Author 
Message 
mathlover science forum beginner
Joined: 04 May 2005
Posts: 13

Posted: Wed Jul 19, 2006 11:30 pm Post subject:
A Combinatorics/Graph Theory Question



Hi every body,
There is a problem I have exposed to but, though being badly in need of
an answer, I have not yet been able to solve it. I am not quite sure if
it is better classified as a graph theory problem or a combinatorial
one; anyway, here is the problem:
Assume we have a bipartite graph with X and Y as its two parts. X has
"n" vertices and Y has C(k,n) vertices where "k" is a natural number
less than "n" and by C(k,n) I denote the number of kelement subsets of
an nelement set. The edges of the graph are formed as below: we
correspond each kelement subset of X with a vertex in Y and put an
edge between that vertex of Y and each member of the corresponding
kelement subset of X.
Now it is clear that for every vertex of X there are C(k1, n1)
vertices in Y that have an edge to that vertex. That is every vertex in
X has exactly C(k1, n1) number of neighbors in Y. Now the problem is
as follows: assuming "r" is a natural number not larger than C(k1,
n1) (I am specially interested in the case r=2) determine the minimum
number "p" (or at least a nontrivial upper bound on it) such that for
every pelement subset, like S, of Y the following property holds: for
every node in X, like "v", it has at least "r" neighbors which are
members of S.
Any help or clues are greatly appreciated.
Thanks. 

Back to top 


Brian M. Scott science forum Guru
Joined: 10 May 2005
Posts: 332

Posted: Thu Jul 20, 2006 8:32 pm Post subject:
Re: A Combinatorics/Graph Theory Question



On 19 Jul 2006 16:30:16 0700, mathlover
<immathlover@yahoo.com> wrote in
<news:1153351816.624884.222030@i3g2000cwc.googlegroups.com>
in alt.math.undergrad:
Quote:  Hi every body,
There is a problem I have exposed to but, though being
badly in need of an answer, I have not yet been able to
solve it. I am not quite sure if it is better classified
as a graph theory problem or a combinatorial one; anyway,
here is the problem:
Assume we have a bipartite graph with X and Y as its two
parts. X has "n" vertices

X has n vertices, not "n" vertices: when you enclose the
letter in quotation marks, you're talking about the symbol
itself, not what it represents. The same goes for all of
the other places where you've used quotation marks in this
way.
Quote:  and Y has C(k,n) vertices where "k" is a natural number
less than "n" and by C(k,n) I denote the number of
kelement subsets of an nelement set.

That's backwards from the standard oneline notation, in
which n choose k is written C(n, k).
Quote:  The edges of the graph are formed as below: we correspond
each kelement subset of X with a vertex in Y and put an
edge between that vertex of Y and each member of the
corresponding kelement subset of X.
Now it is clear that for every vertex of X there are
C(k1, n1)

C(n1, k1) here and later.
Quote:  vertices in Y that have an edge to that
vertex. That is every vertex in X has exactly C(k1, n1)
number of neighbors in Y. Now the problem is as follows:
assuming "r" is a natural number not larger than C(k1,
n1) (I am specially interested in the case r=2)
determine the minimum number "p" (or at least a
nontrivial upper bound on it) such that for every
pelement subset, like S, of Y the following property
holds: for every node in X, like "v", it has at least "r"
neighbors which are members of S.

So X is an arbitrary set with cardinality n. Without loss
of generality Y = [X]^k, where [X]^n is a standard notation
for the set of subsets of X of cardinality k. For each x in
X and y in Y, {x, y} is an edge of the graph G iff x in y.
For each x in X, deg(x) = C(n1, k1), and for each y in Y,
deg(y) = k.
Say that a subset S of Y is rdense if each x in X has at
least r neighbors in S. You want to know the minimum
cardinality of an rdense subset of Y; I'll denote this by
p(n, k, r).
Quote:  Any help or clues are greatly appreciated.

Here's a start.
Suppose that S is rdense. Then each element of X belongs
to at least r members of S, so S >= nr/k.
Now suppose that n is a multiple of k^r, say n = m*k^r.
Then without loss of generality we can take X to be
{(i_0, ..., i_r) : 1 <= i_0 <= m, 1 <= i_j <= k for j =
1, ..., r}. For each x = (i_0, ..., i_r) in X and j with
1 <= j <= r let y(x, j) be the set of points in X that
differ from x in at most the jth coordinate; clearly
x in y(x, j) in Y. Moreover, if 1 <= j' <= r with j' != j,
then y(x, j') != y(x, j), so {y(x, j) : 1 <= j <= r} is a
collection of r members of Y, each of which contains x. Let
S = {y(x, j) : x in X & 1 <= j <= r}; we've just seen that S
is rdense. Finally, S = m*r*k^(r1) = nr/k, and it
follows that p(n, k, r) = nr/k when n is a multiple of k^r.
To answer the question completely, therefore, it suffices to
determine p(n, k, r) when n < k^r. I suspect that it's
ceil(nr/k), where ceil is the ceiling function (i.e.,
ceil(x) is the least integer greater than or equal to x),
but I've not had time to think about it seriously.
Brian 

Back to top 


Google


Back to top 



The time now is Fri Oct 20, 2017 4:02 pm  All times are GMT

