Crazy math problems.

Looks like Java... I hate Java.

 

its actually a mutated C that runs in a java based compiler

I brute forced #9 in 1.1865 seconds

Code:

A Pythagorean triplet is a set of three natural numbers, a  b  c, for which,

a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Code:

function [ansA,ansB,ansC] = PythagoreanTriplet()
for c = 1000:-1:1
    for b = c:-1:1
        for a = b:-1:1
            if a + b + c == 1000
                if a^2 + b^2 == c^2
                    ansA = a;
                    ansB = b;
                    ansC = c;
                    return;
                end
            end
        end
    end
end

ans =

Spoiler

31875000 (200*375*425)

 

I figured out #18 in 0.350815 seconds:

Code:

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.

[COLOR="Red"]3[/COLOR]
[COLOR="red"]7[/COLOR] 4
2 [COLOR="red"]4[/COLOR] 6
8 5 [COLOR="red"]9[/COLOR] 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

using the code:

Code:

[SPOILER]function pathSum = MaxTreePath(Tree, node, depth)%start with node = 1; depth = 1
if node > numel(Tree)
    pathSum = 0;
else
    pathSum = max(MaxTreePath(Tree,depth+node,depth+1),MaxTreePath(Tree,depth+1+node,depth+1)) + Tree(node);
end
end[/SPOILER]

and sending it an array of all of the elements in the triangle (read left right, top down)

This was a quick dynamic programming approach that I learned in my algorithm's class last quarter, its efficient, but it does check every path

Edit: i figured out a more efficient algorithm that runs in O(n) time and only calculates the length of the last row paths, ill code it up tomorrow and see if it works on #67


Edit #2:
did #18 in 0.001072 seconds and #67 in 0.002109 seconds using the following code:

Code:

[SPOILER]function pathSum = MaxTreePath(Tree)
dim = size(Tree,2);
for i = (dim-1):-1:1
    for j = 1:i
        Tree(i,j) = Tree(i,j) + max(Tree(i+1,j),Tree(i+1,j+1));
    end
end
pathSum = Tree(1);
end[/SPOILER]

where i had the tree stored in a matrix (which better fits the input given to me)