FilterHN

I think the summary at the beginning of your first video is misleading; it's not a way to "trade space for time", at least not in an arbitrary program. The real statement is a bit odder to wrap one's head around -- "every problem solvable in t time on a multitape Turing machine is also solvable in close to √t space".

For a Turing machine that already solves a problem in n time and √n space (in other words, a lot of them!), it doesn't say anything.

▲

LegionMammal978

2 months ago

[-]

When you convert a generic Turing machine into a Tree Evaluation instance, you end up with square-root space with respect to the original runtime t, but the new runtime will be far, far slower. IME, with these types of circuit reductions, the runtime typically becomes exponential in the space required, which is just about 'as long as possible'.

If we're being pedantic, it's trading time for the space guarantee.

▲

dang

2 months ago

[-]

Related. Others?

For algorithms, a little memory outweighs a lot of time - https://news.ycombinator.com/item?id=44055347 - May 2025 (139 comments)

▲

krackers

2 months ago

[-]

Dupe of today's https://news.ycombinator.com/item?id=44212861 ?

▲

dang

2 months ago

[-]

It looks like that posted just a little later (by the same submitter) so I've put a copy of that link at the top.

▲

wewewedxfgdf

2 months ago

[-]

What are the practical implications and use cases of this?

Is it something like some sort of reverse compiler which creates super efficient code by analyzing the inverse flow of code or something?

▲

throwaway81523

2 months ago

[-]

From Februray 2025 fwiw. Same result there have been multiple articles here about. I wonder how it would work for Haskell programs (no mutable memory).

▲

HappMacDonald

2 months ago

[-]

I'd view "no mutable memory" as misleading, because immutable languages can still create a new variable and forget an old one which has the same memory footprint as mutating one variable.

Obvious example: the flickering stack frame of tail call elimination.

▲

curtisf

2 months ago

[-]

Haskell has genuine mutable memory, through State and IO.

But even without it, you can emulate mutation in a pure language by threading a "heap" parameter through everything.

There's only at most a log factor of extra space and time required in most computing models to "update" a persistent map (though I'm not sure the best way to encode persistent maps directly in Turing machine tapes, which is the model this result is specifically about)

▲

slicktux

2 months ago

[-]

It’s nice to see what a little bitwise manipulation can do(XOR)! Low level programming is always fun!