On 22 Jun 2001, John Steinberger wrote:

> Hi all,

>

> I have recently stumbled into an unknotting algorithm of a very simple

> diagrammatic kind, but I have been unable to prove it will always

> unknot a projection of the unknot. It would be really exciting though

> if it did because the algorithm runs in O(n^3) (n = the number of

> crossings in the projection) whereas the algorithms developped so far

> for unknotting run in exponential times that are not real-life

> implementable. Also I should mention that my algorithm has happily

> weathered the supposedly nasty unknot projections that discouraged

> people in the past from using diagrammatic approaches (some of these

> can be seen at http://f2.org/maths/kt/unknoteq.html).

>

> The algorithm runs like this: define a (p,q) move on a knot projection

> to be a move taking a p-crossing over(or under-)strand to a position

> where it has q crossings (still all over or under, accordingly). For

> instance a -I Reidemeister move (which removes a single loop) is a

> (1,0) move, acting on the overstrand of a single crossing. My

> conjecture is that a non-trivial unknot projection always contains a

> (p,q) move with p>q (which we denote by a '-(p,q)' move). And my

> discovery is that when present in a diagram, these moves can be

> located very quickly--in O(n^2) or less.

>

> I've been thinking about it for a while but I cannot either prove or

> disprove the algorithm--even in much simpler cases (such as requiring

> the unknot's Seifert surface be a flattish, untwisted pancake) I have

> not been able to prove a -(p,q) move is always present. Which is

> either discouraging for the conjecture or just means that I have

> watched one too many Pepsi commercials.

I made this conjecture a million years ago, along with what I

regarded as the obvious generalization, to an algorithm for finding

all the projections of an arbitrary knot that have minimal crossing

number. Let me describe the latter. My terms are (p,q)-PASS for

this move:

--|----##### -------*****

--|----#####------- to -------*****---|---

--|----***** -------*****

where there are p horizontal strings on the left, and q on the right,

(this one is a (3,1)-OVERPASS - there's also a (p,q)-UNDERPASS).

There's also a (p,q)-TWIST, which takes you from

__ __ ______ ______ __ __

\ / ##### ##### \ /

\ ##### to ##### \

__/ \__#####______ ______#####__/ \__

(this one's a (2,2)-TWIST - in general there'd be p strings on the

left, with a single twist before the move and no twist after, and q on

the right, with a single twist after and none before).

The conjecture was originally that if you start from any projection

of the knot (in which term I include links) and apply all possible

(p,q)-passes and (p,q)-twists for which p >= q, then you'll get all

minimal projections.

My evidence for this conjecture was pretty overwhelming, since I

used them to prepare my knot tables. However, the above conjecture is

false. I disproved it first by finding a new move - the WRISTWATCH

move, which was needed to related two diagrams for a certain 10-crossing

link, and which I then realised was even needed for unlinks! I wasn't

too worried by this, since it seemed that the conjecture could be cured

by adding these wristwatch moves.

Let me briefly describe them. Imagine an arm wearing a wristwatch:

/\ /\ /\

/ / /

#######

-------#######-------

-------#######-------

-------#######-------

#######

\ \ \

\/ \/ \/

whose armband consists of a number of strings going around the arm,

while there are also p strings going along the surface of the arm to

the left of the watch, and q similar strings to the right.

Now suppose one's wearing two such watches, one "outside" the

other, and initially to the left of it. Then the wristwatch move

slides this watch rightawrds along the arm, over the other one.

The modified conjecture stated that all minimal projections were

obtainable from an given projection by applying just those passes,

twists and wristwatch moves that never increased the crossing number.

Of course it was slight less credible than the previous one had been,

because it wasn't at all obvious that there wouldn't be more "new"

moves like the wristwatch one, so when I wrote my paper I refrained

from publicly making this conjecture. There's a sort of sotto voce

reference to it, though in a sentence when I say that a certain

algorithm had been applied to all the knots in the table, as evidence

that forms I said were distinct reall were.

However, that conjecture was wrong. There were two 10-crossing

knots in the table, which had been thought to be distinct ever since

Tait first enumerated knots in the 19th century, and which I listed as

distinct since the above algorithm didn't distinguish them, but which

were later found by Perko to be the same. They are related by yet

another "new" move studied by him, to which I gave the name

'Perkolation". I won't attempt to describe it here. A few other

moves were found later, that seemed to be independent of the passes,

twists, watch-moves and Perkolation, and I know longer believe that

there's any algorithm of this kind for finding all minimal

presentations - in other words, I think there will be an infinity

of new and essentially different such moves.

Strictly speaking, this says nothing about the unknot version of

the conjecture. However, my (very well-informed) guess is that will

be false too. I know some intriguing unlinks that disprove certain

natural generalizations, and my guess is that there will be some

unknots in the 20-30 crossing range that disprove this one. But

they're very hard to find!

John Conwa