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Suppose one has a link diagram of the unknot, and applies random Reidemeister moves until the unknot is reached. Surely it requires an exponential number of moves, exponential in, say, the crossing number of the original diagram? The 2001 Hass-Lagarias paper, "The number of Reidemeister moves needed for unknotting," established an exponential upper bound on the number of moves needed, but I am not finding a result on the expected number of random moves needed. I would like affirmation that not only is it hard, but one would not easily stumble into a solution, because then it would not truly be hard! (This in the spirit of Gower's much more substantive MO question, "Are there any very hard unknots?")

A reference would be appreciated! Thanks!

Edit: Apologies for the flawed question (thanks to Ryan Budney for clarifying it). I had in mind the expected number of random moves to reach the unknot from a random (in some sense!) diagram of the unknot.

Answered. The question has been answered in the comments by Theo Johnson-Freyd and Ori Gurel-Gurevich: the expected number of moves is $\infty$! As Ori put it,

for any starting diagram of the unknot, there is a positive probability of never unknotting it.

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I'm a little confused as to why you say "surely" as it's does take an exponential number of moves in general. Moreover, I would guess that if you only make random moves, it's highly unlikely you will ever find a route to the unknot -- it seems more likely that you would be endlessly lost in diagram space. –  Ryan Budney Oct 9 '11 at 1:12
    
@Ryan: Perhaps the cases that require an exponential number of moves are rare? You say, "in general." That is what I would like quantified. And, Yes, it is highly unlikely to wander into the unknot. But has this been proved? Perhaps it follows from Hass-Lagarias in a way I am not seeing... –  Joseph O'Rourke Oct 9 '11 at 1:22
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Here's one reason to think that the answer to your question is $\infty$. Let's work with knots with more than, say, two crossings (as two-crossing knots are not hard to untie). Then there are plenty of strands that with some isotoping can be made parallel. Ok, so among other moves you have access to are the Reidemester-II moves, one direction of which increases the crossing number. If you are not careful, it is easy to use these to include into your diagram space a lattice of large dimension. And you know that it's easy to get lost in them: a drunk bird never arrives home. –  Theo Johnson-Freyd Oct 9 '11 at 1:33
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@Joseph: I think I don't entirely understand your question. There are concrete knot diagrams for which you have to make a large number of Reidemeister moves to simplify them to standard unknot diagrams. So perhaps I've misplaced the quantifier when you use the word "expected" -- are you choosing your knot randomly, as well as choosing the Reidemeister moves randomly? –  Ryan Budney Oct 9 '11 at 2:04
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Theo is correct that the diagram space contains lattices of any dimension. Therefore, for any starting diagram of the unknot, there is a positive probability of never unknotting it. (BTW, you can also do it with type I moves). I would just like to add that even if the diagram space was recurrent, the expected number of moves to unknotting would still be infinite, as is the case for simple random walk on any infinite graph. –  Ori Gurel-Gurevich Oct 9 '11 at 16:42
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This question has been fully answered (the expected number of moves is $\infty$), as detailed in an addendum to the question. I place this community-wiki "answer" here so I can accept it (It let me!) and so (hopefully!) prevent the MO software-bot from re-asking the question.

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I found an answer of sorts in the paper, "Mean unknotting times of random knots and embeddings," by Yao-ban Chan, Aleksander L Owczarek, Andrew Rechnitzer, and Gordon Slade (Journal of Statistical Mechanics: Theory and Experiment, Volume 2007, May 2007.) Here is the beginning of their Abstract:

We study mean unknotting times of knots and knot embeddings by crossing reversals, in a problem motivated by DNA entanglement. Using self-avoiding polygons (SAPs) and self-avoiding polygon trails (SAPTs) we prove that the mean unknotting time grows exponentially in the length of the SAPT and at least exponentially with the length of the SAP.

Their SAPs are on a 3D lattice; their SAPTs are on a 2D lattice; see below. Interesting that they did not establish an upper bound for SAPs.
SAPT

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This paper deals with another question. The "unknotting" they refer to is achieved by reversing crossings, not by Reidemeister's moves. In particular, the diagram is never changed, so the space is finite size. –  Ori Gurel-Gurevich Oct 10 '11 at 1:29
    
Thanks, Ori, this clears up my confusion! –  Joseph O'Rourke Oct 10 '11 at 11:49
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