problem18 (Maximum path sum I)

problem18 (Maximum path sum I)

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23


NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

In Python:
# Maximum path sum I
import time

startTime = time.time()
f = open("./PE18_number.txt", 'r')
dataList = [int(num) for num in f.read().split()]
f.close()
maxValue = 0
for i in range(0, 16384):
    product = dataList[0]
    orderGroup = 1
    indexSubGroup = 0
    for j in range(0, 14):
        if i % 2 == 0:
            orderGroup += 1
            indexSubGroup = indexSubGroup + 0
            product += dataList[int(orderGroup * (orderGroup - 1) / 2) + indexSubGroup]
            i //= 2
        else:
            orderGroup += 1
            indexSubGroup = indexSubGroup + 1
            product += dataList[int(orderGroup * (orderGroup - 1) / 2) + indexSubGroup]
            i //= 2
    if product > maxValue:
        maxValue = product
print(maxValue)
print(time.time() - startTime, "seconds")
Run time: 0.2011110782623291 seconds



In Java:
//Euler18 Maximum path sum I

package project_euler11_20;


import java.io.FileReader;
import java.io.BufferedReader;
import java.io.IOException;
import java.util.ArrayList;


public class Euler18 {

	public static void main(String[] args) throws IOException {
		long startTime = System.currentTimeMillis();
		int maxValue = 0;
		int orderGroup;
		int indexSubGroup;
		int product;
		double num;
		String[] data;
		ArrayList<Integer> dataList = new ArrayList<Integer>();
		BufferedReader br = new BufferedReader(new FileReader("./Euler18_number.txt"));

		while(true) {
			String line = br.readLine();
			if (line == null) break;
			data = line.split(" ");
			for (int i = 0; i < data.length; i++) {
				dataList.add(Integer.parseInt(data[i]));
			}
		}
		br.close();
		for (int i = 0; i < Math.pow(2, 14); i++) {
			product = dataList.get(0);
			orderGroup = 1;
			indexSubGroup = 0;
			num = i;
			for (int j = 0; j < 14; j++) {
				if (num % 2 == 0) {
					orderGroup++;
					product += dataList.get(orderGroup*(orderGroup - 1) / 2 + indexSubGroup);
					num = Math.floor(num / 2);
				}
				else {
					orderGroup++;
					indexSubGroup++;
					product += dataList.get(orderGroup*(orderGroup - 1) / 2 + indexSubGroup);
					num = Math.floor(num / 2);
				}
			}
			if (product > maxValue) {
				maxValue = product;
			}
		}
		System.out.println(maxValue);
		long endTime = System.currentTimeMillis();
		System.out.println((double)(endTime - startTime)/(double)1000 + "seconds");
	}
}
Run time: 0.035seconds


solution: 1074