## Saturday, April 12, 2008

### Project Euler Problem 18

This time we will solve euler problem 18 by semi-bruteforce solution.
Problem description:
--------------------------------------------------------------------------------------
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 5
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
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Solution in C# is more or less short and fast (calculates answer in 1,1 ms):

using System;class Program{    static long MaxTotalFromTriangle()    {        long[][] triangle = new long[15][];        long max = 0;        triangle[0] = new long[] { 75 };        triangle[1] = new long[] { 95, 64 };        triangle[2] = new long[] { 17, 47, 82 };        triangle[3] = new long[] { 18, 35, 87, 10 };        triangle[4] = new long[] { 20, 04, 82, 47, 65 };        triangle[5] = new long[] { 19, 01, 23, 75, 03, 34 };        triangle[6] = new long[] { 88, 02, 77, 73, 07, 63, 67 };        triangle[7] = new long[] { 99, 65, 04, 28, 06, 16, 70, 92 };        triangle[8] = new long[] { 41, 41, 26, 56, 83, 40, 80, 70, 33 };        triangle[9] = new long[] { 41, 48, 72, 33, 47, 32, 37, 16, 94, 29 };        triangle[10] = new long[] { 53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14 };        triangle[11] = new long[] { 70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57 };        triangle[12] = new long[] { 91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48 };        triangle[13] = new long[] { 63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31 };        triangle[14] = new long[] { 04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23 };        for (int i = 1; i < triangle.Length; i++)        {            for (int j = 0; j < triangle[i].Length; j++)            {                // Accumulating maximum total                if (j == 0)                {                    triangle[i][j] += triangle[i - 1][j];                }                else if (j == triangle[i].Length - 1)                {                    triangle[i][j] += triangle[i - 1][triangle[i - 1].Length - 1];                }                else                {                    triangle[i][j] += Math.Max(triangle[i - 1][j], triangle[i - 1][j - 1]);                }                // Finding maximum from last row                if (i == triangle.Length - 1)                {                    if (triangle[i][j] > max) max = triangle[i][j];                }            }        }        return max;    }    static void Main(string[] args)    {        System.Diagnostics.Stopwatch sw = new System.Diagnostics.Stopwatch();        sw.Start();        long ans = MaxTotalFromTriangle();        sw.Stop();        Console.WriteLine("Total sum {0}; cpu time {1} ms", ans, sw.Elapsed.TotalMilliseconds);        Console.ReadLine();    } }

Also this solution is universal and can calculate bigger triangles (lets say 100 rows triangles and more) with a reasonable time.

Have fun!!

#### 1 comment:

1. Hi,
I thought you might be interested in the solution to Problem 18 that I've written up, using LINQ and the Aggregate method.

Sam

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