Subst in combstruct

[Ralf Hemmecke]

It seems there were some errors so that the following mail did not made
it to the list. :-(
Martin has already answered, but I would like to record that mail.

Ralf

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Subject: Subst in combstruct
From: Ralf Hemmecke <ralf-***@public.gmane.org>
Date: Sat, 17 Mar 2007 21:55:10 +0100
To: aldor-combinat-devel <aldor-combinat-devel-5NWGOfrQmneRv+***@public.gmane.org>

I was trying out Maple-combstruct.

Forgetting that I could not say P=Subst(Set(Atom),Set(Atom,card>0) (or
rather with the first and second argument exchanged) to get set
partitions, I am still puzzled about the number of unlabeled structures.
It seems that for them the two unlabled partitions

3, 1, Set(Set(A), Set(A, A))
3, 4, Set(Set(A, A), Set(A))

are different things (which is also reflected in the counting series.
Maybe I haven't read the documentation clearly, but I would not have
expected this.

BTW, the help page combstruct-specifications says

labeled structure:
F=Set(Set(Z,card>=1)) set partitions

unlabeled structure
A=Set(Sequence(Z,card>=1)) integer partition

That somehow doesn't look like integer partitions are isomorphism types
of set partitions.

Any comments?

Nicolas, you mentioned that MC simulates combstruct. Does MC also have
such a strange behaviour with respect to substitution?

Ralf

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for i from 0 to 5 do
print(count([P,sys,labeled], size=i), count([P,sys,unlabeled],

size=i));

od;

0, 0
1, 1
2, 2
5, 4
15, 8
52, 16

for i from 1 to 3 do n:=0:for s in allstructs([P,sys,labeled],

size=i) do n:=n+1:print(i,n,s) od;od;
n := 0
1, 1, Set(Set(A[1]))
n := 0
2, 1, Set(Set(A[1]), Set(A[2]))
2, 2, Set(Set(A[1], A[2]))
n := 0
3, 1, Set(Set(A[3]), Set(A[1], A[2]))
3, 2, Set(Set(A[1], A[3], A[2]))
3, 3, Set(Set(A[1], A[3]), Set(A[2]))
3, 4, Set(Set(A[1]), Set(A[2]), Set(A[3]))
3, 5, Set(Set(A[1]), Set(A[3], A[2]))

for i from 1 to 3 do n:=0:for s in allstructs([P,sys,unlabeled],

size=i) do n:=n+1:print(i,n,s) od;od;
n := 0
1, 1, Set(Set(A))
n := 0
2, 1, Set(Set(A), Set(A))
2, 2, Set(Set(A, A))
n := 0
3, 1, Set(Set(A), Set(A, A))
3, 2, Set(Set(A, A, A))
3, 3, Set(Set(A), Set(A), Set(A))
3, 4, Set(Set(A, A), Set(A))


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