I did problems 11 and 15 this morning. I had been trying to think of a clever way to do problem 11, but could only come up with a brute force approach
require "matrix"
matrix_str = <<MATRIX_STR
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 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 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
MATRIX_STR
rows = matrix_str.split("\n")
rows = rows.inject([]) { |rows,row| rows << row.split.inject([]) { |num_row, str| num_row << str.to_i } }
matrix = Matrix.rows(rows)
def sum_up(matrix, x, y)
return 0 if y < 3
matrix[x,y] * matrix[x,y-1] * matrix[x,y-2] + matrix[x,y-3]
end
def sum_down(matrix, x, y)
col_size_adj_for_zero_start = matrix.column_size - 1
return 0 if y > (col_size_adj_for_zero_start - 3)
matrix[x,y] * matrix[x,y+1] * matrix[x,y+2] + matrix[x,y+3]
end
def sum_right(matrix, x, y)
row_size_adj_for_zero_start = matrix.row_size - 1
return 0 if x > (row_size_adj_for_zero_start - 3)
matrix[x,y] * matrix[x+1,y] * matrix[x+2,y] + matrix[x+3,y]
end
def sum_left(matrix, x, y)
return 0 if x < 3
matrix[x,y] * matrix[x-1,y] * matrix[x-2,y] + matrix[x-3,y]
end
def sum_diag_north_west(matrix, x, y)
return 0 if x < 3 || y < 3
matrix[x,y] * matrix[x-1,y-1] * matrix[x-2, y-2] * matrix[x-3, y-3]
end
def sum_diag_north_east(matrix, x, y)
col_size_adj_for_zero_start = matrix.column_size - 1
return 0 if x < 3 || y > (col_size_adj_for_zero_start - 3)
matrix[x,y] * matrix[x-1,y+1] * matrix[x-2, y+2] * matrix[x-3, y+3]
end
def sum_diag_south_east(matrix, x, y)
col_size_adj_for_zero_start = matrix.column_size - 1
row_size_adj_for_zero_start = matrix.row_size - 1
return 0 if x > (row_size_adj_for_zero_start - 3) || y > (col_size_adj_for_zero_start - 3)
matrix[x,y] * matrix[x+1,y+1] * matrix[x+2, y+2] * matrix[x+3, y+3]
end
def sum_diag_south_west(matrix, x, y)
row_size_adj_for_zero_start = matrix.row_size - 1
return 0 if x > (row_size_adj_for_zero_start - 3) || y < 3
matrix[x,y] * matrix[x+1,y-1] * matrix[x+2, y-2] * matrix[x+3, y-3]
end
max = 0
1.upto(matrix.row_size-1) do |x|
1.upto(matrix.column_size-1) do |y|
temp = sum_up(matrix, x, y)
max = temp if (max < temp)
temp = sum_down(matrix, x, y)
max = temp if (max < temp)
temp = sum_left(matrix, x, y)
max = temp if (max < temp)
temp = sum_right(matrix, x, y)
max = temp if (max < temp)
temp = sum_diag_north_west(matrix, x, y)
max = temp if (max < temp)
temp = sum_diag_north_east(matrix, x, y)
max = temp if (max < temp)
temp = sum_diag_south_east(matrix, x, y)
max = temp if (max < temp)
temp = sum_diag_south_west(matrix, x, y)
max = temp if (max < temp)
end
end
puts "max=#{max}"
Now don't get me wrong, this is terrible Ruby code, but I had been more focussed on trying to work out a good mathematical way and ended up with this. I'll probably rewrite it to make better use of Ruby over the weekend, also maybe port it to scala?
Problem 15 was far more fruitful on the math approach. I'm glad that when I first saw this problem my initial reaction wasn't to think algorithmically, and rather mathematically. Aaron and I had talked about that recently, and he was right that as a programmer one tends to think of solutions to these problems in code. So this was somewhat of a litmus test to see if my maths and physics instructors brainwashing was still in there somewhere.
The problem is as follows:
Starting in the top left corner of a 2
2 grid, there are 6 routes (without backtracking) to the bottom right corner. How many routes are there through a 2020 grid?My gut reaction was that this could be solved using permutations. I'll be honest, it did take me getting a pencil and paper plus a few references to my discrete maths book; but, happily I sorted out that the problem was just a case of permutations of indistinguishable objects.
My thinking was as follows, if I use the top-left corner as the origin and have the x-axis extend to the right and the y-axis extend down,since there is no backtracking, the path will consist of 10 moves in the positive x direction and 10 moves in the positive y direction. I tested my idea with the 2x2 grid, which gives the following six paths xxyy, xyxy, xyyx, yyxx, yxyx, yxxy. This essentially is just taking xxyy and seeing how many permutations there are of it, scaled up of course we're just looking for permutations of 20 x's and 20 y's. The simple formula for this is
n! / (n1! n2!)
where n is the total number of objects, n1 is the number of objects type 1 (x) and n2 is the number of objects of type 2 (y).
