Not my problem sort

0x4E4F@infosec.pub · 2 年前

xmunk@sh.itjust.works · 2 年前

Guaranteed to sort the list in nearly instantaneous time and with absolutely no downsides that are capable of objecting.

frezik@midwest.social · 2 年前

You still have to check that it’s sorted, which is O(n).

We’ll also assume that destroying the universe takes constant time.

BatmanAoD@programming.dev · 2 年前

In the universe where the list is sorted, it doesn’t actually matter how long the destruction takes!

groet@feddit.org · 2 年前

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.

MBM@lemmings.world · 2 年前

Eh, trillion is a constant

FiskFisk33@startrek.website · 2 年前

amortized O(0)

YonderEpochs@lemmy.world · 2 年前

We’ll also assume that destroying the universe takes constant time.

Well yeah just delete the pointer to it!

PoolloverNathan@programming.dev · 2 年前

universe.take()

vithigar@lemmy.ca · 2 年前

Except you missed a bug in the “check if it’s sorted” code and it ends up destroying every universe.

db2@lemmy.world · 2 年前

There’s a bug in it now, that’s why we’re still here.

frezik@midwest.social · 2 年前

deleted by creator