How to solve the 2 glass balls puzzle programmatically? [closed]

Question

There is a popular puzzle about a 100-story building and two glass balls. I see the solution and now I wonder if I can solve the puzzle programmatically.

The trivial programmatic solution is a full search (I believe I can code it using backtracking). Is there any better programmatic solution ? Can I use dynamic programming to solve the puzzle?

Answer 1

I am sorry but the previous answer just does no justice to this question. Best possible time is sqrt of N. Someone will attempt to refute this with logn but i am sorry, its just not the case.

// my prentend breaking function
function itBreaks(level) {
    return level > 36;
}

function search(maxLevel) {
    var sqrtN = Math.floor(Math.sqrt(maxLevel));
    var i = 0;
    for (;i < maxLevel; i += sqrtN) {
        if (itBreaks(i)) {
            break;
        }
    }

    for (i -= sqrtN; i < maxLevel; i++) {
        if (itBreaks(i)) {
            return i - 1;
        }
    }
}

Answer 2

This is similar to the egg dropping puzzle. I'll give you the basic strategy in dynamic programming. Relate this closely with your question.

# include <stdio.h>
# include <limits.h>

// A utility function to get maximum of two integers
int max(int a, int b) { return (a > b)? a: b; }

/* Function to get minimum number of trails needed in worst
case with n eggs and k floors */
int eggDrop(int n, int k)
{
/* A 2D table where entery eggFloor[i][j] will represent minimum
   number of trials needed for i eggs and j floors. */
int eggFloor[n+1][k+1];
int res;
int i, j, x;

// We need one trial for one floor and0 trials for 0 floors
for (i = 1; i <= n; i++)
{
    eggFloor[i][1] = 1;
    eggFloor[i][0] = 0;
}

// We always need j trials for one egg and j floors.
for (j = 1; j <= k; j++)
    eggFloor[1][j] = j;

// Fill rest of the entries in table using optimal substructure
// property
for (i = 2; i <= n; i++)
{
    for (j = 2; j <= k; j++)
    {
        eggFloor[i][j] = INT_MAX;
        for (x = 1; x <= j; x++)
        {
            res = 1 + max(eggFloor[i-1][x-1], eggFloor[i][j-x]);
            if (res < eggFloor[i][j])
                eggFloor[i][j] = res;
        }
    }
}

// eggFloor[n][k] holds the result
return eggFloor[n][k];
}

/* Driver program to test to pront printDups*/
int main()
{
    int n = 2, k = 36;
    printf ("\nMinimum number of trials in worst case with %d eggs and %d floors is %d \n", n, k, eggDrop(n, k));
    return 0;
}

Output: Minimum number of trials in worst case with 2 eggs and 36 floors is 8