Not my problem sort0x4E4F@infosec.pub to Programmer Humor@programming.dev – 736 points – 1 weeks ago71Post a CommentPreviewYou are viewing a single commentView all commentsShow the parent commentGuaranteed to sort the list in nearly instantaneous time and with absolutely no downsides that are capable of objecting.You still have to check that it's sorted, which is O(n). We'll also assume that destroying the universe takes constant time.In the universe where the list is sorted, it doesn't actually matter how long the destruction takes!It actually takes a few trillion years but its fine because we just stop considering the "failed" universes because they will be gone soon™ anyway.Eh, trillion is a constantamortized O(0) We'll also assume that destroying the universe takes constant time. Well yeah just delete the pointer to it!universe.take()Except you missed a bug in the "check if it's sorted" code and it ends up destroying every universe.There's a bug in it now, that's why we're still here.1 more...
Guaranteed to sort the list in nearly instantaneous time and with absolutely no downsides that are capable of objecting.You still have to check that it's sorted, which is O(n). We'll also assume that destroying the universe takes constant time.In the universe where the list is sorted, it doesn't actually matter how long the destruction takes!It actually takes a few trillion years but its fine because we just stop considering the "failed" universes because they will be gone soon™ anyway.Eh, trillion is a constantamortized O(0) We'll also assume that destroying the universe takes constant time. Well yeah just delete the pointer to it!universe.take()Except you missed a bug in the "check if it's sorted" code and it ends up destroying every universe.There's a bug in it now, that's why we're still here.1 more...
You still have to check that it's sorted, which is O(n). We'll also assume that destroying the universe takes constant time.In the universe where the list is sorted, it doesn't actually matter how long the destruction takes!It actually takes a few trillion years but its fine because we just stop considering the "failed" universes because they will be gone soon™ anyway.Eh, trillion is a constantamortized O(0) We'll also assume that destroying the universe takes constant time. Well yeah just delete the pointer to it!universe.take()
In the universe where the list is sorted, it doesn't actually matter how long the destruction takes!It actually takes a few trillion years but its fine because we just stop considering the "failed" universes because they will be gone soon™ anyway.Eh, trillion is a constantamortized O(0)
It actually takes a few trillion years but its fine because we just stop considering the "failed" universes because they will be gone soon™ anyway.Eh, trillion is a constant
We'll also assume that destroying the universe takes constant time. Well yeah just delete the pointer to it!universe.take()
Except you missed a bug in the "check if it's sorted" code and it ends up destroying every universe.There's a bug in it now, that's why we're still here.
Guaranteed to sort the list in nearly instantaneous time and with absolutely no downsides that are capable of objecting.
You still have to check that it's sorted, which is O(n).
We'll also assume that destroying the universe takes constant time.
In the universe where the list is sorted, it doesn't actually matter how long the destruction takes!
It actually takes a few trillion years but its fine because we just stop considering the "failed" universes because they will be gone soon™ anyway.
Eh, trillion is a constant
amortized O(0)
Well yeah just delete the pointer to it!
universe.take()Except you missed a bug in the "check if it's sorted" code and it ends up destroying every universe.
There's a bug in it now, that's why we're still here.