Braids, Cables, and CellsRepresenting Art and Craft with Mathematics and Computer

Science

Joshua Holden

Rose-Hulman Institute of Technologyhttp://www.rose-hulman.edu/~holden

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“Knotwork” has been used in visual arts for manycenturies.

Left: Detail from Roman mosaic at Woodchester, c. 325 CE

Right: Detail from the “Book of Kells”, c. 800 CE

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“Knotwork” has been used in visual arts for manycenturies.

Left: by A. Reed Mihaloew, Right: by Christian Mercat

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In knitting, raised knot-like designs are known as“cables”.

Left: Design by Meredith Morioka, knitted by Lana Holden

Right: Design by Julie Levy, knitted by Lana Holden

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“Cables” can also be done in crochet.

Both: Designed and crocheted by Lisa Naskrent5 / 43

A somewhat similar effect is given by “travelingeyelets” in knitted lace.

From Barbara Walker’s Charted Knitting Designs

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And of course, there are many actual weavingpatterns.

Left: 2/2 twill weave, woven by Sarah, a.k.a. Aranel

Right: “Noonday Sun” pattern, woven by Peggy Brennan(Cherokee Nation)

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Vandermonde was interested in the mathematicalstudy of knots and braids.

From “Remarques sur les problèmes de situation”, 1771

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So was Gauss.

From Page 283 of Gauss’s Handbuch 7, c. 1825?

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Today, braids are studied from the perspectives oftopology and group theory.

Two equal braids (Wikipedia)

I Two braids which are the same except for “pulling thestrands” are considered equal.

I All strands are required to move from bottom to top.

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You can make braids into a group by “multiplying”them.

× =

Multiplying braids (Wikipedia)

I You multiply two braids by stacking them and thensimplifying.

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A Cellular Automaton is a mathematical constructmodeling a system evolving in time.

I Finite number of cells in a regular gridI Finite number of states that a cell can be inI Each cell has a well-defined finite neighborhoodI Time moves in discrete stepsI State of each cell at time t is determined by the states of

its neighbors at time t − 1I Each cell uses the same rule

The “von Neumann neighborhood”(Wikibooks)

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“The Game of Life” is an example you might know.I Invented by John ConwayI Grid is two-dimensionalI Two states, “live” and “dead”I Neighborhood is the eight cells which are directly

horizontally, vertically, or diagonally adjacentI Any live cell with two or three live neighbors stays live.

I Any other live cell dies.I Any dead cell with exactly three live neighbors becomes a

live cell.

I Any other dead cell stays dead.

(Figures by Paul Callahan, from www.math.com)13 / 43

Example: A “Pulsar”

(Wikipedia)

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Example: A “Pulsar”

(Wikipedia)

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Example: A “Pulsar”

(Wikipedia)

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Example: A “Pulsar”

(Wikipedia)

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Example: A “Pulsar”

(Wikipedia)

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Example: A “Pulsar”

(Wikipedia)

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Example: A “Pulsar”

(Wikipedia)

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Another well-known class of automata are the“Elementary” Cellular Automata.

I Popularized by Stephen Wolfram (A New Kind of Science)I Grid is one-dimensionalI Two states, “white” and “black”I Neighborhood includes self and one cell on each side

“Rule 30” (Mathworld)

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Example: “Rule 90”I Second dimension is used for “time”I Produces the Sierpinski triangle fractalI An “additive” rule

(Mathworld)

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Cellular automata can exhibit aperiodic behavior.

Conjecture (Wolfram, 1984)The sequence of colors produced by the cell at the center ofRule 30 is aperiodic.

I This sequence is used by the pseudorandom numbergenerator in the program Mathematica.

I The center and right portions of Rule 30 appear to havesome of the characteristics of “chaotic” systems.

Theorem (Jen, 1986 and 1990)

(a) At most one cell of Rule 30 produces a periodic sequenceof colors.

(b) The sequence of color pairs produced by any two adjacentcells of Rule 30 is aperiodic.

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Rule 30

(Mathworld)

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Cellular automata are also “computationally universal”.

Theorem (Cook, 1994+)Rule 110 can be used to simulate any Turing machine.

This is important because of the widely accepted:

Church-Turing ThesisAnything that can be computed by an algorithm can becomputed by some Turing machine.

And for complexity geeks:

Theorem (Neary and Woods, 2006)Rule 110 can be used to simulate any polynomial time Turingmachine in polynomial time. (I.e., it is “P-complete”.)

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Rule 110 on a Single Cell Input

(Mathworld)

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How is this possible?

1. Use Rule 110 to simulate a “cyclic tag system”.

A cyclic tag system has:

I A data stringI A cyclic list of “production rules”

To perform a computation:I If the first data symbol is 1, add the production rule to the

end of the data string. If the first data symbol is 0 donothing.

I Delete the first data symbol.I Move to the next production rule.I Repeat until the data string is empty.

2. Show that any Turing machine can be simulated by a cyclictag system.

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Cyclic Tag System Example

Production rules Data string

010 11001000 10010101111 001010000010 01010000000 10100001111 010000000010 10000000

.... . .

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To simulate a cyclic tag system with Rule 110, youneed:

I a representation of the data string (stationary)I a representation of the production rules (left-moving)I “clock pulses” (right-moving)

(Wikipedia)23 / 43

Rule 110 Performing (Part of) a Computation

(Wikipedia)

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CAs have been used in fiber arts before.

Left: Designed and crocheted by Jake Wildstrom

Right: Knitted by Pamela Upright, after Debbie New25 / 43

Each of our cells will store 4 bits of information in 8states.

upright slanted

no strands

left only

right only

both

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The neighborhood will be a “brick wall” neighborhood.

(Time moves from bottom to top, like a knitting pattern.)

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The CA rule system can actually be thought of as foursimpler CAs. The first two just control whether strandsare present or not.

no left left

no right

right

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The CA rule system can actually be thought of as foursimpler CAs. The first two just control whether strandsare present or not.

no left left

no right

right

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The third CA controls whether the strands are uprightor diagonal, specified by a numbered rule.

“Turning Rule 39”30 / 43

And the fourth CA controls which strand is on top if thestrands cross, also specified by a numbered rule.

“Crossing Rule 39”31 / 43

There are several possible choices for what to do atthe edges of the grid.

I Make the grid infinite?I Have a special kind of state for edge cells?I Make the grid cylindrical? (“Periodic boundary conditions”)I Reflect cells at the edges? (Where to put the axis?)

I have so far only implemented the cylindrical case.

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The rules can produce fractal patterns, . . .

Rules 68 and 033 / 43

. . . weaving patterns, . . .

Left: Rules 0 and 47, Right: Rules 0 and 448

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. . . traditional braids, . . .

’

Left: Wikipedia, Right: Rules 333 and 39

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. . . slightly less traditional braids, . . .

Left: backstrapweaving.wordpress.com

Right: Rules 333 and 9936 / 43

. . . and other sorts of “cable” patterns.

Left: Rules 47 and 0, Right: Rules 201 and 39

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If all strands are present and only one rule is active,previously known results on “elementary” CA’s apply.

Rules 68 and 0 give the same result as Wolfram’s Rule 90

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Since the width is finite, the pattern must eventuallyrepeat. For a given width, how long can a repeat be?

PropositionAssume only the crossing rule is active. For a given width m, norepeat can be longer than m2m − 2m rows.

Proof.

After 2m rows, all of the strandshave returned to their originalpositions. The only question iswhich strand of each crossingis on top.

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Since the width is finite, the pattern must eventuallyrepeat. For a given width, how long can a repeat be?

PropositionAssume only the crossing rule is active. For a given width m, norepeat can be longer than m2m − 2m rows.

Proof.

If there are m crossings thenthere are 2m possiblearrangements of the crossingsbut only 2 different ways therow can be shifted. So themaximum repeat is the lcm of anumber ≤ 2m and anumber ≤ 2.

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Since the width is finite, the pattern must eventuallyrepeat. For a given width, how long can a repeat be?

PropositionAssume only the crossing rule is active. For a given width m, norepeat can be longer than m2m − 2m rows.

Proof.

If there are m − 1 crossings,then there are 2m−1 possiblearrangements but 2m differentshifts, so the maximum repeatis the lcm of a number ≤ 2m−1

and a number ≤ 2m.

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Since the width is finite, the pattern must eventuallyrepeat. For a given width, how long can a repeat be?

PropositionAssume only the crossing rule is active. For a given widthm ≥ 2k , the maximum repeat is at least lcm(2k+1,2m) rowslong.

Proof.

Consider the starting row withone single strand and m − 1crossings. Crossing Rule 100(which is additive) acts on thiswith a repeat (modulo cyclicshift) of 2k+1 if m > 2k . Thecyclic shift gives the 2m.

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Since the width is finite, the pattern must eventuallyrepeat. For a given width, how long can a repeat be?

PropositionAssume only the crossing rule is active. For a given widthm ≥ 2k , the maximum repeat is at least lcm(2k+1,2m) rowslong.

RemarkFor m ≤ 5, this is sharp. Form = 23, m crossings andCrossing Rule 257 (which isalso additive) does better.

For large m, neither the upperbound above nor this lowerbound seems especially likelyto be sharp.

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If only one or two strands are present then themaximum length of a repeat can be determined.

PropositionAssume only one or two strands arepresent. For a given width m ≤ 5, themaximum repeat is (2m)(2m + 1) rows long.

Proof.This is achieved by Turning Rule 97 and twostrands.

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There is much future work to be done.

I If all strands are present and both rules are active, then wehave two “elementary” CA’s where one can “overwrite” theother.

I The length of a maximum repeat in other cases is open.I What is the computational complexity of predicting things

that the CA might do?I More work can be done with different edge conditions.I Which braids can be represented? (In the sense of braid

groups)I Which rules are “reversible”?I Two-dimensional grids with time as the third dimension

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Thanks for listening!

“Barolo”, designed and knitted by Lana Holden

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