ProjectEuler 411

[Panphobia]

As my exams get closer I am trying to keep away from project euler, but I cant resist trying the latest one, http://projecteuler.net/problem=411 , basically what the question is saying is if you are given n what is the maximum path from (0,0) to (n,n) when x,y cannot decrease, and (2^i mod n, 3^i mod n) are the x and y values 0 <= i <= 2n, so in their example they have 11 distinct points, when n is 22, how is this possible, is this because some of the x points are repeated throughout getting the x's and y's, if so is there a way to keep track of the values already calculated without looking through the whole list to check for duplicates, for efficiency, i am guessing it can be solved using a dfs, but that would take too long to run

[d310]

I actually got in the top 100 for this one!
A good place to start is looking at the maximum number of points; what is an acceptable O time for an algorithm given that many points?
If you want to keep track of all the unique points, you could use a TreeSet in Java.

[Panphobia]

I have the way to get distinct points, all I need is to get the maximum number of points, problem is finding an efficient algorithm that doesnt take days to run

[d310]

Try sorting your points, that's a good starting point.

[Panphobia]

Thats what I was thinking of but I dont know which value to sort them by, x or y?

[d310]

It doesn't really matter, as long as you sort by either the x or y coordinate, you will remove one dimension, and break down a two dimensional problem into a one dimensional problem (which is much easier to handle).

[d310]

For a path to be valid, the x and y coordinates have to be non-decreasing. If we sort by one coordinate we have one guaranteed coordinate that is non-decreasing, so all we have to do is...

[Panphobia]

The way I generate the distinct points, it takes forever, is there a much more efficient way than this?

code:

out:         for (int i = 0; i <= twon; ++i) {             long x = (long) Math.pow(2, i) % n;             long y = (long) Math.pow(3, i) % n;             for (int q = 0; q < node.size(); ++q) {                 if (node.get(q).x == x) {                     continue out;                 }             }             node.add(new Node(x, y));         }

[d310]

O(n^2) isn't going to cut it at 20 million points.
I already recommended a TreeSet which is O(n log n)

[Panphobia]

Ok I got the points generating quickly enough

code:

for (int i = 0; i <= twon; ++i) {             long x = (long) Math.pow(2, i) % n;             long y = (long) Math.pow(3, i) % n;             a=nodeX.size();             nodeX.add(x);             if(a!=nodeX.size())nodeY.add(y);         }

now on to the tricky part, finding an algorithm that actually runs at a decent speed