Perfect shuffle
A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:
You are encouraged to solve this task according to the task description, using any language you may know.
- 7♠ 8♠ 9♠ J♠ Q♠ K♠→7♠ 8♠ 9♠
J♠ Q♠ K♠→7♠ J♠ 8♠ Q♠ 9♠ K♠
When you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:
original: 1 2 3 4 5 6 7 8
after 1st shuffle: 1 5 2 6 3 7 4 8
after 2nd shuffle: 1 3 5 7 2 4 6 8
after 3rd shuffle: 1 2 3 4 5 6 7 8
The Task
- Write a function that can perform a perfect shuffle on an even-sized list of values.
- Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under "Test Cases" below.
- You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all "cards" are unique within each deck.
- Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.
Test Cases
input (deck size) output (number of shuffles required) 8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
11l
<lang 11l>F flatten(lst)
[Int] r
L(sublst) lst
L(i) sublst
r [+]= i
R r
F magic_shuffle(deck)
V half = deck.len I/ 2 R flatten(zip(deck[0 .< half], deck[half ..]))
F after_how_many_is_equal(start, end)
V deck = magic_shuffle(start)
V counter = 1
L deck != end
deck = magic_shuffle(deck)
counter++
R counter
print(‘Length of the deck of cards | Perfect shuffles needed to obtain the same deck back’) L(length) (8, 24, 52, 100, 1020, 1024, 10000)
V deck = Array(0 .< length) V shuffles_needed = after_how_many_is_equal(deck, deck) print(‘#<5 | #.’.format(length, shuffles_needed))</lang>
- Output:
Length of the deck of cards | Perfect shuffles needed to obtain the same deck back 8 | 3 24 | 11 52 | 8 100 | 30 1020 | 1018 1024 | 10 10000 | 300
Action!
Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU. <lang Action!>DEFINE MAXDECK="5000"
PROC Order(INT ARRAY deck INT count)
INT i
FOR i=0 TO count-1 DO deck(i)=i OD
RETURN
BYTE FUNC IsOrdered(INT ARRAY deck INT count)
INT i
FOR i=0 TO count-1
DO
IF deck(i)#i THEN
RETURN (0)
FI
OD
RETURN (1)
PROC Shuffle(INT ARRAY src,dst INT count)
INT i,i1,i2
i=0 i1=0 i2=count RSH 1 WHILE i<count DO dst(i)=src(i1) i==+1 i1==+1 dst(i)=src(i2) i==+1 i2==+1 OD
RETURN
PROC Test(INT ARRAY deck,deck2 INT count)
INT ARRAY tmp INT n
Order(deck,count)
n=0
DO
Shuffle(deck,deck2,count)
tmp=deck deck=deck2 deck2=tmp
n==+1
Poke(77,0) ;turn off the attract mode
PrintF("%I cards -> %I iterations%E",count,n)
Put(28) ;move cursor up
UNTIL IsOrdered(deck,count)
OD
PutE()
RETURN
PROC Main()
INT ARRAY deck(MAXDECK),deck2(MAXDECK) INT ARRAY counts=[8 24 52 100 1020 1024 MAXDECK] INT i
FOR i=0 TO 6 DO Test(deck,deck2,counts(i)) OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
8 cards -> 3 iterations 24 cards -> 11 iterations 52 cards -> 8 iterations 100 cards -> 30 iterations 1020 cards -> 1018 iterations 1024 cards -> 10 iterations 5000 cards -> 357 iterations
Ada
<lang ada>with ada.text_io;use ada.text_io;
procedure perfect_shuffle is
function count_shuffle (half_size : Positive) return Positive is
subtype index is Natural range 0..2 * half_size - 1;
subtype index_that_move is index range index'first+1..index'last-1;
type deck is array (index) of index;
initial, d, next : deck;
count : Natural := 1;
begin
for i in index loop initial (i) := i; end loop;
d := initial;
loop
for i in index_that_move loop
next (i) := (if d (i) mod 2 = 0 then d(i)/2 else d(i)/2 + half_size);
end loop;
exit when next (index_that_move)= initial(index_that_move);
d := next;
count := count + 1;
end loop;
return count;
end count_shuffle;
test : array (Positive range <>) of Positive := (8, 24, 52, 100, 1020, 1024, 10_000);
begin
for size of test loop
put_line ("For" & size'img & " cards, there are "& count_shuffle (size / 2)'img & " shuffles needed.");
end loop;
end perfect_shuffle;</lang>
- Output:
For 8 cards, there are 3 shuffles needed. For 24 cards, there are 11 shuffles needed. For 52 cards, there are 8 shuffles needed. For 100 cards, there are 30 shuffles needed. For 1020 cards, there are 1018 shuffles needed. For 1024 cards, there are 10 shuffles needed. For 10000 cards, there are 300 shuffles needed.
ALGOL 68
<lang algol68># returns an array of the specified length, initialised to an ascending sequence of integers # OP DECK = ( INT length )[]INT:
BEGIN
[ 1 : length ]INT result;
FOR i TO UPB result DO result[ i ] := i OD;
result
END # DECK # ;
- in-place shuffles the deck as per the task requirements #
- LWB deck is assumed to be 1 #
PROC shuffle = ( REF[]INT deck )VOID:
BEGIN
[ 1 : UPB deck ]INT result;
INT left pos := 1;
INT right pos := ( UPB deck OVER 2 ) + 1;
FOR i FROM 2 BY 2 TO UPB result DO
result[ left pos ] := deck[ i - 1 ];
result[ right pos ] := deck[ i ];
left pos +:= 1;
right pos +:= 1
OD;
FOR i TO UPB deck DO deck[ i ] := result[ i ] OD
END # SHUFFLE # ;
- compares two integer arrays for equality #
OP = = ( []INT a, b )BOOL:
IF LWB a /= LWB b OR UPB a /= UPB b
THEN # the arrays have different bounds #
FALSE
ELSE
BOOL result := TRUE;
FOR i FROM LWB a TO UPB a WHILE result := a[ i ] = b[ i ] DO SKIP OD;
result
FI # = # ;
- compares two integer arrays for inequality #
OP /= = ( []INT a, b )BOOL: NOT ( a = b );
- returns the number of shuffles required to return a deck of the specified length #
- back to its original state #
PROC count shuffles = ( INT length )INT:
BEGIN
[] INT original deck = DECK length;
[ 1 : length ]INT shuffled deck := original deck;
INT count := 1;
WHILE shuffle( shuffled deck );
shuffled deck /= original deck
DO
count +:= 1
OD;
count
END # count shuffles # ;
- test the shuffling #
[]INT lengths = ( 8, 24, 52, 100, 1020, 1024, 10 000 ); FOR l FROM LWB lengths TO UPB lengths DO
print( ( whole( lengths[ l ], -8 ) + ": " + whole( count shuffles( lengths[ l ] ), -6 ), newline ) )
OD</lang>
- Output:
8: 3
24: 11
52: 8
100: 30
1020: 1018
1024: 10
10000: 300
APL
<lang apl>faro ← ∊∘⍉(2,2÷⍨≢)⍴⊢ count ← {⍺←⍵ ⋄ ⍺≡r←⍺⍺ ⍵:1 ⋄ 1+⍺∇r} (⊢,[1.5] (faro count ⍳)¨) 8 24 52 100 1020 1024 10000</lang>
- Output:
8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
Arturo
<lang rebol>perfectShuffle: function [deckSize][
deck: 1..deckSize original: new deck halfDeck: deckSize/2
i: 1
while [true][
deck: flatten combine first.n: halfDeck deck last.n: halfDeck deck
if deck = original -> return i
i: i+1
]
]
loop [8 24 52 100 1020 1024 10000] 's ->
print [
pad.right join @["Perfect shuffles required for deck size " to :string s ":"] 48
perfectShuffle s
]</lang>
- Output:
Perfect shuffles required for deck size 8: 3 Perfect shuffles required for deck size 24: 11 Perfect shuffles required for deck size 52: 8 Perfect shuffles required for deck size 100: 30 Perfect shuffles required for deck size 1020: 1018 Perfect shuffles required for deck size 1024: 10 Perfect shuffles required for deck size 10000: 300
AutoHotkey
<lang AutoHotkey>Shuffle(cards){ n := cards.MaxIndex()/2, res := [] loop % n res.push(cards[A_Index]), res.push(cards[round(A_Index + n)]) return res }</lang> Examples:<lang AutoHotkey>test := [8, 24, 52, 100, 1020, 1024, 10000] for each, val in test { cards := [], original:=rep:="" loop, % val cards.push(A_Index), original .= (original?", ":"") A_Index while (res <> original) { res := "" for k, v in (cards := Shuffle(cards)) res .= (res?", ":"") v rep := A_Index } result .= val "`t" rep "`n" } MsgBox % result return</lang>
Outputs:
8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
C
<lang c>/* ===> INCLUDES <============================================================*/
- include <stdlib.h>
- include <stdio.h>
- include <string.h>
/* ===> CONSTANTS <===========================================================*/
- define N_DECKS 7
const int kDecks[N_DECKS] = { 8, 24, 52, 100, 1020, 1024, 10000 };
/* ===> FUNCTION PROTOTYPES <=================================================*/ int CreateDeck( int **deck, int nCards ); void InitDeck( int *deck, int nCards ); int DuplicateDeck( int **dest, const int *orig, int nCards ); int InitedDeck( int *deck, int nCards ); int ShuffleDeck( int *deck, int nCards ); void FreeDeck( int **deck );
/* ===> FUNCTION DEFINITIONS <================================================*/
int main() {
int i, nCards, nShuffles; int *deck = NULL;
for( i=0; i<N_DECKS; ++i ) {
nCards = kDecks[i];
if( !CreateDeck(&deck,nCards) ) {
fprintf( stderr, "Error: malloc() failed!\n" );
return 1;
}
InitDeck( deck, nCards );
nShuffles = 0;
do {
ShuffleDeck( deck, nCards );
++nShuffles;
} while( !InitedDeck(deck,nCards) );
printf( "Cards count: %d, shuffles required: %d.\n", nCards, nShuffles );
FreeDeck( &deck ); }
return 0;
}
int CreateDeck( int **deck, int nCards ) {
int *tmp = NULL;
if( deck != NULL )
tmp = malloc( nCards*sizeof(*tmp) );
return tmp!=NULL ? (*deck=tmp)!=NULL : 0; /* (?success) (:failure) */
}
void InitDeck( int *deck, int nCards ) {
if( deck != NULL ) {
int i;
for( i=0; i<nCards; ++i )
deck[i] = i;
}
}
int DuplicateDeck( int **dest, const int *orig, int nCards ) {
if( orig != NULL && CreateDeck(dest,nCards) ) {
memcpy( *dest, orig, nCards*sizeof(*orig) );
return 1; /* success */
}
else {
return 0; /* failure */
}
}
int InitedDeck( int *deck, int nCards ) {
int i;
for( i=0; i<nCards; ++i )
if( deck[i] != i )
return 0; /* not inited */
return 1; /* inited */
}
int ShuffleDeck( int *deck, int nCards ) {
int *copy = NULL;
if( DuplicateDeck(©,deck,nCards) ) {
int i, j;
for( i=j=0; i<nCards/2; ++i, j+=2 ) {
deck[j] = copy[i];
deck[j+1] = copy[i+nCards/2];
}
FreeDeck( © );
return 1; /* success */
}
else {
return 0; /* failure */
}
}
void FreeDeck( int **deck ) {
if( *deck != NULL ) {
free( *deck );
*deck = NULL;
}
} </lang>
- Output:
Cards count: 8, shuffles required: 3. Cards count: 24, shuffles required: 11. Cards count: 52, shuffles required: 8. Cards count: 100, shuffles required: 30. Cards count: 1020, shuffles required: 1018. Cards count: 1024, shuffles required: 10. Cards count: 10000, shuffles required: 300. Press "Enter" to quit...
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
public static class PerfectShuffle {
static void Main()
{
foreach (int input in new [] {8, 24, 52, 100, 1020, 1024, 10000}) {
int[] numbers = Enumerable.Range(1, input).ToArray();
Console.WriteLine($"{input} cards: {ShuffleThrough(numbers).Count()}");
}
IEnumerable<T[]> ShuffleThrough<T>(T[] original) {
T[] copy = (T[])original.Clone();
do {
yield return copy = Shuffle(copy);
} while (!Enumerable.SequenceEqual(original, copy));
}
}
public static T[] Shuffle<T>(T[] array) {
if (array.Length % 2 != 0) throw new ArgumentException("Length must be even.");
int half = array.Length / 2;
T[] result = new T[array.Length];
for (int t = 0, l = 0, r = half; l < half; t+=2, l++, r++) {
result[t] = array[l];
result[t+1] = array[r];
}
return result;
}
}</lang>
- Output:
8 cards: 3 24 cards: 11 52 cards: 8 100 cards: 30 1020 cards: 1018 1024 cards: 10 10000 cards: 300
C++
<lang cpp>
- include <iostream>
- include <algorithm>
- include <vector>
int pShuffle( int t ) {
std::vector<int> v, o, r;
for( int x = 0; x < t; x++ ) {
o.push_back( x + 1 );
}
r = o; int t2 = t / 2 - 1, c = 1;
while( true ) {
v = r;
r.clear();
for( int x = t2; x > -1; x-- ) {
r.push_back( v[x + t2 + 1] );
r.push_back( v[x] );
}
std::reverse( r.begin(), r.end() );
if( std::equal( o.begin(), o.end(), r.begin() ) ) return c;
c++;
}
}
int main() {
int s[] = { 8, 24, 52, 100, 1020, 1024, 10000 };
for( int x = 0; x < 7; x++ ) {
std::cout << "Cards count: " << s[x] << ", shuffles required: ";
std::cout << pShuffle( s[x] ) << ".\n";
}
return 0;
} </lang>
- Output:
Cards count: 8, shuffles required: 3. Cards count: 24, shuffles required: 11. Cards count: 52, shuffles required: 8. Cards count: 100, shuffles required: 30. Cards count: 1020, shuffles required: 1018. Cards count: 1024, shuffles required: 10. Cards count: 10000, shuffles required: 300.
Clojure
<lang clojure>(defn perfect-shuffle [deck]
(let [half (split-at (/ (count deck) 2) deck)] (interleave (first half) (last half))))
(defn solve [deck-size]
(let [original (range deck-size)
trials (drop 1 (iterate perfect-shuffle original))
predicate #(= original %)]
(println (format "%5s: %s" deck-size
(inc (some identity (map-indexed (fn [i x] (when (predicate x) i)) trials)))))))
(map solve [8 24 52 100 1020 1024 10000])</lang>
- Output:
8: 3 24: 11 52: 8 100: 30 1020: 1018 1024: 10 10000: 300
Common Lisp
<lang lisp>(defun perfect-shuffle (deck)
(let* ((half (floor (length deck) 2))
(left (subseq deck 0 half))
(right (nthcdr half deck)))
(mapcan #'list left right)))
(defun solve (deck-size)
(loop with original = (loop for n from 1 to deck-size collect n)
for trials from 1
for deck = original then shuffled
for shuffled = (perfect-shuffle deck)
until (equal shuffled original)
finally (format t "~5D: ~4D~%" deck-size trials)))
(solve 8) (solve 24) (solve 52) (solve 100) (solve 1020) (solve 1024) (solve 10000)</lang>
- Output:
8: 3 24: 11 52: 8 100: 30 1020: 1018 1024: 10 10000: 300
D
<lang D>import std.stdio;
void main() {
auto sizes = [8, 24, 52, 100, 1020, 1024, 10_000];
foreach(s; sizes) {
writefln("%5s : %5s", s, perfectShuffle(s));
}
}
int perfectShuffle(int size) {
import std.exception : enforce; enforce(size%2==0);
import std.algorithm : copy, equal; import std.range; int[] orig = iota(0, size).array;
int[] process; process.length = size; copy(orig, process);
for(int count=1; true; count++) {
process = roundRobin(process[0..$/2], process[$/2..$]).array;
if (equal(orig, process)) {
return count;
}
}
assert(false, "How did this get here?");
}</lang>
- Output:
8 : 3 24 : 11 52 : 8 100 : 30 1020 : 1018 1024 : 10 10000 : 300
Delphi
<lang Delphi> program Perfect_shuffle;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
type
TDeck = record Cards: TArray<Integer>; Len: Integer; constructor Create(deckSize: Integer); overload; constructor Create(deck: TDeck); overload; procedure shuffleDeck(); class operator Equal(a, b: TDeck): boolean; function ShufflesRequired: Integer; procedure Assign(a: TDeck); end;
{ TDeck }
procedure TDeck.Assign(a: TDeck); begin
Len := a.Len; Cards := copy(a.Cards, 0, len);
end;
constructor TDeck.Create(deckSize: Integer); begin
if deckSize < 1 then
raise Exception.Create('Error: Deck size must have above zero');
if Odd(deckSize) then
raise Exception.Create('Error: Deck size must be even');
SetLength(Cards, deckSize); Len := deckSize;
for var i := 0 to High(Cards) do Cards[i] := i;
end;
constructor TDeck.Create(deck: TDeck); begin
Assign(deck);
end;
class operator TDeck.Equal(a, b: TDeck): boolean; begin
if a.len <> b.len then
raise Exception.Create('Error: Decks arent equally sized');
if a.Len = 0 then exit(True);
for var i := 0 to a.Len - 1 do
if a.Cards[i] <> b.Cards[i] then
exit(False);
Result := True;
end;
procedure TDeck.shuffleDeck; var
tmp: TArray<Integer>;
begin
SetLength(tmp, len); for var i := 0 to len div 2 - 1 do begin tmp[i * 2] := Cards[i]; tmp[i * 2 + 1] := Cards[len div 2 + i]; end; Cards := copy(tmp, 0, len);
end;
function TDeck.ShufflesRequired: Integer; var
ref: TDeck;
begin
Result := 1; ref := TDeck.Create(self); shuffleDeck; while not (self = ref) do begin shuffleDeck; inc(Result); end;
end;
const
cases: TArray<Integer> = [8, 24, 52, 100, 1020, 1024, 10000];
begin
for var size in cases do
begin
var deck := TDeck.Create(size);
writeln(format('Cards count: %d, shuffles required: %d', [size, deck.ShufflesRequired]));
end;
readln;
end.</lang>
Dyalect
<lang dyalect>func shuffle(arr) {
if arr.len() % 2 != 0 {
throw "Length must be even."
}
var half = arr.len() / 2
var result = Array.empty(size: arr.len())
var (t, l, r) = (0, 0, half)
while l < half {
result[t] = arr[l]
result[t+1] = arr[r]
l += 1
r += 1
t += 2
}
result
}
func arrayEqual(xs, ys) {
if xs.len() != ys.len() {
return false
}
for i in xs.indices() {
if xs[i] != ys[i] {
return false
}
}
return true
}
func shuffleThrough(original) {
var copy = original.clone()
while true {
copy = shuffle(copy)
yield copy
if arrayEqual(original, copy) {
break
}
}
}
for input in yields { 8, 24, 52, 100, 1020, 1024, 10000} {
var numbers = [1..input]
print("\(input) cards: \(shuffleThrough(numbers).len())");
}</lang>
- Output:
8 cards: 3 24 cards: 11 52 cards: 8 100 cards: 30 1020 cards: 1018 1024 cards: 10 10000 cards: 300
EchoLisp
<lang lisp>
- shuffler
- a permutation vector which interleaves both halves of deck
(define (make-shuffler n) (let ((s (make-vector n))) (for ((i (in-range 0 n 2))) (vector-set! s i (/ i 2))) (for ((i (in-range 0 n 2))) (vector-set! s (1+ i) (+ (/ n 2) (vector-ref s i)))) s))
- output
- (n . # of shuffles needed to go back)
(define (magic-shuffle n) (when (odd? n) (error "magic-shuffle:odd input" n)) (let [(deck (list->vector (iota n))) ;; (0 1 ... n-1) (dock (list->vector (iota n))) ;; keep trace or init deck (shuffler (make-shuffler n))]
(cons n (1+ (for/sum ((i Infinity)) ; (in-naturals missing in EchoLisp v2.9) (vector-permute! deck shuffler) ;; permutes in place #:break (eqv? deck dock) ;; compare to first 1))))) </lang>
- Output:
<lang lisp> map magic-shuffle '(8 24 52 100 1020 1024 10000))
→ ((8 . 3) (24 . 11) (52 . 8) (100 . 30) (1020 . 1018) (1024 . 10) (10000 . 300))
- Let's look in the On-line Encyclopedia of Integer Sequences
- Given a list of numbers, the (oeis ...) function looks for a sequence
(lib 'web) Lib: web.lib loaded. map magic-shuffle (range 2 18 2))
→ ((2 . 1) (4 . 2) (6 . 4) (8 . 3) (10 . 6) (12 . 10) (14 . 12) (16 . 4))
(oeis '(1 2 4 3 6 10 12 4)) → Sequence A002326 found </lang>
Elixir
<lang elixir>defmodule Perfect do
def shuffle(n) do
start = Enum.to_list(1..n)
m = div(n, 2)
shuffle(start, magic_shuffle(start, m), m, 1)
end
defp shuffle(start, start, _, step), do: step
defp shuffle(start, deck, m, step) do
shuffle(start, magic_shuffle(deck, m), m, step+1)
end
defp magic_shuffle(deck, len) do
{left, right} = Enum.split(deck, len)
Enum.zip(left, right)
|> Enum.map(&Tuple.to_list/1)
|> List.flatten
end
end
Enum.each([8, 24, 52, 100, 1020, 1024, 10000], fn n ->
step = Perfect.shuffle(n)
IO.puts "#{n} : #{step}"
end)</lang>
- Output:
8 : 3 24 : 11 52 : 8 100 : 30 1020 : 1018 1024 : 10 10000 : 300
F#
<lang fsharp> let perfectShuffle xs =
let h = (List.length xs) / 2 xs |> List.mapi (fun i x->(if i<h then i * 2 else ((i-h) * 2) + 1), x) |> List.sortBy fst |> List.map snd
let orderCount n =
let xs = [1..n] let rec spin count ys = if xs=ys then count else ys |> perfectShuffle |> spin (count + 1) xs |> perfectShuffle |> spin 1
[ 8; 24; 52; 100; 1020; 1024; 10000 ] |> List.iter (fun n->n |> orderCount |> printfn "%d %d" n) </lang>
- Output:
8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
Factor
<lang factor>USING: arrays formatting kernel math prettyprint sequences sequences.merged ; IN: rosetta-code.perfect-shuffle
CONSTANT: test-cases { 8 24 52 100 1020 1024 10000 }
- shuffle ( seq -- seq' ) halves 2merge ;
- shuffle-count ( n -- m )
<iota> >array 0 swap dup [ 2dup = ] [ shuffle [ 1 + ] 2dip ] do until 2drop ;
"Deck size" "Number of shuffles required" "%-11s %-11s\n" printf test-cases [ dup shuffle-count "%-11d %-11d\n" printf ] each</lang>
- Output:
Deck size Number of shuffles required 8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
Fortran
<lang fortran>MODULE PERFECT_SHUFFLE
IMPLICIT NONE
CONTAINS
! Shuffle the deck/array of integers
FUNCTION SHUFFLE(NUM_ARR)
INTEGER, DIMENSION(:), INTENT(IN) :: NUM_ARR
INTEGER, DIMENSION(SIZE(NUM_ARR)) :: SHUFFLE
INTEGER :: I, IDX
IF (MOD(SIZE(NUM_ARR), 2) .NE. 0) THEN
WRITE(*,*) "ERROR: SIZE OF DECK MUST BE EVEN NUMBER"
CALL EXIT(1)
END IF
IDX = 1
DO I=1, SIZE(NUM_ARR)/2
SHUFFLE(IDX) = NUM_ARR(I)
SHUFFLE(IDX+1) = NUM_ARR(SIZE(NUM_ARR)/2+I)
IDX = IDX + 2
END DO
END FUNCTION SHUFFLE
! Compare two arrays element by element
FUNCTION COMPARE_ARRAYS(ARRAY_1, ARRAY_2)
INTEGER, DIMENSION(:) :: ARRAY_1, ARRAY_2
LOGICAL :: COMPARE_ARRAYS
INTEGER :: I
DO I=1,SIZE(ARRAY_1)
IF (ARRAY_1(I) .NE. ARRAY_2(I)) THEN
COMPARE_ARRAYS = .FALSE.
RETURN
END IF
END DO
COMPARE_ARRAYS = .TRUE. END FUNCTION COMPARE_ARRAYS
! Generate a deck/array of consecutive integers
FUNCTION GEN_DECK(DECK_SIZE)
INTEGER, INTENT(IN) :: DECK_SIZE
INTEGER, DIMENSION(DECK_SIZE) :: GEN_DECK
INTEGER :: I
GEN_DECK = (/(I, I=1,DECK_SIZE)/) END FUNCTION GEN_DECK
END MODULE PERFECT_SHUFFLE
! Program to demonstrate the perfect shuffle algorithm ! for various deck sizes PROGRAM DEMO_PERFECT_SHUFFLE
USE PERFECT_SHUFFLE IMPLICIT NONE
INTEGER, PARAMETER, DIMENSION(7) :: DECK_SIZES = (/8, 24, 52, 100, 1020, 1024, 10000/) INTEGER, DIMENSION(:), ALLOCATABLE :: DECK, SHUFFLED INTEGER :: I, COUNTER
WRITE(*,'(A, A, A)') "input (deck size)", " | ", "output (number of shuffles required)"
WRITE(*,*) REPEAT("-", 55)
DO I=1, SIZE(DECK_SIZES)
IF (I .GT. 1) THEN
DEALLOCATE(DECK)
DEALLOCATE(SHUFFLED)
END IF
ALLOCATE(DECK(DECK_SIZES(I)))
ALLOCATE(SHUFFLED(DECK_SIZES(I)))
DECK = GEN_DECK(DECK_SIZES(I))
SHUFFLED = SHUFFLE(DECK)
COUNTER = 1
DO WHILE (.NOT. COMPARE_ARRAYS(DECK, SHUFFLED))
SHUFFLED = SHUFFLE(SHUFFLED)
COUNTER = COUNTER + 1
END DO
WRITE(*,'(I17, A, I35)') DECK_SIZES(I), " | ", COUNTER END DO
END PROGRAM DEMO_PERFECT_SHUFFLE</lang>
input (deck size) | output (number of shuffles required)
-------------------------------------------------------
8 | 3
24 | 11
52 | 8
100 | 30
1020 | 1018
1024 | 10
10000 | 300
FreeBASIC
<lang freebasic>function is_in_order( d() as uinteger ) as boolean
'tests if a deck is in order
for i as uinteger = lbound(d) to ubound(d)-1
if d(i) > d(i+1) then return false
next i
return true
end function
sub init_deck( d() as uinteger )
for i as uinteger = 1 to ubound(d)
d(i) = i
next i
return
end sub
sub shuffle( d() as uinteger )
'does a faro shuffle of the deck
dim as integer n = ubound(d), i
dim as integer b( 1 to n )
for i = 1 to n/2
b(2*i-1) = d(i)
b(2*i) = d(n/2+i)
next i
for i = 1 to n
d(i) = b(i)
next i
return
end sub
function shufs_needed( size as integer ) as uinteger
dim as uinteger d(1 to size), s = 0
init_deck(d())
do
shuffle(d())
s+=1
if is_in_order(d()) then exit do
loop
return s
end function
dim as uinteger tests(1 to 7) = {8, 24, 52, 100, 1020, 1024, 10000}, i for i = 1 to 7
print tests(i);" cards require "; shufs_needed(tests(i)); " shuffles."
next i</lang>
- Output:
8 cards require 3 shuffles. 24 cards require 11 shuffles. 52 cards require 8 shuffles. 100 cards require 30 shuffles. 1020 cards require 1018 shuffles. 1024 cards require 10 shuffles. 10000 cards require 300 shuffles.
Go
<lang go>package main
import "fmt"
type Deck struct { Cards []int length int }
func NewDeck(deckSize int) (res *Deck){ if deckSize % 2 != 0{ panic("Deck size must be even") } res = new(Deck) res.Cards = make([]int, deckSize) res.length = deckSize for i,_ := range res.Cards{ res.Cards[i] = i } return } func (d *Deck)shuffleDeck(){ tmp := make([]int,d.length) for i := 0;i <d.length/2;i++ { tmp[i*2] = d.Cards[i] tmp[i*2+1] = d.Cards[d.length / 2 + i] } d.Cards = tmp } func (d *Deck) isEqualTo(c Deck) (res bool) { if d.length != c.length { panic("Decks aren't equally sized") } res = true for i, v := range d.Cards{ if v != c.Cards[i] { res = false } } return }
func main(){
for _,v := range []int{8,24,52,100,1020,1024,10000} {
fmt.Printf("Cards count: %d, shuffles required: %d\n",v,ShufflesRequired(v))
}
}
func ShufflesRequired(deckSize int)(res int){ deck := NewDeck(deckSize) Ref := *deck deck.shuffleDeck() res++ for ;!deck.isEqualTo(Ref);deck.shuffleDeck(){ res++ } return }</lang>
- Output:
Cards count: 8, shuffles required: 3 Cards count: 24, shuffles required: 11 Cards count: 52, shuffles required: 8 Cards count: 100, shuffles required: 30 Cards count: 1020, shuffles required: 1018 Cards count: 1024, shuffles required: 10 Cards count: 10000, shuffles required: 300
Haskell
<lang Haskell>shuffle :: [a] -> [a] shuffle lst = let (a,b) = splitAt (length lst `div` 2) lst
in foldMap (\(x,y) -> [x,y]) $ zip a b
findCycle :: Eq a => (a -> a) -> a -> [a] findCycle f x = takeWhile (/= x) $ iterate f (f x)
main = mapM_ report [ 8, 24, 52, 100, 1020, 1024, 10000 ]
where
report n = putStrLn ("deck of " ++ show n ++ " cards: "
++ show (countSuffles n) ++ " shuffles!")
countSuffles n = 1 + length (findCycle shuffle [1..n])</lang>
- Output:
deck of 8 cards: 3 shuffles! deck of 24 cards: 11 shuffles! deck of 52 cards: 8 shuffles! deck of 100 cards: 30 shuffles! deck of 1020 cards: 1018 shuffles! deck of 1024 cards: 10 shuffles! deck of 10000 cards: 300 shuffles!
J
The shuffle routine:
<lang J> shuf=: /: $ /:@$ 0 1"_</lang>
Here, the phrase ($ $ 0 1"_) would generate a sequence of 0s and 1s the same length as the argument sequence:
<lang J> ($ $ 0 1"_) 'abcdef' 0 1 0 1 0 1</lang>
And we can use grade up (/:) to find the indices which would sort the argument sequence so that the values in the positions corresponding to our generated zeros would come before the values in the positions corresponding to our ones.
<lang J> /: ($ $ 0 1"_) 'abcdef' 0 2 4 1 3 5</lang>
But we can use grade up again to find what would have been the original permutation (grade up is a self inverting function for this domain).
<lang J> /:/: ($ $ 0 1"_) 'abcdef' 0 3 1 4 2 5</lang>
And, that means it can also sort the original sequence into that order:
<lang J> shuf 'abcdef' adbecf
shuf 'abcdefgh'
aebfcgdh</lang>
And this will work for sequences of arbitrary length.
(The rest of the implementation of shuf is pure syntactic sugar - you can use J's dissect and trace facilities to see the details if you are trying to learn the language.)
Meanwhile, the cycle length routine could look like this:
<lang J> shuflen=: [: *./ #@>@C.@shuf@i.</lang>
Here, we first generate a list of integers of the required length in their natural order. We then reorder them using our shuf function, find the cycles which result, find the lengths of each of these cycles then find the least common multiple of those lengths.
So here is the task example (with most of the middle trimmed out to avoid crashing the rosettacode wiki implementation):
<lang J> shuflen"0 }.2*i.5000 1 2 4 3 6 10 12 4 8 18 6 11 20 18 28 5 10 12 36 12 20 14 12 23 21 8 52 20 18 ... 4278 816 222 1332 384</lang>
Task example:
<lang J> ('deck size';'required shuffles'),(; shuflen)&> 8 24 52 100 1020 1024 10000 ┌─────────┬─────────────────┐ │deck size│required shuffles│ ├─────────┼─────────────────┤ │8 │3 │ ├─────────┼─────────────────┤ │24 │11 │ ├─────────┼─────────────────┤ │52 │8 │ ├─────────┼─────────────────┤ │100 │30 │ ├─────────┼─────────────────┤ │1020 │1018 │ ├─────────┼─────────────────┤ │1024 │10 │ ├─────────┼─────────────────┤ │10000 │300 │ └─────────┴─────────────────┘</lang>
Note that the implementation of shuf defines a behavior for odd length "decks". Experimentation shows that cycle length for an odd length deck is often the same as the cycle length for an even length deck which is one "card" longer.
Java
<lang java>import java.util.Arrays; import java.util.stream.IntStream;
public class PerfectShuffle {
public static void main(String[] args) {
int[] sizes = {8, 24, 52, 100, 1020, 1024, 10_000};
for (int size : sizes)
System.out.printf("%5d : %5d%n", size, perfectShuffle(size));
}
static int perfectShuffle(int size) {
if (size % 2 != 0)
throw new IllegalArgumentException("size must be even");
int half = size / 2;
int[] a = IntStream.range(0, size).toArray();
int[] original = a.clone();
int[] aa = new int[size];
for (int count = 1; true; count++) {
System.arraycopy(a, 0, aa, 0, size);
for (int i = 0; i < half; i++) {
a[2 * i] = aa[i];
a[2 * i + 1] = aa[i + half];
}
if (Arrays.equals(a, original))
return count;
}
}
}</lang>
8 : 3 24 : 11 52 : 8 100 : 30 1020 : 1018 1024 : 10 10000 : 300
JavaScript
ES6
<lang JavaScript>(() => {
'use strict';
// shuffleCycleLength :: Int -> Int
const shuffleCycleLength = deckSize =>
firstCycle(shuffle, range(1, deckSize))
.all.length;
// shuffle :: [a] -> [a]
const shuffle = xs =>
concat(zip.apply(null, splitAt(div(length(xs), 2), xs)));
// firstycle :: Eq a => (a -> a) -> a -> [a]
const firstCycle = (f, x) =>
until(
m => EqArray(x, m.current),
m => {
const fx = f(m.current);
return {
current: fx,
all: m.all.concat([fx])
};
}, {
current: f(x),
all: [x]
}
);
// Two arrays equal ?
// EqArray :: [a] -> [b] -> Bool
const EqArray = (xs, ys) => {
const [nx, ny] = [xs.length, ys.length];
return nx === ny ? (
nx > 0 ? (
xs[0] === ys[0] && EqArray(xs.slice(1), ys.slice(1))
) : true
) : false;
};
// GENERIC FUNCTIONS
// zip :: [a] -> [b] -> [(a,b)]
const zip = (xs, ys) =>
xs.slice(0, Math.min(xs.length, ys.length))
.map((x, i) => [x, ys[i]]);
// concat :: a -> [a] const concat = xs => [].concat.apply([], xs);
// splitAt :: Int -> [a] -> ([a],[a]) const splitAt = (n, xs) => [xs.slice(0, n), xs.slice(n)];
// div :: Num -> Num -> Int const div = (x, y) => Math.floor(x / y);
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
const go = x => p(x) ? x : go(f(x));
return go(x);
}
// range :: Int -> Int -> [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// length :: [a] -> Int // length :: Text -> Int const length = xs => xs.length;
// maximumBy :: (a -> a -> Ordering) -> [a] -> a
const maximumBy = (f, xs) =>
xs.reduce((a, x) => a === undefined ? x : (
f(x, a) > 0 ? x : a
), undefined);
// transpose :: a -> a const transpose = xs => xs[0].map((_, iCol) => xs.map((row) => row[iCol]));
// show :: a -> String const show = x => JSON.stringify(x, null, 2);
// replicateS :: Int -> String -> String
const replicateS = (n, s) => {
let v = s,
o = ;
if (n < 1) return o;
while (n > 1) {
if (n & 1) o = o.concat(v);
n >>= 1;
v = v.concat(v);
}
return o.concat(v);
};
// justifyRight :: Int -> Char -> Text -> Text
const justifyRight = (n, cFiller, strText) =>
n > strText.length ? (
(replicateS(n, cFiller) + strText)
.slice(-n)
) : strText;
// TEST
return transpose(transpose([
['Deck', 'Shuffles']
].concat(
[8, 24, 52, 100, 1020, 1024, 10000]
.map(n => [n.toString(), shuffleCycleLength(n)
.toString()
])))
.map(col => { // Right-justified number columns
const width = length(
maximumBy((a, b) => length(a) - length(b), col)
) + 2;
return col.map(x => justifyRight(width, ' ', x));
}))
.map(row => row.join())
.join('\n');
})();</lang>
- Output:
Deck Shuffles
8 3
24 11
52 8
100 30
1020 1018
1024 10
10000 300
jq
A small point of interest in the following is the `recurrence` function as it is generic. <lang jq>def perfect_shuffle:
. as $a
| if (length % 2) == 1 then "cannot perform perfect shuffle on odd-length array" | error
else (length / 2) as $mid
| reduce range(0; $mid) as $i (null;
.[2*$i] = $a[$i]
| .[2*$i + 1] = $a[$mid+$i] )
end;
- How many iterations of f are required to get back to . ?
def recurrence(f):
def r:
# input: [$init, $current, $count]
(.[1]|f) as $next
| if .[0] == $next then .[-1] + 1
else [.[0], $next, .[-1]+1] | r
end;
[., ., 0] | r;
def count_perfect_shuffles:
[range(0;.)] | recurrence(perfect_shuffle);
(8, 24, 52, 100, 1020, 1024, 10000, 100000) | [., count_perfect_shuffles]</lang>
- Output:
[8,3] [24,11] [52,8] [100,30] [1020,1018] [1024,10] [10000,300] [100000,540]
Julia
<lang julia>using Printf
function perfect_shuffle(a::Array)::Array
if isodd(length(a)) error("cannot perform perfect shuffle on odd-length array") end
rst = zeros(a)
mid = div(length(a), 2)
for i in 1:mid
rst[2i-1], rst[2i] = a[i], a[mid+i]
end
return rst
end
function count_perfect_shuffles(decksize::Int)::Int
a = collect(1:decksize)
b, c = perfect_shuffle(a), 1
while a != b
b = perfect_shuffle(b)
c += 1
end
return c
end
println(" Deck n.Shuffles") for i in (8, 24, 52, 100, 1020, 1024, 10000, 100000)
count = count_perfect_shuffles(i)
@printf("%7i%7i\n", i, count)
end</lang>
- Output:
Deck n.Shuffles
8 3
24 11
52 8
100 30
1020 1018
1024 10
10000 300
100000 540
Kotlin
<lang scala>// version 1.1.2
fun areSame(a: IntArray, b: IntArray): Boolean {
for (i in 0 until a.size) if (a[i] != b[i]) return false return true
}
fun perfectShuffle(a: IntArray): IntArray {
var b = IntArray(a.size)
val hSize = a.size / 2
for (i in 0 until hSize) b[i * 2] = a[i]
var j = 1
for (i in hSize until a.size) {
b[j] = a[i]
j += 2
}
return b
}
fun countShuffles(a: IntArray): Int {
require(a.size >= 2 && a.size % 2 == 0)
var b = a
var count = 0
while (true) {
val c = perfectShuffle(b)
count++
if (areSame(a, c)) return count
b = c
}
}
fun main(args: Array<String>) {
println("Deck size Num shuffles")
println("--------- ------------")
val sizes = intArrayOf(8, 24, 52, 100, 1020, 1024, 10000)
for (size in sizes) {
val a = IntArray(size) { it }
val count = countShuffles(a)
println("${"%-9d".format(size)} $count")
}
}</lang>
- Output:
Deck size Num shuffles --------- ------------ 8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
Lua
<lang Lua>-- Perform weave shuffle function shuffle (cards)
local pile1, pile2 = {}, {}
for card = 1, #cards / 2 do table.insert(pile1, cards[card]) end
for card = (#cards / 2) + 1, #cards do table.insert(pile2, cards[card]) end
cards = {}
for card = 1, #pile1 do
table.insert(cards, pile1[card])
table.insert(cards, pile2[card])
end
return cards
end
-- Return boolean indicating whether or not the cards are in order function inOrder (cards)
for k, v in pairs(cards) do
if k ~= v then return false end
end
return true
end
-- Count the number of shuffles needed before the cards are in order again function countShuffles (deckSize)
local deck, count = {}, 0
for i = 1, deckSize do deck[i] = i end
repeat
deck = shuffle(deck)
count = count + 1
until inOrder(deck)
return count
end
-- Main procedure local testCases = {8, 24, 52, 100, 1020, 1024, 10000} print("Input", "Output") for _, case in pairs(testCases) do print(case, countShuffles(case)) end</lang>
- Output:
Input Output 8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
Mathematica/Wolfram Language
<lang Mathematica>shuffle[deck_] := Apply[Riffle, TakeDrop[deck, Length[deck]/2]]; shuffleCount[n_] := Block[{count=0}, NestWhile[shuffle, shuffle[Range[n]], (count++; OrderedQ[#] )&];count]; Map[shuffleCount, {8, 24, 52, 100, 1020, 1024, 10000}]</lang>
- Output:
{3, 11, 8, 30, 1018, 10, 300}
MATLAB
PerfectShuffle.m: <lang matlab>function [New]=PerfectShuffle(Nitems, Nturns)
if mod(Nitems,2)==0 %only if even number
X=1:Nitems; %define deck
for c=1:Nturns %defines one shuffle
X=reshape(X,Nitems/2,2)'; %split the deck in two and stack halves
X=X(:)'; %mix the halves
end
New=X; %result of multiple shufflings
end</lang>
Main: <lang matlab>Result=[]; %vector to store results Q=[8, 24, 52, 100, 1020, 1024, 10000]; %queries for n=Q %for each query
Same=0; %initialize comparison
T=0; %initialize number of shuffles
while ~Same %while the result is not the original query
T=T+1; %one more shuffle
R=PerfectShuffle(n,T); %result of shuffling the query
Same=~(any(R-(1:n))); %same vector as the query
end %when getting the same vector
Result=[Result;T]; %collect results
end disp([Q', Result])</lang>
- Output:
8 3
24 11
52 8
100 30
1020 1018
1024 10
10000 300
Modula-2
<lang modula2>MODULE PerfectShuffle; FROM FormatString IMPORT FormatString; FROM Storage IMPORT ALLOCATE,DEALLOCATE; FROM SYSTEM IMPORT ADDRESS,TSIZE; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteCard(c : CARDINAL); VAR buf : ARRAY[0..15] OF CHAR; BEGIN
FormatString("%c", buf, c);
WriteString(buf)
END WriteCard;
PROCEDURE Init(VAR arr : ARRAY OF INTEGER); VAR i : CARDINAL; BEGIN
FOR i:=0 TO HIGH(arr) DO
arr[i] := i + 1
END
END Init;
PROCEDURE PerfectShuffle(VAR arr : ARRAY OF INTEGER);
PROCEDURE Inner(ti : CARDINAL);
VAR
tv : INTEGER;
tp,tn,n : CARDINAL;
BEGIN
n := HIGH(arr);
tn := ti;
tv := arr[ti];
REPEAT
tp := tn;
IF tp MOD 2 = 0 THEN
tn := tp / 2
ELSE
tn := (n+1)/2+tp/2
END;
arr[tp] := arr[tn];
UNTIL tn = ti;
arr[tp] := tv
END Inner;
VAR
done : BOOLEAN; i,c : CARDINAL;
BEGIN
c := 0; Init(arr);
REPEAT
i := 1;
WHILE i <= (HIGH(arr)/2) DO
Inner(i);
INC(i,2)
END;
INC(c);
done := TRUE;
FOR i:=0 TO HIGH(arr) DO
IF arr[i] # INT(i+1) THEN
done := FALSE;
BREAK
END
END
UNTIL done;
WriteCard(HIGH(arr)+1);
WriteString(": ");
WriteCard(c);
WriteLn
END PerfectShuffle;
(* Main *) VAR
v8 : ARRAY[1..8] OF INTEGER;
v24 : ARRAY[1..24] OF INTEGER;
v52 : ARRAY[1..52] OF INTEGER;
v100 : ARRAY[1..100] OF INTEGER;
v1020 : ARRAY[1..1020] OF INTEGER;
v1024 : ARRAY[1..1024] OF INTEGER;
v10000 : ARRAY[1..10000] OF INTEGER;
BEGIN
PerfectShuffle(v8); PerfectShuffle(v24); PerfectShuffle(v52); PerfectShuffle(v100); PerfectShuffle(v1020); PerfectShuffle(v1024); PerfectShuffle(v10000);
ReadChar
END PerfectShuffle.</lang>
- Output:
8: 3 24: 11 52: 8 100: 30 1020: 1018 1024: 10 10000: 300
Nim
<lang Nim>import sequtils, strutils
proc newValList(size: Positive): seq[int] =
if (size and 1) != 0: raise newException(ValueError, "size must be even.") result = toSeq(1..size)
func shuffled(list: seq[int]): seq[int] =
result.setLen(list.len) let half = list.len div 2 for i in 0..<half: result[2 * i] = list[i] result[2 * i + 1] = list[half + i]
for size in [8, 24, 52, 100, 1020, 1024, 10000]:
let initList = newValList(size)
var valList = initList
var count = 0
while true:
inc count
valList = shuffled(valList)
if valList == initList:
break
echo ($size).align(5), ": ", ($count).align(4)</lang>
- Output:
8: 3 24: 11 52: 8 100: 30 1020: 1018 1024: 10 10000: 300
Oforth
<lang oforth>: shuffle(l) l size 2 / dup l left swap l right zip expand ;
- nbShuffles(l) 1 l while( shuffle dup l <> ) [ 1 under+ ] drop ;</lang>
- Output:
>[ 8, 24, 52, 100, 1020, 1024, 10000 ] map(#[ seq nbShuffles ]) . [3, 11, 8, 30, 1018, 10, 300] ok
PARI/GP
<lang parigp>magic(v)=vector(#v,i,v[if(i%2,1,#v/2)+i\2]); shuffles_slow(n)=my(v=[1..n],o=v,s=1);while((v=magic(v))!=o,s++);s; shuffles(n)=znorder(Mod(2,n-1)); vector(5000,n,shuffles_slow(2*n))</lang>
- Output:
%1 = [1, 2, 4, 3, 6, 10, 12, 4, 8, 18, 6, 11, 20, 18, 28, 5, 10, 12, 36, 12, 20, 14, 12, 23, 21, 8, 52, 20, 18, 58, 60, 6, 12, 66, 22, 35, 9, 20, 30, 39, 54 , 82, 8, 28, 11, 12, 10, 36, 48, 30, 100, 51, 12, 106, 36, 36, 28, 44, 12, 24, 1 10, 20, 100, 7, 14, 130, 18, 36, 68, 138, 46, 60, 28, 42, 148, 15, 24, 20, 52, 5 2, 33, 162, 20, 83, 156, 18, 172, 60, 58, 178, 180, 60, 36, 40, 18, 95, 96, 12, 196, 99, 66, 84, 20, 66, 90, 210, 70, 28, 15, 18, 24, 37, 60, 226, 76, 30, 29, 9 2, 78, 119, 24, 162, 84, 36, 82, 50, 110, 8, 16, 36, 84, 131, 52, 22, 268, 135, 12, 20, 92, 30, 70, 94, 36, 60, 136, 48, 292, 116, 90, 132, 42, 100, 60, 102, 10 2, 155, 156, 12, 316, 140, 106, 72, 60, 36, 69, 30, 36, 132, 21, 28, 10, 147, 44 , 346, 348, 36, 88, 140, 24, 179, 342, 110, 36, 183, 60, 156, 372, 100, 84, 378, 14, 191, 60, 42, 388, 88, 130, 156, 44, 18, 200, 60, 108, 180, 204, 68, 174, 16 4, 138, 418, 420, 138, 40, 60, 60, 43, 72, 28, 198, 73, 42, 442, 44, 148, 224, 2 0, 30, 12, 76, 72, 460, 231, 20, 466, 66, 52, 70, 180, 156, 239, 36, 66, 48, 243 , 162, 490, 56, 60, 105, 166, 166, 251, 100, 156, 508, 9, 18, 204, 230, 172, 260 , 522, 60, 40, 253, 174, 60, 212, 178, 210, 540, 180, 36, 546, 60, 252, 39, 36, 556, 84, 40, 562, 28, 54, 284, 114, 190, 220, 144, 96, 246, 260, 12, 586, 90, 19 6, 148, 24, 198, 299, 25, 66, 220, 303, 84, 276, 612, 20, 154, 618, 198, 33, 500 , 90, 72, 45, 210, 28, 84, 210, 64, 214, 28, 323, 290, 30, 652, 260, 18, 658, 66 0, 24, 36, 308, 74, 60, 48, 180, 676, 48, 226, 22, 68, 76, 156, 230, 30, 276, 40 , 58, 700, 36, 92, 300, 708, 78, 55, 60, 238, 359, 51, 24, 140, 121, 486, 56, 24 4, 84, 330, 246, 36, 371, 148, 246, 318, 375, 50, 60, 756, 110, 380, 36, 24, 348 , 384, 16, 772, 20, 36, 180, 70, 252, 52, 786, 262, 84, 60, 52, 796, 184, 66, 90 , 132, 268, 404, 270, 270, 324, 126, 12, 820, 411, 20, 826, 828, 92, 168, 332, 9 0, 419, 812, 70, 156, 330, 94, 396, 852, 36, 428, 858, 60, 431, 172, 136, 390, 1 32, 48, 300, 876, 292, 55, 882, 116, 443, 21, 270, 414, 356, 132, 140, 104,[+++]
(By default gp won't show more than 25 lines of output, though an arbitrary amount can be printed or written to a file; use print, write, or default(lines, 100) to show more.)
Perl
<lang perl>use List::Util qw(all);
sub perfect_shuffle {
my $mid = @_ / 2;
map { @_[$_, $_ + $mid] } 0..($mid - 1);
}
for my $size (8, 24, 52, 100, 1020, 1024, 10000) {
my @shuffled = my @deck = 1 .. $size;
my $n = 0;
do { $n++; @shuffled = perfect_shuffle(@shuffled) }
until all { $shuffled[$_] == $deck[$_] } 0..$#shuffled;
printf "%5d cards: %4d\n", $size, $n;
}</lang>
- Output:
8 cards: 3 24 cards: 11 52 cards: 8 100 cards: 30 1020 cards: 1018 1024 cards: 10 10000 cards: 300
Phix
<lang Phix>function perfect_shuffle(sequence deck) integer mp = length(deck)/2 sequence res = deck
integer k = 1
for i=1 to mp do
res[k] = deck[i] k += 1
res[k] = deck[i+mp] k += 1
end for
return res
end function
constant testsizes = {8, 24, 52, 100, 1020, 1024, 10000} for i=1 to length(testsizes) do
sequence deck = tagset(testsizes[i])
sequence work = perfect_shuffle(deck)
integer count = 1
while work!=deck do
work = perfect_shuffle(work)
count += 1
end while
printf(1,"%5d cards: %4d\n", {testsizes[i],count})
end for</lang>
- Output:
8 cards: 3 24 cards: 11 52 cards: 8 100 cards: 30 1020 cards: 1018 1024 cards: 10 10000 cards: 300
PicoLisp
<lang PicoLisp>(de perfectShuffle (Lst)
(mapcan '((B A) (list A B))
(cdr (nth Lst (/ (length Lst) 2)))
Lst ) )
(for N (8 24 52 100 1020 1024 10000)
(let (Lst (range 1 N) L Lst Cnt 1)
(until (= Lst (setq L (perfectShuffle L)))
(inc 'Cnt) )
(tab (5 6) N Cnt) ) )</lang>
Output:
8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
Python
<lang python> import doctest import random
def flatten(lst):
""" >>> flatten([[3,2],[1,2]]) [3, 2, 1, 2] """ return [i for sublst in lst for i in sublst]
def magic_shuffle(deck):
""" >>> magic_shuffle([1,2,3,4]) [1, 3, 2, 4] """ half = len(deck) // 2 return flatten(zip(deck[:half], deck[half:]))
def after_how_many_is_equal(shuffle_type,start,end):
""" >>> after_how_many_is_equal(magic_shuffle,[1,2,3,4],[1,2,3,4]) 2 """
start = shuffle_type(start)
counter = 1
while start != end:
start = shuffle_type(start)
counter += 1
return counter
def main():
doctest.testmod()
print("Length of the deck of cards | Perfect shuffles needed to obtain the same deck back")
for length in (8, 24, 52, 100, 1020, 1024, 10000):
deck = list(range(length))
shuffles_needed = after_how_many_is_equal(magic_shuffle,deck,deck)
print("{} | {}".format(length,shuffles_needed))
if __name__ == "__main__":
main()
</lang> More functional version of the same code: <lang python> """ Brute force solution for the Perfect Shuffle problem. See http://oeis.org/A002326 for possible improvements """ from functools import partial from itertools import chain from operator import eq from typing import (Callable,
Iterable,
Iterator,
List,
TypeVar)
T = TypeVar('T')
def main():
print("Deck length | Shuffles ")
for length in (8, 24, 52, 100, 1020, 1024, 10000):
deck = list(range(length))
shuffles_needed = spin_number(deck, shuffle)
print(f"{length:<11} | {shuffles_needed}")
def shuffle(deck: List[T]) -> List[T]:
"""[1, 2, 3, 4] -> [1, 3, 2, 4]""" half = len(deck) // 2 return list(chain.from_iterable(zip(deck[:half], deck[half:])))
def spin_number(source: T,
function: Callable[[T], T]) -> int:
"""
Applies given function to the source
until the result becomes equal to it,
returns the number of calls
"""
is_equal_source = partial(eq, source)
spins = repeat_call(function, source)
return next_index(is_equal_source,
spins,
start=1)
def repeat_call(function: Callable[[T], T],
value: T) -> Iterator[T]:
"""(f, x) -> f(x), f(f(x)), f(f(f(x))), ..."""
while True:
value = function(value)
yield value
def next_index(predicate: Callable[[T], bool],
iterable: Iterable[T],
start: int = 0) -> int:
"""
Returns index of the first element of the iterable
satisfying given condition
"""
for index, item in enumerate(iterable, start=start):
if predicate(item):
return index
if __name__ == "__main__":
main()
</lang>
- Output:
Deck length | Shuffles 8 | 3 24 | 11 52 | 8 100 | 30 1020 | 1018 1024 | 10 10000 | 300
Reversed shuffle or just calculate how many shuffles are needed: <lang python>def mul_ord2(n): # directly calculate how many shuffles are needed to restore # initial order: 2^o mod(n-1) == 1 if n == 2: return 1
n,t,o = n-1,2,1 while t != 1: t,o = (t*2)%n,o+1 return o
def shuffles(n): a,c = list(range(n)), 0 b = a
while True: # Reverse shuffle; a[i] can be taken as the current # position of the card with value i. This is faster. a = a[0:n:2] + a[1:n:2] c += 1 if b == a: break return c
for n in range(2, 10000, 2): #print(n, mul_ord2(n)) print(n, shuffles(n))</lang>
Quackery
<lang Quackery> [ [] swap
times [ i^ join ] ] is deck ( n --> [ )
[ dup size 2 / split swap
witheach
[ swap i^ 2 * stuff ] ] is weave ( [ --> [ )
[ 0 swap
deck dup
[ rot 1+ unrot
weave 2dup = until ]
2drop ] is shuffles ( n --> n )
' [ 8 24 52 100 1020 1024 10000 ]
witheach
[ say "A deck of " dup echo say " cards needs " shuffles echo say " shuffles." cr ]</lang>
- Output:
A deck of 8 cards needs 3 shuffles. A deck of 24 cards needs 11 shuffles. A deck of 52 cards needs 8 shuffles. A deck of 100 cards needs 30 shuffles. A deck of 1020 cards needs 1018 shuffles. A deck of 1024 cards needs 10 shuffles. A deck of 10000 cards needs 300 shuffles.
R
Matrix solution
<lang rsplus>wave.shuffle <- function(n) {
deck <- 1:n ## create the original deck
new.deck <- c(matrix(data = deck, ncol = 2, byrow = TRUE)) ## shuffle the deck once
counter <- 1 ## track the number of loops
## defining a loop that shuffles the new deck until identical with the original one
## and in the same time increses the counter with 1 per loop
while (!identical(deck, new.deck)) { ## logical condition
new.deck <- c(matrix(data = new.deck, ncol = 2, byrow = TRUE)) ## shuffle
counter <- counter + 1 ## add 1 to the number of loops
}
return(counter) ## final result - total number of loops until the condition is met
} test.values <- c(8, 24, 52, 100, 1020, 1024, 10000) ## the set of the test values test <- sapply(test.values, wave.shuffle) ## apply the wave.shuffle function on each element names(test) <- test.values ## name the result test ## print the result out</lang>
- Output:
> test
8 24 52 100 1020 1024 10000
3 11 8 30 1018 10 300
Sequence solution
The previous solution exploits R's matrix construction; This solution exploits its array indexing. <lang rsplus>#A strict reading of the task description says that we need a function that does exactly one shuffle. pShuffle <- function(deck) {
n <- length(deck)#Assumed even (as in task description). new <- array(n)#Maybe not as general as it could be, but the task said to use whatever was convenient. new[seq(from = 1, to = n, by = 2)] <- deck[seq(from = 1, to = n/2, by = 1)] new[seq(from = 2, to = n, by = 2)] <- deck[seq(from = 1 + n/2, to = n, by = 1)] new
}
task2 <- function(deck) {
new <- deck
count <- 0
repeat
{
new <- pShuffle(new)
count <- count + 1
if(all(new == deck)) break
}
cat("It takes", count, "shuffles of a deck of size", length(deck), "to return to the original deck.","\n")
invisible(count)#For the unit tests. The task wanted this printed so we only return it invisibly.
}
- Tests - All done in one line.
mapply(function(x, y) task2(1:x) == y, c(8, 24, 52, 100, 1020, 1024, 10000), c(3, 11, 8, 30, 1018, 10, 300))</lang>
- Output:
> mapply(function(x,y) task2(1:x)==y,c(8,24,52,100,1020,1024,10000),c(3,11,8,30,1018,10,300)) It takes 3 shuffles of a deck of size 8 to return to the original deck. It takes 11 shuffles of a deck of size 24 to return to the original deck. It takes 8 shuffles of a deck of size 52 to return to the original deck. It takes 30 shuffles of a deck of size 100 to return to the original deck. It takes 1018 shuffles of a deck of size 1020 to return to the original deck. It takes 10 shuffles of a deck of size 1024 to return to the original deck. It takes 300 shuffles of a deck of size 10000 to return to the original deck. [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE
Racket
<lang racket>#lang racket/base (require racket/list)
(define (perfect-shuffle l)
(define-values (as bs) (split-at l (/ (length l) 2))) (foldr (λ (a b d) (list* a b d)) null as bs))
(define (perfect-shuffles-needed n)
(define-values (_ rv)
(for/fold ((d (perfect-shuffle (range n))) (i 1))
((_ (in-naturals))
#:break (apply < d))
(values (perfect-shuffle d) (add1 i))))
rv)
(module+ test
(require rackunit) (check-equal? (perfect-shuffle '(1 2 3 4)) '(1 3 2 4)) (define (test-perfect-shuffles-needed n e) (define psn (perfect-shuffles-needed n)) (printf "Deck size:\t~a\tShuffles needed:\t~a\t(~a)~%" n psn e) (check-equal? psn e))
(for-each test-perfect-shuffles-needed
'(8 24 52 100 1020 1024 10000)
'(3 11 8 30 1018 10 300)))</lang>
- Output:
Deck size: 8 Shuffles needed: 3 (3) Deck size: 24 Shuffles needed: 11 (11) Deck size: 52 Shuffles needed: 8 (8) Deck size: 100 Shuffles needed: 30 (30) Deck size: 1020 Shuffles needed: 1018 (1018) Deck size: 1024 Shuffles needed: 10 (10) Deck size: 10000 Shuffles needed: 300 (300)
Raku
(formerly Perl 6)
<lang perl6>sub perfect-shuffle (@deck) {
my $mid = @deck / 2; flat @deck[0 ..^ $mid] Z @deck[$mid .. *];
}
for 8, 24, 52, 100, 1020, 1024, 10000 -> $size {
my @deck = ^$size;
my $n;
repeat until [<] @deck {
$n++;
@deck = perfect-shuffle @deck;
}
printf "%5d cards: %4d\n", $size, $n;
}</lang>
- Output:
8 cards: 3 24 cards: 11 52 cards: 8 100 cards: 30 1020 cards: 1018 1024 cards: 10 10000 cards: 300
REXX
unoptimized
<lang rexx>/*REXX program performs a "perfect shuffle" for a number of even numbered decks. */ parse arg X /*optional list of test cases from C.L.*/ if X= then X=8 24 52 100 1020 1024 10000 /*Not specified? Then use the default.*/ w=length(word(X, words(X))) /*used for right─aligning the numbers. */
do j=1 for words(X); y=word(X,j) /*use numbers in the test suite (list).*/
do k=1 for y; @.k=k; end /*k*/ /*generate a deck to be used (shuffled)*/
do t=1 until eq(); call magic; end /*t*/ /*shuffle until before equals after.*/
say 'deck size:' right(y,w)"," right(t,w) 'perfect shuffles.' end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ eq: do ?=1 for y; if @.?\==? then return 0; end; return 1 /*──────────────────────────────────────────────────────────────────────────────────────*/ magic: z=0 /*set the Z pointer (used as index).*/
h=y%2 /*get the half─way (midpoint) pointer. */
do s=1 for h; z=z+1; h=h+1 /*traipse through the card deck pips. */
!.z=@.s; z=z+1 /*assign left half; then bump pointer. */
!.z=@.h /* " right " */
end /*s*/ /*perform a perfect shuffle of the deck*/
do r=1 for y; @.r=!.r; end /*re─assign to the original card deck. */
return</lang>
output (abbreviated) when using the default input:
deck size: 8, 3 perfect shuffles. deck size: 24, 11 perfect shuffles. deck size: 52, 8 perfect shuffles. deck size: 100, 30 perfect shuffles. deck size: 1020, 1018 perfect shuffles. deck size: 1024, 10 perfect shuffles. deck size: 10000, 300 perfect shuffles.
optimized
This REXX version takes advantage that the 1st and last cards of the deck don't change. <lang rexx>/*REXX program does a "perfect shuffle" for a number of even numbered decks. */ parse arg X /*optional list of test cases from C.L.*/ if X= then X=8 24 52 100 1020 1024 10000 /*Not specified? Use default.*/ w=length(word(X, words(X))) /*used for right─aligning the numbers. */
do j=1 for words(X); y=word(X,j) /*use numbers in the test suite (list).*/
do k=1 for y; @.k=k; end /*generate a deck to be shuffled (used)*/
do t=1 until eq(); call magic; end /*shuffle until before equals after.*/
say 'deck size:' right(y,w)"," right(t,w) 'perfect shuffles.' end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ eq: do ?=1 for y; if @.?\==? then return 0; end; return 1 /*──────────────────────────────────────────────────────────────────────────────────────*/ magic: z=1; h=y%2 /*H is (half─way) pointer.*/
do L=3 by 2 for h-1; z=z+1; !.L=@.z; end /*assign left half of deck.*/
do R=2 by 2 for h-1; h=h+1; !.R=@.h; end /* " right " " " */
do a=2 for y-2; @.a=!.a; end /*re─assign──►original deck*/
return</lang>
output is the same as the 1st version.
Ruby
<lang ruby>def perfect_shuffle(deck_size = 52)
deck = (1..deck_size).to_a original = deck.dup half = deck_size / 2 1.step do |i| deck = deck.first(half).zip(deck.last(half)).flatten return i if deck == original end
end
[8, 24, 52, 100, 1020, 1024, 10000].each {|i| puts "Perfect shuffles required for deck size #{i}: #{perfect_shuffle(i)}"} </lang>
- Output:
Perfect shuffles required for deck size 8: 3 Perfect shuffles required for deck size 24: 11 Perfect shuffles required for deck size 52: 8 Perfect shuffles required for deck size 100: 30 Perfect shuffles required for deck size 1020: 1018 Perfect shuffles required for deck size 1024: 10 Perfect shuffles required for deck size 10000: 300
Rust
<lang Rust>extern crate itertools;
fn shuffle<T>(mut deck: Vec<T>) -> Vec<T> {
let index = deck.len() / 2; let right_half = deck.split_off(index); itertools::interleave(deck, right_half).collect()
}
fn main() {
for &size in &[8, 24, 52, 100, 1020, 1024, 10_000] {
let original_deck: Vec<_> = (0..size).collect();
let mut deck = original_deck.clone();
let mut iterations = 0;
loop {
deck = shuffle(deck);
iterations += 1;
if deck == original_deck {
break;
}
}
println!("{: >5}: {: >4}", size, iterations);
}
}</lang>
- Output:
8: 3 24: 11 52: 8 100: 30 1020: 1018 1024: 10 10000: 300
Scala
Imperative, Quick, dirty and ugly
- Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<lang Scala>object PerfectShuffle extends App {
private def sizes = Seq(8, 24, 52, 100, 1020, 1024, 10000)
private def perfectShuffle(size: Int): Int = {
require(size % 2 == 0, "Card deck must be even")
val (half, a) = (size / 2, Array.range(0, size))
val original = a.clone
var count = 1
while (true) {
val aa = a.clone
for (i <- 0 until half) {
a(2 * i) = aa(i)
a(2 * i + 1) = aa(i + half)
}
if (a.deep == original.deep) return count
count += 1
}
0
}
for (size <- sizes) println(f"$size%5d : ${perfectShuffle(size)}%5d")
}</lang>
Scilab
<lang>function New=PerfectShuffle(Nitems,Nturns)
if modulo(Nitems,2)==0 then
X=1:Nitems;
for c=1:Nturns
X=matrix(X,Nitems/2,2)';
X=X(:);
end
New=X';
end
endfunction
Result=[]; Q=[8, 24, 52, 100, 1020, 1024, 10000]; for n=Q
Same=0;
T=0;
Compare=[];
while ~Same
T=T+1;
R=PerfectShuffle(n,T);
Compare = find(R-(1:n));
if Compare == [] then
Same = 1;
end
end
Result=[Result;T];
end disp([Q', Result])</lang>
- Output:
8. 3. 24. 11. 52. 8. 100. 30. 1020. 1018. 1024. 10. 10000. 300.
Sidef
<lang ruby>func perfect_shuffle(deck) {
deck/2 -> zip.flat
}
[8, 24, 52, 100, 1020, 1024, 10000].each { |size|
var deck = @(1..size) var shuffled = deck
var n = (1..Inf -> lazy.first {
(shuffled = perfect_shuffle(shuffled)) == deck
})
printf("%5d cards: %4d\n", size, n)
}</lang>
- Output:
8 cards: 3 24 cards: 11 52 cards: 8 100 cards: 30 1020 cards: 1018 1024 cards: 10 10000 cards: 300
Swift
<lang swift>func perfectShuffle<T>(_ arr: [T]) -> [T]? {
guard arr.count & 1 == 0 else {
return nil
}
let half = arr.count / 2 var res = [T]()
for i in 0..<half {
res.append(arr[i])
res.append(arr[i + half])
}
return res
}
let decks = [
Array(1...8), Array(1...24), Array(1...52), Array(1...100), Array(1...1020), Array(1...1024), Array(1...10000)
]
for deck in decks {
var shuffled = deck var shuffles = 0
repeat {
shuffled = perfectShuffle(shuffled)!
shuffles += 1
} while shuffled != deck
print("Deck of \(shuffled.count) took \(shuffles) shuffles to get back to original order")
}</lang>
- Output:
Deck of 8 took 3 shuffles to get back to original order Deck of 24 took 11 shuffles to get back to original order Deck of 52 took 8 shuffles to get back to original order Deck of 100 took 30 shuffles to get back to original order Deck of 1020 took 1018 shuffles to get back to original order Deck of 1024 took 10 shuffles to get back to original order Deck of 10000 took 300 shuffles to get back to original order
Tcl
Using tcltest to include an executable test case ..
<lang Tcl>namespace eval shuffle {
proc perfect {deck} {
if {[llength $deck]%2} {
return -code error "Deck must be of even length!"
}
set split [expr {[llength $deck]/2}]
set top [lrange $deck 0 $split-1]
set btm [lrange $deck $split end]
foreach a $top b $btm {
lappend res $a $b
}
return $res
}
proc cycle_length {transform deck} {
set d $deck
while 1 {
set d [$transform $d]
incr i
if {$d eq $deck} {return $i}
}
return $i
}
proc range {a {b ""}} {
if {$b eq ""} {
set b $a; set a 0
}
set res {}
while {$a < $b} {
lappend res $a
incr a
}
return $res
}
}
set ::argv {} package require tcltest tcltest::test "Test perfect shuffle cycles" {} -body {
lmap size {8 24 52 100 1020 1024 10000} {
shuffle::cycle_length perfect [range $size]
}
} -result {3 11 8 30 1018 10 300}</lang>
VBA
<lang vb>Option Explicit
Sub Main() Dim T, Arr, X As Long, C As Long
Arr = Array(8, 24, 52, 100, 1020, 1024, 10000)
For X = LBound(Arr) To UBound(Arr)
C = 0
Call PerfectShuffle(T, CLng(Arr(X)), C)
Debug.Print Right(String(19, " ") & "For " & Arr(X) & " cards => ", 19) & C & " shuffles needed."
Erase T
Next
End Sub
Private Sub PerfectShuffle(tb, NbCards As Long, Count As Long) Dim arr1, arr2, StrInit As String, StrTest As String
tb = CreateArray(1, NbCards)
StrInit = Join(tb, " ")
Do
Count = Count + 1
Call DivideArr(tb, arr1, arr2)
tb = RemakeArray(arr1, arr2)
StrTest = Join(tb, " ")
Loop While StrTest <> StrInit
End Sub
Private Function CreateArray(First As Long, Length As Long) As String() Dim i As Long, T() As String, C As Long
If IsEven(Length) Then
ReDim T(Length - 1)
For i = First To First + Length - 1
T(C) = i
C = C + 1
Next i
CreateArray = T
End If
End Function
Private Sub DivideArr(A, B, C) Dim i As Long
B = A
ReDim Preserve B(UBound(A) \ 2)
ReDim C(UBound(B))
For i = LBound(C) To UBound(C)
C(i) = A(i + UBound(B) + 1)
Next
End Sub
Private Function RemakeArray(A1, A2) As String() Dim i As Long, T() As String, C As Long
ReDim T((UBound(A2) * 2) + 1)
For i = LBound(T) To UBound(T) - 1 Step 2
T(i) = A1(C)
T(i + 1) = A2(C)
C = C + 1
Next
RemakeArray = T
End Function
Private Function IsEven(Number As Long) As Boolean
IsEven = (Number Mod 2 = 0)
End Function</lang>
- Output:
For 8 cards => 3 shuffles needed. For 24 cards => 11 shuffles needed. For 52 cards => 8 shuffles needed. For 100 cards => 30 shuffles needed. For 1020 cards => 1018 shuffles needed. For 1024 cards => 10 shuffles needed. For 10000 cards => 300 shuffles needed.
Wren
<lang ecmascript>import "/fmt" for Fmt
var areSame = Fn.new { |a, b|
for (i in 0...a.count) if (a[i] != b[i]) return false return true
}
var perfectShuffle = Fn.new { |a|
var n = a.count
var b = List.filled(n, 0)
var hSize = (n/2).floor
for (i in 0...hSize) b[i * 2] = a[i]
var j = 1
for (i in hSize...n) {
b[j] = a[i]
j = j + 2
}
return b
}
var countShuffles = Fn.new { |a|
var n = a.count
if (n < 2 || n % 2 == 1) Fiber.abort("Array must be even-sized and non-empty.")
var b = a
var count = 0
while (true) {
var c = perfectShuffle.call(b)
count = count + 1
if (areSame.call(a, c)) return count
b = c
}
}
System.print("Deck size Num shuffles") System.print("--------- ------------") var sizes = [8, 24, 52, 100, 1020, 1024, 10000] for (size in sizes) {
var a = List.filled(size, 0)
for (i in 1...size) a[i] = i
var count = countShuffles.call(a)
Fmt.print("$-9d $d", size, count)
}</lang>
- Output:
Deck size Num shuffles --------- ------------ 8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
XPL0
<lang XPL0>int Deck(10000), Deck0(10000); int Cases, Count, Test, Size, I;
proc Shuffle; \Do perfect shuffle of Deck0 into Deck int DeckLeft, DeckRight; int I; [DeckLeft:= Deck0; DeckRight:= Deck0 + Size*4/2; \4 bytes per integer for I:= 0 to Size-1 do
Deck(I):= if I&1 then DeckRight(I/2)
else DeckLeft(I/2);
];
[Cases:= [8, 24, 52, 100, 1020, 1024, 10000]; for Test:= 0 to 7-1 do
[Size:= Cases(Test);
for I:= 0 to Size-1 do Deck(I):= I;
Count:= 0;
repeat for I:= 0 to Size-1 do Deck0(I):= Deck(I);
Shuffle;
Count:= Count+1;
for I:= 0 to Size-1 do
if Deck(I) # I then I:= Size;
until I = Size; \equal starting configuration
IntOut(0, Size); ChOut(0, 9\tab\); IntOut(0, Count); CrLf(0);
];
]</lang>
- Output:
8 3 24 11 52 8 100 30 1020 1018 1024 10 10000 300
zkl
<lang zkl>fcn perfectShuffle(numCards){
deck,shuffle,n,N:=numCards.pump(List),deck,0,numCards/2;
do{ shuffle=shuffle[0,N].zip(shuffle[N,*]).flatten(); n+=1 }
while(deck!=shuffle);
n
} foreach n in (T(8,24,52,100,1020,1024,10000)){
println("%5d : %d".fmt(n,perfectShuffle(n)));
}</lang>
- Output:
8 : 3 24 : 11 52 : 8 100 : 30 1020 : 1018 1024 : 10 10000 : 300