algorithmics in braid groups juan gonzález-meneses universidad de sevilla jornada temática...

Algorithmics in braid groups

Juan González-Meneses

Universidad de Sevilla

Jornada Temática Interdisciplinar de laRed Española de Topología

Barcelona , October 22 – 23, 2010.

Trenzas

Seville. June 13-17, 2011

www.imus.us.es/ACT/braids2011

Computing in braid groups Garside structure


1

2345

E. Artin (1925)



Monoid of positive braids:

Induces a lattice order:

Garside elementGarside element

Simple elements = Positive prefixes of

1

G

P



Garside group

Dehornoy-Paris (1999)

And some otherproperties…

Lattice of simple elements in the braid group B4.



Word problem Left normal form


Garside (1969), Deligne (1972), Adyan (1984), Elrifai-Morton (1988), Thurston (1992).

Every braid x can be written in Left normal form:

Simple elements

Every product must be left-weighted.

Biggest possible simple element in any such writing of this product.Biggest possible simple element in any such writing of this product.

Artin (1925)



a s t

In general: Given simple elements

a b

simple simple

Not left-weighted

left-weighted

We call this procedure a local sliding applied to a b .



Computation of a left normal form, given a product of simple elements:

Apply all possible local slidings,until all consecutive factors are left weighted

Left normal form:

Maximal power of Minimal number of factors.

(canonical length)

Conjugacy problem


Charney: (1992) Artin-Tits groups of spherical type are biautomatic.

Garside (1969)

ElRifai-Morton (1988)

Birman-Ko-Lee (1998)

Franco-GM (2003)

Gebhardt (2005)

Birman-Gebhardt-GM (2008)

Gebhardt-GM (2009)

Conjugacy problem The main idea


Given an element x, compute the set of simplest conjugates of x.

(in some sense)

x and y are conjugate , their corresponding sets coincide. x and y are conjugate , their corresponding sets coincide.

Conjugacy problem Cyclic sliding


Can x be simplified by a conjugation?

For simplicity, we will assume that there is no power of :

Consecutive factors are left-weighted. What about x5 and x1 ?

Up to conjugacy, we can consider that x5 and x1 are consecutiveUp to conjugacy, we can consider that x5 and x1 are consecutive



x2

x1 x 5

x4 x3

x2

x1 x 5

x4 x3

t x2

x 5

x4 x3



s Cyclic sliding:

x2

x 5 s

x4 x3

t



x2

x 5 s

x4 x3

t Cyclic sliding: The resulting braid is notIn left normal form

The resulting braid is notIn left normal form

We compute its left normal form, and it may become simplerWe compute its left normal form, and it may become simpler

Iterate…

Conjugacy problem Sliding circuits


xsliding x1

sliding x2 x3

sliding …

Sliding circuit of x.Sliding circuit of x.

SC(x) Set of sliding circuits in the

Conjugacy class of x.

Set of sliding circuits in the

Conjugacy class of x.

Computer experiments Random braids


The solution to the word problem is very efficient.

The solution to the conjugacy problem seems to be very efficient.

(taking braids at random)

At random?At random?

Many authors do computer experiments with “random braids”.


Quotes from papers on braid-cryptography:


“A” chooses a random secret braid…“A” chooses a random secret braid…

Ko, Lee, Cheon, Han, Kang, Park.

…we took 5000 random pairs of positive braids in Bn of length l (using Artin presentation)

…we took 5000 random pairs of positive braids in Bn of length l (using Artin presentation)

Franco, González-Meneses.

…the elements ai and bi are products of 10 random Artin generators.…the elements ai and bi are products of 10 random Artin generators.

Garber, Kaplan, Teicher, Tsaban,Vishne.

Random elements for these tests were obtained as follows. We choose independent random simple elements A1,A2, ... until len(A1/ Am) = r.

Random elements for these tests were obtained as follows. We choose independent random simple elements A1,A2, ... until len(A1/ Am) = r.

Gebhardt.

None of these procedures generates truly random braids!


We will focus on the positive braid monoid:


Length of a (positive) braid = Word lengthLength of a (positive) braid = Word length

We want to generate a random positive braid of length k.

What if we take the product of k randomly chosen generators? What if we take the product of k randomly chosen generators?



There is only one word representing the braid

There are two words representing the braid

The probability of obtaining is twice the probability of obtaining

This becomes more dramatic as n and k grow.

There are 16 words representing the braid

There is only one word representing the braid

The probability of obtaining is 16 times the probability of obtaining

Generating random braids Lex-representative


How do we generate braids with the same probability?

Given a braid , its Lex-representative, w, is the (lexicographically) smallest word representing .

Given a braid , its Lex-representative, w, is the (lexicographically) smallest word representing .

{ Braids of length k } { Lex-representatives of length k }Bij.

Generating random braids Braid tree


Example, in B4+ :

The situation can be described using a rooted tree.

Root.

The only braid of length 0 in B4+.



Example, in B4+ :


3 braids of length 1 in B4+.



Example, in B4+ :


8 braids of length 2 in B4+. The word is not there,

as the Lex-representativeof is .



Example, in B4+ :


19 braids of length 3 in B4+. There is no subword ,

neither , nor .

Generating random braids Generation procedure


To generate a random braid:

1) Compute the number of leaves of the tree: 19

2) Choose a random number between 1 and 19: 16

3) Find the braid corresponding to the 16th leaf: In polynomial time!In polynomial time!

Warning: the tree is exponentially big!Warning: the tree is exponentially big!

Counting braids of given length Bronfman method


x n,k = Number of braids in Bn+ of length k.

Recursive method to compute the growth function of Bn+

(want to compute this number)

Bronfman, 2001:

We wil use another method: simpler and faster.We wil use another method: simpler and faster.

Counting braids of given length Our method


Example, in Bn+ :

= # Braids: Cannot start with

# Braids: Cannot start with

+ # Braids: # Braids:



This yields an easy recursive formula:

+

+

+

-

-

=



+

+

+

-

=

Example, in B4+ :

1 1 1 1

1233



+

+

+

-

=

Example, in B4+ :

1 1 1 1

1233

2588



+

+

+

-

=

Example, in B4+ :

1 1 1 1

1233

2588

4111919



Example, in B4+ :

1 1 1 1

1233

2588

4111919

8244343+

+

+

-

=



Example, in B4+ :

1 1 1 1

1233

2588

4111919

8244343

16519494+

+

+

-

=



Example, in B4+ :

1 1 1 1

1233

2588

4111919

8244343

16519494

33108202202 …

What are we computing?What are we computing?

+

+

+

-

=

123

258

41119



123

258

41119

The first column of our table contains x n,1 , x n,2 , x n,3 ,…

Computing k rows, we obtain x n,k.Computing k rows, we obtain x n,k.

Generating random braids Generation procedure


To generate a random braid:

1) Compute the number of leaves of the tree: 19

2) Choose a random number between 1 and 19: 16

3) Find the braid corresponding to the 16th leaf: NextNext

Finding the rth braid of length k Hanging leaves


Given a vertex w, suppose we can compute, in polynomial time,

the number of leaves hanging from w:

Given a vertex w, suppose we can compute, in polynomial time,

the number of leaves hanging from w:

Consider the graph of height k as before.

Finding the rth braid of length k The procedure


We want to compute the 16th braid:

The first letter is .The first letter is .

The second letter is .The second letter is .

The third letter is .The third letter is .

Computing pending leaves Forbidden prefixes


Computing at most (n-2)k times the hanging leaves of a vertex,

one computes the rth braid of length k.

Computing at most (n-2)k times the hanging leaves of a vertex,

one computes the rth braid of length k.

How to compute the hanging leaves?

{ Leaves hanging from w } = { Lex-representatives starting with w }

Lemma: (GM, 2010) The number of leaves hanging from w is equal to

the number of braids, of length k-|w|, which cannot start with someforbidden prefixes.

Lemma: (GM, 2010) The number of leaves hanging from w is equal to

the number of braids, of length k-|w|, which cannot start with someforbidden prefixes.



Example:

Forbidden prefixes for the word

Denote

Know how to compute



By the inclusion-exclusion principle, we can compute



Good news: Every summand is equal to § xn,t, for some t· k.

These are the elements in the first column of our table.

But we can treat the same way all elements of the form a1Ç a2Ç Ç as

such that: ² Have the same length.

² Coincide in other technical questions, regarding permutation of strands.

Bad news: There are exponentially many summands.

Generating random positive braids The result


As a consequence:

Theorem: (GM, 2010) There is a procedure to generate a random positive braid

in Bn+ of length k, whose time and space complexity is a polynomial in n and k.

Theorem: (GM, 2010) There is a procedure to generate a random positive braid

in Bn+ of length k, whose time and space complexity is a polynomial in n and k.

One can compute in ploynomial space and time with respect to n and k.

)

Every braid can be written, in a unique way, as:

Generating random braids From positive braids to braids


where is positive, and cannot start with .

There are xn,k positive braids of length k,

xn,k-|| of them can start with .1 1 1 1

1233

2588

4111919

8244343

16519494

33108202202Hence:

It is easy to show that

Generating random braids The result


Theorem: (GM, 2010) There is a partial algorithm to generate a random braid

in Bn, where | | = k, whose time and space complexity is a polynomial in n and k,

and whose rate of failure is < 2-n(n -1)/2.

Theorem: (GM, 2010) There is a partial algorithm to generate a random braid

in Bn, where | | = k, whose time and space complexity is a polynomial in n and k,

and whose rate of failure is < 2-n(n -1)/2.

Procedure: Choose a random .

Compute its left normal form.

If can start with , discard it and start again.

Probability: <

Software to compute with braid groups


² www.personal.us.es/meneses

² www.indiana.edu/~knotinfo/

² GAP3

² MAGMA

(C++ source code)

(web applet)

(CHEVIE package)

algorithmics in braid groups juan gonzález-meneses universidad de sevilla jornada temática...

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main idea juan gonzlezmeneses

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random elements

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resulting braid