Programming Praxis

4 Responses to “Mind-Boggling Card Trick”

1. Informatmiago said

   September 4, 2018 at 11:52 AM

   This is indeed a magic trick:

   The first phase is not random at all, because the operation ensures that there are the same number of black in the black stack as there are reds in the red stack. Discading the seen cards, ensures that there will remain the exact same number of cards (either of the same color, or the opposite color) to be moved to the other stack.

   The second part is just for the spectacle, since it’s an operation that does not change that property (if two cards of the same color are exchanged, then no change; if cards of different colors are exchanged, then either it increments both counts (if the right color leaves each stack), or it decrements both counts (if red leaves the red stack and black leaves the black stack). Randomness does not matter, you could choose the cards you want, as long as you exchange the same number from both stacks :-)
2. Rutger said

   September 4, 2018 at 1:00 PM

   This code almost reads like the assignment to me ;)

   from random import shuffle
   from random import randint
   
   deck = 26*['black'] + 26*['red']
   shuffle(deck)
   
   red_stack = []
   black_stack = []
   
   while deck:
   	item = deck.pop()
   	if item == 'black':
   		black_stack.append(deck.pop())
   	else:
   		red_stack.append(deck.pop())
   
   shuffle(black_stack)
   shuffle(red_stack)
   swap = randint(1, min(len(black_stack), len(red_stack)))
   black_stack, red_stack = red_stack[:swap]+black_stack[swap:], black_stack[:swap]+red_stack[swap:]
   
   print(black_stack.count('black'))
   print(red_stack.count('red'))
   
   
3. Globules said

   September 9, 2018 at 5:49 AM

   A Haskell version.

   import Control.Monad.Random (RandT, evalRandT, getRandomR, liftIO)
   import Data.Tuple.Extra (both)
   import Data.List (partition)
   import System.Random.Shuffle (shuffleM)
   import System.Random.TF.Gen (TFGen)
   import System.Random.TF.Init (initTFGen)
   import Text.Printf (PrintfArg, formatArg, formatString, printf)
   
   data Card = Black | Red deriving (Eq, Show)
   
   orderedDeck :: [Card]
   orderedDeck = replicate 26 Black ++ replicate 26 Red
   
   -- Look at the first card of the list.  If it's black add the next card to the
   -- black stack, otherwise add it to the red stack.  Discard the first card.
   -- Repeat this until the list has been exhausted.  Return the black and red
   -- stacks.  If the list contains an odd number of cards the last one will simply
   -- be discarded.
   makeStacks :: [Card] -> ([Card], [Card])
   makeStacks = both (map snd) . partition ((== Black) . fst) . pairs
     where pairs (x:y:zs) = (x, y) : pairs zs
           pairs _        = []
   
   -- The result of randomly swapping elements between two lists.  (We don't bother
   -- retaining the relative order of the original elements.)
   randomSwap :: ([a], [a]) -> RandC ([a], [a])
   randomSwap (xs, ys) = do n <- getRandomR (0, (length xs `min` length ys) - 1)
                            (xs1, xs2) <- splitShuffle n xs
                            (ys1, ys2) <- splitShuffle n ys
                            return (ys1 ++ xs2, xs1 ++ ys2)
     where splitShuffle n zs = splitAt n <$> shuffleM zs
   
   -- Simulate the card trick described in the exercise.
   cardTrick :: RandC ()
   cardTrick = do deck <- shuffleM orderedDeck
                  (blackStack, redStack) <- randomSwap (makeStacks deck)
                  printNumberOf Black blackStack
                  printNumberOf Red   redStack
   
   main :: IO ()
   main = initTFGen >>= evalRandT cardTrick
   
   ---------------------------------- Utilities ----------------------------------
   
   type RandC a = RandT TFGen IO a
   
   instance PrintfArg Card where
     formatArg c = formatString (show c)
   
   printNumberOf :: Card -> [Card] -> RandC ()
   printNumberOf c = liftIO . printf "%s cards in %s stack: %d\n" c c
                            . length . filter (== c)
   

   $ for i in $(seq 1 10); do ./cardtrick; done
   Black cards in Black stack: 4
   Red cards in Red stack: 4
   Black cards in Black stack: 6
   Red cards in Red stack: 6
   Black cards in Black stack: 8
   Red cards in Red stack: 8
   Black cards in Black stack: 5
   Red cards in Red stack: 5
   Black cards in Black stack: 9
   Red cards in Red stack: 9
   Black cards in Black stack: 10
   Red cards in Red stack: 10
   Black cards in Black stack: 6
   Red cards in Red stack: 6
   Black cards in Black stack: 7
   Red cards in Red stack: 7
   Black cards in Black stack: 6
   Red cards in Red stack: 6
   Black cards in Black stack: 6
   Red cards in Red stack: 6
   
4. Daniel said

   September 11, 2018 at 12:55 AM

   Here’s a solution in Python.

   from collections import deque
   import random
   
   deck = [0] * 26 + [1] * 26
   random.shuffle(deck)
   
   stacks = [deque(), deque()]
   
   for top, bottom in zip(deck[0::2], deck[1::2]):
       stacks[top].append(bottom)
   
   for stack in stacks:
       random.shuffle(stack)
   min_len = min(len(stack) for stack in stacks)
   num_swaps = random.randint(0, min_len - 1)
   for idx in range(num_swaps):
       stacks[0].appendleft(stacks[1].pop())
       stacks[1].appendleft(stacks[0].pop())
   
   assert stacks[0].count(0) == stacks[1].count(1)
   

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