# Project Euler #11 in Ruby and some Iterations

Question:

Given a 20x20 Grid of integers, find the greatest product of four adjacent numbers. (up, down, diagonal, reverse diagonal).

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 7107 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 0547 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 8335 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 6857 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

Example, above 87 x 97 x 94 x 89 = 70600674


I wrote a procedural approach to solve the problem. It solved the problem and gave out the correct answer, just that I was not very satisfied with the clumsiness of the code. My intention was to make this piece of code more dynamic so that 1) grids of various sizes (e.g. 30x30), and length of the candidates (e.g. instead of 4, make it product of 5 adjacent numbers) can be solved by this code.

1) So first thing you will notice is I write loops like this:

width = data.length       # length of first row of array
height = data.length         # length of array (height)
length = 4                   # 4 adjacent numbers

(0..height-1).each do |i|
(0..width-length).each do |j|
numbers_array = []
(0..length-1).each do |l|
# # Do things here
end
end
end


which I personally think is pretty unreadable. I wish there is a way so I don't have to write all these loops.

2) For the strategies, I came up with the following code which I think is pretty neat, but since there are four different strategies (i.e. up, down, diagonal /, and diagonal ). I had to write this code four times, which violates the DRY principle. So I wonder how I can improve this logic by using a more elegant approach

numbers_array << data[i][j+l].to_i # i, j, l are iterations, where l represent the length of the adjacent numbers, while i, j are positions in the grid.
product = numbers_array.inject(:*)

numbers_array << data[i+l][j].to_i # vertical
numbers_array << data[i+l][j+l].to_i # diagonal \
numbers_array << data[i-l][j+l].to_i # diagaonal /


May be I should encapsulate the four strategies in a def and call it accordingly. Any thoughts?

Below is the complete code for reference. :)

def load_data
data = []
data << line.delete("\n").split(/ /)
end
data
end

width = data.length
height = data.length
length = 4
max_product = [0,0,0,'']

# Horizontal
(0..height-1).each do |i|
(0..width-length).each do |j|
numbers_array = []
(0..length-1).each do |l|
numbers_array << data[i][j+l].to_i
end
product = numbers_array.inject(:*)
puts "Horizontal:[#{i}, #{j}] - #{numbers_array * " x "} = #{product}"
max_product = product > max_product ? i : max_product
max_product = product > max_product ? j : max_product
max_product = product > max_product ? 'Horizontal' : max_product
max_product = product > max_product ? product : max_product
end
end

# Vertical
(0..height-length).each do |i|
(0..width-1).each do |j|
numbers_array = []
(0..length-1).each do |l|
numbers_array << data[i+l][j].to_i # want to make this into a def but since data[i][j] is complicated to put into a def.
end
product = numbers_array.inject(:*)
puts "Vertical:[#{i}, #{j}] - #{numbers_array * " x "} = #{product}"
max_product = product > max_product ? i : max_product
max_product = product > max_product ? j : max_product
max_product = product > max_product ? 'Vertical' : max_product
max_product = product > max_product ? product : max_product
end
end

# Diagonal \
(0..height-length).each do |i|
(0..width-length).each do |j|
numbers_array = []
(0..length-1).each do |l|
numbers_array << data[i+l][j+l].to_i
end
product = numbers_array.inject(:*)
puts "Diagonal \:[#{i}, #{j}] - #{numbers_array * " x "} = #{product}"
max_product = product > max_product ? i : max_product
max_product = product > max_product ? j : max_product
max_product = product > max_product ? 'Diagonal \\' : max_product
max_product = product > max_product ? product : max_product
end
end

# Diagonal /
(length-1..height-1).each do |i|
(0..width-length).each do |j|
numbers_array = []
(0..length-1).each do |l|
numbers_array << data[i-l][j+l].to_i
end
product = numbers_array.inject(:*)
puts "Diagonal \:[#{i}, #{j}] - #{numbers_array * " x "} = #{product}"
max_product = product > max_product ? i : max_product
max_product = product > max_product ? i : max_product
max_product = product > max_product ? j : max_product
max_product = product > max_product ? 'Diagonal /' : max_product
max_product = product > max_product ? product : max_product
end
end

puts max_product.inspect


So you can easily calculate additional answers by simply changing the length now~

• For length 5, max product is: Diagonal :[15, 3] - 87 x 97 x 94 x 89 x 47 = 3318231678
• For length 6, max product is: Horizontal:[17, 10] - 99 x 69 x 82 x 67 x 59 x 85 = 188210512710

• For length 7, max product is: Horizontal:[17, 10] - 99 x 69 x 82 x 67 x 59 x 85 x 74 = 13927577940540 ...etc

• One minor improvement I saw now is to use 0...height instead of 0..height-1 – Chris Yeung Apr 4 '15 at 13:45

Firstly, to load the data, I'd do this:

grid = File.readlines("some-file").map do |line|
line.chomp.split.map(&:to_i)
end


that gets you a grid of integers with minimum hassle.

For the horizontal products there's each_cons (each consecutive), reduce (aka inject), and max to help you (all part of the Enumerable module). Using them, finding the maximum horizontal product is simple:

max_horizontal = grid.map do |row|
row.each_cons(4).map { |group| group.reduce(&:*) }.max
end.max


Now, to find the vertical maximum, we only need to transpose the grid and do the exact same thing. Thankfully, there's Array#transpose which does exactly what it says on the tin. So we can wrap the above in a method, and call it once with the plain grid, and once with the transposed grid to get the horizontal and vertical maxima:

def max_linear_product(grid)
grid.map do |row|
row.each_cons(4).map { |group| group.reduce(&:*) }.max
end.max
end

max_horizontal = max_linear_product(grid)
max_vertical = max_linear_product(grid.transpose)


You can of course add an extra argument to the method instead of the hard-coded 4.

The diagonals require a bit more fiddling. But again, we can use one method, and just feed it two different versions of the grid to find the two different diagonal maxima.

Now, we know we can skip a few rows and columns, since we need 4 numbers for each product. So we won't gain anything by starting our diagonal grouping in the last 3 rows, or in the last three columns of each row.

I came up with a method like this (which isn't too pretty):

def max_diagonal_product(grid)
grid[0..-4].each_with_index.flat_map do |row, y|
row[0..-4].each_with_index.map do |_, x|
4.times.reduce(1) { |product, i| product * grid[y + i][x + i] }
end
end.max
end


(Again, the hard-coded 4 can be replaced with an argument.)

To find the top-left to bottom-right diagonal maximum, we just feed the plain grid to that method:

max_diagonal = max_diagonal_product(grid)


To find the reverse diagonal's maximum, we just need to feed it the grid with each row reversed:

max_reverse_diagonal = max_diagonal_product(grid.map(&:reverse))


Finally, to find the maximum of all of those, you can of course just do:

[max_horizontal, max_vertical, max_diagonal, max_reverse_diagonal].max


All together:

def max_linear_product(grid)
grid.map do |row|
row.each_cons(4).map { |group| group.reduce(&:*) }.max
end.max
end

def max_diagonal_product(grid)
grid[0..-4].each_with_index.flat_map do |row, y|
row[0..-4].each_with_index.map do |_, x|
4.times.reduce(1) { |product, i| product * grid[y + i][x + i] }
end
end.max
end

line.chomp.split.map(&:to_i)
end

max_horizontal = max_linear_product(grid)
max_vertical = max_linear_product(grid.transpose)
max_diagonal = max_diagonal_product(grid)
max_reverse_diagonal = max_diagonal_product(grid.map(&:reverse))

solution = [max_horizontal, max_vertical, max_diagonal, max_reverse_diagonal].max


At least, that's one way to do it.

Another, more low-level way, could be to run through the grid once, getting horizontal, vertical and diagonal groups all in one go:

groups = grid.each_with_index.flat_map do |row, y|
row[0..-4].each_with_index.flat_map do |_, x|
groups = []
groups << 4.times.map { |i| grid[y][x + i] } # horizontal
if y <= grid.count - 4
groups << 4.times.map { |i| grid[y + i][x] } # vertical
groups << 4.times.map { |i| grid[y + i][x + i] } # diagonal top-left/bottom-right
groups << 4.times.map { |i| grid[y + i][x + 3 - i] } # opposite diagonal
end
groups
end
end

maximum = groups.map { |group| group.reduce(&:*) }.max