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Re: How many concave quadrilaterals are in a 3 by n array?
Posted:
Jan 26, 2014 9:38 PM


The formula works perfectly and it prevents me from bruteforcing: For odd n use P(n) = (11n^4  30n^3 + 16n^2 18n +21)/8. For even n use P(n) = (11n^4  30n^3 + 16n^2)/8.
But I cannot understand why. For example, I would use this approach with a n x m grid. Where are the mparts in that formula (should be constant 3)?
Are there any hints to the basics of the mentioned approach?



