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Algorithmics in braid groups

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1 Algorithmics in braid groups
Juan González-Meneses Universidad de Sevilla Trenzas Jornada Temática Interdisciplinar de la Red Española de Topología Barcelona , October 22 – 23,

2 Seville. June 13-17, 2011

3 Computing in braid groups
Garside structure 1 2 3 4 5 E. Artin (1925) Juan González-Meneses

4 Computing in braid groups
Garside structure Monoid of positive braids: Induces a lattice order: Garside element Simple elements = Positive prefixes of  Juan González-Meneses

5 Computing in braid groups
Garside structure Garside group Dehornoy-Paris (1999) P 1 G And some other properties… Juan González-Meneses

6 Computing in braid groups
Garside structure Lattice of simple elements in the braid group B4. Juan González-Meneses

7 Word problem Left normal form Artin (1925)
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. Juan González-Meneses 7

8 a b a s t Word problem Left normal form
In general: Given simple elements a b Not left-weighted a s t left-weighted simple simple We call this procedure a local sliding applied to a b Juan González-Meneses 8

9 Word problem Left normal form
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 D. Minimal number of factors. Juan González-Meneses (canonical length) 9

10 Conjugacy problem Garside (1969) ElRifai-Morton (1988)
Birman-Ko-Lee (1998) Franco-GM (2003) Gebhardt (2005) Birman-Gebhardt-GM (2008) Gebhardt-GM (2009) Charney: (1992) Artin-Tits groups of spherical type are biautomatic. Juan González-Meneses 10

11 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. Juan González-Meneses 11

12 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 consecutive Juan González-Meneses 12

13 Conjugacy problem Cyclic sliding x5 x1 x4 x2 x3 Juan González-Meneses
13

14 Conjugacy problem s t t Cyclic sliding x2 x5 s x4 x3 x2 x1 x5 x4 x3 x5
Juan González-Meneses 14

15 The resulting braid is not
Conjugacy problem Cyclic sliding x2 x5 s x4 x3 t The resulting braid is not In left normal form Cyclic sliding: We compute its left normal form, and it may become simpler Iterate… Juan González-Meneses 15

16 Conjugacy problem SC(x) x x1 x2 x3 Sliding circuits …
Sliding circuit of x. SC(x) Set of sliding circuits in the Conjugacy class of x. Juan González-Meneses 16

17 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? Many authors do computer experiments with “random braids”. Juan González-Meneses

18 Computer experiments “A” chooses a random secret braid… Random braids
Quotes from papers on braid-cryptography: “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) Franco, González-Meneses. …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. Gebhardt. Juan González-Meneses None of these procedures generates truly random braids!

19 Computer experiments Random braids
We will focus on the positive braid monoid: Length 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? Juan González-Meneses

20 Computer experiments Random braids
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 There is only one word representing the braid There are 16 words representing the braid The probability of obtaining is 16 times the probability of obtaining This becomes more dramatic as n and k grow. Juan González-Meneses

21 Generating random braids
Lex-representative How do we generate braids with the same probability? Given a braid a, its Lex-representative, w(a ), is the (lexicographically) smallest word representing a. Bij. { Braids of length k } { Lex-representatives of length k } Juan González-Meneses

22 Generating random braids
Braid tree The situation can be described using a rooted tree. Example, in B4+ : Root. The only braid of length 0 in B4+. Juan González-Meneses

23 Generating random braids
Braid tree The situation can be described using a rooted tree. Example, in B4+ : 3 braids of length 1 in B4+. Juan González-Meneses

24 Generating random braids
Braid tree The situation can be described using a rooted tree. Example, in B4+ : 8 braids of length 2 in B4+. The word is not there, as the Lex-representative of is Juan González-Meneses

25 Generating random braids
Braid tree The situation can be described using a rooted tree. Example, in B4+ : 19 braids of length 3 in B4+. There is no subword , neither , nor Juan González-Meneses

26 Generating random braids
Braid tree The situation can be described using a rooted tree. Example, in B4+ : 19 braids of length 3 in B4+. There is no subword , neither , nor Juan González-Meneses

27 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! Juan González-Meneses Warning: the tree is exponentially big!

28 Counting braids of given length
Bronfman method x n,k = Number of braids in Bn+ of length k. (want to compute this number) Bronfman, 2001: Recursive method to compute the growth function of Bn+ We wil use another method: simpler and faster. Juan González-Meneses

29    Counting braids of given length Our method Example, in Bn+ :
Cannot start with = # Braids: # Braids: Cannot start with Cannot start with Juan González-Meneses

30    Counting braids of given length Our method Example, in Bn+ :
Cannot start with = # Braids: Cannot start with Cannot start with # Braids: Juan González-Meneses

31 +    Counting braids of given length Our method Example, in Bn+ :
Cannot start with = # Braids: Cannot start with + # Braids: Cannot start with Juan González-Meneses

32 +     Counting braids of given length Our method Example, in Bn+ :
Cannot start with Cannot start with + # Braids: # Braids: Juan González-Meneses

33 +     Counting braids of given length Our method Example, in Bn+ :
Cannot start with = # Braids: Cannot start with + # Braids: # Braids: Juan González-Meneses

34 Counting braids of given length
Our method This yields an easy recursive formula: + - = Juan González-Meneses

35 Counting braids of given length
Our method + - = Example, in B4+ : 1 3 3 2 1 Juan González-Meneses

36 Counting braids of given length
Our method Example, in B4+ : + - = 1 3 3 2 1 8 8 5 2 Juan González-Meneses

37 Counting braids of given length
Our method Example, in B4+ : + - = 1 3 3 2 1 8 8 5 2 19 19 11 4 Juan González-Meneses

38 Counting braids of given length
Our method Example, in B4+ : + - = 1 3 3 2 1 8 8 5 2 19 19 11 4 43 43 24 8 Juan González-Meneses

39 Counting braids of given length
Our method Example, in B4+ : + - = 1 3 3 2 1 8 8 5 2 19 19 11 4 43 43 24 8 94 94 51 16 Juan González-Meneses

40 … Counting braids of given length Our method Example, in B4+ : 1 3 1 2
- = 1 3 1 2 3 5 8 4 11 19 3 2 1 8 8 5 2 19 19 11 4 43 43 24 8 94 94 51 16 202 202 108 33 What are we computing? Juan González-Meneses

41 Counting braids of given length
Our method 1 2 3 5 8 4 11 19 The first column of our table contains x n,1 , x n,2 , x n,3 ,… Computing k rows, we obtain x n,k. Juan González-Meneses

42 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: Next Juan González-Meneses

43 Finding the rth braid of length k
Hanging leaves Consider the graph of height k as before. Given a vertex w, suppose we can compute, in polynomial time, the number of leaves hanging from w: Juan González-Meneses

44 Finding the rth braid of length k
The procedure We want to compute the 16th braid: The first letter is The second letter is The third letter is Juan González-Meneses

45 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. 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 some forbidden prefixes. Juan González-Meneses

46 Computing pending leaves
Forbidden prefixes Example: The number of leaves hanging from 5, is equal to the number of braids, of length k-1, that cannot start with: Forbidden prefixes Lemma: (GM, 2010) For every w, ending with j, one just needs at most n -1 forbidden prefixes, of a very particular kind: The forbidden prefixes of wi are obtained from those of w, in time O(n). Juan González-Meneses

47 Computing pending leaves
Forbidden prefixes Example: Forbidden prefixes for the word Denote Know how to compute Juan González-Meneses

48 Computing pending leaves
Forbidden prefixes By the inclusion-exclusion principle, we can compute Juan González-Meneses

49 Computing pending leaves
Forbidden prefixes By the inclusion-exclusion principle, we can compute Juan González-Meneses

50 Computing pending leaves
Forbidden prefixes Good news: Every summand is equal to § xn,t, for some t· k. These are the elements in the first column of our table. Bad news: There are exponentially many summands. 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. Juan González-Meneses

51 ) Generating random positive braids The result As a consequence:
One can compute in ploynomial space and time with respect to 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. Juan González-Meneses

52 Generating random braids
From positive braids to braids Every braid can be written, in a unique way, as: where  is positive, and cannot start with . There are xn,k positive braids of length k, 1 2 3 5 8 4 11 19 24 43 16 51 94 33 108 202 xn,k-|| of them can start with . It is easy to show that Hence: Juan González-Meneses

53 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. Procedure: Choose a random . Compute its left normal form. If  can start with , discard it and start again. Probability: < Juan González-Meneses

54 Open problems ² Generate random positive braids of given canonical length. ² Generate random braids of given length, or given canonical length. ² Count elements in Artin monoids of finite type. (without using Saito’s or Albenque-Nadeau’s formulae for growth functions) ² Count elements in Artin groups of finite type. (without using Finite State Automata to compute growth functions) Charney, 1995 Juan González-Meneses

55 Software to compute with braid groups
² (C++ source code) ² (web applet) ² GAP3 (CHEVIE package) ² MAGMA Juan González-Meneses


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