TCSS 321 Discrete Mathematics

Question:

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be a fixed positive integer.

>=

Hall’s condition is satisfied.]

Hall’s theorem can be proved by using the generalized Menger’s theorem. (See Theorem 2.9)

Consider the following:

>= k

(1) Show that G contains stars on the k + 1 vertex vertices that saturate.

A star on G is a graph with k vertex points of degree 1 that are joined to a vertex in degree k.

Do you think that G is an N-vertex graph of maximum degree?

(G) with no vertex of degree 0.

Let’s take as an example a 5-regular graph, which is a graph in which all vertices possess degree 5, but that does not have a 1-factor.

(Graduate Exercise) Use Tutte’s Theorem to prove Hall’s Theorem.

Answer:

Be sure to use LaTeX to write your answers.

Help with creating your HW file can be found at hw-template.tex

You can also create a separate page for each solution.

You can use the command pagebreak to force page break.

be a fixed positive integer.

Consider that for each set A, we have

– d. [hint: convert G into a graph that satisfies Hall’s condition.]

>=

Add d vertices to B. Each one should be connected to every vertice in A.

Next, let G0 be a new graph.

>=

>= (

/2 – d edges from G. Hence G contains a matching of size

Hall’s theorem can be proved by using the generalized Menger’s theorem. (See Theorem 2.9)

>=

G is a bipartite diagram with vertex classes X, Y.

Add two new vertices a, b, to G and join a all elements of X and b all elements Y.

Let G0 represent the graph that was obtained this way.

Let C be a set vertices which separate a from b at G0.

=

>=

independent paths between a and b, this paths induce a matching in G.

Consider the following:

>= k

Show that G contains stars on k+1 vertices which saturate a collection.

A star on G is a graph with k vertex points of degree 1 that are joined to a vertex in degree k.

Let G be a graph bipartite with bipartition in V1, V2, and let M represent the maximum match of G (Hall’s condition).

You can then denote U the set M that is not saturated vertices in the V1, and Z the set M-alternating paths connecting vertices to U.

Then, set S=Z/V1 & T=Z/V2, and as in the half-theorem, we can see that every vertex of T is M saturated and?

(s) =T so G contains a collection star-shaped vertices on k+1 that saturate. A graph with k vertexes of degree 1 joined to a vertex at k is called a star graph on k+1.

Do you think that G is an N-vertex graph of maximum degree?

(G) with no vertex of degree 0.

G is an nvertex graph with the maximum degree.

If (G) has no vertex of the degree 0, then the upper limit is immediately and clearly sharp.

We use induction to find the size m of a graph connected to verify the lower limit. If m = 2, the lower bound follows.

The lower bound is assumed to hold for connected graphs of positive sizes less than k. Let k>=2 be the k-value and let G be a connected diagram of order n with a size k+1.

If G has a cycl edge e, then

b1(1)(G) >= or b1(1)(G-e >= ; otherwise, G is a tree.

If G=K1, n-1 then G contains

If GK1, N-1, is G, then G contains an edge (e) such that (Ge-e) has two nontrivial elements G1 and G2.

We have the following:

This means that b1(1) = b1(1) >= b1(2) =

Let’s take as an example a 5-regular graph, which is a graph in which all vertices possess degree 5, but that does not have a 1-factor.

Solution

Hall Theorem – A bipartite graph with partition (A,B) has a matching of

A =S,?N(S),?

T V (H).

and H has a 1-factor (H is graph obtained from G by adding one vertex to Y if V (G) is odd and then adding the edges of a clique (=a full graph) on the vertices of Y ) it follows that G satisfies Hall’s condition.

Assuming H is a 1-factor, (i.e.

Let M be the edges that occur with vertices in X when H has a 1-factor (i.e.

=

Therefore, the edges M are still edges in H and match every vertex of X to some vertex of Y.

While there may be some vertices that are not matched by Y, H constructed the graph that enables us to complete our matching.

Let us suppose that G meets Hall’s condition for subsets SXX, and let T V(H).

vertices left – each a connected component.

>=

Because Y 6 T is a clique, there is one connected component of H T that contains all the vertex Y T. Let S equal X V (B).

<= is even we are therefore done. or Particularly, we've done it again. >=

and another matching of size

References

5-regular graphs have a 3-colorable, positive probability.

Algorithms: ESA 2008 (Brodal & Leonardi, eds.

Random graphs: Wiley, New York.

M (2010).

The threshold values, stability analysis and high Q asymptotic for coloring problems on random graphs, Physics.

Rev.

Random K-Satisfiability. From an Analytic solution to a new efficient algorithm.

Rev.

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