Project Euler Problem# 276 - Primitive Triangles
algorithm, c, math, performance
Solution
It's good that you're profiling before optimizing, but the fact that a lot of time in the `gcd` function doesn't necessarily mean you need to (or can) make it faster, rather that you're calling it too often. :-) Here's a hint — an algorithmic improvement that will improve the running time by orders of magnitude, instead of merely an improvement in the implementation.
You're currently counting just the primitive triangles individually. Instead, ask yourself this: can you efficiently count all triangles (not necessarily primitive) with perimeter a+b+c=n? (Running time O(n) will do – your current algorithm is Ω(n3).) Having done that, which triangles have you over-counted? For example, how many triangles are there with p dividing the sides? (Hint: a+b+c=n <=> (a/p) + (b/p) + (c/p) = n/p.) And so on.
Edit: After solving the problem and checking the thread on Project Euler, I found there are some other nice approaches to this problem, but the above approach is most common, and it works. For the first part, you can count it directly (some people have done it; it's certainly doable), or you may find this additional hint/trick helpful:
- Suppose 1 ≤ a ≤ b ≤ c, and a+b+c = n. Let p = b+c-a = n-2a, let q = c+a-b = n-2b, and let r = a+b-c = n-2c. Then 1 ≤ r ≤ q ≤ p, and p+q+r = a+b+c = n. Also, p,q,r all have the same parity as n.
- Conversely, for any 1 ≤ r ≤ q ≤ p with p+q+r = n with all three having the same parity (as n), let a = (q+r)/2, let b = (r+p)/2, c = (p+q)/2. Then 1 ≤ a ≤ b ≤ c and a+b+c=n. Further, these are integers and form a triangle (check).
- So the number of integer triangles (a,b,c) with perimeter n is exactly the number of partitions of n into parts (p,q,r) of same parity. You may find this easier to count.
Additionally/alternatively, try directly relating this number T(n) to smaller numbers T(n-k), to get a recurrence relation.
(Of course, there are also some very simple formulae you can find if you Google, but what's the fun in that?)
Problem
I tried to solve problem# 276 from Project Euler, but without success so far. Consider the triangles with integer sides a, b and c with a ≤ b ≤ c. An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1. How many primitive integer sided triangles exist with a perimeter not exceeding 10 000 000? The bottleneck in my code is the `GCD` function. It almost takes 90% of execution time for my sample input. I would love to hear hints, comments on how to improve the solution... My solution is written in C, but it is quit simple: ``` #include <stdio.h> #include <math.h> #define sides 3 // This is where my program spends most of its time size_t bi_gcd (size_t a, size_t b); size_t tri_gcd (size_t a, size_t b, size_t c); size_t count_primitive_triangles (size_t maximum_perimeter); int main() { printf( "primitive_triangles = %lu \n", count_primitive_triangles(1000) ); } size_t bi_gcd (size_t a, size_t b) { size_t t; while(b != 0) { t = b; b = a % b; a = t; } return a; } size_t tri_gcd (size_t a, size_t b, size_t c) { return bi_gcd(a, bi_gcd(b, c)); } size_t count_primitive_triangles (size_t max_perimeter) { size_t count = 0; // number of primitive triangles size_t a, b, c; // sides of the triangle // the following are the bounds of each side size_t // because b >= a && c >= b ==> max of a // is when a+b+c > 10,000,000 // b == a (at least) // c == b (at least) // ==> a+b+c >= 10,000,000 // ==> 3*a >= 10,000,000 // ==> a= 10,000,000/3 a_limit = max_perimeter/sides, b_limit, c_limit; for (a = 1; a <= a_limit; ++a) { // because c >= b && a+b+c <= 10,000,000 // ==> 2*b + a = 10,000,000 // ==> 2*b = 10,000,000 - a // ==> b = (10,000,000 - a)/2 for (b = a, b_limit = (max_perimeter-a)/2; b <= b_limit; ++b) { // The triangle inequality: // a+b > c (for a triangle to be valid!) // ==> c < a+b for (c = b, c_limit = a+b; c < c_limit; ++c) { if (tri_gcd(a, b, c) == 1) ++count; } } } return count; } ```