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Egg Drop Experiment: Number of Drops

/*************************************************************************** * Author: Isai Damier * Title: EggDropPuzzle * Project: geekviewpoint * Package: algorithms.dynamic.programming * * Description: * While eggs are fragile, there are heights from which a fallen egg will * not break. If you are given a measuring tape and a carton of eggs, can * you devise a strategy for finding the breaking height of the eggs using * the fewest trials possible? * A few constraints: * 1] If you drop an egg and it didn't break, you can reuse it: * it's as good as new. * 2] The eggs are completely interchangeable: same chemical * and mechanical structure. * 3] You are guaranteed the eggs will break within the range of the * measuring tape, that is 1 <= B <= h, where B is the breaking height * and h is the length of the measuring tape. * * Solution approach: * 1] If you have zero eggs, then the problem is unsolvable. * 2] If you have one egg, the problem becomes trivial: you simply must * try all the heights starting with the lowest. If the tape is 15 * centimeters tall, you must start at 1cm, then 2cm, then so on. If * after reaching 5cm you decide to skip to 7cm and the egg breaks, you * won't know if it would have broken at 6cm. * 3] If you have two eggs, things get a bit tricky. You must drop the * first egg from a height x such that if it breaks, you can apply * step 2 the minimum number of times; and if it does not break, you * can re-apply step 3 the minimum number of times. * * Formulation: * Let P(n,h) means you have n eggs and a tape of height h. Then dropping * 1 egg from some height x (1<=x<=h) has two possible effects: * a) if it breaks, then the problem becomes P(n-1,x-1): * You lost 1 egg and must test all the heights below x. * b) if it does not break, the problem becomes P(n, h-x): * you lost no egg but must test all the heights above x. * * On the bright side, whether the egg breaks or not, "you will have * reduced the problem to a smaller problem whose solution you can use * toward solving the original problem." The quoted text is the description * of dynamic programming. Therefore we will use dynamic programming * to solve the problem. * * So how do you select height x? Use brute force: try all heights from 1 * to h, then pick the one which gives the minimum worst case drops: * * for(int x=1; x<h; x++) * best = Math.min(best, Math.max(P(n-1,x-1),P(n,h-x))) * * Why Math.max(P(n-1,x-1),P(n,h-x))? because you are looking for the worst * case drop of each x. Once you get all the worst case drops, then you * can choose the smallest one. * * Program: * P(int e, int h) { * if(1 >= e) return h;//base case * if(1 >= h) return 1;//base case * int best = h+1;//h+1 is infinity * for(int x=1; x < h; x++) { * best = Math.min(best, Math.max(P(e-1,x-1),P(e,h-x))); * } * return best+1; * } **************************************************************************/ package algorithms.dynamic.programming; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class EggDropPuzzle { /************************************************************************* * Function: numberOfDrops * @param * e: number of eggs * h: length to test * * Description: Find the fewest number of drops necessary to * determine the breaking height of the given eggs. * * Technical Details: * While program P above is sufficient to solve the problem, it is quite * inefficient. It is generally good instinct to suspect any dynamic * programming algorithm that does not use memoization. Memoization * simply means using a memo to remember subproblems you have already * solved so you don't waste time solving them again. The following * algorithm augments memoization to P for a more efficient solution. * * 1] create a table to memoize results. * 2] call recursive eggDrop with e, h, and the table to * get the optimum number of drops necessary. ************************************************************************/ public int numberOfDrops(int e, int h) { int[][] memo = new int[e + 1][h]; for (int x = 0; x <= e; x++) { for (int y = 0; y < h; y++) { memo[x][y] = Integer.MAX_VALUE;// infinity could also be s+1 } } return numberOfDrops(e, h, memo); }//numberOfDrops /****************************************************************** * Function: numberOfDrops * @param * e: number of eggs * h: length to test * memo: table to memoize results to subproblems. * * Description: This is the recursive portion of the eggDrop algorithm. * * Technical Details: * Remember the guarantee that the eggs will break within the * range of the measuring tape (constraint #3). * * 1] if there is only one egg, then the worse case is to * test each grade from 1 to h. So just return h. * 2] if there is only one grade to test (i.e. h=1), then it * must be the optimal solution. Simply return 1. * 3] initialize variable 'best' to infinity. * 4] for each height x from 1 to h * a) using either the memoized or the computed * value, get the greater of numberOfDrops(e-1,i-1) * and numberOfDrops(e,h-i) * b) if the value in part-a is less than 'best', assign * it to best. * 5] memoize the result as best+1. Plus 1 recognizes the * present height, since the for loop excluded it. * 6] return the result as best+1. ****************************************************************/ private int numberOfDrops(int e, int h, int[][] memo) { if (1 >= e) { return h; } if (1 >= h) { return 1; } int best = h + 1;// infinity for (int a, b, x = 1; x < h; x++) { if ((a = memo[e - 1][x - 1]) == Integer.MAX_VALUE) { a = numberOfDrops(e - 1, x - 1, memo); } if ((b = memo[e][h - x]) == Integer.MAX_VALUE) { b = numberOfDrops(e, h - x, memo); } best = Math.min(best, Math.max(a, b)); } if (h < memo[0].length) { memo[e][h] = best + 1; } return best + 1; }//numberOfDrops }

package algorithms.dynamic.programming; import java.util.ArrayList; import java.util.Arrays; import java.util.List; import org.junit.Test; import static org.junit.Assert.*; public class EggDropPuzzleTest { /** * Test of numberOfDrops method, of class EggDropPuzzle. */ @Test public void testNumberOfDrops() { System.out.println("numberOfDrops"); EggDropPuzzle eggDrops = new EggDropPuzzle(); int[][] result = {{1, 1, 1, 2}, {2, 3, 2, 2}, {4, 6, 3, 2}, {7, 10, 4, 2}, {11, 15, 5, 2}, {16, 21, 6, 2}, {22, 28, 7, 2}, {29, 36, 8, 2}, {37, 45, 9, 2}, {46, 55, 10, 2}, {56, 66, 11, 2}, {67, 78, 12, 2}, {79, 91, 13, 2}, {92, 105, 14, 2}, {106, 120, 15, 2}, {121, 136, 16, 2}, {137, 153, 17, 2}, {154, 171, 18, 2}, {172, 190, 19, 2}, {191, 200, 20, 2}, {16, 31, 5, 5}, {32, 62, 6, 5}, {63, 119, 7, 5}, {120, 153, 8, 5}, {32, 63, 6, 8}, {64, 127, 7, 8}, {128, 182, 8, 8}}; for (int[] R : result) { for (int s = R[0]; s <= R[1]; s++) { int x = eggDrops.numberOfDrops(R[3], s); if (R[2] != x) { fail(" Floor " + R[0] + ": expected " + R[2] + " found " + x); } } } } }