## Project Euler Problem #102

### Nov 20, 2009

Read the details of the problem here

**Summary**

For how many triangles in the text file does the interior contain the origin?

**Solution**

This question had me wondering for a while as to what the best approach to take might be. In the end I went for a simple approach that just relied on the fact that the interior angles from any point within a triangle to the vertexes would add up to 360° or 2 * Pi radians. To calculate the angles at the origin I relied on some things I’d learned when I used to write computer games as a hobby.

The dot product of two 2D vectors, U and V, is equivalent to the following two formulae:

U•V = U_{0}V_{0} + U_{1}V_{1}

U•V = |U| * |V| * cos θ

Where |U| is the magnitude of the vector U, i.e. sqrt( U_{0}² + U_{1}² )

So this gives: cos θ = ( U_{0}V_{0} + U_{1}V_{1} ) / ( |U| * |V| )

The coded solution is then quite trivial and just calculates the angle at the origin subtended by vectors out to the pairs of vertices for each line segment.

def answer = 0 new File("triangles.txt").eachLine { def (x0, y0, x1, y1, x2, y2) = it.split(",")*.toInteger() def ( a, b, c ) = [ [ x0, y0 ], [ x1, y1 ], [ x2, y2 ] ] def theta = 0.0 [ [ a, b ], [ a, c ], [ b, c ] ].each { def ( u, v ) = it def dot = u[0] * v[0] + u[1] * v[1] def ul = Math.sqrt(u[0] * u[0] + u[1] * u[1]) def vl = Math.sqrt(v[0] * v[0] + v[1] * v[1]) theta += Math.acos(dot / (ul * vl)) } if (Math.abs(theta - 2 * Math.PI) < 1E-7) answer++ }

It ran in 0.96 seconds and wasn’t difficult to write once I’d decided what approach to take! Note the importance of using an epsilon value (1E-7) when comparing floating-point numbers as you can’t rely on an exact equality.

There are probably quicker ways to do this (in fact, I know there are) but this approach was nice & straightforward to code off the top of my head and ran in an acceptable time.

**Conclusion**

Groovy makes reading a text file and parsing it into appropriate data structures quite simple. It didn’t really make any difference to the rest of the solution apart from perhaps the nice syntax on the <list of sides>.each{} loop as it was mostly just straightforward maths.