Just to elucidate my problem, because I think I'm doing a dismal job of
explaining it.
Keep a4 and v4 as defined previously (alternating group and klien 4 group)
Consider
a4 C."(1 _) v4
This gives an almost expected result.
But it becomes suspicous when I observe each of the 12 sets of 4 permutations
(not been nubbed yet). ALL of them contain 0 1 2 3, i.e. they all contain the
identity, which I think shouldn't be. In fact, they are all the same 4
permutations, with rows rearranged, whereas I expected 12 sets of 4
permutations, which can be partitioned into 3 separate sets, after nubbing.
Then I try
result1 =: (1{a4) C."(1 _) v4 NB. multiply the 2nd permutaiton in a4 by V4
This gives
0 1 2 32 3 0 13 2 1 0
1 0 3 2
Which I think is wrong. Well, unexpected anyway.
So, next I multiply 2nd permutation in A4 by the first permutation in V4 (which
is the identity permutation).
1{a4 C."(1 _) (0{ v4)
0 2 3 1
This is what I expected. And actually I believe this should be the first
permutation in result1. I cannot understand why result1's ifrst permutation is
the identity, 0 1 2 3.
The root of my problem is, why the discrepancy?
To add to my problem, if I do the multiplication the other way around:
v4 C."(_ 1) a4
I get what I expect: 12 sets of permutations that can be, after removing
duplicates, reduced to 3 sets, i.e. the right cosets of V4 in A4.
> From: [email protected]
> To: [email protected]
> Date: Sun, 13 Jul 2014 07:38:38 +0100
> Subject: Re: [Jprogramming] I just don't C. it.
>
> Thanks,
> I'm worried your lcosets and rcosets are actually the same verb, given that
> the arguments are swapped.
> The idea is that left multiplying (C.) everything by v4 and right multiplying
> everything by v4 will give individually different result, because A4 is not
> commutative, but will partition A4 into the same sets.
>
> Does not the "~" swap a4 and v4 in rcosets, so that it is effectively the
> same as lcosets?
> If so, this is not what I intended.
> a4 lcosets v4 should be the results of doing " for all a in a4, v in v4,
> a*v", where * can be considered as C.
> whereas,a4 rcosets v4 should be the results of doing the same but v*a,
> instead of a*v.
>
> Both your lcosets and rcosets verbs seem to do the same, v C. a.
>
> Unless, that is, my reading of your verbs is wrong.
> Regards.
>
> > From: [email protected]
> > Date: Sat, 12 Jul 2014 22:41:38 -0400
> > To: [email protected]
> > Subject: Re: [Jprogramming] I just don't C. it.
> >
> > I've reduced your notation slightly - playing with alternate expressions
> > helps me understand.
> > Sym=: A.&i.~ !
> > Alt=: ({~ 1 I.@:= C.!.2)@:Sym
> > rcosets=: ~.@:(/:~"2)@:(C."_ 1~)
> > a4=: Alt 4
> > v4=: 0 7 16 23 A.i.4
> > lcosets=: ~.@:(/:~"2)@:(C."_ 1)
> >
> > (a4 rcosets v4) -: (v4 lcosets a4)
> > 1
> >
> > Note that I played with sub expressions (and sometimes looked at result
> > shapes instead of results) while playing with this.
> >
> > It seems obvious now that I think I have things straight, but originally it
> > was not. But let's see if you agree.
> >
> > Thanks,
> >
> > --
> > Raul
> >
> >
> >
> > On Sat, Jul 12, 2014 at 9:16 PM, Jon Hough <[email protected]> wrote:
> >
> > > Thanks for the reply, but my problem still persists.
> > > Please consider the following verbs:
> > >
> > > Sym =: (i. @: !) A. i.
> > >
> > >
> > >
> > >
> > > Alt =: ( (I. @: (1&=) @: (C.!.2)) { ] ) @: Sym
> > >
> > >
> > >
> > >
> > > a4 =. Alt 4
> > >
> > >
> > > a4 is the alternating group on 4 letters.
> > > Now I'll sloppily create v4
> > >
> > >
> > >
> > >
> > > v4 =. 4 4 $ 0 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0
> > >
> > >
> > > v4 (klien four group) is a normal subgroup of a4. This means the left
> > > cosets and right cosets of v4 in a4 should be the same.
> > >
> > >
> > > i.e.
> > >
> > >
> > > if I multiply ( use C.) all elements of a4 with v4, and then nub out
> > > duplicates, a4 will be partitioned into 3 sets of size 4.
> > >
> > >
> > > here is the right coset verb:
> > >
> > >
> > > rcosets =: ~.@:(/:~"2)@:( ] C."_ 1 [ ) NB. a4 should be on left, v4 on
> > > right.
> > >
> > >
> > > Yes the verb is weird, I swap x and y when using C. So we have v4 C."(_ 1)
> > > a4 , and then nub duplicates. I get
> > >
> > >
> > > a4 rcosets v4
> > >
> > >
> > >
> > >
> > > 0 1 2 3
> > > 1 0 3 2
> > > 2 3 0 1
> > > 3 2 1 0
> > >
> > > 0 2 3 1
> > > 1 3 2 0
> > > 2 0 1 3
> > > 3 1 0 2
> > >
> > >
> > > 0 3 1 2
> > > 1 2 0 3
> > > 2 1 3 0
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > 3 0 2 1
> > >
> > >
> > > i.e. I get 3 sets of 4 permutations. Exactly what I wanted.
> > > What I cannot do, is find the equivalent verb for getting left cosets.
> > > i.e.
> > > I need a verb that does:
> > > a4 C."(r1 r2) v4 NB. what ranks r1, r2?
> > > and then nub the resulting duplicate sets ~.@:(/:~"2
> > > And the answer SHOULD be the same as the right cosets.
> > > So, if I try
> > >
> > > lcosets =: ~.@:(/:~"2) @: ( C."(1 _) ) NB. superfluous brackets, I know
> > >
> > >
> > > The reuslt of
> > >
> > >
> > > a4 lcosets v4
> > >
> > >
> > > is simply
> > >
> > >
> > > 0 1 2 31 0 3 22 3 0 1
> > >
> > >
> > >
> > >
> > > 3 2 1 0
> > >
> > >
> > > i.e. it's just v4 again. This makes no sense to me.
> > >
> > >
> > >
> > >
> > > > Date: Sat, 12 Jul 2014 07:24:32 -0400
> > > > From: [email protected]
> > > > To: [email protected]
> > > > Subject: Re: [Jprogramming] I just don't C. it.
> > > >
> > > > The explanation may be the rank of C, as you ask.
> > > >
> > > > C. b. 0
> > > > 1 1 _
> > > > 0 2 3 1 C."1 v
> > > > 0 2 3 1
> > > > 1 3 2 0
> > > > 2 0 1 3
> > > > 3 1 0 2
> > > >
> > > >
> > > >
> > > >
> > > > --
> > > > (B=)
> > > > ----------------------------------------------------------------------
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> > >
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