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.
- Related tasks
360 Assembly
The program uses two ASSIST macro (XDECO,XPRNT) to keep the code as short as possible. <lang 360asm>* Topswops optimized 12/07/2016 TOPSWOPS CSECT
USING TOPSWOPS,R13 base register B 72(R15) skip savearea DC 17F'0' savearea STM R14,R12,12(R13) prolog ST R13,4(R15) " <- ST R15,8(R13) " -> LR R13,R15 " addressability MVC N,=F'1' n=1
LOOPN L R4,N n; do n=1 to 10 ===-------------==*
C R4,=F'10' " * BH ELOOPN . * MVC P(40),PINIT p=pinit MVC COUNTM,=F'0' countm=0
REPEAT MVC CARDS(40),P cards=p -------------------------+
SR R11,R11 count=0 |
WHILE CLC CARDS,=F'1' do while cards(1)^=1 ---------+
BE EWHILE . | MVC M,CARDS m=cards(1) L R2,M m SRA R2,1 m/2 ST R2,MD2 md2=m/2 L R3,M @card(mm)=m SLA R3,2 *4 LA R3,CARDS-4(R3) @card(mm) LA R2,CARDS @card(i)=0 LA R6,1 i=1
LOOPI C R6,MD2 do i=1 to m/2 -------------+
BH ELOOPI . | L R0,0(R2) swap r0=cards(i) MVC 0(4,R2),0(R3) swap cards(i)=cards(mm) ST R0,0(R3) swap cards(mm)=r0 AH R2,=H'4' @card(i)=@card(i)+4 SH R3,=H'4' @card(mm)=@card(mm)-4 LA R6,1(R6) i=i+1 | B LOOPI ----------------------------+
ELOOPI LA R11,1(R11) count=count+1 |
B WHILE -------------------------------+
EWHILE C R11,COUNTM if count>countm
BNH NOTGT then ST R11,COUNTM countm=count
NOTGT BAL R14,NEXTPERM call nextperm
LTR R0,R0 until nextperm=0 | BNZ REPEAT ---------------------------------+ L R1,N n XDECO R1,XDEC edit n MVC PG(2),XDEC+10 output n MVI PG+2,C':' output ':' L R1,COUNTM countm XDECO R1,XDEC edit countm MVC PG+3(4),XDEC+8 output countm XPRNT PG,L'PG print buffer L R1,N n * LA R1,1(R1) +1 * ST R1,N n=n+1 * B LOOPN ===------------------------------==*
ELOOPN L R13,4(0,R13) epilog
LM R14,R12,12(R13) " restore XR R15,R15 " rc=0 BR R14 exit
PINIT DC F'1',F'2',F'3',F'4',F'5',F'6',F'7',F'8',F'9',F'10' CARDS DS 10F cards P DS 10F p COUNTM DS F countm M DS F m N DS F n MD2 DS F m/2 PG DC CL20' ' buffer XDEC DS CL12 temp
- ------- ---- nextperm ----------{-----------------------------------
NEXTPERM L R9,N nn=n
SR R8,R8 jj=0 LR R7,R9 nn BCTR R7,0 j=nn-1 LTR R7,R7 if j=0 BZ ELOOPJ1 then skip do loop
LOOPJ1 LR R1,R7 do j=nn-1 to 1 by -1; j ----+
SLA R1,2 . | L R2,P-4(R1) p(j) C R2,P(R1) if p(j)<p(j+1) BNL PJGEPJP then LR R8,R7 jj=j B ELOOPJ1 leave j |
PJGEPJP BCT R7,LOOPJ1 j=j-1 ---------------------+ ELOOPJ1 LA R7,1(R8) j=jj+1 LOOPJ2 CR R7,R9 do j=jj+1 while j<nn ------+
BNL ELOOPJ2 . | LR R2,R7 j SLA R2,2 . LR R3,R9 nn SLA R3,2 . L R0,P-4(R2) swap p(j),p(nn) L R1,P-4(R3) " ST R0,P-4(R3) " ST R1,P-4(R2) " BCTR R9,0 nn=nn-1 LA R7,1(R7) j=j+1 | B LOOPJ2 ----------------------------+
ELOOPJ2 LTR R8,R8 if jj=0
BNZ JJNE0 then LA R0,0 return(0) BR R14 "
JJNE0 LA R7,1(R8) j=jj+1
LR R2,R7 j SLA R2,2 r@p(j) LR R3,R8 jj SLA R3,2 r@p(jj)
LOOPJ3 L R0,P-4(R2) p(j) ----------------------+
C R0,P-4(R3) do j=jj+1 while p(j)<p(jj) | BNL ELOOPJ3 LA R2,4(R2) r@p(j)=r@p(j)+4 LA R7,1(R7) j=j+1 | B LOOPJ3 ----------------------------+
ELOOPJ3 L R1,P-4(R3) swap p(j),p(jj)
ST R0,P-4(R3) " ST R1,P-4(R2) " LA R0,1 return(1) BR R14 ---------------}----------------------------------- YREGS END TOPSWOPS</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
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>
C++
<lang cpp>
- include <iostream>
- include <vector>
- include <numeric>
- include <algorithm>
int topswops(int n) {
std::vector<int> list(n); std::iota(std::begin(list), std::end(list), 1); int max_steps = 0; do { auto temp_list = list; for (int steps = 1; temp_list[0] != 1; ++steps) { std::reverse(std::begin(temp_list), std::begin(temp_list) + temp_list[0]); if (steps > max_steps) max_steps = steps; } } while (std::next_permutation(std::begin(list), std::end(list))); return max_steps;
}
int main() {
for (int i = 1; i <= 10; ++i) { std::cout << i << ": " << topswops(i) << std::endl; } return 0;
}</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
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.typecons;
__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; staticIota!(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; staticIota!(0, n)) // Copy. deck2[i] = deck[i];
d++; foreach (immutable i; staticIota!(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; staticIota!(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 (immutable i; staticIota!(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.
Eiffel
<lang Eiffel> class TOPSWOPS
create make
feature
make (n: INTEGER) -- Topswop game. local perm, ar: ARRAY [INTEGER] tcount, count: INTEGER do create perm_sol.make_empty create solution.make_empty across 1 |..| n as c loop create ar.make_filled (0, 1, c.item) across 1 |..| c.item as d loop ar [d.item] := d.item end permute (ar, 1) across 1 |..| perm_sol.count as e loop tcount := 0 from until perm_sol.at (e.item).at (1) = 1 loop perm_sol.at (e.item) := reverse_array (perm_sol.at (e.item)) tcount := tcount + 1 end if tcount > count then count := tcount end end solution.force (count, c.item) end end
solution: ARRAY [INTEGER]
feature {NONE}
perm_sol: ARRAY [ARRAY [INTEGER]]
reverse_array (ar: ARRAY [INTEGER]): ARRAY [INTEGER] -- Array with 'ar[1]' elements reversed. require ar_not_void: ar /= Void local i, j: INTEGER do create Result.make_empty Result.deep_copy (ar) from i := 1 j := ar [1] until i > j loop Result [i] := ar [j] Result [j] := ar [i] i := i + 1 j := j - 1 end ensure same_elements: across ar as a all Result.has (a.item) end end
permute (a: ARRAY [INTEGER]; k: INTEGER) -- All permutations of array 'a' stored in perm_sol. require ar_not_void: a.count >= 1 k_valid_index: k > 0 local i, t: INTEGER temp: ARRAY [INTEGER] do create temp.make_empty if k = a.count then across a as ar loop temp.force (ar.item, temp.count + 1) end perm_sol.force (temp, perm_sol.count + 1) else from i := k until i > a.count loop t := a [k] a [k] := a [i] a [i] := t permute (a, k + 1) t := a [k] a [k] := a [i] a [i] := t i := i + 1 end end end
end </lang> Test: <lang Eiffel> class APPLICATION
create make
feature
make do create topswop.make (10) across topswop.solution as t loop io.put_string (t.item.out + "%N") end end
topswop: TOPSWOPS
end </lang>
- Output:
0 1 2 4 7 10 16 22 30 38
Elixir
<lang elixir>defmodule Topswops do
def get_1_first( [1 | _t] ), do: 0 def get_1_first( list ), do: 1 + get_1_first( swap(list) ) defp swap( [n | _t]=list ) do {swaps, remains} = Enum.split( list, n ) Enum.reverse( swaps, remains ) end def task do IO.puts "N\ttopswaps" Enum.map(1..10, fn n -> {n, permute(Enum.to_list(1..n))} end) |> Enum.map(fn {n, n_permutations} -> {n, get_1_first_many(n_permutations)} end) |> Enum.map(fn {n, n_swops} -> {n, Enum.max(n_swops)} end) |> Enum.each(fn {n, max} -> IO.puts "#{n}\t#{max}" end) end def get_1_first_many( n_permutations ), do: (for x <- n_permutations, do: get_1_first(x)) defp permute([]), do: [[]] defp permute(list), do: for x <- list, y <- permute(list -- [x]), do: [x|y]
end
Topswops.task</lang>
- Output:
N topswaps 1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
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
Factor
<lang factor>USING: formatting kernel math math.combinatorics math.order math.ranges sequences ; FROM: sequences.private => exchange-unsafe ; IN: rosetta-code.topswops
! Reverse a subsequence in-place from 0 to n.
- head-reverse! ( seq n -- seq' )
dupd [ 2/ ] [ ] bi rot [ [ over - 1 - ] dip exchange-unsafe ] 2curry each-integer ;
! Reverse the elements in seq according to the first element.
- swop ( seq -- seq' ) dup first head-reverse! ;
! Determine the number of swops until 1 is the head.
- #swops ( seq -- n )
0 swap [ dup first 1 = ] [ [ 1 + ] [ swop ] bi* ] until drop ;
! Determine the maximum number of swops for a given length.
- topswops ( n -- max )
[1,b] <permutations> [ #swops ] [ max ] map-reduce ;
- main ( -- )
10 [1,b] [ dup topswops "%2d: %2d\n" printf ] each ;
MAIN: main</lang>
- Output:
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
Searching permutations
<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_ (putStrLn . ((++) <$> show <*> (":\t" ++) . show . topswops)) [1 .. 10]</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Searching derangements
Alternate version
Uses only permutations with all elements out of place.
<lang Haskell>import Data.List (permutations, inits)
import Control.Arrow (first)
derangements :: [Int] -> Int derangements = (\x -> filter (and . zipWith (/=) x)) <*> permutations
topswop :: Int -> [a] -> [a] topswop x xs = uncurry (++) (first reverse (splitAt x xs))
topswopIter :: [Int] -> Int topswopIter = takeWhile ((/= 1) . head) . iterate (topswop =<< head)
swops :: [Int] -> [Int] swops = fmap (length . topswopIter) . derangements
topSwops :: [Int] -> [(Int, Int)] topSwops = zip [1 ..] . fmap (maximum . (0 :) . swops) . tail . inits
main :: IO () main = mapM_ print $ take 10 $ topSwops [1 ..]</lang> Output
(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
jq
The following uses permutations and is therefore impractical for n>10 or so.
Infrastructure: <lang jq># "while" as defined here is included in recent versions (>1.4) of jq: def until(cond; next):
def _until: if cond then . else (next|_until) end; _until;
- Generate a stream of permutations of [1, ... n].
- This implementation uses arity-0 filters for speed.
def permutations:
# Given a single array, insert generates a stream by inserting (length+1) at different positions def insert: # state: [m, array] .[0] as $m | (1+(.[1]|length)) as $n | .[1] | if $m >= 0 then (.[0:$m] + [$n] + .[$m:]), ([$m-1, .] | insert) else empty end;
if .==0 then [] elif . == 1 then [1] else . as $n | ($n-1) | permutations | [$n-1, .] | insert end;</lang>
Topswops: <lang jq># Input: a permutation; output: an integer def flips:
# state: [i, array] [0, .] | until( .[1][0] == 1; .[1] as $p | $p[0] as $p0
| [.[0] + 1, ($p[:$p0] | reverse) + $p[$p0:] ] )
| .[0];
- input: n, the number of items
def fannkuch:
reduce permutations as $p (0; [., ($p|flips) ] | max);</lang>
Example: <lang jq>range(1; 11) | [., fannkuch ]</lang>
- Output:
<lang sh>$ jq -n -c -f topswops.jq [1,0] [2,1] [3,2] [4,4] [5,7] [6,10] [7,16] [8,22] [9,30] [10,38]</lang>
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
Kotlin
<lang scala>// version 1.1.2
val best = IntArray(32)
fun trySwaps(deck: IntArray, f: Int, d: Int, n: Int) {
if (d > best[n]) best[n] = d for (i in n - 1 downTo 0) { if (deck[i] == -1 || deck[i] == i) break if (d + best[i] <= best[n]) return } val deck2 = deck.copyOf() for (i in 1 until n) { val k = 1 shl i if (deck2[i] == -1) { if ((f and k) != 0) continue } else if (deck2[i] != i) continue deck2[0] = i for (j in i - 1 downTo 0) deck2[i - j] = deck[j] trySwaps(deck2, f or k, d + 1, n) }
}
fun topswops(n: Int): Int {
require(n > 0 && n < best.size) best[n] = 0 val deck0 = IntArray(n + 1) for (i in 1 until n) deck0[i] = -1 trySwaps(deck0, 1, 0, n) return best[n]
}
fun main(args: Array<String>) {
for (i in 1..10) println("${"%2d".format(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
Lua
<lang Lua>-- Return an iterator to produce every permutation of list function permute (list)
local function perm (list, n) if n == 0 then coroutine.yield(list) end for i = 1, n do list[i], list[n] = list[n], list[i] perm(list, n - 1) list[i], list[n] = list[n], list[i] end end return coroutine.wrap(function() perm(list, #list) end)
end
-- Perform one topswop round on table t function swap (t)
local new, limit = {}, t[1] for i = 1, #t do if i <= limit then new[i] = t[limit - i + 1] else new[i] = t[i] end end return new
end
-- Find the most swaps needed for any starting permutation of n cards function topswops (n)
local numTab, highest, count = {}, 0 for i = 1, n do numTab[i] = i end for numList in permute(numTab) do count = 0 while numList[1] ~= 1 do numList = swap(numList) count = count + 1 end if count > highest then highest = count end end return highest
end
-- Main procedure for i = 1, 10 do print(i, topswops(i)) end</lang>
- Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
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
Nim
<lang nim>import strformat
const maxBest = 32 var best: array[maxBest, int]
proc trySwaps(deck: seq[int], f, d, n: int) =
if d > best[n]: best[n] = d for i in countdown(n - 1, 0): if deck[i] == -1 or deck[i] == i: break if d + best[i] <= best[n]: return var deck2 = deck for i in 1..<n: var k = 1 shl i if deck2[i] == -1: if (f and k) != 0: continue elif deck2[i] != i: continue deck2[0] = i for j in countdown(i - 1, 0): deck2[i - j] = deck[j] trySwaps(deck2, f or k, d + 1, n)
proc topswops(n: int): int =
assert(n > 0 and n < maxBest) best[n] = 0 var deck0 = newSeq[int](n + 1) for i in 1..<n: deck0[i] = -1 trySwaps(deck0, 1, 0, n) best[n]
for i in 1..10:
echo &"{i:2}: {topswops(i):2}"</lang>
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
PARI/GP
Naive solution: <lang parigp>flip(v:vec)={
my(t=v[1]+1); if (t==2, return(0)); for(i=1,t\2, [v[t-i],v[i]]=[v[i],v[t-i]]); 1+flip(v)
} topswops(n)={
my(mx); for(i=0,n!-1, mx=max(flip(Vecsmall(numtoperm(n,i))),mx) ); mx;
} vector(10,n,topswops(n))</lang>
- Output:
%1 = [0, 1, 2, 4, 7, 10, 16, 22, 30, 38]
An efficient solution would use PARI, following the C solution.
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
Phix
Originally contributed by Jason Gade as part of the Euphoria version of the Great Computer Language Shootout benchmarks. <lang Phix>function fannkuch(integer n) sequence start = tagset(n),
perm, perm1 = start, count = start
integer maxFlipsCount = 0, r = n+1 integer perm0, flipsCount, k, k2, j, j2
while 1 do while r!=1 do count[r-1] = r r -= 1 end while if not (perm1[1]=1 or perm1[n]=n) then perm = perm1 flipsCount = 0 k = perm[1] while k!=1 do k2 = floor((k+1)/2) perm = reverse(perm[1..k]) & perm[k+1..n] flipsCount += 1 k = perm[1] end while if flipsCount>maxFlipsCount then maxFlipsCount = flipsCount end if end if -- Use incremental change to generate another permutation while 1 do if r>n then return maxFlipsCount end if perm0 = perm1[1] j2 = 1 while j2<r do j = j2+1 perm1[j2] = perm1[j] j2 = j end while perm1[r] = perm0 count[r] = count[r]-1 if count[r]>1 then exit else r += 1 end if end while end while
end function -- fannkuch
for i=1 to 10 do
? fannkuch(i)
end for</lang>
- Output:
0 1 2 4 7 10 16 22 30 38
PicoLisp
<lang PicoLisp>(de fannkuch (N)
(let (Lst (range 1 N) L Lst Max) (recur (L) # Permute (if (cdr L) (do (length L) (recurse (cdr L)) (rot L) ) (zero N) # For each permutation (for (P (copy Lst) (> (car P) 1) (flip P (car P))) (inc 'N) ) (setq Max (max N Max)) ) ) Max ) )
(for I 10
(println I (fannkuch I)) )</lang>
Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
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 perml = range(0, n), count = list(n), perm = list(n)
loop: 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): (n, maxf) say 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
n = argv(1) number if (n<1): n=10. fannkuch(n) </lang>
Output follows that of Raku and Python, ~2.5x faster than perl5
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
R
Using iterpc package for optimization <lang R> topswops <- function(x){
i <- 0 while(x[1] != 1){ first <- x[1] if(first == length(x)){ x <- rev(x) } else{ x <- c(x[first:1], x[(first+1):length(x)]) } i <- i + 1 } return(i)
}
library(iterpc)
result <- NULL
for(i in 1:10){
I <- iterpc(i, labels = 1:i, ordered = T) A <- getall(I) A <- data.frame(A) A$flips <- apply(A, 1, topswops) result <- rbind(result, c(i, max(A$flips)))
} </lang>
Output:
[,1] [,2] [1,] 1 0 [2,] 2 1 [3,] 3 2 [4,] 4 4 [5,] 5 7 [6,] 6 10 [7,] 7 16 [8,] 8 22 [9,] 9 30 [10,] 10 38
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
Raku
(formerly Perl 6) <lang perl6>sub swops(@a is copy) {
my int $count = 0; until @a[0] == 1 { @a[ ^@a[0] ] .= reverse; ++$count; } $count
}
sub topswops($n) { max (1..$n).permutations.race.map: *.&swops }
say "$_ {topswops $_}" for 1 .. 10;</lang>
- Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
REXX
The decks function is a modified permSets (permutation sets) subroutine,
and is optimized somewhat to take advantage by eliminating one-swop "decks".
<lang rexx>/*REXX program generates N decks of numbered cards and finds the maximum "swops". */
parse arg things .; if things== then things=10
do n=1 for things; #=decks(n, n) /*create a (things) number of "decks". */ mx= (n\==1) /*handle the case of a one-card deck.*/ do i=1 for #; p=swops(!.i) /*compute the SWOPS for this iteration.*/ if p>mx then mx=p /*This a new maximum? Use a new max. */ end /*i*/ say '──────── maximum swops for a deck of' right(n,2) ' cards is' right(mx,4) end /*n*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ decks: procedure expose !.; parse arg x,y,,$ @. /* X things taken Y at a time. */
#=0; call .decks 1 /* [↑] initialize $ & @. to null.*/ return # /*return number of permutations (decks)*/
.decks: procedure expose !. @. x y $ #; parse arg ?
if ?>y then do; _=@.1; do j=2 for y-1; _=_ @.j; end /*j*/; #=#+1; !.#=_ end else do; qm=? - 1 if ?==1 then qs=2 /*don't use 1-swops that start with 1 */ else if @.1==? then qs=2 /*skip the 1-swops: 3 x 1 x ···*/ else qs=1 do q=qs to x /*build the permutations recursively. */ do k=1 for qm; if @.k==q then iterate q end /*k*/ @.?=q; call .decks ? + 1 end /*q*/ end return
/*──────────────────────────────────────────────────────────────────────────────────────*/
swops: parse arg z; do u=1; parse var z t .; if \datatype(t, 'W') then t=x2d(t)
if word(z, t)==1 then return u /*found unity at T. */ do h=10 to things; if pos(h, z)==0 then iterate z=changestr(h, z, d2x(h) ) /* [↑] any H's in Z?*/ end /*h*/ z=reverse( subword(z, 1, t) ) subword(z, t + 1) end /*u*/</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>
Scala
<lang Scala>object Fannkuch extends App {
def fannkuchen(l: List[Int], n: Int, i: Int, acc: Int): Int = { def flips(l: List[Int]): Int = (l: @unchecked) match { case 1 :: ls => 0 case (n :: ls) => val splitted = l.splitAt(n) flips(splitted._2.reverse_:::(splitted._1)) + 1 }
def rotateLeft(l: List[Int]) = l match { case Nil => List() case x :: xs => xs ::: List(x) }
if (i >= n) acc else { if (n == 1) acc.max(flips(l)) else { val split = l.splitAt(n) fannkuchen(rotateLeft(split._1) ::: split._2, n, i + 1, fannkuchen(l, n - 1, 0, acc)) } } } // def fannkuchen(
val result = (1 to 10).map(i => (i, fannkuchen(List.range(1, i + 1), i, 0, 0))) println("Computing results...") result.foreach(x => println(s"Pfannkuchen(${x._1})\t= ${x._2}")) assert(result == Vector((1, 0), (2, 1), (3, 2), (4, 4), (5, 7), (6, 10), (7, 16), (8, 22), (9, 30), (10, 38)), "Bad results") println(s"Successfully completed without errors. [total ${scala.compat.Platform.currentTime - executionStart} ms]")
}</lang>
- Output:
Computing results... 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 Successfully completed without errors. [total 7401 ms] Process finished with exit code 0
Tcl
Probably an integer overflow at n=10.
<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(_).pump(List,"copy",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