How do I identify the set of words of given length in matrix generators

asked 2016-04-02 01:30:25 +0100

j0equ1nn
69 ●2 ●8 ●14

updated 2016-04-02 01:31:00 +0100

I should mention that I am a beginner with Sage so please bear with me. Also this question regards a small sub-step of what this question is about: http://ask.sagemath.org/question/3293... But I expect the current issue would be much more common.

Take a small set of matrices, let's say 3 of them: m1, m2, m3. Let G be the group generated by these matrices. I want to be able to work with words of a given length in these generators. Let Gn be the words of length n in G. Then for example

G1 = (m1,m2,m3,m1^(-1),m2^(-1),m3^(-1)).

Is there a pre-built module for working with words of length n for n arbitrary? If not, I'm sure someone can find improvements on the approach I am taking. For instance to get G2, I have been doing this:

G2 = []
for g in G1:
    for h in G1:
        if g*h not in G2:
            G2.append(g*h)
        if h*g not in G2:
            G2.append(h*g)

This gives me everything in G2, and avoids repitition, but does not extend well to Gn.

To get G3 I have been doing this:

G3 = []
for g in G2:
    for h in G1:
        if g*h not in G2:
            if g*h not in G1:
                G2.append(g*h)
        if h*g not in G2:
            if h*g not in G1:
            G2.append(h*g)

If I want to go to G4 in this way, I need to add a line that makes sure my new addition isn't in G3, and I need to keep adding new lines at each increase in word length. I would like to define a function which outputs all words of length n, but as you can imagine from my approach above that is not working out well.

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Like that ?

sage: Words(['a','b'],5).list()

FrédéricC ( 2016-04-02 12:00:49 +0100 )edit

@FredéricC I did not know of that command for generating words, but there is a big difference between words in the free group <a,b> and reduced words in a matrix group. I'm interested in reducing run-time by avoiding repetition of equivalent words, and also being able to efficiently search through the words generated.

j0equ1nn ( 2016-04-04 01:30:44 +0100 )edit

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answered 2016-04-02 16:49:17 +0100

nbruin
4153 ●3 ●45 ●91

You can of course first generate the words, but if the group generated by your generators is far from free on those generators, you will be generating a lot of extraneous entries. In addition M in <List> is much slower than M in <Set>, and sets would take care of repetitions automatically.

Matrices and sets are a little inconvenient together, because matrices are by default "mutable" and it's a bad idea to put mutable objects in a set (if an element were to mutate into one that is equal to another one, it would need to be taken out!), but with a little helper function we can fix that.

def immutable(a):
  a.set_immutable()
  return a

m1=immutable(matrix(ZZ,2,2,[1,1,0,1]))
m2=immutable(matrix(ZZ,2,2,[0,1,-1,0]))
V={m1,m2,immutable(m1^(-1)),immutable(m2^(-1))}

G=[V]
for i in [1..5]:
    G.append( {immutable(m*g) for m in G[0] for g in G[-1]})

Note that for every i, a new Gi is computed. Furthermore, in python, negative indices mean from the end, so G[-1] is the last element in the list G, i.e., the last computed list G[i-1]. List indexing in python is 0-based, so G[0] is the first entry, i.e., V.

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This is great! I had improved my code a little from what I posted, but still a run-time long enough to go grocery shopping before getting the data. This approach by contract even runs in Sage cell online in a few seconds.

j0equ1nn ( 2016-04-03 06:07:44 +0100 )edit

Something strange about this though. We are not creating a set, but a list of sets, right? Because of this there is in fact a lot of repetition. For instance, every time a matrix can be written as a non-reduced word of length L in the generators, it will be included in the (L-1)th set in the list. Is there an easy modification that will put all these sets together, perhaps improving run-time further?

j0equ1nn ( 2016-04-03 07:02:48 +0100 )edit

Do you mean that you only want to keep track of the union of the G[i] ? Something like

R=set(V)
W=V
for i in [1..5]:
    W= {immutable(w*v) for w in W for v in V}
    R.update(W)

This might slightly reduce runtime, because memory use is a little lower, but we're still doing the same multiplication work.

I see one possible improvement: if you've just obtained an element by multiplying with v, you shouldn't multiply if with v^(-1) the next time around. That would at most bring you an improvement factor of (#V - 1)/(#V) and you'd have more overhead in administration, so I'm not sure it's worth it.

nbruin ( 2016-04-03 18:02:34 +0100 )edit

I want to be able to search for words satisfying certain properties, up to some chosen length. I can do that in the previous version, but I must search through each item in the list of sets G and I get a lot of repetition. It would be okay to have G be a list of sets by reduced word length (meaning if something simplifies to a shorter word, it doesn't get included in a new set). It would also be okay to have G just be a set of all words up to the given length. With the new version you've given I get the error "local variable 'W' referenced before assignment," which makes no sense to me since W=V, and V is the set of generators and their inverses.

j0equ1nn ( 2016-04-04 00:35:01 +0100 )edit

One solution I have is just, use your original answer, then in my function that searches through G I can tell it to make the collected output into a set before returning it. This will not reduce run-time, but does save me the trouble of looking through the output for distinct elements. BTW I really appreciate your help with this! There are many things that just take me forever being inexperienced and you've already saved me a lot of time.

j0equ1nn ( 2016-04-04 01:14:03 +0100 )edit

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