The guidelines for Project Euler say that

Each problem has been designed according to a “one-minute rule”, which means that although it may take several hours to design a successful algorithm with more difficult problems, an efficient implementation will allow a solution to be obtained on a modestly powered computer in less than one minute.

which makes me think that my solution to problem 15,

Starting in the top left corner of a 22 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 2020 grid?

which has been running since Wednesday, is less than optimal (makes me think I don’t quite deserve the 9% genius rating). I am sure it is going to find the right answer and I calculated that it would take three days. When it is done, I’ll try to get it down to a minute. That’s part of the fun.

Since #15 is taking such a long time, I skipped to some later problems. #18 was fun:

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

My recursive solution below the fold (no peeking until you have solved it).

``` input = "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" @triangle = input.split(/\n/).collect { |line| line.split(/\s/).collect{ |number| number.to_i } } def total row, col total = @triangle[row][col] return total if row == @triangle.size - 1 left = total(row+1, col) right = total(row+1, col+1) return total + right if right > left return total + left end before = Time.now puts total(0,0) after = Time.now puts "time=#{after - before}" ```