Timeline for Singly lossy integers: Concatenated sequences missing a single element
Current License: CC BY-SA 3.0
12 events
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| Jun 17, 2020 at 9:04 | history | edited | CommunityBot |
Commonmark migration
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| Feb 22, 2016 at 19:36 | comment | added | Michael Klein |
@mbomb007 Quick calculation shows: if a(n) is the nth term in the series, calculating the n+1th term should take less than O(a(n)^4) time, because allUpToN takes up to O(n^4) time (n for i, for j, for k, and for filter/concatInt). I'm not sure how fast the series grows so I don't have a more explicit bound at the moment.
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| Feb 22, 2016 at 17:40 | comment | added | mbomb007 | What's the big oh of your program? | |
| Feb 21, 2016 at 4:57 | comment | added | Michael Klein | @mbomb007 Yeah, it took over an hour to get to the 14th entry on my laptop (compiled) | |
| Feb 21, 2016 at 4:53 | comment | added | mbomb007 | Wow, it doesn't get very far before timing out. :D | |
| Feb 20, 2016 at 10:10 | comment | added | Michael Klein |
@mbomb007 Best I've found so far is this, though it needs main=print$ that GHCi does not. GHC.io runs out of memory and TryHaskell.org's feature set is too limited.
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| Feb 19, 2016 at 21:55 | comment | added | mbomb007 | Can you link to an interpreter where I can run this? I pasted it into TryHaskell.org and it didn't work. | |
| Feb 18, 2016 at 23:46 | history | edited | Michael Klein | CC BY-SA 3.0 |
nimi's suggestion
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| Feb 18, 2016 at 23:15 | comment | added | Michael Klein |
@mbomb007 With Integer, it will continue until you run out of memory (or patience). It'll continue with Int, but will start giving wrong answers once it overflows (> 2^29-1).
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| Feb 18, 2016 at 23:13 | comment | added | mbomb007 |
Is this infinite, or does it take n?
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| Feb 18, 2016 at 23:04 | history | edited | Michael Klein | CC BY-SA 3.0 |
golfed more + explanation
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| Feb 18, 2016 at 22:56 | history | answered | Michael Klein | CC BY-SA 3.0 |