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 0849 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 0081 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 6552 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 9122 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 8024 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 5032 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 7067 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 2124 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 7221 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 9578 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 9216 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 5786 56 00 48 35 71[89]07 05 44 44 37 44 60 21 58 51 54 17 5819 80 81 68 05[94]47 69 28 73 92 13 86 52 17 77 04 89 55 4004 52 08 83[97]35 99 16 07 97 57 32 16 26 26 79 33 27 98 6688 36 68[87]57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 6904 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 3620 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 1620 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 5401 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48Example, 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[0].length       # length of first row of arrayheight = 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     endendwhich 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 approachnumbers_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 # verticalnumbers_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 = []    File.readlines('11-resource').each_with_index do |line|        data << line.delete(\n).split(/ /)    end    dataenddata = load_datawidth = data[0].lengthheight = data.lengthlength = 4max_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[1] = product > max_product[0] ? i : max_product[1]        max_product[2] = product > max_product[0] ? j : max_product[2]        max_product[3] = product > max_product[0] ? 'Horizontal' : max_product[3]        max_product[0] = product > max_product[0] ? product : max_product[0]    endend# 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[1] = product > max_product[0] ? i : max_product[1]        max_product[2] = product > max_product[0] ? j : max_product[2]        max_product[3] = product > max_product[0] ? 'Vertical' : max_product[3]        max_product[0] = product > max_product[0] ? product : max_product[0]    endend# 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[1] = product > max_product[0] ? i : max_product[1]        max_product[2] = product > max_product[0] ? j : max_product[2]        max_product[3] = product > max_product[0] ? 'Diagonal \\' : max_product[3]        max_product[0] = product > max_product[0] ? product : max_product[0]    endend# 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[1] = product > max_product[0] ? i : max_product[1]        max_product[1] = product > max_product[0] ? i : max_product[1]        max_product[2] = product > max_product[0] ? j : max_product[2]        max_product[3] = product > max_product[0] ? 'Diagonal /' : max_product[3]        max_product[0] = product > max_product[0] ? product : max_product[0]    end endputs max_product.inspectSo 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 = 3318231678For length 6, max product is: Horizontal:[17, 10] - 99 x 69 x 82 x 67 x 59 x 85 = 188210512710For length 7, max product is: Horizontal:[17, 10] - 99 x 69 x 82 x 67 x 59 x 85 x 74 = 13927577940540 ...etc
Project Euler #11 in Ruby and some Iterations
ruby;programming challenge