Re: Does this problem have a name?
- From: "Ted Hopp" <ted@xxxxxxxxxxxxxxx>
- Date: Thu, 27 Mar 2008 12:52:19 -0400
You're correct. Only the "if" part makes sense. Thanks for pointing this
out. I'm still looking for some way to efficiently find a minimal valid
partition pair.
Ted Hopp
"Chip Eastham" <hardmath@xxxxxxxxx> wrote in message
news:64ed8262-0e79-4fc7-a988-075451856220@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 26, 8:31 pm, "Hopp Family" <ada...@xxxxxxxxx> wrote:
Given: Finite sets D and S, and a function f : D x D -> S.
Definition: Let a "valid partition pair" be two (not necessarily
distinct)
partitions U and V of D such that there exists a function g : U x V -> S
where g(u,v) = f(d,e) iff d is an element of u and e is an element of v.
(Letting U and V be the identity partition shows that there always exists
at
least one valid partition pair.)
I believe there is a problem with your claim that
the "identity" partitions always give a "valid
partition pair" (relative to map f:DxD --> S).
Consider S = D = {0,1} as Z/2Z, and define the
natural map f(x,y) = x+y mod 2. According to
your definition, for U = V = {{0},{1}} to be a
valid partition pair, we would need g:UxV --> S
such that:
g(u,v) = f(d,e) if and only if d in u & e in v
Clearly g({0},{0}) and g({1},{1}) would both
have to be f(0,0) = f(1,1) = 0 mod 2 in order
to meet the "if" part of the definition above,
but this contradicts the "only if" part since
0 is not in {1} and 1 is not in {0}.
Perhaps you meant to write "if" rather than
"iff" in your definition of a valid partition
pair? In that case the partitions U,V can be
considered "quotient spaces" of the underlying
sets D and D (which happen to be equal, but
this is not an essential feature of the
construction.
regards, chip
.
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