/***************************************************************************
* 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);
}
}
}
}
}
