by Hew Wolff

An *orthogonal discrete knot* is a path through the three-dimensional lattice of
integer-valued points and orthogonal edges between them, which never
visits the same point twice and ends where it starts.
Two knots are *equivalent* if, when they are realized with pieces of stretchy string, we can move one around to look like the other.
(Also, we consider mirror images to be the same knot.)
Naturally we
are interested in knots which are not *trivial*, that is, equivalent
to a simple circle.

Here are some questions about making these knots as small as possible. There are links to the answers below, but often I don't know the answer. If you do, please contact me through my home page.

**What's the shortest knot?** Here we are measuring the
knot by
the length of its path. The picture above
shows a knot of length 64 which turns out to be equivalent
to the trefoil knot.
64 is certainly not the best we can do, though.

- Can you find a nontrivial knot of length 24? (See below.)
- Is there a shorter one? I'm pretty sure that the length must be at least 20 (see below).

**What are the minimal bounding boxes?** The knot above has a
bounding box of size 3 by 3 by 3, which I will write as (3, 3,
3). But again, we can do better than that.

- Can you find a version of the trefoil knot which fits in the box (1, 3, 3)? I believe that this bounding box is minimal, that is, reducing its size in any dimension will not allow any equivalent knot to fit. Can you show this? (See below.)
- Can you fit a trefoil knot into (1, 2,
*c*) for some*c*? (See below.) If so, what's the minimal*c*? What other minimal bounding boxes does the trefoil knot have? - What other knots can fit into (1, 2,
*c*) for some*c*? Just for fun, can you fit in equivalents of the cinquefoil knot, the figure-eight knot, and Buddhism's "glorious endless knot"? (See below.)

**What about links?** A link is just like a knot, but with more
than one component. For example, the Borromean
rings form a link with three components.

- Find the shortest nontrivial link with two components (see below).
- Fit an equivalent link into the box (1, 2, 3) (see below). What are all its minimal bounding boxes?
- Find a different link with two components that fits into (1, 3, 3) (see below).
- Do the Borromean rings fit into (1, 2,
*c*) for some*c*? (See below.)

**I claim** that any nontrivial knot *K* must have length at least 20.
I'll use some basic knot theory such as you can find in Rolfsen's *Knots and Links*.

First of all, I claim it's enough to show that the bounding box
*B* must be reasonably large. Suppose its size is (*a*,
*b*,
*c*), with *a* <= *b* <= *c*. The projection of
*K* onto the *x*-axis is a path ranging over a distance of
*a* and returning to its starting point, so that path must have
length at least 2*a*. If the path has length 2*a*, then it
has only one local maximum. But the bridge index, and therefore the
number of local maxima, of *K* must be at least 2 because *K* is
nontrivial. So the path has length at least 2*a* + 2. Applying the same argument to
the other two axes, the total length of *K* must be at least 2(*a*
+
*b* + *c*) + 6. So we only need to show that
*a* +
*b* + *c* is at least 7.

Take *L* to be intersection of *K* with the interior of
*B*. *L* is a union of disjoint open arcs. If all these arcs
are unknotted, then there must be at least two of them, again because
the bridge number of *K* is at least 2. So certainly a box of
the form (0, *b*, *c*) will not work, because its interior allows no
arcs at all. Similarly, (1, 1, *c*) will not work. (1, 2, 2) or
(2, 2, 2) allows only one unknotted arc, so that won't work either. So all we need to do is show that (1, 2, 3) won't work; then
we know that *a* +
*b* + *c* is at least 7, and we're done.

So suppose the bounding box is (1, 2, 3). From above, both of the
interior edges must be included in *K*: Look at the intersection
*J* of *K* with the near side of *B*. *J* is a
union of disjoint closed arcs. Now, we can assume that *K* has
minimal length among equivalent knots. Then an arc in *J* cannot
have length 1, since that would make a U-turn path in *K*:
There can't be any of these U-turns, since we can replace one
with a single edge to shorten *K*. There are similar problems if an arc
in *J* has length 2:

So all arcs have length at least 3, which means *K* must include arcs that look like one of the following (up to symmetry):

It's easy to see that in each of these cases, avoiding U-turns and
dead ends brings us to a link (in fact, the simplest two-component
link, 2^{2}_{1}
in this table). For example: So there is no nontrivial knot in
(1, 2, 3), and we are done. **Q.E.D.**

In fact the trefoil fits into the box (1, 3, 3), so *a* +
*b* + *c* = 7 is as small as we can get. Here's a
symmetrical trefoil of length 32 and also one of length 26:

It's clear from the proof above that this is a minimal
bounding box for the trefoil.

What are the shortest equivalents of other simple knots, like the figure-eight knot?

It's surprising how many knots fit into (1, 2, *c*) for some *c*. For example, here are the trefoil and the figure-eight knot:

Call a link *narrow*
if it fits in (1, 2, *c*).
Then **I claim** that all knots of bridge index 2 are narrow.
Here's a sketch of a proof. Given a braid on 4 strands, we can make a link by attaching adjacent pairs of strands at each end. Any knot with bridge index 2 can be obtained in this way, as follows. Take a drawing of the knot with 2 local maxima, pull each local maximum to the top and cut there, and do the same with the local minima at the bottom, forming a braid.

So all we need to do is show that all braids on 4 strands are narrow. Here are two examples of braids:
The first one exchanges a pair of adjacent strands, and the second one cycles all the strands. By composing these, we can also construct a braid which exchanges any pair of adjacent strands. This means we can generate the entire braid group, so we're done. **Q.E.D.**

In particular, since the knots listed in the question above are all 2-bridge knots (see Bar-Natan's table), they are all narrow.

Is the obvious converse also true? That is, does any narrow knot have bridge index at most 2? (Note that only the unknot has bridge index less than 2, so 2-bridge knots are really the only interesting case.) Not quite: I believe, for example, that the connected sum of two trefoil knots is narrow but has bridge number 3.

In fact, **I claim** that a link is narrow if and only if it is a connected
sum of 2-bridge links. Here's another sketchy proof.

To clarify a few things: I'm not sure whether bridge index is normally used for proper links (links with more than one component), but here I simply mean the smallest number of local maxima. A 2-bridge proper link must have only two components, each with bridge index 1 and therefore unknotted. The connected sum of two links is defined up to equivalence by the components you choose to connect. Finally, in order to cover the trivial case of the unknot, I'll say that it's the connected sum of no links.

So suppose *L* is a narrow link.
Look at the intersections of *L* with the planes *c* =
*t*, where *t* ranges from minus infinity to infinity. We
may perturb *L* slightly to put it in general position, so that
the intersection *I(t)* is finite for each *t*. The size
|*I(t)*| changes by +2 at each local minimum and -2 at each local
maximum, so it is even except at a finite number of points. If |*I(t)*| ever
reaches 6, then part of *L* must look like , and it's easy to see that there
must be a U-turn eventually. We may assume that *L* was
initially a shortest equivalent, so it has no U-turns.

This means that
the sequence of values of |*I(t)*|, ignoring the local extrema,
looks like this: 0, 2, then the sequence (4, 2) occurring zero or more
times, then finally 0 again.
If we snip the sequence whenever a 2 occurs in the middle of it, we
see that *L* is the connected sum of a series of links, each with
the sequence (0, 2, 4, 2, 0). Each of these links has 2 local maxima,
and therefore has bridge index 2 (as claimed) or is the unknot.

For the converse, suppose *L* is a connected sum
*L*_{0} # *L*_{1} # ... of 2-bridge links.
From the proof above, we can realize each *L*_{i}
as a discrete braid in
(1, 2, *c*),
with the strands joined together in two pairs at
each end of the braid.
Also, note that in the case of a 2-component link, each end of the
braid has one pair of strands from each component. This means that if
we put the *L*_{0} braid on top of the
*L*_{1} braid, we can perform the connected sum by
connecting a pair of strands from the bottom of *L*_{0}
with a pair of strands at the top of *L*_{1}. Repeating
this for each *L*_{i}, we have constructed a
narrow equivalent of *L*. **Q.E.D.**

In particular, since the Borromean rings has three components and is prime, it is not narrow.

Moving on to links in general, the shortest example has length 16, and is unique up to symmetry:
It's easy to see this, because any link component of length 6 is not
large enough for another link component to pass through it, so the
best we can do is to have two components of length 8. This link also
fits into (1, 2, 3) as shown above. Here's
another nice symmetrical link, 4^{2}_{1}:

Is this length minimal? My guess is yes.

There's some interesting related work in
molecular biology,
wordplay (allowing diagonal steps in the lattice),
and random walks. Holden's *Orderly Tangles* has related examples.