Jane Street Puzzles

https://www.janestreet.com/puzzles/current-puzzle/

by dmmalam • 8 years ago

117 14 8 years ago

kej• 8 years ago

These are pretty good puzzles, the full archive is at https://www.janestreet.com/puzzles/archive/

Interesting. For a mainly numerical superset of similar puzzles, I guess that https://projecteuler.net/ still is the best known collection.

gre• 8 years ago

IBM posts monthly puzzles as well.

http://www.research.ibm.com/haifa/ponderthis/index.shtml

http://www.research.ibm.com/haifa/ponderthis/challenges/July...

21• 8 years ago

Nope: https://www.janestreet.com/puzzles/turn-based-strategy-game/

extra__tofu• 8 years ago

   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   ...          ...

summing up #non-congruent tris for each n I see

so I think that answer may be 2^(n+1) - 1.

Could be wrong because I forgot what congruent really means.

tylerhou• 8 years ago

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.

dilatedmind2• 8 years ago

for the first 8 im getting 1,3,6,14,28,60,120,248

yantrams4• 8 years ago

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.

knappa• 8 years ago

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.

mikhailfranco• 8 years ago

Not sure why, but that is unexpected.

And the proof is so simple and elegant.

Jane Street Puzzles

Related Stories