Throwing cats out of windows

Question

Imagine you\'re in a tall building with a cat. The cat can survive a fall out of a low story window, but will die if thrown from a high floor. How can you figure out the lon

Answer 1

You can easily write a little DP (dynamic programming) for the general case of n floors and m cats.

The main formula, a[n][m] = min(max(a[k - 1][m - 1], a[n - k][m]) + 1) : for each k in 1..n, should be self-explanatory:

• If first cat is thrown from k-th floor and dies, we now have k - 1 floors to check (all below k) and m - 1 cats (a[k - 1][m - 1]).
• If cat survives, there're n - k floors left (all floors above k) and still m cats.
• The worst case of two should be chosen, hence max.
• + 1 comes from the fact that we've just used one attempt (regardless of whether cat has survived or not).
• We try every possible floor to find the best result, hence min(f(k)) : for k in 1..n.

It agrees with Google result from Gaurav Saxena's link for (100, 2).

int n = 100; // number of floors
int m = 20; // number of cats
int INFINITY = 1000000;

int[][] a = new int[n + 1][m + 1];
for (int i = 1; i <= n; ++i) {
    // no cats - no game
    a[i][0] = INFINITY;
}

for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
        // i floors, j cats
        a[i][j] = INFINITY;

        for (int k = 1; k <= i; ++k) {
            // try throw first cat from k-th floor
            int result = Math.max(a[k - 1][j - 1], a[i - k][j]) + 1;
            a[i][j] = Math.min(a[i][j], result);
        }
    }
}

System.out.println(a[n][m]);

You can easily find strategy (how to throw first cat), if you save best k in another array.

There's also a faster solution, not involving O(n^3) computations, but I'm a bit sleepy already.

edit
Oh yeah, I remember where I saw this problem before.