Comprehensive Exam (8/19/24)

The Tanglenomicon

Zachary Bryhtan, Nicholas Connolly, Isabel Darcy, Ethan Rooke, Joseph Starr*

Mathematics Department at The University of Iowa

Knots

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https://www.knotplot.com/

The natural question

How many knots?

Knot Tables

Lord Kelvin’s vortex theory of the atom

Atoms are knotted vortices in the æther.

By Hand

  • 1860’s Tait computes knots up to 7 crossings
    • 15 knots
  • 1870’s Tait, Kirkman, and Little compute knots up to 10 crossings
    • Takes about 25 years
    • 250 knots
  • 1960’s Conway computes knots up to 11 crossings
    • “A few hours”
    • 802 knots

By Computer

  • 1980’s Dowker and Thistlethwaite compute up to 13 crossings
    • First using a computer
    • 12,966 knots
  • 1990’s Hoste, Thistlethwaite, and Weeks compute up to 16 crossings
    • Computer runtime on the order of weeks
    • 1,701,936 knots
  • 2020’s Burton computes up to 19 crossings
    • 350 Million knots

KnotInfo

C. Livingston and A. H. Moore, KnotInfo: Table of Knot Invariants, https://knotinfo.math.indiana.edu/, today’s date (eg. August 24, 2023)

Conway

How did Conway compute 25 years of work in "a few hours"?

Tangles

“We define a tangle as a portion of a knot diagram from which there emerge just 4 arcs pointing in the compass directions NW, NE, SW, SE.” - Conway, J.H.

Conway, J.H. “An Enumeration of Knots and Links, and Some of Their Algebraic Properties.” In Computational Problems in Abstract Algebra, 329-58. Elsevier, 1970. https://doi.org/10.1016/B978-0-08-012975-4.50034-5

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NWNESWSE
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Basic Operations

Operation $+$

$+$
$=$
$=$
$=$
$2$

Operation $\vee$

$\vee$
$=$
$=$
$=$
$\frac{1}{2}$

The Tanglenomicon

A table of two string tangles

(up to fixed boundary)

Building up

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Where we are

Rational Tangles

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$\begin{aligned}\to&\ \LP 3 \vee \frac{1}{2}\RP + 2\\&\\ \to&\ [3\ 2\ 2]\end{aligned}$
$\begin{aligned}\to&\ \LP 1 \vee \frac{1}{3}\RP + 1\\&\\ \to&\ [1\ 3\ 1]\end{aligned}$
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$\begin{aligned}\to& \frac{1}{4} + 1\,\,\\&\\ \to&\ [4\ 1]\end{aligned}$

Generation

For any $N$ an obvious twist vector is the twist vector of all $1$s $$[1\ 1\ 1\ \cdots\ 1]$$ Noting that when we write this sequence, we have $N-1$ spaces.

If we choose to place a $+$ instead of the left most space we get $$[1+1\ 1\ \cdots\ 1]=[2\ 1\ \cdots\ 1]$$ we’re free to make this choice for each space

this gives $N-1$ choices between ‘$+$’ and space $$[1\square 1\square 1\square\cdots\square1]$$ letting us generate twist vectors by simply counting from $0\to 2^{N-1}$.

Twist Vectors for $N=5$
$$\begin{array}{|l|l|l|l|} \hline [1\ 1\ 1\ 1\ 1]\ &\ [1\ 1\ 1\ 2]\ &\ [1\ 1\ 2\ 1]\ &\ [1\ 1\ 3]\\\hline [1\ 2\ 1\ 1]\ &\ [1\ 2\ 2]\ &\ [1\ 3\ 1]\ &\ [1\ 4]\\\hline [2\ 1\ 1\ 1]\ &\ [2\ 1\ 2]\ &\ [2\ 2\ 1]\ &\ [2\ 3]\\\hline [3\ 1\ 1]\ &\ [3\ 2]\ &\ [4\ 1]\ &\ [5]\\\hline \end{array}$$

Programmatic Description

stateDiagram-v2 direction LR state if_done <<choice>> State_i: i=0 State_ipp: i++ state "Construct TV from i as a bitfield" as tv_calc{ state "tmplt=i;j=0;cnt=N" as State_temp State_jpp: j++ State_cntmm: cnt-- State_sum_tv: TV[j]++ State_rsh: tmplt=tmplt>>1 state if_lsb <<choice>> state if_cnteo <<choice>> State_store_tv: Store TV [*] --> State_temp State_temp --> if_cnteo if_cnteo--> State_cntmm: if cnt>0 if_cnteo--> State_store_tv: if cnt==0 State_store_tv --> [*] State_cntmm -->if_lsb if_lsb -->State_sum_tv: if (tmplt & 0x01u)==1u State_sum_tv --> State_rsh if_lsb -->State_jpp: if (tmplt & 0x01u)==0u State_jpp --> State_rsh State_rsh --> if_cnteo } [*] --> State_i State_i --> if_done if_done --> tv_calc: if i < 2**(N-1) tv_calc --> State_ipp State_ipp --> if_done if_done --> [*]: if i == 2**(N-1)
i tmplt cnt j tv
0 0000 5 0 [1,1,1,1,1]
0 0000 4 1 [1,1,1,1,1]
0 0000 3 2 [1,1,1,1,1]
0 0000 2 3 [1,1,1,1,1]
0 0000 1 4 [1,1,1,1,1]
0 0000 0 4 [1,1,1,1,1]
i tmplt cnt j tv
6 0110 5 4 [1,1,1,1,1]
6 0011 4 1 [1,1,1,1,1]
6 0001 3 1 [1,2,1,1,1]
6 0000 2 1 [1,3,1,1,1]
6 0000 1 2 [1,3,1,1,1]
6 0000 0 2 [1,3,1,1,1]
i tmplt cnt j tv
7 0111 5 0 [1,1,1,1,1]
7 0011 4 0 [2,1,1,1,1]
7 0001 3 0 [3,1,1,1,1]
7 0000 2 0 [4,1,1,1,1]
7 0000 1 1 [4,1,1,1,1]
7 0000 0 1 [4,1,1,1,1]

Canonical Twist Vectors

We can write a canonical twist vector by taking the odd length vectors (appending $0$ where needed).

Canonical Twist Vectors for $N=5$
$$\begin{array}{|l|l|l|l|} \hline [1\ 1\ 1\ 1\ 1]\ &\ [1\ 1\ 1\ 2\ 0]\ &\ [1\ 1\ 2\ 1\ 0]\ &\ [1\ 1\ 3]\\\hline [1\ 2\ 1\ 1\ 0]\ &\ [1\ 2\ 2]\ &\ [1\ 3\ 1]\ &\ [1\ 4\ 0]\\\hline [2\ 1\ 1\ 1\ 0]\ &\ [2\ 1\ 2]\ &\ [2\ 2\ 1]\ &\ [2\ 3\ 0]\\\hline [3\ 1\ 1]\ &\ [3\ 2\ 0]\ &\ [4\ 1\ 0]\ &\ [5]\\\hline \end{array}$$

Computations

Rational Number (continued fraction)

The rational number for a twist vector is computed by taking the twist vector as a finite continued fraction, that is: $$\LB a\ b\ c\RB=c+\frac{1}{b+\frac{1}{a}}$$

Twist Vector to rational number
$$\ =\LB 3\ 2\ 2\RB=2+\frac{1}{2+\frac{1}{3}}=\frac{17}{7}$$
  • J.R. Goldman, L.H. Kauffman, Rational Tangles, Advances in Applied Math., 18 (1997), 300-332.
  • Conway, J.H. “An Enumeration of Knots and Links, and Some of Their Algebraic Properties.” In Computational Problems in Abstract Algebra, 329-58. Elsevier, 1970. https://doi.org/10.1016/B978-0-08-012975-4.50034-5

To play with twist vectors and continued fractions, visit: