Perfect shuffle

From Rosetta Code
Revision as of 20:48, 23 April 2015 by rosettacode>CRGreathouse (GP)
Perfect shuffle is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A deck can be divided in two halves:

   a b c d e f -> a b c | d e f

Perfect shuffling means taking one card from the first half, one card for the second half, and so on:

   a d b e c f

Perfect shuffling can be done only with an even number of cards.

A remarkable feature of perfect shuffling is that the deck goes back to the start after a number that is fixed for every n (where n is the number of cards)

  • Implement a magic shuffle function
  • Print the number of perfect shuffles needed to go back to start for decks from 2 to 10,000 cards (by steps of 2), determined by using the magic shuffle function

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.)

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 range(2,10**4,2):
       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> 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>