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
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contain the most consonants
- Find words which contains more than 3 vowels
- Find words whose first and last three letters are equal
- Find words with alternating vowels and consonants
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 bottles of beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
- Reference
- The Minimal Superpermutation Problem. by Nathaniel Johnston.
- oeis A180632 gives 0-5 as 0, 1, 3, 9, 33, 153. 6 is thought to be 872.
- Superpermutations - Numberphile. A video
- Superpermutations: the maths problem solved by 4chan - Standupmaths. A video of recent (2018) mathematical progress.
- New Superpermutations Discovered! Standupmaths & Numberphile.
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))- 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++
#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];
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 .. $] = trialAdd[];
if (trialPerm in toFind) {
result ~= trialAdd;
toFind.remove(trialPerm);
break MIDDLE;
}
}
}
}
return result;
}
void main() {
foreach (immutable n; 1 .. 8)
n.superpermutation.length.writeln;
}
- Output:
1 3 9 35 164 932 6247
Using the ldc2 compiler with n=10, it finds the result string of length 4_235_533 in less than 9 seconds.
Faster Version
From the C version with some improvements.
import std.stdio, std.range, std.algorithm, std.ascii;
enum uint nMax = 12;
__gshared char[] superperm;
__gshared uint pos;
__gshared uint[nMax] count;
/// factSum(n) = 1! + 2! + ... + n!
uint factSum(in uint n) pure nothrow @nogc @safe {
return iota(1, n + 1).map!(m => reduce!q{ a * b }(1u, iota(1, m + 1))).sum;
}
bool r(in uint n) nothrow @nogc {
if (!n)
return false;
immutable c = superperm[pos - n];
if (!--count[n]) {
count[n] = n;
if (!r(n - 1))
return false;
}
superperm[pos++] = c;
return true;
}
void superPerm(in uint n) nothrow {
static immutable chars = digits ~ uppercase;
static assert(chars.length >= nMax);
pos = n;
superperm.length = factSum(n);
foreach (immutable i; 0 .. n + 1)
count[i] = i;
foreach (immutable i; 1 .. n + 1)
superperm[i - 1] = chars[i];
while (r(n)) {}
}
void main() {
foreach (immutable n; 0 .. nMax) {
superPerm(n);
writef("superPerm(%2d) len = %d", n, superperm.length);
// Use -version=doPrint to see the string itself.
version (doPrint) write(": ", superperm);
writeln;
}
}
- Output:
superPerm( 0) len = 0 superPerm( 1) len = 1 superPerm( 2) len = 3 superPerm( 3) len = 9 superPerm( 4) len = 33 superPerm( 5) len = 153 superPerm( 6) len = 873 superPerm( 7) len = 5913 superPerm( 8) len = 46233 superPerm( 9) len = 409113 superPerm(10) len = 4037913 superPerm(11) len = 43954713
Run-time: about 0.40 seconds.
Delphi
program Superpermutation_minimisation;
{$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.
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)
- 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
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
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
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
Runs in about 1/4 second.
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 $(length(s)) ", (verbose ? String(s) : ""))
end
end
testsuper(nmax)
- Output:
Superperm(0) has length 0 Superperm(1) has length 1 Superperm(2) has length 3 Superperm(3) has length 9 Superperm(4) has length 33 Superperm(5) has length 153 Superperm(6) has length 873 Superperm(7) has length 5913 Superperm(8) has length 46233 Superperm(9) has length 409113 Superperm(10) has length 4037913 Superperm(11) has length 43954713
Kotlin
// version 1.1.2
const val MAX = 12
var sp = CharArray(0)
val count = IntArray(MAX)
var pos = 0
fun factSum(n: Int): Int {
var s = 0
var x = 0
var f = 1
while (x < n) {
f *= ++x
s += f
}
return s
}
fun r(n: Int): Boolean {
if (n == 0) return false
val c = sp[pos - n]
if (--count[n] == 0) {
count[n] = n
if (!r(n - 1)) return false
}
sp[pos++] = c
return true
}
fun superPerm(n: Int) {
pos = n
val len = factSum(n)
if (len > 0) sp = CharArray(len)
for (i in 0..n) count[i] = i
for (i in 1..n) sp[i - 1] = '0' + i
while (r(n)) {}
}
fun main(args: Array<String>) {
for (n in 0 until MAX) {
superPerm(n)
println("superPerm(${"%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
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)}")
- 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
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.
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
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
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])):
#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])))
- 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
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
#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)
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
Rust
const MAX: usize = 12;
// 1! + 2! + ... + n!
fn fact_sum(n: usize) -> usize {
let mut sum = 0_usize;
let mut fact = 1_usize;
let mut x = 0_usize;
while x < n {
x += 1;
fact *= x;
sum += fact;
}
return sum;
}
fn r(n: usize, sup: &mut Vec<u8>, pos: &mut usize, cnt: &mut Vec<usize>) -> bool {
if n == 0 {
return false;
}
let c = sup[*pos - n];
cnt[n] -= 1;
if cnt[n] == 0 {
cnt[n] = n;
if !r(n - 1, sup, pos, cnt) {
return false;
}
}
sup[*pos] = c;
*pos += 1;
return true;
}
fn superperm(n: usize, sup: &mut Vec<u8>, pos: &mut usize, cnt: &mut Vec<usize>) -> usize {
*pos = n;
let new_len = fact_sum(n);
sup.resize(new_len, 0);
for i in 0..=n {
cnt[i] = i;
}
for i in 1..=n {
sup[i - 1] = (i as u8) + ('0' as u8);
}
while r(n, sup, pos, cnt) {}
return sup.len();
}
fn main() {
let sup: &mut Vec<u8> = &mut [0_u8; 0].to_vec();
let pos: &mut usize = &mut 0_usize;
let cnt: &mut Vec<usize> = &mut [0_usize; MAX].to_vec();
for n in 0..MAX {
println!("superperm of {:<2} has a length of {}", n, superperm(n, sup, pos, cnt));
}
}
- Output:
superperm of 0 has a length of 0 superperm of 1 has a length of 1 superperm of 2 has a length of 3 superperm of 3 has a length of 9 superperm of 4 has a length of 33 superperm of 5 has a length of 153 superperm of 6 has a length of 873 superperm of 7 has a length of 5913 superperm of 8 has a length of 46233 superperm of 9 has a length of 409113 superperm of 10 has a length of 4037913 superperm of 11 has a length of 43954713
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
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
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
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
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