctPerms($)>]>)> do $! end shouldSwap templates findPerms when <$@distinctPerms::length..> do $@distinctPerms ! otherwise def index: $; $index..$@distinctPerms::length -> shouldSwap&{start: $index} -> \( @findPerms: $; @distinctPerms([$, $index]): $@distinctPerms([$index, $])...; $index + 1 -> findPerms ! \) ! @distinctPerms([$@findPerms, $index..~$@findPerms]): $@distinctPerms($index..$@findPerms)...; end findPerms @: $ -> \[i](1..$ -> $i !\); 1 -> findPerms !
end distinctPerms
def alpha: ['ABC'...]; [[2,3,1] -> distinctPerms -> '$alpha($)...;' ] -> !OUT::write </lang>
- Output:
[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCAB, ABBCBA, ABCABB, ABCBAB, ABCBBA, ACABBB, ACBABB, ACBBAB, ACBBBA, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCAB, BABCBA, BACABB, BACBAB, BACBBA, BBAABC, BBAACB, BBABAC, BBABCA, BBACAB, BBACBA, BBBAAC, BBBACA, BBBCAA, BBCAAB, BBCABA, BBCBAA, BCAABB, BCABAB, BCABBA, BCBAAB, BCBABA, BCBBAA, CAABBB, CABABB, CABBAB, CABBBA, CBAABB, CBABAB, CBABBA, CBBAAB, CBBABA, CBBBAA]
Wren
<lang ecmascript>import "./perm" for Perm
var createList = Fn.new { |nums, charSet|
var chars = [] for (i in 0...nums.count) { for (j in 0...nums[i]) chars.add(charSet[i]) } return chars
}
var nums = [2, 1] var a = createList.call(nums, "12") System.print(Perm.listDistinct(a).map { |p| p.join() }.toList) System.print()
nums = [2, 3, 1] a = createList.call(nums, "123") System.print(Perm.listDistinct(a).map { |p| p.join() }.toList) System.print()
a = createList.call(nums, "ABC") System.print(Perm.listDistinct(a).map { |p| p.join() }.toList)</lang>
- Output:
[112, 121, 211] [112223, 112232, 112322, 113222, 121223, 121232, 121322, 122123, 122132, 122213, 122231, 122321, 122312, 123221, 123212, 123122, 132221, 132212, 132122, 131222, 211223, 211232, 211322, 212123, 212132, 212213, 212231, 212321, 212312, 213221, 213212, 213122, 221123, 221132, 221213, 221231, 221321, 221312, 222113, 222131, 222311, 223121, 223112, 223211, 231221, 231212, 231122, 232121, 232112, 232211, 312221, 312212, 312122, 311222, 321221, 321212, 321122, 322121, 322112, 322211] [AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCBA, ABBCAB, ABCBBA, ABCBAB, ABCABB, ACBBBA, ACBBAB, ACBABB, ACABBB, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCBA, BABCAB, BACBBA, BACBAB, BACABB, BBAABC, BBAACB, BBABAC, BBABCA, BBACBA, BBACAB, BBBAAC, BBBACA, BBBCAA, BBCABA, BBCAAB, BBCBAA, BCABBA, BCABAB, BCAABB, BCBABA, BCBAAB, BCBBAA, CABBBA, CABBAB, CABABB, CAABBB, CBABBA, CBABAB, CBAABB, CBBABA, CBBAAB, CBBBAA]
zkl
<lang zkl> // eg ( (2,3,1), "ABC" ) == permute "A","A","B","B","B","C" and remove duplicates
// --> ( "AABBBC", "AABBCB" .. ) // this gets ugly lots sooner than it should
fcn permutationsWithSomeIdenticalElements(ns,abcs){
ns.zipWith(fcn(n,c){ List.createLong(n,c) },abcs).flatten() : # (3,"B")-->("B","B,"B") Utils.Helpers.permute(_) : Utils.Helpers.listUnique(_) .apply("concat") // ("A","A","B","B","B","C")-->"AABBCB"
}</lang> <lang zkl>permutationsWithSomeIdenticalElements(T(2,1),"123").println(); permutationsWithSomeIdenticalElements(T(2,1),L("\u2192","\u2191")).concat(" ").println();
z:=permutationsWithSomeIdenticalElements(T(2,3,1),"ABC"); println(z.len()); z.pump(Void,T(Void.Read,9,False), // print rows of ten items fcn{ vm.arglist.concat(" ").println() });</lang>
- Output:
L("112","121","211") →→↑ →↑→ ↑→→ 60 AABBBC AABBCB AABCBB AACBBB ACABBB CAABBB CABABB ACBABB ABCABB ABACBB ABABCB ABABBC BAABBC BAABCB BAACBB BACABB BCAABB CBAABB ABBABC ABBACB ABBCAB ABCBAB ACBBAB CABBAB BABABC BABACB BABCAB BACBAB BCABAB CBABAB CBBAAB BCBAAB BBCAAB BBACAB BBAACB BBAABC CBABBA BCABBA BACBBA BABCBA BABBCA BABBAC CBBABA BCBABA BBCABA BBACBA BBABCA BBABAC ABBBAC ABBBCA ABBCBA ABCBBA ACBBBA CABBBA BBBAAC BBBACA BBBCAA BBCBAA BCBBAA CBBBAA