algorithm for project euler problem no 18
- by Valentino Ru
Problem number 18 from Project Euler's site is as follows:
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)
The formulation of this problems does not make clear if
the "Traversor" is greedy, meaning that he always choosed the child with be higher value
the maximum of every single walkthrough is asked
The NOTE says, that it is possible to solve this problem by trying every route. This means to me, that is is also possible without!
This leads to my actual question: Assumed that not the greedy one is the max, is there any algorithm that finds the max walkthrough value without trying every route and that doesn't act like the greedy algorithm?
I implemented an algorithm in Java, putting the values first in a node structure, then applying the greedy algorithm. The result, however, is cosidered as wrong by Project Euler.
sum = 0;
void findWay(Node node){
sum += node.value;
if(node.nodeLeft != null && node.nodeRight != null){
if(node.nodeLeft.value > node.nodeRight.value){
findWay(node.nodeLeft);
}else{
findWay(node.nodeRight);
}
}
}
