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 word Pfannkuchen meaning pancake.
- Related tasks
Contents
- 1 11l
- 2 360 Assembly
- 3 Ada
- 4 AutoHotkey
- 5 C
- 6 C++
- 7 D
- 8 Eiffel
- 9 Elixir
- 10 Erlang
- 11 Factor
- 12 Fortran
- 13 Go
- 14 Haskell
- 15 Icon and Unicon
- 16 J
- 17 Java
- 18 jq
- 19 Julia
- 20 Kotlin
- 21 Lua
- 22 Mathematica
- 23 Nim
- 24 PARI/GP
- 25 Perl
- 26 Phix
- 27 PicoLisp
- 28 PL/I
- 29 Potion
- 30 Python
- 31 R
- 32 Racket
- 33 Raku
- 34 REXX
- 35 Ruby
- 36 Rust
- 37 Scala
- 38 Tcl
- 39 Wren
- 40 XPL0
- 41 zkl
11l[edit]
V best = [0] * 16
F try_swaps(&deck, f, =s, d, n)
I d > :best[n]
:best[n] = d
V i = 0
V k = 1 << s
L s != 0
k >>= 1
s--
I deck[s] == -1 | deck[s] == s
L.break
i [|]= k
I (i [&] f) == i & d + :best[s] <= :best[n]
R d
s++
V deck2 = copy(deck)
k = 1
L(i2) 1 .< s
k <<= 1
I deck2[i2] == -1
I (f [&] k) != 0
L.continue
E I deck2[i2] != i2
L.continue
deck[i2] = i2
L(j) 0 .. i2
deck2[j] = deck[i2 - j]
try_swaps(&deck2, f [|] k, s, 1 + d, n)
F topswops(n)
:best[n] = 0
V deck0 = [-1] * 16
deck0[0] = 0
try_swaps(&deck0, 1, n, 0, n)
R :best[n]
L(i) 1..12
print(‘#2: #.’.format(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: 63
360 Assembly[edit]
The program uses two ASSIST macro (XDECO,XPRNT) to keep the code as short as possible.
* 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)[email protected](i)+4
SH R3,=H'4' @card(mm)[email protected](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 [email protected](j)
LR R3,R8 jj
SLA R3,2 [email protected](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) [email protected](j)[email protected](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
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Ada[edit]
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]].
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
AutoHotkey[edit]
Topswops(Obj, n){Examples:
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
}
Cards := [2, 4, 1, 3]Outputs:
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")
}
2 4 1 3 4 2 1 3 3 1 2 4 2 1 3 4 1 2 3 4
C[edit]
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;
}
C++[edit]
#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;
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
D[edit]
Permutations generator from: http://rosettacode.org/wiki/Permutations#Faster_Lazy_Version
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);
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
D: Faster Version[edit]
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());
}
- 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[edit]
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
Test:
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
- Output:
0 1 2 4 7 10 16 22 30 38
Elixir[edit]
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
- Output:
N topswaps 1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
Erlang[edit]
This code is using the permutation code by someone else. Thank you.
-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].
- 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[edit]
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
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Fortran[edit]
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
Go[edit]
// 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
Haskell[edit]
Searching permutations[edit]
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]
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Searching derangements[edit]
Alternate version
Uses only permutations with all elements out of place.
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 ..]
Output
(1,0) (2,1) (3,2) (4,4) (5,7) (6,10) (7,16) (8,22) (9,30) (10,38)
Icon and Unicon[edit]
This doesn't compile in Icon only because of the use of list comprehension to build the original list of 1..n values.
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
Sample run:
->topswop 1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38 ->
J[edit]
Solution:swops =: ((|[email protected]:{. , }.)~ {.)^: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 + ! >./@:(#@[email protected] >:)&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.
Java[edit]
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
jq[edit]
The following uses permutations and is therefore impractical for n>10 or so.
Infrastructure:
# "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;
Topswops:
# 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);
Example:
range(1; 11) | [., fannkuch ]
- Output:
$ 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]
Julia[edit]
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
Kotlin[edit]
// 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)}")
}
- Output:
1 : 0 2 : 1 3 : 2 4 : 4 5 : 7 6 : 10 7 : 16 8 : 22 9 : 30 10 : 38
Lua[edit]
-- 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
- Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
Mathematica[edit]
An exhaustive search of all possible permutations is done
flip[a_] :=
Block[{a1 = [email protected]},
If[a1 == [email protected], Reverse[a],
Join[Reverse[a[[;; a1]]], a[[a1 + 1 ;;]]]]]
swaps[a_] := [email protected][flip, a] - 2
Print[#, ": ", Max[swaps /@ Permutations[[email protected]#]]] & /@ Range[10];
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Nim[edit]
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}"
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
PARI/GP[edit]
Naive solution:
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))
- Output:
%1 = [0, 1, 2, 4, 7, 10, 16, 22, 30, 38]
An efficient solution would use PARI, following the C solution.
Perl[edit]
Recursive backtracking solution, starting with the final state and going backwards.
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;
- 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[edit]
Originally contributed by Jason Gade as part of the Euphoria version of the Great Computer Language Shootout benchmarks.
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
- Output:
0 1 2 4 7 10 16 22 30 38
PicoLisp[edit]
(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)) )
Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
PL/I[edit]
(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;
1: 1 2: 1 3: 2 4: 2 5: 4 6: 2 7: 1 8: 9 9: 16 10: 1
Potion[edit]
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)
Output follows that of Raku and Python, ~2.5x faster than perl5
Python[edit]
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
>>>
Python: Faster Version[edit]
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
R[edit]
Using iterpc package for optimization
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)))
}
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[edit]
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
Raku[edit]
(formerly Perl 6)
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 $_}" for 1 .. 10;
- Output:
1 0 2 1 3 2 4 4 5 7 6 10 7 16 8 22 9 30 10 38
Alternately, using string manipulation instead. Much faster, though honestly, still not very fast.
sub swops($a is copy) {
my int $count = 0;
while (my \l = $a.ord) > 1 {
$a = $a.substr(0, l).flip ~ $a.substr(l);
++$count;
}
$count
}
sub topswops($n) { max (1..$n).permutations.map: { .chrs.join.&swops } }
say "$_ {topswops $_}" for 1 .. 10;
Same output
REXX[edit]
The decks function is a modified permSets (permutation sets) subroutine,
and is optimized somewhat to take advantage by eliminating one-swop "decks".
/*REXX program generates N decks of numbered cards and finds the maximum "swops". */
parse arg things .; if things=='' then things= 10
do n=1 for things; #= decks(n, n) /*create a (things) number of "decks". */
mx= n\==1 /*handle the case of a one-card deck.*/
do i=1 for #; p= swops(!.i) /*compute the SWOPS for this iteration.*/
if p>mx then mx= p /*This a new maximum? Use a new max. */
end /*i*/
say '──────── maximum swops for a deck of' right(n,2) ' cards is' right(mx,4)
end /*n*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
decks: procedure expose !.; parse arg x,y,,$ @. /* 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; [email protected].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*/
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[edit]
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
- 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
Rust[edit]
use itertools::Itertools;
fn solve(deck: &[usize]) -> usize {
let mut counter = 0_usize;
let mut shuffle = deck.to_vec();
loop {
let p0 = shuffle[0];
if p0 == 1 {
break;
}
shuffle[..p0].reverse();
counter += 1;
}
counter
}
// this is a naive method which tries all permutations and works up to ~12 cards
fn topswops(number: usize) -> usize {
(1..=number)
.permutations(number)
.fold(0_usize, |mut acc, p| {
let steps = solve(&p);
if steps > acc {
acc = steps;
}
acc
})
}
fn main() {
(1_usize..=10).for_each(|x| println!("{}: {}", x, topswops(x)));
}
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
Scala[edit]
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]")
}
- 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[edit]
Probably an integer overflow at n=10.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
Wren[edit]
import "/fmt" for Fmt
var maxn = 10
var maxl = 50
var steps = Fn.new { |n|
var a = List.filled(maxl, null)
var b = List.filled(maxl, null)
var x = List.filled(maxl, 0)
for (i in 0...maxl) {
a[i] = List.filled(maxn + 1, 0)
b[i] = List.filled(maxn + 1, 0)
}
a[0][0] = 1
var m = 0
var l = 0
while (true) {
x[l] = x[l] + 1
var k = x[l]
var cont = false
if (k >= n) {
if (l <= 0) break
l = l - 1
cont = true
} else if (a[l][k] == 0) {
if (b[l][k+1] != 0) cont = true
} else if (a[l][k] != k + 1) {
cont = true
}
if (!cont) {
a[l+1] = a[l].toList
var j = 1
while (j <= k) {
a[l+1][j] = a[l][k-j]
j = j + 1
}
b[l+1] = b[l].toList
a[l+1][0] = k + 1
b[l+1][k+1] = 1
if (l > m - 1) {
m = l + 1
}
l = l + 1
x[l] = 0
}
}
return m
}
for (i in 1..maxn) Fmt.print("$2d: $d", i, steps.call(i))
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
XPL0[edit]
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
XPL0: Faster Version[edit]
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
zkl[edit]
Slow version
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)) }
- Output:
1: 0 2: 1 3: 2 4: 4 5: 7 6: 10 7: 16 8: 22 9: 30 10: 38
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