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141 points baruchel | 5 comments | | HN request time: 0.74s | source
1. viraptor ◴[] No.45124051[source]
Is this something people have been actively trying to disprove? The example provided seems to not be hard to bruteforce - given it's only 5 moves. Does anyone know why there's no older counter example? (Or am I totally underestimating how the number of options explodes in 5 moves?)
replies(3): >>45124262 #>>45124346 #>>45129087 #
2. adgjlsfhk1 ◴[] No.45124262[source]
I think this is a combination of things.

1: knot theory is somewhat obscure. it generally only comes up in undergrad in a topology class for a week or two so there aren't a ton of people interested

2. It's 5 cuts on a joining of 2 knots with 6 crossings. it's brute forcable, but not trivially (i.e. you have to code it up and possibly wait a while)

3. for conjectures that feel intuitively true more effort goes into finding the proof than looking for a counterexample that feels unlikely to exist.

3. lilyball ◴[] No.45124346[source]
It's not just 5 moves. It's 5 crossing changes (which don't change the number of crossings, they just change the order of the strings in a crossing). Unknotting also involves moving the strings around to add or remove crossings, without performing crossing changes (if you take a loop and twist it into a figure eight, you've moved the strings and created a crossing but you haven't cut the strings and performed a crossing change).

If you look at the preprint paper, the knot it starts with has 14 crossings, but they actually move the strings around to end up with 20 crossings prior to performing the first 2 crossing changes in the unknotting sequence. So the potential space for moves here is actually rather large.

replies(1): >>45124425 #
4. viraptor ◴[] No.45124425[source]
> crossing changes (which don't change the number of crossings

Ok, that explains the search space explosion. Thanks for explaining!

5. cottonseed ◴[] No.45129087[source]
You cannot bruteforce this. Exhibiting a unknotting of K with n moves only gives you an upper bound u(K) <= n. Proving u(K) = n is an entirely different matter.