6 ms·
Jane Street Puzzles
- kej 8y agoThese are pretty good puzzles, the full archive is at https://www.janestreet.com/puzzles/archive/ https://www.janestreet.com/puzzles/archive/
- jschulenklopper 8y agoInteresting. For a mainly numerical superset of similar puzzles, I guess that https://projecteuler.net/ https://projecteuler.net/ still is the best known collection.
- gre 8y agoIBM posts monthly puzzles as well. http://www.research.ibm.com/haifa/ponderthis/index.shtml http://www.research.ibm.com/haifa/ponderthis/index.shtml http://www.research.ibm.com/haifa/ponderthis/challenges/July2018.html http://www.research.ibm.com/haifa/ponderthis/challenges/July...
- 21 8y agoNope: https://www.janestreet.com/puzzles/turn-based-strategy-game/ https://www.janestreet.com/puzzles/turn-based-strategy-game/
- extra__tofu 8y agon=1, A=2: h b #tri #non-congruent tri 1 4 b-1 roundup(#tri/2) 2 2 b-1 roundup(#tri/2) . . . n=4, A=16: h b #tri #non-congruent tri 1 2A/h b-1 roundup(#tri/2) 2 2A/h b-1 roundup(#tri/2) 4 ... ... 8 ... 16 ... ... summing up #non-congruent tris for each n I see n tri 1 3 2 7 3 15 4 31 so I think that answer may be 2^(n+1) - 1. Could be wrong because I forgot what congruent really means.
- tylerhou 8y agoYou're undercounting some triangles and missing the acute condition. For n = 4, there is more than one triangle with base 4 and height 4, for example: [(0, 0), (4, 0), (1, 4)]; [(0, 0), (4, 0), (2, 4)]; (and [(0, 0), (4, 0), (3, 4)] but it's congruent to the first so we don't count it). Furthermore, a triangle with base 16 and height 1 might be non-acute.
- deleted 8y ago[deleted]
- dilatedmind 8y agofor the first 8 im getting 1,3,6,14,28,60,120,248
- yantrams 8y agoJust a hunch but I think you might be skipping isosceles triangles. And possibly adding right angled triangles because of floating-point arithmetic.
- dilatedmind 8y agohow about 3,6,14,28,60,120,248,496?
- yantrams 8y agoNope :( The sequence starts with 1 because there is just one triangle with an area of 2 that satisfies the conditions. I ended up with a sequence similar to your previous one when I forgot to take the absolute value of area -- x1(y2-y3) + x2(y3-y1) + x3*(y1-y2))/2 . That could be a possibility in case you were using the same formula. In your first result - 1, 3, 6, 14.... , first two are right, third over-counted and rest are undercounted.
- yantrams 8y agoListed the solutions for n = 3 and 4 here. Could help you with debugging- https://imgur.com/a/fHS4tNd https://imgur.com/a/fHS4tNd
- yantrams 8y agoClose enough but wrong :) I went the same route as you did and realized my mistake after a while. Base and altitude values can be fractional/irrational too. PS : I bruteforced my way to the answer.
- deleted 8y ago[deleted]
- knappa 8y agoI haven't thought it through yet, but Pick's theorem ( https://en.wikipedia.org/wiki/Pick%27s_theorem https://en.wikipedia.org/wiki/Pick%27s_theorem ) might be useful for the current puzzle.
- deleted 8y ago[deleted]
- mikhailfranco 8y agoNot sure why, but that is unexpected. And the proof is so simple and elegant.