Timeline for Calculating Collatz Cousins
Current License: CC BY-SA 3.0
13 events
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| May 29, 2015 at 6:54 | history | edited | Sp3000 | CC BY-SA 3.0 |
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| May 29, 2015 at 6:48 | comment | added | Sp3000 | @isaacg Oh wow, I'm slightly ashamed I didn't notice that :P | |
| May 29, 2015 at 6:42 | comment | added | isaacg |
I think the ~- is unnecessary because you're using integer division.
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| May 29, 2015 at 4:47 | comment | added | Sp3000 | @DLosc Ahaha thanks. I still swear there's got to be a better short-circuiting way though... | |
| May 29, 2015 at 4:47 | history | edited | Sp3000 | CC BY-SA 3.0 |
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| May 29, 2015 at 4:35 | comment | added | DLosc |
Nice! Here's 2 more bytes off: f=lambda n,k=1:sorted([k][n:]or(k>4==k%6and f(n-1,~-k/3)or[])+f(n-1,k*2))
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| May 29, 2015 at 3:06 | history | edited | Sp3000 | CC BY-SA 3.0 |
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| May 29, 2015 at 2:55 | history | edited | Sp3000 | CC BY-SA 3.0 |
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| May 29, 2015 at 2:38 | history | edited | Sp3000 | CC BY-SA 3.0 |
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| May 29, 2015 at 2:30 | comment | added | Sp3000 | @DLosc I liked your idea and made it one better :) | |
| May 29, 2015 at 2:29 | history | edited | Sp3000 | CC BY-SA 3.0 |
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| May 29, 2015 at 2:00 | comment | added | DLosc |
Interesting--my original solution is very similar, but (taking a couple optimizations from yours) comes out 2 bytes shorter: f=lambda d,n=1:d and sorted(sum((c(d-1,x)for x in[n*2]+[~-n/3][:n>4==n%6]),[]))or[n]. It's less efficient with the function calls but still does n = 30 in under a second.
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| May 29, 2015 at 1:44 | history | answered | Sp3000 | CC BY-SA 3.0 |