Download presentation

Presentation is loading. Please wait.

Published byLeanna Gillian Modified over 2 years ago

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

2
Seville. June 13-17, 2011

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

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

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

6
Lattice of simple elements in the braid group B 4. Juan González-Meneses Computing in braid groups Garside structure

7
Word problem Left normal form Juan González-Meneses 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)

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

9
Word problem Left normal form Juan González-Meneses 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)

10
Conjugacy problem Juan González-Meneses 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)

11
Conjugacy problem The main idea Juan González-Meneses Given an element x, compute the set of simplest conjugates of x. (in some sense) x and y are conjugate, their corresponding sets coincide.

12
Conjugacy problem Cyclic sliding Juan González-Meneses 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 x 5 and x 1 ? Up to conjugacy, we can consider that x 5 and x 1 are consecutive

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

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

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

16
Conjugacy problem Sliding circuits Juan González-Meneses x sliding x1x1 x2x2 x3x3 … 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.

17
Computer experiments Random braids Juan González-Meneses 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.

18
Computer experiments Random braids Quotes from papers on braid-cryptography: Juan González-Meneses A chooses a random secret braid… Ko, Lee, Cheon, Han, Kang, Park. …we took 5000 random pairs of positive braids in B n of length l (using Artin presentation) Franco, González-Meneses. …the elements a i and b i 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 A 1, A 2,... until len( A 1 / A m ) = r. Gebhardt. None of these procedures generates truly random braids!

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

20
Computer experiments Random braids Juan González-Meneses 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

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

22
Generating random braids Braid tree Juan González-Meneses Example, in B 4 + : The situation can be described using a rooted tree. Root. The only braid of length 0 in B 4 +.

23
Generating random braids Braid tree Juan González-Meneses Example, in B 4 + : The situation can be described using a rooted tree. 3 braids of length 1 in B 4 +.

24
Generating random braids Braid tree Juan González-Meneses Example, in B 4 + : The situation can be described using a rooted tree. 8 braids of length 2 in B 4 +. The word is not there, as the Lex-representative of is.

25
Generating random braids Braid tree Juan González-Meneses Example, in B 4 + : The situation can be described using a rooted tree. 19 braids of length 3 in B 4 +. There is no subword, neither, nor.

26
Generating random braids Braid tree Juan González-Meneses Example, in B 4 + : The situation can be described using a rooted tree. 19 braids of length 3 in B 4 +. There is no subword, neither, nor.

27
Generating random braids Generation procedure Juan González-Meneses 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! Warning: the tree is exponentially big!

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

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

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

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

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

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

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

35
Counting braids of given length Our method Juan González-Meneses = Example, in B 4 + :

36
Counting braids of given length Our method Juan González-Meneses = Example, in B 4 + :

37
Counting braids of given length Our method Juan González-Meneses = Example, in B 4 + :

38
Counting braids of given length Our method Juan González-Meneses Example, in B 4 + : =

39
Counting braids of given length Our method Juan González-Meneses Example, in B 4 + : =

40
Counting braids of given length Our method Juan González-Meneses Example, in B 4 + : … What are we computing? =

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

42
Generating random braids Generation procedure Juan González-Meneses 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

43
Finding the r th braid of length k Hanging leaves Juan González-Meneses 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.

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

45
Computing pending leaves Forbidden prefixes Juan González-Meneses 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.

46
Computing pending leaves Forbidden prefixes Juan González-Meneses 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 ).

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

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

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

50
Computing pending leaves Forbidden prefixes Juan González-Meneses Good news: Every summand is equal to § x n,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 a 1 Ç a 2 Ç Ç a s such that: ² Have the same length. ² Coincide in other technical questions, regarding permutation of strands. Bad news: There are exponentially many summands.

51
Generating random positive braids The result Juan González-Meneses As a consequence: Theorem: (GM, 2010) There is a procedure to generate a random positive braid in B n + 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. )

52
Every braid can be written, in a unique way, as: Generating random braids From positive braids to braids Juan González-Meneses where is positive, and cannot start with. There are x n,k positive braids of length k, x n,k-| | of them can start with Hence: It is easy to show that

53
Generating random braids The result Juan González-Meneses Theorem: (GM, 2010) There is a partial algorithm to generate a random braid in B n, 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: <

54
Open problems Juan González-Meneses 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 Finite State Automata to compute growth functions) Charney, 1995 (without using Saitos or Albenque-Nadeaus formulae for growth functions) ² Count elements in Artin groups of finite type.

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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google