Perfect shuffle

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

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>