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# Project Euler | Maximum Sum Traversing Top To Bottom In A Triangle

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The 18th and the 67th problems in Project Euler are one and the same. The only difference is in the input test case. The problem 18 has a smaller input and 67 has a large input. For the explanation purpose, I’ll consider problem 18. The code without any modification can be extended to 67th problem.

Given a triangle of numbers, the objective is to determine the maximum sum along paths descending from the apex of the triangle to the base. Consider the example:

         3      7     4   2     4     68     5     9     3

The maximum sum occurs along the path 3–7–4–9.

The problem required us to determine the maximum sum for any given triangle. Also an interesting thing is mentioned about the problem, in case you were considering a brute force method of checking every path:

It is not possible to try every route to solve this problem (where the base of the triangle has 100 numbers), as there are 2⁹⁹ altogether! If you could check one trillion (10¹²) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it.

Being forewarned, I never once gave a thought to the brute force solution.

This problem was not as tricky as they tried to sound it. The solution is quite easy to see. I’ll run you through the idea as it formed in my mind. I used the top-to-bottom approach for forming the idea and coming up with the solution, and then implemented the solution in bottom-to-top sweep. I hope you have realized that a greedy solution will not work here.

Consider the triangle given above. I’ll work out the maximum path for it from top to bottom. We start with 3. We may choose either 7 or 4. To help make this choice, consider the two triangles with apex 7 and the other with 4. So we have two smaller triangles of height 3 each. Say the maximum length for the triangle with apex 7 is x and with apex 4 is y. If x > y, we choose 7 as the next on the path, else 4.

Thus, at every step we form two triangle of height 1 smaller, and choose the apex which has the larger sum. Implemented as it is will give us the recursive solution. To get an iterative solution to this problem, one could traverse from bottom-to-top. The bottom-to-top approach is also simple. It is greedy in approach.

Consider the penultimate row, specifically the triangle with 2 as apex and 8, 5 as the base. The maximum sum in this triangle is 2+8 = 10. Replace 2 with 10. Do the same for all the triangles formed with the penultimate layer as the apex. Thus our triangle reduces to

       3    7     410     13     15

You know what to do next. Keep applying the same till the apex of the original triangle has the maximum sum.

I have implemented the bottom-to-top method so as to avoid recursion. Here is my code:

import java.io.BufferedReader;import java.io.FileReader;import java.io.IOException;public class Problem_18 {	private static final int heightOfTree = 100;	private final String fileName = "problem_67.in";	private int[][] tree;	public int maxValue() throws IOException {		readTree();		for (int i = Problem_18.heightOfTree - 2; i >= 0; i--)			for (int j = 0; j <= i; j++)				tree[i][j] += tree[i + 1][j] > tree[i + 1][j + 1] ? tree[i + 1][j] : tree[i + 1][j + 1];		return tree[0][0];	}	private void readTree() throws IOException {		tree = new int[Problem_18.heightOfTree][];		BufferedReader bufferedReader = new BufferedReader(new FileReader(fileName));		for (int i = 0; i < Problem_18.heightOfTree; i++) {			tree[i] = new int[i + 1];			String[] values = bufferedReader.readLine().split(" ");			for (int j = 0; j <= i; j++)				tree[i][j] = Integer.parseInt(values[j]);		}	}	public static void main(String[] args) throws IOException {		System.out.println(new Problem_18().maxValue());	}}