description=%{
    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

    NOTE: As there are only 16384 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)

}

reflexion=%{
    use bottom to top method

     z
    x y chose max(x,y) path for z, new triangle = z + max(x,y)
}

data_test=%{3
7 4
2 4 6
8 5 9 3}
data=%{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=data.split("\n").collect! { |line| line.split(" ").collect! {|n| n.to_i} }.reverse!

p triangle


def max(x,y)
    return x if x>y
    return y
end

nbline=0
triangle.each do |line|
    if nbline==0
        nbline +=1
        next
    end
    col=0
    line.each do |n|
        triangle[nbline][col] = 
            n + max( triangle[nbline-1][col], 
                     triangle[nbline-1][col+1] )
        col +=1
    end
    nbline+=1
    puts '==='
    p triangle
end
p triangle
puts triangle[-1][0]