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#1 |
Aug 2002
2×19×223 Posts |
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#2 |
Jun 2016
1910 Posts |
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To help people check their answers without spoiling the puzzle, these were the last two digits of each of my values for n up to 20.
00, 01, 03, 11, 35, 08, 80, 48, 14, 23, 34, 49, 46, 37, 33, 34, 76, 71, 38, 52 |
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#3 |
Jan 2017
2·73 Posts |
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Has anyone here tried to calculate values especially far? I've computed up to 23 (with a fast enough program that calculating the next few values would be realistic if I left it running for a while).
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#4 |
Oct 2017
2×32×7 Posts |
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#5 |
"Kebbaj Reda"
May 2018
Casablanca, Morocco
32×11 Posts |
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#6 |
Jan 2017
100100102 Posts |
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Those values match what I got. There are some things I'd say about the calculation, but they'd be kind of spoilery, so perhaps better left for private messages or after the month is over.
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#7 |
Jun 2016
19 Posts |
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My (fairly naive, single-threaded) code took about two and a half hours to do n=20, and it goes up by a factor of about 4 for each increase in n. By multithreading it I think I could do n=23 or even n=25 in a not totally absurd time, but wondering if other people have faster approaches?
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#8 |
Jan 2017
2×73 Posts |
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Well my code is faster, but not a totally different approach - still proportional to the overall number of polyominoes. I'll send details in PM.
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#9 |
Jul 2015
11 Posts |
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I am seriously struggling with this problem.
![]() If there is an algorithm that efficiently creates polyomino without duplicates, what would it be? The method of creating and saving all polyominos to check for duplicates is easily limited around N=15. Is it correct to save all polyominos of size N-1 to create a new polyomino of size N? I tried to classify polyominos by width size, but the method of saving all polyominos to check for duplicates also runs into a limit as N increases. ![]() |
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