Superpermutation minimisation

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Revision as of 03:09, 9 July 2021 by Alextretyak (talk | contribs) (Added 11l)
Task
Superpermutation minimisation
You are encouraged to solve this task according to the task description, using any language you may know.

A superpermutation of N different characters is a string consisting of an arrangement of multiple copies of those N different characters in which every permutation of those characters can be found as a substring.

For example, representing the characters as A..Z, using N=2 we choose to use the first two characters 'AB'.
The permutations of 'AB' are the two, (i.e. two-factorial), strings: 'AB' and 'BA'.

A too obvious method of generating a superpermutation is to just join all the permutations together forming 'ABBA'.

A little thought will produce the shorter (in fact the shortest) superpermutation of 'ABA' - it contains 'AB' at the beginning and contains 'BA' from the middle to the end.

The "too obvious" method of creation generates a string of length N!*N. Using this as a yardstick, the task is to investigate other methods of generating superpermutations of N from 1-to-7 characters, that never generate larger superpermutations.

Show descriptions and comparisons of algorithms used here, and select the "Best" algorithm as being the one generating shorter superpermutations.

The problem of generating the shortest superpermutation for each N might be NP complete, although the minimal strings for small values of N have been found by brute -force searches.


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences


Reference



11l

Translation of: Kotlin

<lang 11l>-V MAX = 12

[Char] sp V count = [0] * MAX V pos = 0

F factSum(n) 

  V s = 0
  V x = 0
  V f = 1
  L x < n
     f *= ++x
     s += f
  R s

F r(n) 

  I n == 0
     R 0B
  V c = :sp[:pos - n]
  I --:count[n] == 0
     :count[n] = n
     I !r(n - 1)
        R 0B
  :sp[:pos++] = c
  R 1B

F superPerm(n) 

  :pos = n
  V len = factSum(n)
  I len > 0
     :sp = [Char("\0")] * len
  L(i) 0 .. n
     :count[i] = i
  L(i) 1 .. n
     :sp[i - 1] = Char(code' ‘0’.code + i)
  L r(n) {}

L(n) 0 .< MAX 

  superPerm(n)
  print(‘superPerm(#2) len = #.’.format(n, sp.len))</lang>
Output:
superPerm( 0) len = 0
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 33
superPerm( 5) len = 153
superPerm( 6) len = 873
superPerm( 7) len = 5913
superPerm( 8) len = 46233
superPerm( 9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

AWK

<lang AWK>

  1. syntax: GAWK -f SUPERPERMUTATION_MINIMISATION.AWK
  2. converted from C

BEGIN { 

   arr[0] # prevents fatal: attempt to use scalar 'arr' as an array
   limit = 11
   for (n=0; n<=limit; n++) {
     leng = super_perm(n)
     printf("%2d %d ",n,leng)
  1. for (i=0; i<length(arr); i++) { printf(arr[i]) } # un-comment to see the string
     printf("\n")
   }
   exit(0)

} function fact_sum(n, f,s,x) { 

   f = 1
   s = x = 0
   for (;x<n;) {
     f *= ++x
     s += f
   }
   return(s)

} function super_perm(n, i,leng) { 

   delete arr
   pos = n
   leng = fact_sum(n)
   for (i=0; i<leng; i++) {
     arr[i] = ""
   }
   for (i=0; i<=n; i++) {
     cnt[i] = i
   }
   for (i=1; i<=n; i++) {
     arr[i-1] = i + "0"
   }
   while (r(n)) { }
   return(leng)

} function r(n, c) { 

   if (!n) { return(0) }
   c = arr[pos-n]
   if (!--cnt[n]) {
     cnt[n] = n
     if (!r(n-1)) { return(0) }
   }
   arr[pos++] = c
   return(1)

} </lang>

Output:
 0 0
 1 1
 2 3
 3 9
 4 33
 5 153
 6 873
 7 5913
 8 46233
 9 409113
10 4037913
11 43954713

C

Finding a string whose length follows OEIS A007489.   Complexity is the length of output string.   It is known to be not optimal. <lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>
  1. define MAX 12

char *super = 0; int pos, cnt[MAX];

// 1! + 2! + ... + n! int fact_sum(int n) { int s, x, f; for (s = 0, x = 0, f = 1; x < n; f *= ++x, s += f); return s; }

int r(int n) { if (!n) return 0;

char c = super[pos - n]; if (!--cnt[n]) { cnt[n] = n; if (!r(n-1)) return 0; } super[pos++] = c; return 1; }

void superperm(int n) { int i, len;

pos = n; len = fact_sum(n); super = realloc(super, len + 1); super[len] = '\0';

for (i = 0; i <= n; i++) cnt[i] = i; for (i = 1; i <= n; i++) super[i - 1] = i + '0';

while (r(n)); }

int main(void) { int n; for (n = 0; n < MAX; n++) { printf("superperm(%2d) ", n); superperm(n); printf("len = %d", (int)strlen(super)); // uncomment next line to see the string itself // printf(": %s", super); putchar('\n'); }

return 0; }</lang>

Output:
superperm( 0) len = 0
superperm( 1) len = 1
superperm( 2) len = 3
superperm( 3) len = 9
superperm( 4) len = 33
superperm( 5) len = 153
superperm( 6) len = 873
superperm( 7) len = 5913
superperm( 8) len = 46233
superperm( 9) len = 409113
superperm(10) len = 4037913
superperm(11) len = 43954713

C++

Translation of: Kotlin

<lang cpp>#include <array>

  1. include <iostream>
  2. include <vector>

constexpr int MAX = 12;

static std::vector<char> sp; static std::array<int, MAX> count; static int pos = 0;

int factSum(int n) { 

   int s = 0;
   int x = 0;
   int f = 1;
   while (x < n) {
       f *= ++x;
       s += f;
   }
   return s;

}

bool r(int n) { 

   if (n == 0) {
       return false;
   }
   char c = sp[pos - n];
   if (--count[n] == 0) {
       count[n] = n;
       if (!r(n - 1)) {
           return false;
       }
   }
   sp[pos++] = c;
   return true;

}

void superPerm(int n) { 

   pos = n;
   int len = factSum(n);
   if (len > 0) {
       sp.resize(len);
   }
   for (size_t i = 0; i <= n; i++) {
       count[i] = i;
   }
   for (size_t i = 1; i <= n; i++) {
       sp[i - 1] = '0' + i;
   }
   while (r(n)) {}

}

int main() { 

   for (size_t n = 0; n < MAX; n++) {
       superPerm(n);
       std::cout << "superPerm(" << n << ") len = " << sp.size() << '\n';
   }

   return 0;

}</lang>

Output:
superPerm(0) len = 0
superPerm(1) len = 1
superPerm(2) len = 3
superPerm(3) len = 9
superPerm(4) len = 33
superPerm(5) len = 153
superPerm(6) len = 873
superPerm(7) len = 5913
superPerm(8) len = 46233
superPerm(9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

D

The greedy algorithm from the Python entry. This is a little more complex than the Python code because it uses some helper arrays to avoid some allocations inside the loops, to increase performance. <lang d>import std.stdio, std.ascii, std.algorithm, core.memory, permutations2;

/** Uses greedy algorithm of adding another char (or two, or three, ...) until an unseen perm is formed in the last n chars. */ string superpermutation(in uint n) nothrow in { 

   assert(n > 0 && n < uppercase.length);

} out(result) { 

   // It's a superpermutation.
   assert(uppercase[0 .. n].dup.permutations.all!(p => result.canFind(p)));

} body { 

   string result = uppercase[0 .. n];

   bool[const char[]] toFind;
   GC.disable;
   foreach (const perm; result.dup.permutations)
       toFind[perm] = true;
   GC.enable;
   toFind.remove(result);

   auto trialPerm = new char[n];
   auto auxAdd = new char[n];

   while (toFind.length) {
       MIDDLE: foreach (immutable skip; 1 .. n) {
           auxAdd[0 .. skip] = result[$ - n .. $ - n + skip];
           foreach (const trialAdd; auxAdd[0 .. skip].permutations!false) {
               trialPerm[0 .. n - skip] = result[$ + skip - n .. $];
               trialPerm[n - skip .. $] = trialAdd[];
               if (trialPerm in toFind) {
                   result ~= trialAdd;
                   toFind.remove(trialPerm);
                   break MIDDLE;
               }
           }
       }
   }

   return result;

}

void main() { 

   foreach (immutable n; 1 .. 8)
       n.superpermutation.length.writeln;

}</lang>

Output:
1
3
9
35
164
932
6247

Using the ldc2 compiler with n=10, it finds the result string of length 4_235_533 in less than 9 seconds.

Faster Version

Translation of: C

From the C version with some improvements. <lang d>import std.stdio, std.range, std.algorithm, std.ascii;

enum uint nMax = 12;

__gshared char[] superperm; __gshared uint pos; __gshared uint[nMax] count;

/// factSum(n) = 1! + 2! + ... + n! uint factSum(in uint n) pure nothrow @nogc @safe { 

   return iota(1, n + 1).map!(m => reduce!q{ a * b }(1u, iota(1, m + 1))).sum;

}

bool r(in uint n) nothrow @nogc { 

   if (!n)
       return false;

   immutable c = superperm[pos - n];
   if (!--count[n]) {
       count[n] = n;
       if (!r(n - 1))
           return false;
   }
   superperm[pos++] = c;
   return true;

}

void superPerm(in uint n) nothrow { 

   static immutable chars = digits ~ uppercase;
   static assert(chars.length >= nMax);
   pos = n;
   superperm.length = factSum(n);

   foreach (immutable i; 0 .. n + 1)
       count[i] = i;
   foreach (immutable i; 1 .. n + 1)
       superperm[i - 1] = chars[i];

   while (r(n)) {}

}

void main() { 

   foreach (immutable n; 0 .. nMax) {
       superPerm(n);
       writef("superPerm(%2d) len = %d", n, superperm.length);
       // Use -version=doPrint to see the string itself.
       version (doPrint) write(": ", superperm);
       writeln;
   }

}</lang>

Output:
superPerm( 0) len = 0
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 33
superPerm( 5) len = 153
superPerm( 6) len = 873
superPerm( 7) len = 5913
superPerm( 8) len = 46233
superPerm( 9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

Run-time: about 0.40 seconds.

Delphi

Library: System.SysUtils
Translation of: C

<lang Delphi> program Superpermutation_minimisation;

{$APPTYPE CONSOLE}

uses 

 System.SysUtils;

const 

 Max = 12;

var 

 super: ansistring;
 pos: Integer;
 cnt: TArray<Integer>;

function factSum(n: Integer): Uint64; begin 

 var s: Uint64 := 0;
 var f := 1;
 var x := 0;

 while x < n do
 begin
   inc(x);
   f := f * x;
   inc(s, f);
 end;

 Result := s;

end;

function r(n: Integer): Boolean; begin 

 if n = 0 then
   exit(false);

 var c := super[pos - n];

 dec(cnt[n]);

 if cnt[n] = 0 then
 begin
   cnt[n] := n;
   if not r(n - 1) then
     exit(false);
 end;
 super[pos] := c;
 inc(pos);
 result := true;

end;

procedure SuperPerm(n: Integer); begin 

 var pos := n;
 var le: Uint64 := factSum(n);
 SetLength(super, le);

 for var i := 0 to n do
   cnt[i] := i;

 for var i := 1 to n do
   super[i] := ansichar(i + ord('0'));

 while r(n) do
   ;

end;

begin 

 SetLength(cnt, max);

 for var n := 0 to max - 1 do
 begin
   write('superperm(', n: 2, ') ');
   SuperPerm(n);
   writeln('len = ', length(super));
 end;
 {$IFNDEF UNIX}   readln; {$ENDIF}

end.</lang>

Elixir

Translation of: Ruby

<lang elixir>defmodule Superpermutation do 

 def minimisation(1), do: [1]
 def minimisation(n) do
   Enum.chunk(minimisation(n-1), n-1, 1)
   |> Enum.reduce({[],nil}, fn sub,{acc,last} ->
     if Enum.uniq(sub) == sub do
       i = if acc==[], do: 0, else: Enum.find_index(sub, &(&1==last)) + 1
       {acc ++ (Enum.drop(sub,i) ++ [n] ++ sub), List.last(sub)}
     else
       {acc, last}
     end
   end)
   |> elem(0)
 end

end

to_s = fn list -> Enum.map_join(list, &Integer.to_string(&1,16)) end Enum.each(1..8, fn n -> 

 result = Superpermutation.minimisation(n)
 :io.format "~3w: len =~8w : ", [n, length(result)]
 IO.puts if n<5, do: Enum.join(result),
               else: to_s.(Enum.take(result,20)) <> "...." <> to_s.(Enum.slice(result,-20..-1))

end)</lang>

Output:
  1: len =       1 : 1
  2: len =       3 : 121
  3: len =       9 : 123121321
  4: len =      33 : 123412314231243121342132413214321
  5: len =     153 : 12345123415234125341....14352143251432154321
  6: len =     873 : 12345612345162345126....62154326154321654321
  7: len =    5913 : 12345671234561723456....65432716543217654321
  8: len =   46233 : 12345678123456718234....43281765432187654321

FreeBASIC

<lang freebasic>' version 28-06-2018 ' compile with: fbc -s console

Function superpermsize(n As UInteger) As UInteger 

   Dim As UInteger x, y, sum, fac
   For x = 1 To n
       fac = 1
       For y = 1 To x
           fac *= y
       Next
       sum += fac
   Next

   Function = sum

End Function

Function superperm(n As UInteger) As String 

   If n = 1 Then Return "1"

   Dim As String sup_perm = "1", insert
   Dim As String p, q()
   Dim As UInteger a, b, i, l, x

   For x = 2 To n
       insert = IIf(x < 10, Str(x), Chr(x + 55))
       l = Len(sup_perm)
       If l > 1 Then l = Len(sup_perm) - x +2
       ReDim q(l)
       For i = 1 To l
           p = Mid(sup_perm, i, x -1)
           If x > 2 Then
           For a = 0 To Len(p) -2
               For b = a+1 To Len(p) -1
                   If p[a] = p[b] Then Continue For, For, For
               Next
           Next
           End If
           q(i) = p + insert + p
       Next
       sup_perm = q(1)
       For i = 2 To UBound(q)
           a = x -1
           Do
               If Right(sup_perm, a) = Left(q(i), a) Then
                   sup_perm += Mid(q(i), a +1)
                   Exit Do
               End If
               a -= 1
           Loop
       Next
   Next

   Function = sup_perm

End Function

' ------=< MAIN >=------

Dim As String superpermutation Dim As UInteger n

For n = 1 To 10 

   superpermutation = superperm(n)
   Print Using "### ######## ########   "; n; superpermsize(n); Len(superpermutation);
   If n < 5 Then
       Print superpermutation
   Else
       Print
   End If

Next

' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>

Output:
  1        1        1   1
  2        3        3   121
  3        9        9   123121321
  4       33       33   123412314231243121342132413214321
  5      153      153
  6      873      873
  7     5913     5913
  8    46233    46233
  9   409113   409113
 10  4037913  4037913

Go

Translation of: C

<lang go>package main

import "fmt"

const max = 12

var ( 

   super []byte
   pos   int
   cnt   [max]int

)

// 1! + 2! + ... + n! func factSum(n int) int { 

   s := 0
   for x, f := 0, 1; x < n; {
       x++
       f *= x
       s += f
   }
   return s

}

func r(n int) bool { 

   if n == 0 {
       return false
   }
   c := super[pos-n]
   cnt[n]--
   if cnt[n] == 0 {
       cnt[n] = n
       if !r(n - 1) {
           return false
       }
   }
   super[pos] = c
   pos++
   return true

}

func superperm(n int) { 

   pos = n
   le := factSum(n)
   super = make([]byte, le)
   for i := 0; i <= n; i++ {
       cnt[i] = i
   }
   for i := 1; i <= n; i++ {
       super[i-1] = byte(i) + '0'
   }

   for r(n) {
   }

}

func main() { 

   for n := 0; n < max; n++ {
       fmt.Printf("superperm(%2d) ", n)
       superperm(n)
       fmt.Printf("len = %d\n", len(super))
   }

}</lang>

Output:
superperm( 0) len = 0
superperm( 1) len = 1
superperm( 2) len = 3
superperm( 3) len = 9
superperm( 4) len = 33
superperm( 5) len = 153
superperm( 6) len = 873
superperm( 7) len = 5913
superperm( 8) len = 46233
superperm( 9) len = 409113
superperm(10) len = 4037913
superperm(11) len = 43954713

Groovy

Translation of: Java

<lang groovy>import static java.util.stream.IntStream.rangeClosed

class Superpermutation { 

   final static int nMax = 12

   static char[] superperm
   static int pos
   static int[] count = new int[nMax]

   static int factSum(int n) {
       return rangeClosed(1, n)
           .map({ m -> rangeClosed(1, m).reduce(1, { a, b -> a * b }) }).sum()
   }

   static boolean r(int n) {
       if (n == 0) {
           return false
       }

       char c = superperm[pos - n]
       if (--count[n] == 0) {
           count[n] = n
           if (!r(n - 1)) {
               return false
           }
       }
       superperm[pos++] = c
       return true
   }

   static void superPerm(int n) {
       String chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"

       pos = n
       superperm = new char[factSum(n)]

       for (int i = 0; i < n + 1; i++) {
           count[i] = i
       }
       for (int i = 1; i < n + 1; i++) {
           superperm[i - 1] = chars.charAt(i)
       }

       while (r(n)) {
       }
   }

   static void main(String[] args) {
       for (int n = 0; n < nMax; n++) {
           superPerm(n)
           printf("superPerm(%2d) len = %d", n, superperm.length)
           println()
       }
   }

}</lang>

Output:
superPerm( 0) len = 0
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 33
superPerm( 5) len = 153
superPerm( 6) len = 873
superPerm( 7) len = 5913
superPerm( 8) len = 46233
superPerm( 9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

J

If there's an 872 long superpermutation for a six letter alphabet, this is not optimal.

<lang J>approxmin=:3 :0 

 seqs=. y{~(A.&i.~ !)#y
 r=.{.seqs
 seqs=.}.seqs
 while.#seqs do.
   for_n. i.-#y do.
     tail=. (-n){. r
     b=. tail -:"1 n{."1 seqs
     if. 1 e.b do.
       j=. b i.1
       r=. r, n}.j{seqs
       seqs=. (<<<j) { seqs
       break.
     end.
   end.
 end.
 r

)</lang>

Some sequence lengths:

<lang J> (#, #@approxmin)@> (1+i.8) {.&.> <'abcdefghijk' 1 1 2 3 3 9 4 33 5 153 6 873 7 5913 8 46233</lang>

Java

Translation of C via D

Works with: Java version 8

<lang java>import static java.util.stream.IntStream.rangeClosed;

public class Test { 

   final static int nMax = 12;

   static char[] superperm;
   static int pos;
   static int[] count = new int[nMax];

   static int factSum(int n) {
       return rangeClosed(1, n)
               .map(m -> rangeClosed(1, m).reduce(1, (a, b) -> a * b)).sum();
   }

   static boolean r(int n) {
       if (n == 0)
           return false;

       char c = superperm[pos - n];
       if (--count[n] == 0) {
           count[n] = n;
           if (!r(n - 1))
               return false;
       }
       superperm[pos++] = c;
       return true;
   }

   static void superPerm(int n) {
       String chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

       pos = n;
       superperm = new char[factSum(n)];

       for (int i = 0; i < n + 1; i++)
           count[i] = i;
       for (int i = 1; i < n + 1; i++)
           superperm[i - 1] = chars.charAt(i);

       while (r(n)) {
       }
   }

   public static void main(String[] args) {
       for (int n = 0; n < nMax; n++) {
           superPerm(n);
           System.out.printf("superPerm(%2d) len = %d", n, superperm.length);
           System.out.println();
       }
   }

}</lang> 

superPerm( 0) len = 0
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 33
superPerm( 5) len = 153
superPerm( 6) len = 873
superPerm( 7) len = 5913
superPerm( 8) len = 46233
superPerm( 9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

Julia

Translation of: D

Runs in about 1/4 second. <lang julia>const nmax = 12

function r!(n, s, pos, count) 

   if n == 0
       return false
   end
   c = s[pos + 1 - n]
   count[n + 1] -= 1
   if count[n + 1] == 0
       count[n + 1] = n
       if r!(n - 1, s, pos, count) == 0
           return false
       end
   end
   s[pos + 1] = c
   pos += 1
   true

end

function superpermutation(n) 

   count = zeros(nmax)
   pos = n
   superperm = zeros(UInt8, n < 2 ? n : mapreduce(factorial, +, 1:n))
   for i in 0:n-1
       count[i + 1] = i
       superperm[i + 1] = Char(i + '0')
   end
   count[n + 1] = n
   while r!(n, superperm, pos, count) ; end
   superperm

end

function testsuper(N, verbose=false) 

   for i in 0:N-1
       s = superpermutation(i)
       println("Superperm($i) has length $(length(s)) ", (verbose ? String(s) : ""))
   end

end

testsuper(nmax)

</lang>

Output:
Superperm(0) has length 0
Superperm(1) has length 1
Superperm(2) has length 3
Superperm(3) has length 9
Superperm(4) has length 33
Superperm(5) has length 153
Superperm(6) has length 873
Superperm(7) has length 5913
Superperm(8) has length 46233
Superperm(9) has length 409113
Superperm(10) has length 4037913
Superperm(11) has length 43954713

Kotlin

Translation of: C

<lang scala>// version 1.1.2

const val MAX = 12

var sp = CharArray(0) val count = IntArray(MAX) var pos = 0

fun factSum(n: Int): Int { 

   var s = 0
   var x = 0
   var f = 1    
   while (x < n) {
       f *= ++x
       s += f
   }
   return s

}

fun r(n: Int): Boolean { 

   if (n == 0) return false
   val c = sp[pos - n]
   if (--count[n] == 0) {
       count[n] = n
       if (!r(n - 1)) return false
   }
   sp[pos++] = c
   return true

}

fun superPerm(n: Int) { 

   pos = n
   val len = factSum(n)
   if (len > 0) sp = CharArray(len) 
   for (i in 0..n) count[i] = i
   for (i in 1..n) sp[i - 1] = '0' + i 
   while (r(n)) {}

}

fun main(args: Array<String>) { 

   for (n in 0 until MAX) {
       superPerm(n)
       println("superPerm(${"%2d".format(n)}) len = ${sp.size}")       
   }

}</lang>

Output:
superPerm( 0) len = 0
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 33
superPerm( 5) len = 153
superPerm( 6) len = 873
superPerm( 7) len = 5913
superPerm( 8) len = 46233
superPerm( 9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

Mathematica / Wolfram Language

Greedy algorithm: <lang Mathematica>ClearAll[OverlapDistance, ConstructDistances] OverlapDistance[{s1_List, s2_List}] := OverlapDistance[s1, s2] OverlapDistance[s1_List, s2_List] := Module[{overlaprange, overlap, l}, 

 overlaprange = {Min[Length[s1], Length[s2]], 0};
 l = LengthWhile[Range[Sequence @@ overlaprange, -1], Take[s1, -#] =!= Take[s2, #] &];
 overlap = overlaprange1 - l;
 <|"Overlap" -> overlap, "Distance" -> Length[s2] - overlap|>
 ]

ConstructDistances[perms_List] := Module[{sel, OD, fullseq}, 

 OD = BlockMap[OverlapDistance, perms, 2, 1];
 fullseq = 
  Fold[Join[#1, Drop[#22, #21["Overlap"]]] &, 
   First[perms], {OD, Rest[perms]} // Transpose];
 fullseq
 ]

Dynamic[Length[perms]] Do[ 

n = i;
perms = Permutations[Range[n]];
{start, perms} = TakeDrop[perms, 1];
While[Length[perms] > 0,
 last = Last[start];
 dists = 
  Table[<|"Index" -> i, OverlapDistance[last, permsi]|>, {i, 
    Length[perms]}];
 sel = First[TakeSmallestBy[dists, #["Distance"] &, 1]];
 AppendTo[start, perms[[sel["Index"]]]];
 perms = Delete[perms, sel["Index"]];
 ];
Print[{n, Length@ConstructDistances[start]}]
,
{i, 1, 7}

]</lang>

Output:
{1,1}
{2,3}
{3,9}
{4,33}
{5,153}
{6,873}
{7,5913}

Nim

Translation of: Go

<lang nim>import strformat

const MAX = 12

var super: seq[char] = @[] var pos: int var cnt: array[MAX, int]

proc factSum(n: int): int = 

 var s, x = 0
 var f = 1
 while x < n:
   inc x
   f *= x
   inc s, f
 s

proc r(n: int): bool = 

 if n == 0:
   return false
 var c = super[pos - n]
 dec cnt[n]
 if cnt[n] == 0:
   cnt[n] = n
   if not r(n - 1):
     return false
 super[pos] = c
 inc pos
 true

proc superperm(n: int) = 

 pos = n
 var le = factSum(n)
 super.setLen(le)
 for i in 0..n:
   cnt[i] = i
 for i in 1..n:
   super[i-1] = char(i + ord('0'))
 while r(n):
   discard

for n in 0..<MAX: 

 write(stdout, fmt"superperm({n:2})")
 superperm(n)
 writeLine(stdout, fmt" len = {len(super)}")</lang>
Output:
superperm( 0) len = 0
superperm( 1) len = 1
superperm( 2) len = 3
superperm( 3) len = 9
superperm( 4) len = 33
superperm( 5) len = 153
superperm( 6) len = 873
superperm( 7) len = 5913
superperm( 8) len = 46233
superperm( 9) len = 409113
superperm(10) len = 4037913
superperm(11) len = 43954713

Objeck

Translation of: C

<lang objeck>class SuperPermutation { 

 @super : static : Char[];
 @pos : static : Int;
 @cnt : static : Int[];

 function : Main(args : String[]) ~ Nil {
   max := 12;
   @cnt := Int->New[max];
   @super := Char->New[0];

   for(n := 0; n < max; n += 1;) {
     "superperm({$n}) "->Print();
     SuperPerm(n);
     len := @super->Size() - 1;
     "len = {$len}"->PrintLine();
   };
 }

 function : native : FactSum(n : Int) ~ Int {
   s := 0; x := 0; f := 1;
   while(x < n) {
     f *= ++x; s += f;
   };
   return s;
 }

 function : native : R(n : Int) ~ Bool {
   if(n = 0) {
     return false;
    };

   c := @super[@pos - n];
   if(--@cnt[n] = 0) {
     @cnt[n] := n;
     if(<>R(n - 1)) {
       return false;
     };
   };
   @super[@pos++] := c;

   return true;
 }

 function : SuperPerm(n : Int) ~ Nil {
   @pos := n;
   len := FactSum(n);

   tmp := Char->New[len + 1];
   Runtime->Copy(tmp, 0, @super, 0, @super->Size());
   @super := tmp;

   for(i := 0; i <= n; i += 1;) {
     @cnt[i] := i;
   };

   for(i := 1; i <= n; i += 1;) {
     @super[i - 1] := i + '0';
   };
    
   do {
     r := R(n);
   }
   while(r);
 }

} </lang>

Output:
superperm(0) len = 0
superperm(1) len = 1
superperm(2) len = 3
superperm(3) len = 9
superperm(4) len = 33
superperm(5) len = 153
superperm(6) len = 873
superperm(7) len = 5913
superperm(8) len = 46233
superperm(9) len = 409113
superperm(10) len = 4037913
superperm(11) len = 43954713

Perl

This uses a naive method of just concatenating the new permutation to the end (or prepending to the front) if it is not already in the string. Adding to the end is similar to Python's s_perm1() function.

Library: ntheory

<lang perl>use ntheory qw/forperm/; for my $len (1..8) { 

 my($pre, $post, $t) = ("","");
 forperm {
   $t = join "",@_;
   $post .= $t      unless index($post ,$t) >= 0;
   $pre = $t . $pre unless index($pre, $t) >= 0;
 } $len;
 printf "%2d: %8d %8d\n", $len, length($pre), length($post);

}</lang>

Output:
 1:        1        1
 2:        4        4
 3:       12       15
 4:       48       64
 5:      240      325
 6:     1440     1956
 7:    10080    13699
 8:    80640   109600

The permutations are generated in lexicographic order, and it seems prepending them leads to smaller strings than adding to the end. These are still quite a bit larger than the heuristic methods.

Phix

Translation of: C

<lang Phix>constant nMax = 12

atom t0 = time() string superperm sequence count integer pos

function factSum(int n) 

   integer s = 0, f = 1
   for i=1 to n do
       f *= i
       s += f
   end for
   return s

end function

function r(int n) 

   if (n == 0) then return false end if
   integer c = superperm[pos-n+1]
   count[n] -= 1
   if count[n]=0 then
       count[n] = n
       if not r(n-1) then return false end if
   end if
   pos += 1
   superperm[pos] = c
   return true

end function

procedure superPerm(int n) 

   string chars = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[1..n]
   pos = n
   superperm = chars&repeat(' ',factSum(n)-n)
   count = tagset(n)
   while r(n) do end while
   if n=0 then
       if superperm!="" then ?9/0 end if
   elsif n<=9 then
       -- (I estimate it would take at least 5 days to validate 
       --  superPerm(12), feel free to try it on your own time)
       for i=1 to factorial(n) do
           if not match(permute(i,chars),superperm) then ?9/0 end if
       end for
   end if

end procedure

for n=0 to nMax do 

   superPerm(n)
   integer l = length(superperm)
   if l>40 then superperm[20..-20] = "..." end if
   string e = elapsed(time()-t0)
   printf(1,"superPerm(%2d) len = %d  %s (%s)\n", {n, l, superperm, e})

end for</lang>

Output:
superPerm( 0) len = 0   (0s)
superPerm( 1) len = 1  1 (0s)
superPerm( 2) len = 3  121 (0s)
superPerm( 3) len = 9  123121321 (0s)
superPerm( 4) len = 33  123412314231243121342132413214321 (0s)
superPerm( 5) len = 153  1234512341523412534...4352143251432154321 (0s)
superPerm( 6) len = 873  1234561234516234512...2154326154321654321 (0.0s)
superPerm( 7) len = 5913  1234567123456172345...5432716543217654321 (0.7s)
superPerm( 8) len = 46233  1234567812345671823...3281765432187654321 (0.7s)
superPerm( 9) len = 409113  1234567891234567819...9187654321987654321 (0.8s)
superPerm(10) len = 4037913  123456789A123456789...987654321A987654321 (1.2s)
superPerm(11) len = 43954713  123456789AB12345678...87654321BA987654321 (6.5s)
superPerm(12) len = 522956313  123456789ABC1234567...7654321CBA987654321 (1 minute and 09s)

Alternative

Finds the longest overlap, similar to Python's greedy s_perm0 but theoretically more efficient.
I also tried prefixing res with any longer overlap at the start, but it just made things worse.
Uses factSum() from above, and compares that with these results (which are always worse for >3). <lang Phix>procedure superPerm(int n) 

   atom t0 = time()
   string chars = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[1..n]
   integer f = factorial(n)
   sequence perms = repeat("",f)
   for i=1 to f do
       perms[i] = permute(i,chars)
   end for
   string res = perms[$]
   perms = perms[1..$-1]
   while length(perms) do
       integer best = 0, bi = length(perms)
       for i=1 to length(perms) do
           string pi = perms[i]
           integer m = length(res),
                   k = find(res[m],pi)
           for l=k to 1 by -1 do
               if res[m]!=pi[l] then
                   k = 0
                   exit
               end if
               m -= 1
           end for
           if k>best then
               best = k
               bi = i
           end if  
       end for
       if match(perms[bi],res) then
           ?9/0 -- (sanity check)
       else
           res &= perms[bi][best+1..$]
       end if
       perms[bi] = perms[$]
       perms = perms[1..$-1]
   end while
   integer lr = length(res)
   integer fsn = factSum(n)
   string op = {"<","=",">"}[compare(lr,fsn)+2]
   t0 = time()-t0
   string e = iff(t0>1?", "&elapsed(t0):"")
   printf(1,"superPerm(%d) len = %d (%s%d%s)\n",{n,lr,op,fsn,e})

end procedure

for n=1 to 7 do -- (note: 8 takes 65x longer than 7) 

   superPerm(n)

end for</lang>

Output:
superPerm(1) len = 1 (=1)
superPerm(2) len = 3 (=3)
superPerm(3) len = 9 (=9)
superPerm(4) len = 35 (>33)
superPerm(5) len = 162 (>153)
superPerm(6) len = 924 (>873)
superPerm(7) len = 6250 (>5913, 2.5s)
superPerm(8) len = 48703 (>46233, 2 minutes and 43s)

PureBasic

Translation of: C

<lang PureBasic>EnableExplicit

  1. MAX=10

Declare.i fact_sum(n.i) : Declare.i r(n.i) : Declare superperm(n.i) Global pos.i, Dim cnt.i(#MAX), Dim super.s{1}(fact_sum(#MAX))

If OpenConsole() ;- MAIN: Superpermutation_minimisation 

 Define.i n  
 For n=0 To #MAX
   superperm(n) : Print("superperm("+RSet(Str(n),2)+") len = "+LSet(Str(pos),10))   
   If n<=4      : Print(~"\t"+PeekS(@super(),pos)) : EndIf
   PrintN("")
 Next
 Input()

EndIf End ;- END: Superpermutation_minimisation

Procedure.i fact_sum(n.i) 

 Define.i s=0,f=1,x=0
 While x<n : x+1 : f*x : s+f : Wend
 ProcedureReturn s

EndProcedure

Procedure.i r(n.i) 

 If Not n             : ProcedureReturn 0 : EndIf
 Define c.s{1}=super(pos-n)
 cnt(n)-1
 If Not cnt(n)
   cnt(n)=n
   If Not r(n-1)      : ProcedureReturn 0 : EndIf
 EndIf
 super(pos)=c : pos+1 : ProcedureReturn 1

EndProcedure

Procedure superperm(n.i) 

 pos=n
 Define.i len=fact_sum(n),i  
 For i=0 To n : cnt(i)=i              : Next
 For i=1 To n : super(i-1)=Chr('0'+i) : Next
 While r(n)   : Wend

EndProcedure</lang>

Output:
superperm( 0) len = 0         	
superperm( 1) len = 1         	1
superperm( 2) len = 3         	121
superperm( 3) len = 9         	123121321
superperm( 4) len = 33        	123412314231243121342132413214321
superperm( 5) len = 153       
superperm( 6) len = 873       
superperm( 7) len = 5913      
superperm( 8) len = 46233     
superperm( 9) len = 409113    
superperm(10) len = 4037913

Python

<lang python>"Generate a short Superpermutation of n characters A... as a string using various algorithms."


from __future__ import print_function, division

from itertools import permutations from math import factorial import string import datetime import gc


MAXN = 7


def s_perm0(n): 

   """
   Uses greedy algorithm of adding another char (or two, or three, ...)
   until an unseen perm is formed in the last n chars
   """
   allchars = string.ascii_uppercase[:n]
   allperms = [.join(p) for p in permutations(allchars)]
   sp, tofind = allperms[0], set(allperms[1:])
   while tofind:
       for skip in range(1, n):
           for trial_add in (.join(p) for p in permutations(sp[-n:][:skip])):
               #print(sp, skip, trial_add)
               trial_perm = (sp + trial_add)[-n:]
               if trial_perm in tofind:
                   #print(sp, skip, trial_add)
                   sp += trial_add
                   tofind.discard(trial_perm)
                   trial_add = None    # Sentinel
                   break
           if trial_add is None:
               break
   assert all(perm in sp for perm in allperms) # Check it is a superpermutation
   return sp

def s_perm1(n): 

   """
   Uses algorithm of concatenating all perms in order if not already part
   of concatenation.
   """
   allchars = string.ascii_uppercase[:n]
   allperms = [.join(p) for p in sorted(permutations(allchars))]
   perms, sp = allperms[::], 
   while perms:
       nxt = perms.pop()
       if nxt not in sp:
           sp += nxt
   assert all(perm in sp for perm in allperms)
   return sp

def s_perm2(n): 

   """
   Uses algorithm of concatenating all perms in order first-last-nextfirst-
   nextlast... if not already part of concatenation.
   """
   allchars = string.ascii_uppercase[:n]
   allperms = [.join(p) for p in sorted(permutations(allchars))]
   perms, sp = allperms[::], 
   while perms:
       nxt = perms.pop(0)
       if nxt not in sp:
           sp += nxt
       if perms:
           nxt = perms.pop(-1)
           if nxt not in sp:
               sp += nxt
   assert all(perm in sp for perm in allperms)
   return sp

def _s_perm3(n, cmp): 

   """
   Uses algorithm of concatenating all perms in order first,
   next_with_LEASTorMOST_chars_in_same_position_as_last_n_chars, ...
   """
   allchars = string.ascii_uppercase[:n]
   allperms = [.join(p) for p in sorted(permutations(allchars))]
   perms, sp = allperms[::], 
   while perms:
       lastn = sp[-n:]
       nxt = cmp(perms,
                 key=lambda pm:
                   sum((ch1 == ch2) for ch1, ch2 in zip(pm, lastn)))
       perms.remove(nxt)
       if nxt not in sp:
           sp += nxt
   assert all(perm in sp for perm in allperms)
   return sp

def s_perm3_max(n): 

   """
   Uses algorithm of concatenating all perms in order first,
   next_with_MOST_chars_in_same_position_as_last_n_chars, ...
   """
   return _s_perm3(n, max)

def s_perm3_min(n): 

   """
   Uses algorithm of concatenating all perms in order first,
   next_with_LEAST_chars_in_same_position_as_last_n_chars, ...
   """
   return _s_perm3(n, min)


longest = [factorial(n) * n for n in range(MAXN + 1)] weight, runtime = {}, {} print(__doc__) for algo in [s_perm0, s_perm1, s_perm2, s_perm3_max, s_perm3_min]: 

   print('\n###\n### %s\n###' % algo.__name__)
   print(algo.__doc__)
   weight[algo.__name__], runtime[algo.__name__] = 1, datetime.timedelta(0)
   for n in range(1, MAXN + 1):
       gc.collect()
       gc.disable()
       t = datetime.datetime.now()
       sp = algo(n)
       t = datetime.datetime.now() - t
       gc.enable()
       runtime[algo.__name__] += t
       lensp = len(sp)
       wt = (lensp / longest[n]) ** 2
       print('  For N=%i: SP length %5i Max: %5i Weight: %5.2f'
             % (n, lensp, longest[n], wt))
       weight[algo.__name__] *= wt
   weight[algo.__name__] **= 1 / n  # Geometric mean
   weight[algo.__name__] = 1 / weight[algo.__name__]
   print('%*s Overall Weight: %5.2f in %.1f seconds.'
         % (29, , weight[algo.__name__], runtime[algo.__name__].total_seconds()))

print('\n###\n### Algorithms ordered by shortest superpermutations first\n###') print('\n'.join('%12s (%.3f)' % kv for kv in 

               sorted(weight.items(), key=lambda keyvalue: -keyvalue[1])))

print('\n###\n### Algorithms ordered by shortest runtime first\n###') print('\n'.join('%12s (%.3f)' % (k, v.total_seconds()) for k, v in 

               sorted(runtime.items(), key=lambda keyvalue: keyvalue[1])))

</lang>

Output:
Generate a short Superpermutation of n characters A... as a string using various algorithms.

###
### s_perm0
###

    Uses greedy algorithm of adding another char (or two, or three, ...)
    until an unseen perm is formed in the last n chars
    
  For N=1: SP length     1 Max:     1 Weight:  1.00
  For N=2: SP length     3 Max:     4 Weight:  0.56
  For N=3: SP length     9 Max:    18 Weight:  0.25
  For N=4: SP length    35 Max:    96 Weight:  0.13
  For N=5: SP length   164 Max:   600 Weight:  0.07
  For N=6: SP length   932 Max:  4320 Weight:  0.05
  For N=7: SP length  6247 Max: 35280 Weight:  0.03
                              Overall Weight:  6.50 in 0.1 seconds.

###
### s_perm1
###

    Uses algorithm of concatenating all perms in order if not already part
    of concatenation.
    
  For N=1: SP length     1 Max:     1 Weight:  1.00
  For N=2: SP length     4 Max:     4 Weight:  1.00
  For N=3: SP length    15 Max:    18 Weight:  0.69
  For N=4: SP length    64 Max:    96 Weight:  0.44
  For N=5: SP length   325 Max:   600 Weight:  0.29
  For N=6: SP length  1956 Max:  4320 Weight:  0.21
  For N=7: SP length 13699 Max: 35280 Weight:  0.15
                              Overall Weight:  2.32 in 0.1 seconds.

###
### s_perm2
###

    Uses algorithm of concatenating all perms in order first-last-nextfirst-
    nextlast... if not already part of concatenation.
    
  For N=1: SP length     1 Max:     1 Weight:  1.00
  For N=2: SP length     4 Max:     4 Weight:  1.00
  For N=3: SP length    15 Max:    18 Weight:  0.69
  For N=4: SP length    76 Max:    96 Weight:  0.63
  For N=5: SP length   420 Max:   600 Weight:  0.49
  For N=6: SP length  3258 Max:  4320 Weight:  0.57
  For N=7: SP length 24836 Max: 35280 Weight:  0.50
                              Overall Weight:  1.49 in 0.3 seconds.

###
### s_perm3_max
###

    Uses algorithm of concatenating all perms in order first,
    next_with_MOST_chars_in_same_position_as_last_n_chars, ...
    
  For N=1: SP length     1 Max:     1 Weight:  1.00
  For N=2: SP length     4 Max:     4 Weight:  1.00
  For N=3: SP length    15 Max:    18 Weight:  0.69
  For N=4: SP length    56 Max:    96 Weight:  0.34
  For N=5: SP length   250 Max:   600 Weight:  0.17
  For N=6: SP length  1482 Max:  4320 Weight:  0.12
  For N=7: SP length 10164 Max: 35280 Weight:  0.08
                              Overall Weight:  3.06 in 50.2 seconds.

###
### s_perm3_min
###

    Uses algorithm of concatenating all perms in order first,
    next_with_LEAST_chars_in_same_position_as_last_n_chars, ...
    
  For N=1: SP length     1 Max:     1 Weight:  1.00
  For N=2: SP length     4 Max:     4 Weight:  1.00
  For N=3: SP length    15 Max:    18 Weight:  0.69
  For N=4: SP length    88 Max:    96 Weight:  0.84
  For N=5: SP length   540 Max:   600 Weight:  0.81
  For N=6: SP length  3930 Max:  4320 Weight:  0.83
  For N=7: SP length 33117 Max: 35280 Weight:  0.88
                              Overall Weight:  1.16 in 49.8 seconds.

###
### Algorithms ordered by shortest superpermutations first
###
     s_perm0 (6.501)
 s_perm3_max (3.057)
     s_perm1 (2.316)
     s_perm2 (1.494)
 s_perm3_min (1.164)

###
### Algorithms ordered by shortest runtime first
###
     s_perm0 (0.099)
     s_perm1 (0.102)
     s_perm2 (0.347)
 s_perm3_min (49.764)
 s_perm3_max (50.192)

Alternative Version

Translation of: D

<lang python>from array import array from string import ascii_uppercase, digits from operator import mul

try: 

   import psyco
   psyco.full()

except: 

   pass

N_MAX = 12

  1. fact_sum(n) = 1! + 2! + ... + n!

def fact_sum(n): 

   return sum(reduce(mul, xrange(1, m + 1), 1) for m in xrange(1, n + 1))


def r(n, superperm, pos, count): 

   if not n:
       return False

   c = superperm[pos - n]
   count[n] -= 1
   if not count[n]:
       count[n] = n
       if not r(n - 1, superperm, pos, count):
           return False

   superperm[pos] = c
   pos += 1
   return True


def super_perm(n, superperm, pos, count, chars = digits + ascii_uppercase): 

   assert len(chars) >= N_MAX
   pos = n
   superperm += array("c", " ") * (fact_sum(n) - len(superperm))

   for i in xrange(n + 1):
       count[i] = i
   for i in xrange(1, n + 1):
       superperm[i - 1] = chars[i]

   while r(n, superperm, pos, count):
       pass


def main(): 

   superperm = array("c", "")
   pos = 0
   count = array("l", [0]) * N_MAX

   for n in xrange(N_MAX):
       super_perm(n, superperm, pos, count)
       print "Super perm(%2d) len = %d" % (n, len(superperm)),
       #print superperm.tostring(),
       print

main()</lang> It is four times slower than the D entry. The output is about the same as the D entry.

Racket

Translation of: Ruby

<lang racket>#lang racket/base (require racket/list racket/format)

(define (index-of1 x l) (for/first ((i (in-naturals 1)) (m (in-list l)) #:when (equal? m x)) i))

(define (sprprm n) 

 (define n-1 (- n 1))
 (define sp:n-1 (superperm n-1))
 (let loop ((subs (let loop ((sp sp:n-1) (i (- (length sp:n-1) n-1 -1)) (rv null))
                    (cond
                      [(zero? i) (reverse rv)]
                      [else
                       (define sub (take sp n-1))
                       (loop (cdr sp)
                             (- i 1)
                             (if (check-duplicates sub) rv (cons sub rv)))])))
            (ary null))
   (if (null? subs)
       ary
       (let ((sub (car subs)))
         (define i (if (null? ary) 0 (index-of1 (last ary) sub)))
         (loop (cdr subs) (append ary (drop sub i) (list n) sub))))))

(define superperm 

 (let ((hsh (make-hash (list (cons 1 (list 1))))))
   (lambda (n) (hash-ref! hsh n (lambda () (sprprm n))))))


(define (20..20 ary) 

 (if (< (length ary) 41) ary (append (take ary 20) (cons '.. (take-right ary 20)))))

(for* ((n (in-range 1 (add1 8))) (ary (in-value (superperm n)))) 

 (printf "~a: len = ~a : ~a~%" (~a n #:width 3) (~a (length ary) #:width 8) (20..20 ary)))</lang>
Output:
1  : len = 1        : (1)
2  : len = 3        : (1 2 1)
3  : len = 9        : (1 2 3 1 2 1 3 2 1)
4  : len = 33       : (1 2 3 4 1 2 3 1 4 2 3 1 2 4 3 1 2 1 3 4 2 1 3 2 4 1 3 2 1 4 3 2 1)
5  : len = 153      : (1 2 3 4 5 1 2 3 4 1 5 2 3 4 1 2 5 3 4 1 .. 1 4 3 5 2 1 4 3 2 5 1 4 3 2 1 5 4 3 2 1)
6  : len = 873      : (1 2 3 4 5 6 1 2 3 4 5 1 6 2 3 4 5 1 2 6 .. 6 2 1 5 4 3 2 6 1 5 4 3 2 1 6 5 4 3 2 1)
7  : len = 5913     : (1 2 3 4 5 6 7 1 2 3 4 5 6 1 7 2 3 4 5 6 .. 6 5 4 3 2 7 1 6 5 4 3 2 1 7 6 5 4 3 2 1)
8  : len = 46233    : (1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 1 8 2 3 4 .. 4 3 2 8 1 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1)

Raku

(formerly Perl 6)

Translation of: Perl

<lang perl>for 1..8 -> $len { 

 my $pre = my $post = my $t = ;
 for  ('a'..'z')[^$len].permutations -> @p {
    $t = @p.join();
    $post ~= $t        unless index($post, $t);
    $pre   = $t ~ $pre unless index($pre,  $t);
 }
 printf "%1d: %8d %8d\n", $len, $pre.chars, $post.chars;

}</lang>

Output:
1:        1        1
2:        4        4
3:       12       15
4:       48       64
5:      240      325
6:     1440     1956
7:    10080    13699
8:    80640   109600

REXX

version 1

This REXX version just does simple finds for the permutations. <lang rexx>/*REXX program attempts to find better minimizations for computing superpermutations.*/ parse arg cycles . /*obtain optional arguments from the CL*/ if cycles== | cycles=="," then cycles= 7 /*Not specified? Then use the default.*/ 

     do n=0  to  cycles
     #= 0;                           $.=        /*populate the first permutation.      */
             do pop=1  for n;        @.pop= d2x(pop);        $.0= $.0  ||  @.pop
             end  /*pop*/

             do  while aPerm(n, 0)
             if n\==0  then #= #+1;  $.#=
               do j=1  for n;        $.#= $.#  ||  @.j
               end   /*j*/
             end     /*while*/
     z= $.0
     nm= n-1
             do p=1  for #;      if $.j==          then iterate
                                 if pos($.p, z)\==0  then iterate
             parse  var   $.p    h  2  R  1  L  =(n)
             if  left(z, nm)==R  then do;    z= h  ||  z;    iterate;   end
             if right(z,  1)==h  then do;    z= z  ||  R;    iterate;   end
             z= z  ||  $.p
             end   /*p*/                        /* [↑]  more IFs could be added for opt*/

      L= commas( length(z) )
      say 'length of superpermutation('n") ="      right(L, max(length(L), cycles+2) )
      end   /*cycle*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ? /*──────────────────────────────────────────────────────────────────────────────────────*/ aPerm: procedure expose @.; parse arg n,i; nm= n - 1; if n==0 then return 0 

         do k=nm  by -1  for nm; kp=k+1; if @.k<@.kp  then do; i=k; leave; end; end /*k*/
         do j=i+1  while  j<n;  parse value  @.j @.n  with  @.n @.j;    n= n-1; end /*j*/
      if i==0  then return 0
         do m=i+1  while @.m<@.i; end /*m*/;    parse value  @.m  @.i   with   @.i  @.m
      return 1</lang>
output   when using the input:     8 
length of superpermutation(0) =          0
length of superpermutation(1) =          1
length of superpermutation(2) =          2
length of superpermutation(3) =          9
length of superpermutation(4) =         50
length of superpermutation(5) =        302
length of superpermutation(6) =      1,922
length of superpermutation(7) =     13,652
length of superpermutation(8) =    109,538

version 2

<lang rexx>/*REXX program attempts to find better minimizations for computing superpermutations.*/ parse arg cycles . /*obtain optional arguments from the CL*/ if cycles== | cycles=="," then cycles= 7 /*Not specified? Then use the default.*/ 

     do n=0  to  cycles
     #= 0;                          $.=         /*populate the first permutation.      */
             do pop=1  for n;       @.pop= d2x(pop);         $.0= $.0  ||  @.pop
             end     /*pop*/

             do  while aPerm(n,0);  if n\==0  then #= #+1;   $.#=
                do j=1  for n;      $.#= $.#  ||  @.j
                end  /*j*/
             end     /*while*/
     z= $.0
                       c= 0                     /*count of found permutations (so far).*/
         do j=1  while c\==#
         if j>#  then do;  c= c + 1             /*exhausted finds and shortcuts; concat*/
                           z= z  ||  $.j;  $.j=
                           j= 1
                      end
         if $.j==          then iterate       /*Already found? Then ignore this perm.*/
         if pos($.j, z)\==0  then do;  c= c + 1;      $.j=
                                       iterate
                                  end

             do k=n-1  to 1  by -1              /*handle the shortcuts in perm finding.*/
             if substr($.j, k)==left(z, k)  then do;  c= c+1  /*found rightish shortcut*/
                                                      z= left($.j, k-1)  ||  z;     $.j=
                                                      iterate j
                                                 end
             if left($.j, k) ==right(z, k)  then do;  c= c+1 /*found   leftish shortcut*/
                                                      z= z  ||  substr($.j, k+1);   $.j=
                                                      iterate j
                                                 end
             end   /*k*/                        /* [↑]  more IFs could be added for opt*/
          end      /*j*/

      L= commas( length(z) )
      say 'length of superpermutation('n") ="     right(L, max(length(L), cycles+2) )
      end   /*n*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ? /*──────────────────────────────────────────────────────────────────────────────────────*/ aPerm: procedure expose @.; parse arg n,i; nm=n-1; if n==0 then return 0 

          do k=nm  by -1  for nm; kp=k+1; if @.k<@.kp  then do; i=k;leave; end; end /*k*/
          do j=i+1  while  j<n;  parse value  @.j @.n  with  @.n @.j;    n=n-1; end /*j*/
      if i==0  then return 0
          do m=i+1  while @.m<@.i; end /*m*/;   parse value  @.m @.i  with  @.i @.m
      return 1</lang>
output   when using the default input:     7 
length of superpermutation(0) =         0
length of superpermutation(1) =         1
length of superpermutation(2) =         3
length of superpermutation(3) =         9
length of superpermutation(4) =        35
length of superpermutation(5) =       183
length of superpermutation(6) =     1,411
length of superpermutation(7) =    12,137

Ruby

Non Recursive Version

<lang ruby>#A straight forward implementation of N. Johnston's algorithm. I prefer to look at this as 2n+1 where

  1. the second n is first n reversed, and the 1 is always the second symbol. This algorithm will generate
  2. just the left half of the result by setting l to [1,2] and looping from 3 to 6. For the purpose of
  3. this task I am going to start from an empty array and generate the whole strings using just the
  4. rules.
  6. Nigel Galloway: December 16th., 2014

l = [] (1..6).each{|e| 

 a, i = [], e-2
 (0..l.length-e+1).each{|g|
    if not (n = l[g..g+e-2]).uniq!
      a.concat(n[(a[0]? i : 0)..-1]).push(e).concat(n)
      i = e-2
    else
      i -= 1
    end
  }
  a.each{|n| print n}; puts "\n\n"
  l = a

}</lang>

Output:
1

121

123121321

123412314231243121342132413214321

123451234152341253412354123145231425314235142315423124531243512431524312543121345213425134215342135421324513241532413524132541321453214352143251432154321

123456123451623451263451236451234651234156234152634152364152346152341652341256341253641253461253416253412653412356412354612354162354126354123654123145623145263145236145231645231465231425631425361425316425314625314265314235614235164235146235142635142365142315642315462315426315423615423165423124563124536124531624531264531246531243561243516243512643512463512436512431562431526431524631524361524316524312564312546312543612543162543126543121345621345261345216345213645213465213425613425163425136425134625134265134215634215364215346215342615342165342135642135462135426135421635421365421324561324516324513624513264513246513241563241536241532641532461532416532413562413526413524613524163524136524132564132546132541632541362541326541321456321453621453261453216453214653214356214352614352164352146352143652143256143251643251463251436251432651432156432154632154362154326154321654321

Recursive Version

<lang ruby>def superperm(n) 

 return [1] if n==1
 superperm(n-1).each_cons(n-1).with_object([]) do |sub, ary|
   next if sub.uniq!
   i = ary.empty? ? 0 : sub.index(ary.last)+1
   ary.concat(sub[i..-1] + [n] + sub)
 end

end

def to_16(a) a.map{|x| x.to_s(16)}.join end

for n in 1..10 

 ary = superperm(n)
 print "%3d: len =%8d :" % [n, ary.size]
 puts n<5 ? ary.join : to_16(ary.first(20)) + "...." + to_16(ary.last(20))

end</lang>

Output:
 1: len =       1 :1
 2: len =       3 :121
 3: len =       9 :123121321
 4: len =      33 :123412314231243121342132413214321
 5: len =     153 :12345123415234125341....14352143251432154321
 6: len =     873 :12345612345162345126....62154326154321654321
 7: len =    5913 :12345671234561723456....65432716543217654321
 8: len =   46233 :12345678123456718234....43281765432187654321
 9: len =  409113 :12345678912345678192....29187654321987654321
10: len = 4037913 :123456789a1234567891....1987654321a987654321

Scala

<lang Scala>object SuperpermutationMinimisation extends App { 

 val nMax = 12

 @annotation.tailrec
 def factorial(number: Int, acc: Long = 1): Long =
   if (number == 0) acc else factorial(number - 1, acc * number)

 def factSum(n: Int): Long = (1 to n).map(factorial(_)).sum

 for (n <- 0 until nMax) println(f"superPerm($n%2d) len = ${factSum(n)}%d")

}</lang>

Sidef

Translation of: Perl

<lang ruby>for len in (1..8) { 

   var (pre="", post="")
   @^len -> permutations {|*p|
       var t = p.join
       post.append!(t) if !post.contains(t)
       pre.prepend!(t) if !pre.contains(t)
   }
   printf("%2d: %8d %8d\n", len, pre.len, post.len)

}</lang>

Output:
 1:        1        1
 2:        4        4
 3:       12       15
 4:       48       64
 5:      240      325
 6:     1440     1956
 7:    10080    13699
 8:    80640   109600

Wren

Translation of: Kotlin
Library: Wren-fmt

<lang ecmascript>import "/fmt" for Fmt

var max = 12 var sp = [] var count = List.filled(max, 0) var pos = 0

var factSum = Fn.new { |n| 

   var s = 0
   var x = 0
   var f = 1
   while (x < n) {
       x = x + 1
       f = f * x
       s = s + f
   }
   return s

}

var r // recursive r = Fn.new { |n| 

   if (n == 0) return false
   var c = sp[pos - n]
   count[n] = count[n] - 1
   if (count[n] == 0) {
       count[n] = n
       if (!r.call(n - 1)) return false
   }
   sp[pos] = c
   pos = pos + 1
   return true

}

var superPerm = Fn.new { |n| 

   pos = n
   var len = factSum.call(n)
   if (len > 0) sp = List.filled(len, "\0")
   for (i in 0..n) count[i] = i
   if (n > 0) {
       for (i in 1..n) sp[i - 1] = String.fromByte(48 + i)
   }
   while (r.call(n)) {}

}

for (n in 0...max) { 

   superPerm.call(n)
   Fmt.print("superPerm($2d) len = $d", n, sp.count)

}</lang>

Output:
superPerm( 0) len = 0
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 33
superPerm( 5) len = 153
superPerm( 6) len = 873
superPerm( 7) len = 5913
superPerm( 8) len = 46233
superPerm( 9) len = 409113
superPerm(10) len = 4037913
superPerm(11) len = 43954713

zkl

Translation of: C

It crawls ... <lang zkl>const MAX = 12; var super=Data(), pos, cnt; // global state, ick

fcn fact_sum(n){ // -->1! + 2! + ... + n! 

  [1..n].reduce(fcn(s,n){ s + [2..n].reduce('*,1) },0)

}

fcn r(n){ 

  if (not n) return(0);

  c := super[pos - n];
  if (not (cnt[n]-=1)){
     cnt[n] = n;
     if (not r(n-1)) return(0);
  }
  super[pos] = c; pos+=1;
  1

}

fcn superperm(n){ 

  pos = n;
  len := fact_sum(n);
  super.fill(0,len);  // this is pretty close to recalloc()

  cnt = (n+1).pump(List()); //-->(0,1,2,3,..n)
  foreach i in (n){ super[i] = i + 0x31; } //-->"1" ... "123456789:;"
  while (r(n)){}

}

foreach n in (MAX){ 

  superperm(n);
  print("superperm(%2d) len = %d".fmt(n,super.len()));
  // uncomment next line to see the string itself
  //print(": %s".fmt(super.text));
  println();

}</lang>

Output:
superperm( 0) len = 0: 
superperm( 1) len = 1: 1
superperm( 2) len = 3: 121
superperm( 3) len = 9: 123121321
superperm( 4) len = 33: 123412314231243121342132413214321
superperm( 5) len = 153: 123451234152341253412354123145231425314235142315423124531243512431524312543121345213425134215342135421324513241532413524132541321453214352143251432154321
superperm( 6) len = 873
superperm( 7) len = 5913
superperm( 8) len = 46233
superperm( 9) len = 409113
superperm(10) len = 4037913
superperm(11) len = 43954713
Retrieved from "https://rosettacode.org/wiki/Superpermutation_minimisation?oldid=251588"