0x4E4F@infosec.pub to Programmer Humor@programming.devEnglish · 17 days agoNot my problem sortinfosec.pubimagemessage-square67fedilinkarrow-up1732arrow-down114
arrow-up1718arrow-down1imageNot my problem sortinfosec.pub0x4E4F@infosec.pub to Programmer Humor@programming.devEnglish · 17 days agomessage-square67fedilink
minus-squarexmunk@sh.itjust.workslinkfedilinkarrow-up92·16 days agoGuaranteed to sort the list in nearly instantaneous time and with absolutely no downsides that are capable of objecting.
minus-squarefrezik@midwest.sociallinkfedilinkarrow-up46·16 days 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-up40·16 days agoIn the universe where the list is sorted, it doesn’t actually matter how long the destruction takes!
minus-squareBenjaben@lemmy.worldlinkfedilinkarrow-up8·16 days ago We’ll also assume that destroying the universe takes constant time. Well yeah just delete the pointer to it!
minus-squarevithigar@lemmy.calinkfedilinkarrow-up16·16 days agoExcept you missed a bug in the “check if it’s sorted” code and it ends up destroying every universe.
minus-squaredb2@lemmy.worldlinkfedilinkarrow-up6·16 days 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.
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