Timeline for Counting maximal domino placements
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2021 at 19:00 | history | edited | fireflame241 | CC BY-SA 4.0 |
added 47 characters in body
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| Aug 10, 2021 at 10:51 | comment | added | Jakque |
219 bytes by replacing the for loop by a while loop
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| Aug 10, 2021 at 10:17 | comment | added | Jakque |
224 bytes by using 1-(2in u) instead of 2not in u
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| Aug 6, 2021 at 20:03 | history | edited | fireflame241 | CC BY-SA 4.0 |
deleted 188 characters in body
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| Aug 6, 2021 at 11:05 | comment | added | ovs |
And 225 bytes by replacing the any with string-based checks
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| Aug 6, 2021 at 11:00 | comment | added | ovs |
And p=[[0]*w] works just fine, combined: 231 bytes
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| Aug 6, 2021 at 10:40 | comment | added | ovs |
Another idea might be to replace the expression in the any with 2>i<len([*j])%[len([*j])+1,2][i], where len([*j])+1 can be replaced by an upper bound for the length of j.
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| Aug 6, 2021 at 10:33 | comment | added | ovs |
[k%3//2*3,1,2][::k%3+1] saves a byte and I think i<len([*j])%2 works instead of i==0<len([*j])%2?
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| Aug 6, 2021 at 9:18 | history | answered | fireflame241 | CC BY-SA 4.0 |