Topswops
Topswops is a card game created by John Conway in the 1970's.
You are encouraged to solve this task according to the task description, using any language you may know.
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.
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 used [[1]].
<lang Ada>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;</lang>
- Output:
0 1 2 4 7 10 16 22 30 38
AutoHotkey
<lang AutoHotkey>Topswops(Obj, n){ R := [] for i, val in obj{ if (i <=n) res := val (A_Index=1?"":",") res else res .= "," val } Loop, Parse, res, `, R[A_Index]:= A_LoopField return R }</lang> Examples:<lang AutoHotkey>Cards := [2, 4, 1, 3] Res := Print(Cards) while (Cards[1]<>1) { Cards := Topswops(Cards, Cards[1]) Res .= "`n"Print(Cards) } MsgBox % Res
Print(M){ for i, val in M Res .= (A_Index=1?"":"`t") val return Trim(Res,"`n") }</lang>
Outputs:
2 4 1 3 4 2 1 3 3 1 2 4 2 1 3 4 1 2 3 4
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): <lang c>#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; }</lang> 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. <lang c>#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; }</lang>
D
Permutations generator from: http://rosettacode.org/wiki/Permutations#Faster_Lazy_Version
<lang d>import std.stdio, std.algorithm, std.range, permutations2;
int topswops(in int n) pure @safe {
static int flip(int[] xa) pure nothrow @safe @nogc {
if (!xa[0]) return 0;
xa[0 .. xa[0] + 1].reverse();
return 1 + flip(xa);
}
return n.iota.array.permutations.map!flip.reduce!max;
}
void main() {
foreach (immutable i; 1 .. 11)
writeln(i, ": ", i.topswops);
}</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
D: Faster Version
<lang d>import std.stdio, std.typetuple;
template Range(int start, int stop) {
static if (stop <= start)
alias Range = TypeTuple!();
else
alias Range = TypeTuple!(Range!(start, stop - 1), stop - 1);
}
__gshared uint[32] best;
uint topswops(size_t n)() nothrow @nogc {
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 @nogc {
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] = 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());
}</lang>
- 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.
Erlang
This code is using the permutation code by someone else. Thank you. <lang Erlang> -module( topswops ).
-export( [get_1_first/1, swap/1, task/0] ).
get_1_first( [1 | _T] ) -> 0; get_1_first( List ) -> 1 + get_1_first( swap(List) ).
swap( [N | _T]=List ) -> {Swaps, Remains} = lists:split( N, List ), lists:reverse( Swaps ) ++ Remains.
task() -> Permutations = [{X, permute:permute(lists:seq(1, X))} || X <- lists:seq(1, 10)], Swops = [{N, get_1_first_many(N_permutations)} || {N, N_permutations} <- Permutations], Topswops = [{N, lists:max(N_swops)} || {N, N_swops} <- Swops], io:fwrite( "N topswaps~n" ), [io:fwrite("~p ~p~n", [N, Max]) || {N, Max} <- Topswops].
get_1_first_many( N_permutations ) -> [get_1_first(X) || X <- N_permutations]. </lang>
- Output:
42> topswops:task(). N topswaps 1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
Fortran
<lang 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 </lang>
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
Haskell
<lang 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]</lang>
- 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.
<lang Haskell>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</lang> 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)
Icon and Unicon
This doesn't compile in Icon only because of the use of list comprehension to build the original list of 1..n values.
<lang unicon>procedure main()
every n := 1 to 10 do {
ts := 0
every (ts := 0) <:= swop(permute([: 1 to n :]))
write(right(n, 3),": ",right(ts,4))
}
end
procedure swop(A)
count := 0
while A[1] ~= 1 do {
A := reverse(A[1+:A[1]]) ||| A[(A[1]+1):0]
count +:= 1
}
return count
end
procedure permute(A)
if *A <= 1 then return A suspend [(A[1]<->A[i := 1 to *A])] ||| permute(A[2:0])
end</lang>
Sample run:
->topswop 1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38 ->
J
Solution:<lang j> swops =: ((|.@:{. , }.)~ {.)^:a:</lang> Example (from task introduction):<lang j> swops 2 4 1 3 2 4 1 3 4 2 1 3 3 1 2 4 2 1 3 4 1 2 3 4</lang> Example (topswops of all permutations of the integers 1..10):<lang j> (,. _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</lang> Notes: Readers less familiar with array-oriented programming may find an alternate solution written in the structured programming style more accessible.
Java
<lang 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));
}
}</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Julia
Fast, efficient version <lang julia>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</lang>
- 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
Mathematica
An exhaustive search of all possible permutations is done <lang Mathematica>flip[a_] :=
Block[{a1 = First@a},
If[a1 == Length@a, Reverse[a],
Join[Reverse[a;; a1], aa1 + 1 ;;]]]
swaps[a_] := Length@FixedPointList[flip, a] - 2
Print[#, ": ", Max[swaps /@ Permutations[Range@#]]] & /@ Range[10];</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Perl
Recursive backtracking solution, starting with the final state and going backwards. <lang perl> sub next_swop {
my( $max, $level, $p, $d ) = @_;
my $swopped = 0;
for( 2..@$p ){ # find possibilities
my @now = @$p;
if( $_ == $now[$_-1] ) {
splice @now, 0, 0, reverse splice @now, 0, $_;
$swopped = 1;
next_swop( $max, $level+1, \@now, [ @$d ] );
}
}
for( 1..@$d ) { # create possibilities
my @now = @$p;
my $next = shift @$d;
if( not $now[$next-1] ) {
$now[$next-1] = $next;
splice @now, 0, 0, reverse splice @now, 0, $next;
$swopped = 1;
next_swop( $max, $level+1, \@now, [ @$d ] );
}
push @$d, $next;
}
$$max = $level if !$swopped and $level > $$max;
}
sub topswops {
my $n = shift; my @d = 2..$n; my @p = ( 1, (0) x ($n-1) ); my $max = 0; next_swop( \$max, 0, \@p, \@d ); return $max;
}
printf "Maximum swops for %2d cards: %2d\n", $_, topswops $_ for 1..10; </lang>
- Output:
Maximum swops for 1 cards: 0 Maximum swops for 2 cards: 1 Maximum swops for 3 cards: 2 Maximum swops for 4 cards: 4 Maximum swops for 5 cards: 7 Maximum swops for 6 cards: 10 Maximum swops for 7 cards: 16 Maximum swops for 8 cards: 22 Maximum swops for 9 cards: 30 Maximum swops for 10 cards: 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.
PL/I
<lang PL/I> (subscriptrange): topswap: procedure options (main); /* 12 November 2013 */
declare cards(*) fixed (2) controlled, t fixed (2); declare dealt(*) bit(1) controlled; declare (count, i, m, n, c1, c2) fixed binary; declare random builtin;
do n = 1 to 10;
allocate cards(n), dealt(n);
/* Take the n cards, in order ... */
do i = 1 to n; cards(i) = i; end;
/* ... and shuffle them. */
do i = 1 to n;
c1 = random*n+1; c2 = random*n+1;
t = cards(c1); cards(c1) = cards(c2); cards(c2) = t;
end;
/* If '1' is the first card, game is trivial; swap it with another. */
if cards(1) = 1 & n > 1 then
do; t = cards(1); cards(1) = cards(2); cards(2) = t; end;
count = 0;
do until (cards(1) = 1);
/* take the value of the first card, M, and reverse the first M cards. */
m = cards(1);
do i = 1 to m/2;
t = cards(i); cards(i) = cards(m-i+1); cards(m-i+1) = t;
end;
count = count + 1;
end;
put skip edit (n, ':', count) (f(2), a, f(4));
end;
end topswap; </lang>
1: 1 2: 1 3: 2 4: 2 5: 4 6: 2 7: 1 8: 9 9: 16 10: 1
Potion
<lang potion>range = (a, b):
i = 0, l = list(b-a+1) while (a + i <= b): l (i) = a + i++. l.
fannkuch = (n):
flips = 0, maxf = 0, k = 0, m = n - 1, r = n, check = 0 perml = range(0, n), count = list(n), perm = list(n)
loop:
if (check < 30):
perml join print, "\n" print
check++.
while (r != 1):
count (r-1) = r
r--.
if (perml (0) != 0 and perml (m) != m):
flips = 0, i = 1
while (i < n):
perm (i) = perml (i)
i++.
k = perml (0)
loop:
i = 1, j = k - 1
while (i < j):
t = perm (i), perm (i) = perm (j), perm (j) = t
i++, j--.
flips++
j = perm (k), perm (k) = k, k = j
if (k == 0): break.
.
if (flips > maxf): maxf = flips.
.
loop:
if (r == n):
maxf string print, "\n" print
return (maxf).
i = 0, j = perml (0)
while (i < r):
k = i + 1
perml (i) = perml (k)
i = k.
perml (r) = j
j = count (r) - 1
count (r) = j
if (j > 0): break.
r++
_ n
fannkuch(11) </lang>
Output follows that of Perl6 and 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 = 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 >>> </lang>
Python: Faster Version
<lang python>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))</lang>
- 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
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>
- 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))) </lang>
Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
REXX
The deckSets subroutine is a modified permSets (permutation sets) subroutine,
and is optimized somewhat to take advantage by eliminating one-swop "decks".
<lang rexx>/*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 /*_*/</lang>
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
Ruby
<lang ruby>def f1(a)
i = 0 while (a0 = a[0]) > 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</lang>
- Output:
1 : 0 2 : 1 3 : 2 4 : 4 5 : 7 6 : 10 7 : 16 8 : 22 9 : 30 10 : 38
Faster Version
<lang ruby>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</lang>
Tcl
<lang 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</lang>
- Output:
n topswops(n) 1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
XPL0
<lang 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); ];
]</lang>
- Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
XPL0: Faster Version
<lang XPL0>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);
];
]</lang>
- 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
zkl
Slow version <lang zkl>fcn topswops(n){
flip:=fcn(xa){
if (not xa[0]) return(0);
xa.reverse(0,xa[0]+1); // inplace, ~4x faster than making new lists
return(1 + self.fcn(xa));
};
(0).pump(n,List):Utils.Helpers.permute(_).apply("copy").apply(flip).reduce("max");
}
foreach n in ([1 .. 10]){ println(n, ": ", topswops(n)) }</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38