Shuffle a deck of cards perfectly

Write a method to shuffle a deck of cards. It must be a perfect shuffle – in other words, each 52! permutations of the deck has to be equally likely. Assume that you are given a random number generator which is perfect.

My initial thoughts:
We first select a card randomly from the 52 cards. That is probability of 1/52. We then select another card from the rest 51 cards. That is probability of 1/51. So on and so forth. Finally we have a particular permutation with probability of 1/52!, which is expected.

My initial codes:

	public static List<Integer> shuffle() {
		Set<Integer> deck = new HashSet<Integer>();
		// build a deck of cards
		for (int i = 0; i < 52; ++i) {
			deck.add(i + 1);
		}
		List<Integer> shuffledDeck = new ArrayList<Integer>();

		for (int i = 0; i < 52; ++i) {
			int card;
			do {
				card = new Random().nextInt(52) + 1;
			} while (!deck.contains(card));
			deck.remove(card);
			shuffledDeck.add(card);
		}
		return shuffledDeck;
	}

Solution:
Let’s start with a brute force approach: we could randomly selecting items and put them into a new array. We must make sure that we don’t pick the same item twice though by somehow marking the node as dead.

Array:					[1] [2] [3] [4] [5] 
Randomly select 4:		[4] [?] [?] [?] [?] 
Mark element as dead: 	[1] [2] [3] [X] [5]

The tricky part is, how do we mark [4] as dead such that we prevent that element from being picked again? One way to do it is to swap the now-dead [4] with the first element in the array:

Array: 					[1] [2] [3] [4] [5]
Randomly select 4: 		[4] [?] [?] [?] [?]
Swap dead element: 		[X] [2] [3] [1] [5]

Array: 					[X] [2] [3] [1] [5]
Randomly select 3: 		[4] [3] [?] [?] [?]
Swap dead element: 		[X] [X] [2] [1] [5]

By doing it this way, it’s much easier for the algorithm to “know”” that the first k elements are dead than that the third, fourth, nineth, etc elements are dead. We can also optimize this by merging the shuffled array and the original array:

Randomly select 4: 		[4] [2] [3] [1] [5] 
Randomly select 3: 		[4] [3] [2] [1] [5]

This is an easy algorithm to implement iteratively:

	public static void shuffleArray(int[] cards) {
		int temp, index;
		for (int i = 0; i < cards.length; i++) {
			index = (int) (Math.random() * (cards.length - i)) + i;
			temp = cards[i];
			cards[i] = cards[index];
			cards[index] = temp;
		}
	}
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1 Comment (+add yours?)

  1. Gayle Laakmann McDowell
    May 29, 2012 @ 01:19:38

    Hi Runhe,

    I’ve been following your blog for a few months now, and I’ve noticed that while some of the posts are just your attempts at solving the problems, others actually re-type the book’s solutions.

    Could you please:
    (1) Remove the Cracking the Coding Interview tag on the posts
    (2) Refrain from re-posting the book’s solution (which is in fact a violation of copyright law).

    Thanks!
    Gayle

    Reply

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