2000 shuffles

[BMAD]

A shuffle of 2n cards puts the first n cards in the odd positions and the last n cards in the even positions.[For example shuffle (1,2,3,4,5,6) and you get (1,4,2,5,3,6).] Heather has 10 cards, 1-10, Briana has 12 cards, 1-12. Each shuffles her deck 2000 times. "Hey, my deck is back to its original state!"

Who said that, and which card does the other deck have in position #5?

[bonanova]

Briana said it.

There is a 2 in the fifth place of Heather's deck.

s out-shuffles will restore deck a deck of n=2m cards,

where s=ord_n-1(2).

For a 10-card deck, that's 6.

Since 2000(mod 6) = 2, 2000 out-shuffles of Heather's deck is equivalent to 2 out-shuffles.

For a 12-card deck, that's 10.

Since 2000(mod 10) = 0, 2000 out-shuffles restores Briana's deck.

[Anza Power]

public class Class1 {
    public static void main(String[] args) {
        int[] arr1 = {1,2,3,4,5,6,7,8,9,10};
        int[] arr2 = {1,2,3,4,5,6,7,8,9,10,11,12};
        for(int i=0;i<2000;i++){
            doShuffle(arr1);
            doShuffle(arr2);
        }
        System.out.println("arr1 = "+java.util.Arrays.toString(arr1));
        System.out.println("arr2 = "+java.util.Arrays.toString(arr2));
    }
    private static void doShuffle(int[] arr){
        int[] temp = java.util.Arrays.copyOf(arr, arr.length);
        for(int i=0;i<arr.length;i++){
            arr[i] = temp[i/2 + (i%2)*(arr.length/2)];
        }
    }
}

Result:

arr1 = [1, 8, 6, 4, 2, 9, 7, 5, 3, 10]

arr2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

[DeathStone0000]

The one who has her cards in order is Briana.

The fifth card in the other deck is 2

By shuffling Heather's cards six times, it goes back to the original state.

So by the 2000th shuffle, it will have the same order as what the deck had after the second shuffle.

Edit: I just realized how tiresome my solution is.

Edited July 19, 2013 by DeathStone0000