Jane Street Puzzles(janestreet.com)
janestreet.com
Jane Street Puzzles
https://www.janestreet.com/puzzles/current-puzzle/
17 comments
These are pretty good puzzles, the full archive is at https://www.janestreet.com/puzzles/archive/
IBM posts monthly puzzles as well.
http://www.research.ibm.com/haifa/ponderthis/index.shtml
http://www.research.ibm.com/haifa/ponderthis/challenges/July...
http://www.research.ibm.com/haifa/ponderthis/index.shtml
http://www.research.ibm.com/haifa/ponderthis/challenges/July...
I haven't thought it through yet, but Pick's theorem ( https://en.wikipedia.org/wiki/Pick%27s_theorem ) might be useful for the current puzzle.
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Not sure why, but that is unexpected.
And the proof is so simple and elegant.
And the proof is so simple and elegant.
Interesting. For a mainly numerical superset of similar puzzles, I guess that https://projecteuler.net/ still is the best known collection.
n=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 ... ...
n tri
1 3
2 7
3 15
4 31
Could be wrong because I forgot what congruent really means.
Close 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.
PS : I bruteforced my way to the answer.
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You'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.
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for the first 8 im getting 1,3,6,14,28,60,120,248
Just a hunch but I think you might be skipping isosceles triangles. And possibly adding right angled triangles because of floating-point arithmetic.
how about 3,6,14,28,60,120,248,496?
Nope :( 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.
Listed the solutions for n = 3 and 4 here. Could help you with debugging- https://imgur.com/a/fHS4tNd