Euler Problem 18 and 67 are exactly the same besides that the data set in the second version is larger than in the first one. In this post, I kill two Eulers with one code.

These problems deal with binary trees, which is a data structure where each node has two children. A practical example of a binary tree is a pedigree chart, where each person or animal has two parents, four grandparents and so on.

## Euler Problem 18 Definition

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

As there are only 16,384 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)

## Solution

This problem seeks a maximum path sum in a binary tree. The brute force method, as indicated in the problem definition, is a very inefficient way to solve this problem. The video visualises the quest for the maximum path, which takes eleven minutes of hypnotic animation.

A more efficient method is to define the maximum path layer by layer, starting at the bottom. The maximum sum of 2+8 or 2+5 is 10, the maximum sum of 4+5 or 4+9 is 13 and the last maximum sum is 15. These numbers are now placed in the next row. This process cycles until only one number is left. This algorithm solves the sample triangle in four steps:

Step 1:

3

7 4

2 4 6

8 5 9 3

Step 2:

3

7 4

10 13 15

Step 3:

3

20 19

Step 4:

23

In the code below, the data is triangle matrix. The variables *rij* (row) and *kol* (column) drive the search for the maximum path. The triangle for Euler Problem 18 is manually created and the triangle for Euler Problem 67 is read from the website.

path.sum <- function(triangle) { for (rij in nrow(triangle):2) { for (kol in 1:(ncol(triangle)-1)) { triangle[rij - 1,kol] <- max(triangle[rij,kol:(kol + 1)]) + triangle[rij - 1, kol] } triangle[rij,] <- NA } return(max(triangle, na.rm = TRUE)) } # Euler Problem 18 triangle <- matrix(ncol = 15, nrow = 15) triangle[1,1] <- 75 triangle[2,1:2] <- c(95, 64) triangle[3,1:3] <- c(17, 47, 82) triangle[4,1:4] <- c(18, 35, 87, 10) triangle[5,1:5] <- c(20, 04, 82, 47, 65) triangle[6,1:6] <- c(19, 01, 23, 75, 03, 34) triangle[7,1:7] <- c(88, 02, 77, 73, 07, 63, 67) triangle[8,1:8] <- c(99, 65, 04, 28, 06, 16, 70, 92) triangle[9,1:9] <- c(41, 41, 26, 56, 83, 40, 80, 70, 33) triangle[10,1:10] <- c(41, 48, 72, 33, 47, 32, 37, 16, 94, 29) triangle[11,1:11] <- c(53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14) triangle[12,1:12] <- c(70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57) triangle[13,1:13] <- c(91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48) triangle[14,1:14] <- c(63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31) triangle[15,1:15] <- c(04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23) answer <- path.sum(triangle) print(answer)

## Euler Problem 67

The solution for problem number 67 is exactly the same. The data is read directly from the Project Euler website.

# Euler Problem 67 triangle.file <- read.delim("https://projecteuler.net/project/resources/p067_triangle.txt", stringsAsFactors = F, header = F) triangle.67 <- matrix(nrow = 100, ncol = 100) for (i in 1:100) { triangle.67[i,1:i] <- as.numeric(unlist(strsplit(triangle.file[i,], " "))) } answer <- path.sum(triangle.67) print(answer)

View the latest version of this code on GitHub.

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Pingback: Euler Problem 18 & 67: Maximum Path Sums – Mubashir Qasim

Hmmm.. for a followup, how about this: if constrained to using the integers 1:N in a pyramid with N locations, what’s the maximum number of winning paths possible (i.e. same max value)?

I am not quite sure what you mean. Are you suggesting that a pyramid with 1:n has more than one maximum path?

1

2, 3

4, 5, 6

7, 8, 9, 10

11, 12, 13, 14 etc.

You can generate these like this:

I am not sure how to test for multiple maximum path sums other than using brute force to create all paths.