burritos, frat boys, and puzzleuzzlement

[mr_bungee]

After solving puzzles for an entire afternoon, and spending the evening making fun of the ridiculous movie that is Tommyknockers, I decided to get some food at 1am. (mmmm...special chicken burrito and i quickly became good friends...) This is the same time that hundreds of frat boys were trying to live out their drunken fantasies and pick up random

Comments

[katiesaur]

2n-3???

[mr_bungee]

Ah, but this doesn't work for n=3, since you can just take the 2 corners to the other side.

[katiesaur]

haha.. oh well. i didn't try very hard so i don't feel too bad ;)

[thedan]

What kind of puzzles were you solving? Prepping for mh2k5? ;)

As far as the puzzle goes, after analyzing cases for small n, I'd be willing to be that the answer is:

k(k+1)+(1/2)(n-2k-1)(n-2k) where k is the closest integer to (n-1)/3

which comes from shifting triangles in the upper corners down an optimal number of rows to make the bottom corners of the new triangle, and then bringing up the bottom to form the top. But I haven't tried (and probably won't try) to prove it.

[mr_bungee]

You bet, and congrats on a very well written mini-hunt! :)

Your solution is correct, though it's a bit simpler to write it as:

(n-1)(n+1)/6 if n mod 3 = 1
n(n+1)/6 otherwise

Hope all is well in the world of dan!

[mr_bungee]

oops, that should be:
(n-1)(n+2)/6 if n mod 3 = 1
n(n+1)/6 otherwise