Topswops

Topswops
You are encouraged to solve this task according to the task description, using any language you may know.

Topswops is a card game created by John Conway in the 1970's.

Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top. A round is composed of reversing the first m cards where m is the value of the topmost card. rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded. For our example the swaps produce:

    [2, 4, 1, 3]    # Initial shuffle
    [4, 2, 1, 3]
    [3, 1, 2, 4]
    [2, 1, 3, 4]
    [1, 2, 3, 4]

For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.

For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.

Task

The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.

Note

Topswops is also known as Fannkuch from the German Pfannkuchen meaning pancake.

Cf.

• Number reversal game
• Sorting algorithms/Pancake sort

C

I don't remember how this code actually works, being an old posting and all, but the results seem right: <lang c>#include <stdio.h>

typedef unsigned char elem;

elem s[16], t[16]; size_t max_n, maxflips;

int flip() { register int i; register elem *x, *y, c;

for (i = 0; i < max_n; i++) if (s[i] == i) return 0;

for (x = t, y = s; i--;) *x++ = *y++;

i = 1; do { for (x = t, y = t + t[0]; x < y; ) c = *x, *x++ = *y, *y-- = c; i++; } while (t[t[0]]);

return i; }

inline void rotate(int n) { elem c; register int i; c = s[0]; for (i = 1; i <= n; i++) s[i-1] = s[i]; s[n] = c; }

/* Tompkin-Paige iterative perm generation */ void tk(int n) { int i = 0, f; maxflips = 0;

elem c[16] = { 0 };

while (i < n) { rotate(i); if (c[i] >= i) { c[i++] = 0; continue; }

c[i]++; i = 1; if (*s) { f = s[s[0]] ? flip() : 1; if (f > maxflips) maxflips = f; } } }

int main(void) { int i; for (max_n = 1; max_n <= 12; max_n++) { for (i = 0; i < max_n; i++) s[i] = i; tk(max_n); printf("Pfannkuchen(%d) = %d\n", max_n, maxflips); }

return 0; }</lang>

Output:

Pfannkuchen(1) = 0
Pfannkuchen(2) = 1
Pfannkuchen(3) = 2
Pfannkuchen(4) = 4
Pfannkuchen(5) = 7
Pfannkuchen(6) = 10
Pfannkuchen(7) = 16
Pfannkuchen(8) = 22
Pfannkuchen(9) = 30
Pfannkuchen(10) = 38
Pfannkuchen(11) = 51
Pfannkuchen(12) = 65

D

This example is incomplete. Missing n in table of output Please ensure that it meets all task requirements and remove this message.

Permutations generator from: http://rosettacode.org/wiki/Permutations#Faster_Lazy_Version <lang d>import std.stdio, std.algorithm, std.range, permutations2;

uint topswops(in uint n) in {

   assert (n > 0 && n < 32);

} body {

   static uint flip(uint[] deck) pure nothrow {
       uint[32] temp = void;
       temp[0 .. deck.length] = deck[];
       uint count = 0;
       for (auto t0 = temp[0]; t0; t0 = temp[0]) {
           temp[0 .. t0 + 1].reverse(); // Slow with DMD.
           count++;
       }
       return count;
   }

   return iota(n).array().permutations!0().map!flip().reduce!max();

}

void main() {

   iota(1, 11).map!topswops().writeln();

}</lang>

Output:

[0, 1, 2, 4, 7, 10, 16, 22, 30, 38]

Go

<lang go>// Adapted from http://www-cs-faculty.stanford.edu/~uno/programs/topswops.w // at Donald Knuth's web site. Algorithm credited there to Pepperdine // and referenced to Mathematical Gazette 73 (1989), 131-133. package main

import "fmt"

const ( // array sizes

   maxn = 10 // max number of cards
   maxl = 50 // upper bound for number of steps

)

func main() {

   for i := 1; i <= maxn; i++ {
       fmt.Printf("%d: %d\n", i, steps(i))
   }

}

func steps(n int) int {

   var a, b [maxl][maxn + 1]int
   var x [maxl]int
   a[0][0] = 1
   var m int
   for l := 0; ; {
       x[l]++
       k := int(x[l])
       if k >= n {
           if l <= 0 {
               break
           }
           l--
           continue
       }
       if a[l][k] == 0 {
           if b[l][k+1] != 0 {
               continue
           }
       } else if a[l][k] != k+1 {
           continue
       }
       a[l+1] = a[l]
       for j := 1; j <= k; j++ {
           a[l+1][j] = a[l][k-j]
       }
       b[l+1] = b[l]
       a[l+1][0] = k + 1
       b[l+1][k+1] = 1
       if l > m-1 {
           m = l + 1
       }
       l++
       x[l] = 0
   }
   return m

}</lang>

Output:

1: 0
2: 1
3: 2
4: 4
5: 7
6: 10
7: 16
8: 22
9: 30
10: 38

Perl 6

<lang perl6>sub postfix:<!>(@a) {

   @a == 1
       ?? [@a]
       !! do for @a -> $a {
               [ $a, @$_ ] for @a.grep(* != $a)!
          }

}

sub swops(@a is copy) {

   my $count = 0;
   until @a[0] == 1 {
       @a[ ^@a[0] ] .= reverse;
       $count++;
   }
   return $count;

} sub topswops($n) { [max] map &swops, (1 .. $n)! }

say "$_ {topswops $_}" for 1 .. 10;</lang>

Output follows that of Python.

Python

This solution uses cards numbered from 0..n-1 and variable p0 is introduced as a speed optimisation <lang python>>>> from itertools import permutations >>> def f1(p): i, p0 = 0, p[0] while p0: i += 1 p0 += 1 p[:p0] = p[:p0][::-1] p0 = p[0] return i

>>> def fannkuch(n): return max(f1(list(p)) for p in permutations(range(n)))

>>> for n in range(1, 11): print(n,fannkuch(n))

1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38 >>> </lang>