Topswops
From Rosetta Code
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.
Contents |
[edit] Ada
This is a straightforward approach that counts the number of swaps for each permutation. To generate all permutations over 1 .. N, for each of N in 1 .. 10, the package Generic_Perm from the Permutations task is ised [[1]].
with Ada.Integer_Text_IO, Generic_Perm;
procedure Topswaps is
function Topswaps(Size: Positive) return Natural is
package Perms is new Generic_Perm(Size);
P: Perms.Permutation;
Done: Boolean;
Max: Natural;
function Swapper_Calls(P: Perms.Permutation) return Natural is
Q: Perms.Permutation := P;
I: Perms.Element := P(1);
begin
if I = 1 then
return 0;
else
for Idx in 1 .. I loop
Q(Idx) := P(I-Idx+1);
end loop;
return 1 + Swapper_Calls(Q);
end if;
end Swapper_Calls;
begin
Perms.Set_To_First(P, Done);
Max:= Swapper_Calls(P);
while not Done loop
Perms.Go_To_Next(P, Done);
Max := natural'Max(Max, Swapper_Calls(P));
end loop;
return Max;
end Topswaps;
begin
for I in 1 .. 10 loop
Ada.Integer_Text_IO.Put(Item => Topswaps(I), Width => 3);
end loop;
end Topswaps;
- Output:
0 1 2 4 7 10 16 22 30 38
[edit] C
An algorithm that doesn't go through all permutations, per Knuth tAoCP 7.2.1.2 exercise 107 (possible bad implementation on my part notwithstanding):
#include <stdio.h>
#include <string.h>
typedef struct { char v[16]; } deck;
typedef unsigned int uint;
uint n, d, best[16];
void tryswaps(deck *a, uint f, uint s) {
# define A a->v
# define B b.v
if (d > best[n]) best[n] = d;
while (1) {
if ((A[s] == s || (A[s] == -1 && !(f & 1U << s)))
&& (d + best[s] >= best[n] || A[s] == -1))
break;
if (d + best[s] <= best[n]) return;
if (!--s) return;
}
d++;
deck b = *a;
for (uint i = 1, k = 2; i <= s; k <<= 1, i++) {
if (A[i] != i && (A[i] != -1 || (f & k)))
continue;
for (uint j = B[0] = i; j--;) B[i - j] = A[j];
tryswaps(&b, f | k, s);
}
d--;
}
int main(void) {
deck x;
memset(&x, -1, sizeof(x));
x.v[0] = 0;
for (n = 1; n < 13; n++) {
tryswaps(&x, 1, n - 1);
printf("%2d: %d\n", n, best[n]);
}
return 0;
}
The code contains critical small loops, which can be manually unrolled for those with OCD. POSIX thread support is useful if you got more than one CPUs.
#define _GNU_SOURCE
#include <stdio.h>
#include <string.h>
#include <pthread.h>
#include <sched.h>
#define MAX_CPUS 8 // increase this if you got more CPUs/cores
typedef struct { char v[16]; } deck;
int n, best[16];
// Update a shared variable by spinlock. Since this program really only
// enters locks dozens of times, a pthread_mutex_lock() would work
// equally fine, but RC already has plenty of examples for that.
#define SWAP_OR_RETRY(var, old, new) \
if (!__sync_bool_compare_and_swap(&(var), old, new)) { \
volatile int spin = 64; \
while (spin--); \
continue; }
void tryswaps(deck *a, int f, int s, int d) {
#define A a->v
#define B b->v
while (best[n] < d) {
int t = best[n];
SWAP_OR_RETRY(best[n], t, d);
}
#define TEST(x) \
case x: if ((A[15-x] == 15-x || (A[15-x] == -1 && !(f & 1<<(15-x)))) \
&& (A[15-x] == -1 || d + best[15-x] >= best[n])) \
break; \
if (d + best[15-x] <= best[n]) return; \
s = 14 - x
switch (15 - s) {
TEST(0); TEST(1); TEST(2); TEST(3); TEST(4);
TEST(5); TEST(6); TEST(7); TEST(8); TEST(9);
TEST(10); TEST(11); TEST(12); TEST(13); TEST(14);
return;
}
#undef TEST
deck *b = a + 1;
*b = *a;
d++;
#define FLIP(x) \
if (A[x] == x || ((A[x] == -1) && !(f & (1<<x)))) { \
B[0] = x; \
for (int j = x; j--; ) B[x-j] = A[j]; \
tryswaps(b, f|(1<<x), s, d); } \
if (s == x) return;
FLIP(1); FLIP(2); FLIP(3); FLIP(4); FLIP(5);
FLIP(6); FLIP(7); FLIP(8); FLIP(9); FLIP(10);
FLIP(11); FLIP(12); FLIP(13); FLIP(14); FLIP(15);
#undef FLIP
}
int num_cpus(void) {
cpu_set_t ct;
sched_getaffinity(0, sizeof(ct), &ct);
int cnt = 0;
for (int i = 0; i < MAX_CPUS; i++)
if (CPU_ISSET(i, &ct))
cnt++;
return cnt;
}
struct work { int id; deck x[256]; } jobs[MAX_CPUS];
int first_swap;
void *thread_start(void *arg) {
struct work *job = arg;
while (1) {
int at = first_swap;
if (at >= n) return 0;
SWAP_OR_RETRY(first_swap, at, at + 1);
memset(job->x, -1, sizeof(deck));
job->x[0].v[at] = 0;
job->x[0].v[0] = at;
tryswaps(job->x, 1 | (1 << at), n - 1, 1);
}
}
int main(void) {
int n_cpus = num_cpus();
for (int i = 0; i < MAX_CPUS; i++)
jobs[i].id = i;
pthread_t tid[MAX_CPUS];
for (n = 2; n <= 14; n++) {
int top = n_cpus;
if (top > n) top = n;
first_swap = 1;
for (int i = 0; i < top; i++)
pthread_create(tid + i, 0, thread_start, jobs + i);
for (int i = 0; i < top; i++)
pthread_join(tid[i], 0);
printf("%2d: %2d\n", n, best[n]);
}
return 0;
}
[edit] D
Permutations generator from: http://rosettacode.org/wiki/Permutations#Faster_Lazy_Version
import std.stdio, std.algorithm, std.range, permutations2;
int topswops(in int n) {
static int flip(int[] xa) pure nothrow {
if (!xa[0]) return 0;
xa[0 .. xa[0] + 1].reverse(); // Slow with DMD.
return 1 + flip(xa);
}
return n.iota.array.permutations.map!flip.reduce!max;
}
void main() {
foreach (immutable i; 1 .. 11)
writeln(i, ": ", i.topswops);
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
[edit] D: Faster Version
import std.stdio, std.typetuple;
template Range(int start, int stop) {
static if (stop <= start)
alias TypeTuple!() Range;
else
alias TypeTuple!(Range!(start, stop - 1), stop - 1) Range;
}
__gshared uint[32] best;
uint topswops(size_t n)() nothrow {
static assert(n > 0 && n < best.length);
size_t d = 0;
alias T = byte;
alias Deck = T[n];
void trySwaps(in ref Deck deck, in uint f) nothrow {
if (d > best[n])
best[n] = d;
foreach_reverse (immutable i; Range!(0, n)) {
if ((deck[i] == i || (deck[i] == -1 && !(f & (1U << i))))
&& (d + best[i] >= best[n] || deck[i] == -1))
break;
if (d + best[i] <= best[n])
return;
}
Deck deck2 = void;
foreach (immutable i; Range!(0, n)) // Copy.
deck2[i] = deck[i];
d++;
foreach (immutable i; Range!(1, n)) {
enum uint k = 1U << i;
if (deck[i] != i && (deck[i] != -1 || (f & k)))
continue;
deck2[0] = cast(T)i;
foreach_reverse (immutable j; Range!(0, i))
deck2[i - j] = deck[j]; // Reverse copy.
trySwaps(deck2, f | k);
}
d--;
}
best[n] = 0;
Deck deck0 = -1;
deck0[0] = 0;
trySwaps(deck0, 1);
return best[n];
}
void main() {
foreach (i; Range!(1, 14))
writefln("%2d: %d", i, topswops!i());
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38 11: 51 12: 65 13: 80
With templates to speed up the computation, using the DMD compiler it's almost as fast as the second C version.
[edit] Fortran
module top
implicit none
contains
recursive function f(x) result(m)
integer :: n, m, x(:),y(size(x)), fst
fst = x(1)
if (fst == 1) then
m = 0
else
y(1:fst) = x(fst:1:-1)
y(fst+1:) = x(fst+1:)
m = 1 + f(y)
end if
end function
recursive function perms(x) result(p)
integer, pointer :: p(:,:), q(:,:)
integer :: x(:), n, k, i
n = size(x)
if (n == 1) then
allocate(p(1,1))
p(1,:) = x
else
q => perms(x(2:n))
k = ubound(q,1)
allocate(p(k*n,n))
p = 0
do i = 1,n
p(1+k*(i-1):k*i,1:i-1) = q(:,1:i-1)
p(1+k*(i-1):k*i,i) = x(1)
p(1+k*(i-1):k*i,i+1:) = q(:,i:)
end do
end if
end function
end module
program topswort
use top
implicit none
integer :: x(10)
integer, pointer :: p(:,:)
integer :: i, j, m
forall(i=1:10)
x(i) = i
end forall
do i = 1,10
p=>perms(x(1:i))
m = 0
do j = 1, ubound(p,1)
m = max(m, f(p(j,:)))
end do
print "(i3,a,i3)", i,": ",m
end do
end program
[edit] 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
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
[edit] Haskell
import Data.List (permutations)
topswops :: Int -> Int
topswops n = maximum $ map tops $ permutations [1 .. n]
where
tops (1 : _) = 0
tops xa@(x : _) = 1 + tops reordered
where
reordered = reverse (take x xa) ++ drop x xa
main = mapM_
(\x -> putStrLn $ show x ++ ":\t" ++ show (topswops x))
[1 .. 10]
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Alternate version
Uses only permutations with all elements out of place.
import Data.List
import Control.Arrow
import Control.Monad
derangements = filter (and . zipWith (/=) [1..] ). permutations
topswop = ((uncurry (++). first reverse).). splitAt
topswopIter = takeWhile((/=1).head). iterate (topswop =<< head)
swops = map (length. topswopIter). derangements
topSwops :: [Int] -> [(Int, Int)]
topSwops = zip [1..]. map (maximum. (0:). swops). tail. inits
Output
*Main> mapM_ print $ take 10 $ topSwops [1..] (1,0) (2,1) (3,2) (4,4) (5,7) (6,10) (7,16) (8,22) (9,30) (10,38)
[edit] J
Solution:swops =: ((|.@:{. , }.)~ {.)^:a:Example (from task introduction):
swops 2 4 1 3Example (topswops of all permutations of the integers 1..10):
2 4 1 3
4 2 1 3
3 1 2 4
2 1 3 4
1 2 3 4
(,. _1 + ! >./@:(#@swops@A. >:)&i. ])&> 1+i.10
1 0
2 1
3 2
4 4
5 7
6 10
7 16
8 22
9 30
10 38
Notes: Readers less familiar with array-oriented programming may find an alternate solution written in the structured programming style more accessible.
[edit] Java
public class Topswops {
static final int maxBest = 32;
static int[] best;
static private void trySwaps(int[] deck, int f, int d, int n) {
if (d > best[n])
best[n] = d;
for (int i = n - 1; i >= 0; i--) {
if (deck[i] == -1 || deck[i] == i)
break;
if (d + best[i] <= best[n])
return;
}
int[] deck2 = deck.clone();
for (int i = 1; i < n; i++) {
final int k = 1 << i;
if (deck2[i] == -1) {
if ((f & k) != 0)
continue;
} else if (deck2[i] != i)
continue;
deck2[0] = i;
for (int j = i - 1; j >= 0; j--)
deck2[i - j] = deck[j]; // Reverse copy.
trySwaps(deck2, f | k, d + 1, n);
}
}
static int topswops(int n) {
assert(n > 0 && n < maxBest);
best[n] = 0;
int[] deck0 = new int[n + 1];
for (int i = 1; i < n; i++)
deck0[i] = -1;
trySwaps(deck0, 1, 0, n);
return best[n];
}
public static void main(String[] args) {
best = new int[maxBest];
for (int i = 1; i < 11; i++)
System.out.println(i + ": " + topswops(i));
}
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
[edit] Julia
Fast, efficient version
function fannkuch(n)
n == 1 && return 0
n == 2 && return 1
p = [1:n]
q = copy(p)
s = copy(p)
sign = 1; maxflips = sum = 0
while true
q0 = p[1]
if q0 != 1
for i = 2:n
q[i] = p[i]
end
flips = 1
while true
qq = q[q0] #??
if qq == 1
sum += sign*flips
flips > maxflips && (maxflips = flips)
break
end
q[q0] = q0
if q0 >= 4
i = 2; j = q0-1
while true
t = q[i]
q[i] = q[j]
q[j] = t
i += 1
j -= 1
i >= j && break
end
end
q0 = qq
flips += 1
end
end
#permute
if sign == 1
t = p[2]
p[2] = p[1]
p[1] = t
sign = -1
else
t = p[2]
p[2] = p[3]
p[3] = t
sign = 1
for i = 3:n
sx = s[i]
if sx != 1
s[i] = sx-1
break
end
i == n && return maxflips
s[i] = i
t = p[1]
for j = 1:i
p[j] = p[j+1]
end
p[i+1] = t
end
end
end
end
- Output:
julia> function main() for i = 1:10 println(fannkuch(i)) end end # methods for generic function main main() at none:2 julia> @time main() 0 1 2 4 7 10 16 22 30 38 elapsed time: 0.299617582 seconds
[edit] Mathematica
An exhaustive search of all possible permutations is done
flip[a_] :=
Block[{a1 = First@a},
If[a1 == Length@a, Reverse[a],
Join[Reverse[a[[;; a1]]], a[[a1 + 1 ;;]]]]]
swaps[a_] := Length@FixedPointList[flip, a] - 2
Print[#, ": ", Max[swaps /@ Permutations[Range@#]]] & /@ Range[10];
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
[edit] Perl 6
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;
Output follows that of Python.
[edit] Python
This solution uses cards numbered from 0..n-1 and variable p0 is introduced as a speed optimisation
>>> from itertools import permutations
>>> def f1(p):
i = 0
while True:
p0 = p[0]
if p0 == 1: break
p[:p0] = p[:p0][::-1]
i += 1
return i
>>> def fannkuch(n):
return max(f1(list(p)) for p in permutations(range(1, n+1)))
>>> 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
>>>
[edit] Python: Faster Version
try:
import psyco
psyco.full()
except ImportError:
pass
best = [0] * 16
def try_swaps(deck, f, s, d, n):
if d > best[n]:
best[n] = d
i = 0
k = 1 << s
while s:
k >>= 1
s -= 1
if deck[s] == -1 or deck[s] == s:
break
i |= k
if (i & f) == i and d + best[s] <= best[n]:
return d
s += 1
deck2 = list(deck)
k = 1
for i2 in xrange(1, s):
k <<= 1
if deck2[i2] == -1:
if f & k: continue
elif deck2[i2] != i2:
continue
deck[i2] = i2
deck2[:i2 + 1] = reversed(deck[:i2 + 1])
try_swaps(deck2, f | k, s, 1 + d, n)
def topswops(n):
best[n] = 0
deck0 = [-1] * 16
deck0[0] = 0
try_swaps(deck0, 1, n, 0, n)
return best[n]
for i in xrange(1, 13):
print "%2d: %d" % (i, topswops(i))
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38 11: 51 12: 65
[edit] Racket
Simple search, only "optimization" is to consider only all-misplaced permutations (as in the alternative Haskell solution), which shaves off around 2 seconds (from ~5).
#lang racket
(define (all-misplaced? l)
(for/and ([x (in-list l)] [n (in-naturals 1)]) (not (= x n))))
(define (topswops n)
(for/fold ([m 0]) ([p (in-permutations (range 1 (add1 n)))]
#:when (all-misplaced? p))
(let loop ([p p] [n 0])
(if (= 1 (car p))
(max n m)
(loop (let loop ([l '()] [r p] [n (car p)])
(if (zero? n) (append l r)
(loop (cons (car r) l) (cdr r) (sub1 n))))
(add1 n))))))
(for ([i (in-range 1 11)]) (printf "~a\t~a\n" i (topswops i)))
Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
[edit] REXX
The deckSets subroutine is a modified permSets (permutation sets) subroutine,
and is optimized somewhat to take advantage by eliminating one-swop "decks".
/*REXX pgm gens N decks of numbered cards and finds the maximum "swops".*/
parse arg things .; if things=='' then things=10; thingsX= things>9
do n=1 for things; #=deckSets(n,n) /*create "decks".*/
mx= n\==1 /*handle case of a one-card deck.*/
do i=1 for #
mx=max(mx,swops(!.i))
end /*i*/
say '──────── maximum swops for a deck of' right(n,2) ' cards is' right(mx,4)
end /*n*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────DECKSETS subroutine─────────────────*/
deckSets: procedure expose !. /*X things taken Y at a time.*/
parse arg x,y,,$ @.; #=0; call .deckset 1 /*set $ & @. to null.*/
return # /*return # permutations (decks).*/
.deckset: procedure expose @. x y $ # !.; parse arg ?
if ?>y then do; _=@.1; do j=2 to y; _=_ @.j; end /*j*/; #=#+1; !.#=_
end
else do
?m=?-1 /*used in the FOR for faster DO.*/
if ?==1 then qs=2 /*¬ use 1-swops that start with 1*/
else do
qs=1
if @.1==? then qs=2 /*skip 1-swops: 3 x 1 x */
end
do q=qs to x /*build permutation recursively. */
do k=1 for ?m; if @.k==q then iterate q; end /*k*/
@.?=q; call .deckset(?+1)
end /*q*/
end
return
/*──────────────────────────────────SWOPS subroutine────────────────────*/
swops: parse arg z; do _=1; t=word(z,1)
if word(z,t)==1 then return _
if thingsX then do h=10 to things
z=changestr(h,z,d2x(h))
end /*h*/
z=reverse(subword(z,1,t)) subword(z,t+1)
if thingsX then do d=10 to things
z=changestr(d2x(d),z,d)
end /*_*/
Some older REXXes don't have a changestr bif, so one is included here ──► CHANGESTR.REX.
output when using the default input
──────── maximum swops for a deck of 1 cards is 0 ──────── maximum swops for a deck of 2 cards is 1 ──────── maximum swops for a deck of 3 cards is 2 ──────── maximum swops for a deck of 4 cards is 4 ──────── maximum swops for a deck of 5 cards is 7 ──────── maximum swops for a deck of 6 cards is 10 ──────── maximum swops for a deck of 7 cards is 16 ──────── maximum swops for a deck of 8 cards is 22 ──────── maximum swops for a deck of 9 cards is 30 ──────── maximum swops for a deck of 10 cards is 38
[edit] Ruby
def f1(a)
i = 0
loop do
a0 = a[0]
break if a0 == 1
a[0...a0] = a[0...a0].reverse
i += 1
end
i
end
def fannkuch(n)
[*1..n].permutation.map{|a| f1(a)}.max
end
for n in 1..10
puts "%2d : %d" % [n, fannkuch(n)]
end
- Output:
1 : 0 2 : 1 3 : 2 4 : 4 5 : 7 6 : 10 7 : 16 8 : 22 9 : 30 10 : 38
Faster Version
def try_swaps(deck, f, d, n)
@best[n] = d if d > @best[n]
(n-1).downto(0) do |i|
break if deck[i] == -1 || deck[i] == i
return if d + @best[i] <= @best[n]
end
deck2 = deck.dup
for i in 1...n
k = 1 << i
if deck2[i] == -1
next if f & k != 0
elsif deck2[i] != i
next
end
deck2[0] = i
deck2[1..i] = deck[0...i].reverse
try_swaps(deck2, f | k, d+1, n)
end
end
def topswops(n)
@best[n] = 0
deck0 = [-1] * (n + 1)
try_swaps(deck0, 1, 0, n)
@best[n]
end
@best = [0] * 16
for i in 1..10
puts "%2d : %d" % [i, topswops(i)]
end
[edit] Tcl
package require struct::list
proc swap {listVar} {
upvar 1 $listVar list
set n [lindex $list 0]
for {set i 0; set j [expr {$n-1}]} {$i<$j} {incr i;incr j -1} {
set tmp [lindex $list $i]
lset list $i [lindex $list $j]
lset list $j $tmp
}
}
proc swaps {list} {
for {set i 0} {[lindex $list 0] > 1} {incr i} {
swap list
}
return $i
}
proc topswops list {
set n 0
::struct::list foreachperm p $list {
set n [expr {max($n,[swaps $p])}]
}
return $n
}
proc topswopsTo n {
puts "n\ttopswops(n)"
for {set i 1} {$i <= $n} {incr i} {
puts $i\t[topswops [lappend list $i]]
}
}
topswopsTo 10
- Output:
n topswops(n) 1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
[edit] XPL0
code ChOut=8, CrLf=9, IntOut=11;
int N, Max, Card1(16), Card2(16);
proc Topswop(D); \Conway's card swopping game
int D; \depth of recursion
int I, J, C, T;
[if D # N then \generate N! permutations of 1..N in Card1
[for I:= 0 to N-1 do
[for J:= 0 to D-1 do \check if object (letter) already used
if Card1(J) = I+1 then J:=100;
if J < 100 then
[Card1(D):= I+1; \card number not used so append it
Topswop(D+1); \recurse next level deeper
];
];
]
else [\determine number of topswops to get card 1 at beginning
for I:= 0 to N-1 do Card2(I):= Card1(I); \make working copy of deck
C:= 0; \initialize swop counter
while Card2(0) # 1 do
[I:= 0; J:= Card2(0)-1;
while I < J do
[T:= Card2(I); Card2(I):= Card2(J); Card2(J):= T;
I:= I+1; J:= J-1;
];
C:= C+1;
];
if C>Max then Max:= C;
];
];
[for N:= 1 to 10 do
[Max:= 0;
Topswop(0);
IntOut(0, N); ChOut(0, ^ ); IntOut(0, Max); CrLf(0);
];
]
- Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
[edit] XPL0: Faster Version
code CrLf=9, IntOut=11, Text=12;
int N, D, Best(16);
proc TrySwaps(A, F, S);
int A, F, S;
int B(16), I, J, K;
[if D > Best(N) then Best(N):= D;
loop [if A(S)=-1 ! A(S)=S then quit;
if D+Best(S) <= Best(N) then return;
if S = 0 then quit;
S:= S-1;
];
D:= D+1;
for I:= 0 to S do B(I):= A(I);
K:= 1;
for I:= 1 to S do
[K:= K<<1;
if B(I)=-1 & (F&K)=0 ! B(I)=I then
[J:= I; B(0):= J;
while J do [J:= J-1; B(I-J):= A(J)];
TrySwaps(B, F!K, S);
];
];
D:= D-1;
];
int I, X(16);
[for I:= 0 to 16-1 do
[X(I):= -1; Best(I):= 0];
X(0):= 0;
for N:= 1 to 13 do
[D:= 0;
TrySwaps(X, 1, N-1);
IntOut(0, N); Text(0, ": "); IntOut(0, Best(N)); CrLf(0);
];
]
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38 11: 51 12: 65 13: 80
