# Topswops

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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.

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.

## 11l

Translation of: Python:_Faster_Version
```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

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)         " ->
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```
Output:
``` 1:   0
2:   1
3:   2
4:   4
5:   7
6:  10
7:  16
8:  22
9:  30
10:  38
```

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

```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
}
```

Examples:

```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")
}
```

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):

```#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 <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;

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;

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++)

for (int i = 0; i < top; i++)

printf("%2d: %2d\n", n, best[n]);
}

return 0;
}
```

## C#

Translation of: Java
```using System;

public class Topswops {
static readonly int maxBest = 32;
static int[] best;

private static 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 = (int[])deck.Clone();
for (int i = 1; i < n; i++) {
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) {
if(n <= 0 || n >= maxBest) throw new ArgumentOutOfRangeException(nameof(n), "n must be greater than 0 and less than 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++)
Console.WriteLine(i + ": " + topswops(i));
}
}
```
Output:
```1: 0
2: 1
3: 2
4: 4
5: 7
6: 10
7: 16
8: 22
9: 30
10: 38

```

## C++

```#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

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

Translation of: C
```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

```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

Translation of: Erlang
```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

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

```
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.

```-module( topswops ).

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.

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

```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

```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
```

## FreeBASIC

Translation of: XPL0:_Faster_Version
```Dim Shared As Byte n, d, best(16)

Sub TrySwaps(A() As Byte, f As Byte, s As Byte)
Dim As Byte B(16), i, j, k
If d > best(n) Then best(n) = d
Do
If A(s) = -1 Or A(s) = s Then Exit Do
If d+best(s) <= best(n) Then Exit Sub
If s = 0 Then Exit Do
s -= 1
Loop

d += 1
For i = 0 To s
B(i) = A(i)
Next

k = 1
For i = 1 To s
k Shl= 1
If B(i) =- 1 AndAlso (f And k) = 0 Or B(i) = i Then
j = i
B(0) = j
While j
j -= 1
B(i-j) = A(j)
Wend
TrySwaps(B(), f Or k, s)
End If
Next
d -= 1
End Sub

Dim As Byte i, X(16)
For i = 0 To 16-1
X(i) = -1
best(i) = 0
Next  i
X(0) = 0

For n = 1 To 13
d = 0
TrySwaps(X(), 1, n-1)
Print Using "##: ##"; n; best(n)
Next n

Sleep
```
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```

## Go

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

import "fmt"

const ( // array sizes
maxn = 10 // max number of cards
maxl = 50 // upper bound for number of steps
)

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

func steps(n int) int {
var a, b [maxl][maxn + 1]int
var x [maxl]int
a[0][0] = 1
var m int
for l := 0; ; {
x[l]++
k := int(x[l])
if k >= n {
if l <= 0 {
break
}
l--
continue
}
if a[l][k] == 0 {
if b[l][k+1] != 0 {
continue
}
} else if a[l][k] != k+1 {
continue
}
a[l+1] = a[l]
for j := 1; j <= k; j++ {
a[l+1][j] = a[l][k-j]
}
b[l+1] = b[l]
a[l+1][0] = k + 1
b[l+1][k+1] = 1
if l > m-1 {
m = l + 1
}
l++
x[l] = 0
}
return m
}
```
Output:
```1: 0
2: 1
3: 2
4: 4
5: 7
6: 10
7: 16
8: 22
9: 30
10: 38
```

#### Searching permutations

```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

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

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

Solution:

```   swops =:  ((|.@:{. , }.)~ {.)^:a:
```

```   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
```

Example (topswops of all permutations of the integers 1..10):

```   (,. _1 + ! >./@:(#@swops@A. >:)&i. ])&> 1+i.10
1  0
2  1
3  2
4  4
5  7
6 10
7 16
8 22
9 30
10 38
```

Notes: Readers less familiar with array-oriented programming may find an alternate solution written in the structured programming style more accessible.

## Java

Translation of: D
```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

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

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

Translation of: Java
```// 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

```-- 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 /Wolfram Language

An exhaustive search of all possible permutations is done

```flip[a_] := Block[{a1 = First@a}, If[a1 == Length@a, Reverse[a], Join[Reverse[a[[;; a1]]], a[[a1 + 1 ;;]]]]]
swaps[a_] := Length@FixedPointList[flip, a] - 2
Print[#, ": ", Max[swaps /@ Permutations[Range@#]]] & /@ Range[10];
```
Output:
```1: 0
2: 1
3: 2
4: 4
5: 7
6: 10
7: 16
8: 22
9: 30
10: 38```

## Nim

Translation of: Java
```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

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

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

Originally contributed by Jason Gade as part of the Euphoria version of the Great Computer Language Shootout benchmarks.

```with javascript_semantics
function fannkuch(integer n)
sequence count = tagset(n),
perm1 = tagset(n)
integer maxFlipsCount = 0, r = n+1
while true do
while r!=1 do
count[r-1] = r
r -= 1
end while
if not (perm1[1]=1 or perm1[n]=n) then
sequence perm = perm1
integer flipsCount = 0,
k = perm[1]
while k!=1 do
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 true do
if r>n then return maxFlipsCount end if
integer perm0 = perm1[1]
perm1[1..r-1] = perm1[2..r]
perm1[r] = perm0
count[r] -= 1
if count[r]>1 then exit end if
r += 1
end while
end while
end function -- fannkuch

atom t0 = time()
for i=1 to iff(platform()=JS?9:10) do
?fannkuch(i)
end for
?elapsed(time()-t0)
```
Output:
```0
1
2
4
7
10
16
22
30
38
"14.1s"
```

It will manage 10 under pwa/p2js but with a blank screen for 38s, so I've capped it to 9 to make it finish in 3s.

## Picat

```go ?=>
member(N,1..10),
Perm = 1..N,
Rev = Perm.reverse(),
Max = 0,
while(Perm != Rev)
next_permutation(Perm),
C = topswops(Perm),
if C > Max then
Max := C
end
end,
printf("%2d: %2d\n",N,Max),
fail,
nl.
go => true.

topswops([]) = 0 => true.
topswops([1]) = 0 => true.
topswops([1|_]) = 0 => true.
topswops(P) = Count =>
Len = P.length,
Count = 0,
while (P[1] > 1)
Pos = P[1],
P := [P[I] : I in 1..Pos].reverse() ++ [P[I] : I in Pos+1..Len],
Count := Count + 1
end.

% Inline
next_permutation(Perm) =>
N = Perm.length,
K = N - 1,
while (Perm[K] > Perm[K+1], K >= 0)
K := K - 1
end,
if K > 0 then
J = N,
while (Perm[K] > Perm[J])  J := J - 1 end,
Tmp := Perm[K],
Perm[K] := Perm[J],
Perm[J] := Tmp,
R = N,
S = K + 1,
while (R > S)
Tmp := Perm[R],
Perm[R] := Perm[S],
Perm[S] := Tmp,
R := R - 1,
S := S + 1
end
end.```
Output:
``` 1:  0
2:  1
3:  2
4:  4
5:  7
6: 10
7: 16
8: 22
9: 30
10: 38```

## 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)) )```

Output:

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

## PL/I

 This example is incorrect. Please fix the code and remove this message.Details: Shown output is incorrect at the very least.
```(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

```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

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

Translation of: C
```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

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

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))))

(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

(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

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;  _=@.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

Translation of: Python
```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

Translation of: Java
```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

```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

Library: 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]")
}
```
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

Library: Tcllib (Package: struct::list)

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

Translation of: Go
Library: Wren-fmt
```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

```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

Translation of: C
```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

Translation of: D

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
```