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breaking out of while, for loops.

asked 10 years ago

anonymous user

Anonymous

updated 10 years ago

vdelecroix gravatar image

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.

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answered 10 years ago

Here are two example functions illustrating breaking out of while and for loops:

def f():
    index = 0
    while True:
        print index
        index += 1
        if index >= 4:
            break

def g():
    for i in range(10):
        if i >= 5:
            continue
        print i
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Thanks! It's useful to me. Using that algorithm, I solved my problem, roughly~. :)

Semin gravatar imageSemin ( 10 years ago )

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Asked: 10 years ago

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Last updated: Nov 22 '14