0x4E4F@infosec.pub to Programmer Humor@programming.devEnglish · 2 years agoNot my problem sortinfosec.pubimagemessage-square48linkfedilinkarrow-up1619arrow-down112
arrow-up1607arrow-down1imageNot my problem sortinfosec.pub0x4E4F@infosec.pub to Programmer Humor@programming.devEnglish · 2 years agomessage-square48linkfedilink
minus-squarexmunk@sh.itjust.workslinkfedilinkarrow-up80·2 years agoGuaranteed to sort the list in nearly instantaneous time and with absolutely no downsides that are capable of objecting.
minus-squarefrezik@midwest.sociallinkfedilinkarrow-up39·2 years agoYou still have to check that it’s sorted, which is O(n). We’ll also assume that destroying the universe takes constant time.
minus-squareBatmanAoD@programming.devlinkfedilinkarrow-up36·2 years agoIn the universe where the list is sorted, it doesn’t actually matter how long the destruction takes!
minus-squareBenjaben@lemmy.worldlinkfedilinkarrow-up8·2 years ago We’ll also assume that destroying the universe takes constant time. Well yeah just delete the pointer to it!
minus-squarevithigar@lemmy.calinkfedilinkarrow-up14·2 years agoExcept you missed a bug in the “check if it’s sorted” code and it ends up destroying every universe.
minus-squaredb2@lemmy.worldlinkfedilinkarrow-up5·2 years agoThere’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!
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.
deleted by creator