Superpermutation minimisation

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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.

Reference

11l

Translation of: Kotlin
```-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))```
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

```# syntax: GAWK -f SUPERPERMUTATION_MINIMISATION.AWK
# 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)
#     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)
}
```
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.

```#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#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;
}
```
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
```#include <array>
#include <iostream>
#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;
}
```
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.

```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];

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

return result;
}

void main() {
foreach (immutable n; 1 .. 8)
n.superpermutation.length.writeln;
}
```
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.

```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;
}
}
```
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```

Delphi

Translation of: C
```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;
end.
```

Elixir

Translation of: Ruby
```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)
```
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

```' 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```
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
```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))
}
}
```
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
```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()
}
}
}
```
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.

```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
)
```

Some sequence lengths:

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

Java

Translation of C via D

Works with: Java version 8
```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();
}
}
}
```
```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

```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)
```
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
```// 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}")
}
}
```
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:

```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 = overlaprange[[1]] - 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[#2[[2]], #2[[1]]["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, perms[[i]]]|>, {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}
]
```
Output:
```{1,1}
{2,3}
{3,9}
{4,33}
{5,153}
{6,873}
{7,5913}```

Nim

Translation of: Go
```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):
for n in 0..<MAX:
write(stdout, fmt"superperm({n:2})")
superperm(n)
writeLine(stdout, fmt" len = {len(super)}")
```
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
```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);
}
}```
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
```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);
}
```
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
```with javascript_semantics
constant nMax = iff(platform()=JS?8:12)
-- Aside: on desktop/Phix, strings can be modified in situ, whereas
--        JavaScript strings are immutable, and the equivalent code
--        in p2js.js ends up doing excessive splitting and splicing
--        hence nMax has to be significantly smaller in a browser.

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<=7 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
```
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).

```with javascript_semantics
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

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), 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
```
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
```EnableExplicit
#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```
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

```"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])):
if trial_perm in tofind:
break
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])))
```
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
```from array import array
from string import ascii_uppercase, digits
from operator import mul

try:
import psyco
psyco.full()
except:
pass

N_MAX = 12

# 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()
```

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/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)))
```
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
```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;
}
```
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.

```/*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
```
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

```/*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
```
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

```#A straight forward implementation of N. Johnston's algorithm. I prefer to look at this as 2n+1 where
#the second n is first n reversed, and the 1 is always the second symbol. This algorithm will generate
#just the left half of the result by setting l to [1,2] and looping from 3 to 6. For the purpose of
#this task I am going to start from an empty array and generate the whole strings using just the
#rules.
#
#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
}
```
Output:
```1

121

123121321

123412314231243121342132413214321

123451234152341253412354123145231425314235142315423124531243512431524312543121345213425134215342135421324513241532413524132541321453214352143251432154321

123456123451623451263451236451234651234156234152634152364152346152341652341256341253641253461253416253412653412356412354612354162354126354123654123145623145263145236145231645231465231425631425361425316425314625314265314235614235164235146235142635142365142315642315462315426315423615423165423124563124536124531624531264531246531243561243516243512643512463512436512431562431526431524631524361524316524312564312546312543612543162543126543121345621345261345216345213645213465213425613425163425136425134625134265134215634215364215346215342615342165342135642135462135426135421635421365421324561324516324513624513264513246513241563241536241532641532461532416532413562413526413524613524163524136524132564132546132541632541362541326541321456321453621453261453216453214653214356214352614352164352146352143652143256143251643251463251436251432651432156432154632154362154326154321654321```

Recursive Version

```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
```
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

```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")

}
```

Sidef

Translation of: Perl
```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)
}
```
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
```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)
}
```
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
```

XPL0

Translation of: C

Works on Raspberry Pi. ReallocMem is not supported in DOS versions.

```include xpllib; \for Print and StrLen

define Maxx = 12;
char Super;
int  Pos, Cnt(Maxx);

func FactSum(N);        \1! + 2! + ... + n!
int  N, S, X, F;
[S:= 0;  X:= 0;  F:= 1;
while X < N do
[X:= X+1;
F:= F*X;
S:= S+F;
];
return S;
];

func R(N);
int  N, C;
[if N = 0 then return false;
C:= Super(Pos - N);
Cnt(N):= Cnt(N)-1;
if Cnt(N) = 0 then
[Cnt(N):= N;
if R(N-1) = 0 then return false;
];
Super(Pos):= C;  Pos:= Pos+1;
return true;
];

proc Superperm(N);
int  N, I, Len;
[Pos:= N;
Len:= FactSum(N);
Super:= ReallocMem(Super, Len+1);
Super(Len):= 0;
for I:= 0 to N do Cnt(I):= I;
for I:= 1 to N do Super(I-1):= I+^0;
while R(N) do [];
];

int N;
[Super:= 0;
for N:= 0 to Maxx-1 do
[Print("Superperm(%d) ", N);
Superperm(N);
Print("len = %d", StrLen(Super));
\Uncomment next line to see the string itself
\Print(": %s", Super);
CrLf(0);
];
]```
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 ...

```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();
}```
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
```