# Revision history [back]

### breaking out of while, for loops.

Hello.

Actually, my problem is to get all new representatives.

But I think it is only an algorithmic problem to me.

My consideration is following:

1. Let A be a group and Rep=[A]. And i=0

2. Make i=i+1 and the list L of all subgroups with index 2^i.

3. Make empty list New.

4. For B in L, if B is not conjugate to any group in Rep then add to Rep and add to New else do nothing.

5. If New is empty then return Rep else make New empty list and go to STEP 2.

How can I do this in SAGE....

I cannot give a simple example. :(

In my problem,

1. A=[1,x,x^2], Rep=[A], i=0 (A is a generator of the rank 3 Z-module in Z[x])

2. i=i+1, L : the list of all submodule of A with index 2^i.

3. New=[]

4. For B in L, [ For C in Rep, t=False, if B~C then t=true. ] If t=False then Rep=Rep+[B] and New=New+[B]

5. If len(New)==0 then return Rep else New=[] and go to STEP 2.

If you have some question I WILL answer as soon as possible. :)

Thanks.

 2 retagged vdelecroix 7147 ●16 ●78 ●155 http://www.labri.fr/pe...

### breaking out of while, for loops.

Hello.

Actually, my problem is to get all new representatives.

But I think it is only an algorithmic problem to me.

My consideration is following:

1. Let A be a group and Rep=[A]. And i=0

2. Make i=i+1 and the list L of all subgroups with index 2^i.

3. Make empty list New.

4. For B in L, if B is not conjugate to any group in Rep then add to Rep and add to New else do nothing.

5. If New is empty then return Rep else make New empty list and go to STEP 2.

How can I do this in SAGE....

I cannot give a simple example. :(

In my problem,

1. A=[1,x,x^2], Rep=[A], i=0 (A is a generator of the rank 3 Z-module in Z[x])

2. i=i+1, L : the list of all submodule of A with index 2^i.

3. New=[]

4. For B in L, [ For C in Rep, t=False, if B~C then t=true. ] If t=False then Rep=Rep+[B] and New=New+[B]

5. If len(New)==0 then return Rep else New=[] and go to STEP 2.

If you have some question I WILL answer as soon as possible. :)

Thanks.