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Remember to prepare your answers in LaTeX. Refer to hw-template.tex for help in preparing your HW file. Also, please create an individual page for each solution. Use the command pagebreak to force page breaks.
1. Let G be a bipartite graph with classes A and B and let d ≤ |A| be a fixed positive integer. Suppose that for every set S ⊂ A we have |N(S)| ≥ |S| − d.Prove that G contains a matching of size |A| − d. [hint: convert G into a graph that
satisfies Hall’s condition.]
2. Using the generalized version of Menger’s theorem (see Theorem 2.9) prove Hall’s theorem.
3. Suppose that instead of Hall’s condition we have the following condition for some positive integer k:

For every vertex subset S ⊆ A we have |N(S)| ≥ k|S|. (1) Show that G contains a collection of stars on k + 1 vertices that saturate A. A star on k + 1 is a graph with k vertices of degree 1 all joined to a vertex of degree k.
4. Prove that if G is an n-vertex graph with maximum degree ?(G) and no vertex of degree 0, then
5. Give an example of a 5-regular graph (i.e., a graph where all vertices have degree 5) that has no 1-factor.
6. (Graduate exercise) Use Tutte’s theorem to prove Hall’s theorem.

## Hall's condition and matching in bipartite graphs

Remember to prepare your answers in LaTeX. Refer to hw-template.tex for help in preparing your HW file. Also, please create an individual page for each solution. Use the command pagebreak to force page breaks.

1. Let G be a bipartite graph with classes A and B and let d ≤ |A| be a fixed positive integer. Suppose that for every set S ⊂ A we have

|N(S)| ≥ |S| − d.

Prove that G contains a matching of size |A| − d. [hint: convert G into a graph that satisfies Hall’s condition.]

Solution.

Let V (G) = A ∪ B be a bipartition of G, with |A| ≥ |B|. Add d new vertices to B, each connected to all vertices in A. Then let G0 be the new graph. Then G0 has |NG0(S)| ≥ |S| for every S ⊂ A (S has at least |S| − d neighbors from G, and is connected to the d new vertices). By Hall’s Theorem, G0 has a matching for A, which has |A| ≥ (|A| + |B|)/2 = |V (G)|/2 edges. At most d of these edges contain a new vertex of G0, which leaves at least |V (G)|/2 − d edges from G. Hence G contains a matching of size |A| − d hence proven.

1. Using the generalized version of Menger’s theorem (see Theorem 2.9) prove Hall’s theorem.

Solution.

Hall’s theorem, that is, |N(S)| ≥ |S| ∀S ⊂ X.

By letting G be a bipartite graph with vertex classes X and Y. We add two new vertices a and b to G, and join a to all elements of X, and b to all elements of Y. Let G0 be the graph obtained in this way.

Let C be a set of vertices separating a from b in G0. Then N(X C) ⊆ Y ∩ C. Since |C| = |C ∩ X| + |C ∩ Y |, we have that |C| ≥ |C ∩ X| + |N(X C)|. By the condition in Hall’s theorem, we have that |N(X C)| ≥ |X C|, so |C| ≥ |C ∩ X| + |X C| = |X|.

Thus, by Menger’s theorem, there are |X| independent paths between a and b, this paths induce a matching in G.

1. Suppose that instead of Hall’s condition we have the following condition for some positive integer k:

For every vertex subset S ⊆ A we have |N(S)| ≥ k|S|. (1)

Show that G contains a collection of stars on k + 1 vertices that saturate A. A star on k + 1 is a graph with k vertices of degree 1 all joined to a vertex of degree k.

Let G be a bipartite graph with bipartition (V1, V2) and let M be a maximum matching of G (Hall’s condition). Then by denoting U the set of M which is unsaturated vertices in V1, and denoting Z the set of all vertices connected by M-alternating paths to vertices of U.

## Converting a graph to satisfy Hall's condition

Set S=Z?V1 and T=Z?V2, then as in the half theorem, we have that every vertex in T is M- saturated and ?(s) =T thus G contains a collection of stars on k + 1 vertices that saturate and a star on k + 1 is a graph with k vertices of degree 1 all joined to a vertex of degree k.

1. Prove that if G is an n-vertex graph with maximum degree ?(G) and no vertex of degree 0, then

If G is an n-vertex graph with maximum degree ? (G) and no vertex of degree 0, then then the upper bound is immediate and clearly sharp. In order to verify the lower bound, we employ induction on the size m of a connected graph: if m ≤ 2 then the lower bound follows.

By assuming that the lower bound holds for all connected graphs of positive sizes not exceeding k, where k≥2 and letting G be a connected graph of order n having a size k+1.

If G has a cycle edge e, then;

β1(G) ≥ β1(G-e) ≥ , otherwise G is a tree.

If G=K1, n-1, then G contains

If G≠K1, n-1, then G contains an edge e such that (G-e) has two nontrivial components G1and G2.

By letting ni denote the order of Gi, i=1, 2 and apply the induction hypothesis to G1 and G2 we have;

β1(G) ≥ β1(G1) ≥ β1(G2) ≥  +

This implies that β1(G) ≥ β1(G1) ≥ β1(G2) =  .

1. Give an example of a 5-regular graph (i.e., a graph where all vertices have degree 5) that has no 1-factor.

Solution

Hall Theorem: A bipartite graph G with partition (A, B) has a matching of;

A ⇔∀S ⊆ A, ?N(S) ? ≥ ?S?

Tutte’s Theorem: A graph G has a 1-factor o(H T) ≤ |T| ∀T ⊂ V (H).

If G has a matching of size |X| 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 that H has a 1-factor (i.e. a perfect matching) and let M be the edges in this matching that are incident with vertices in X. In the construction of the graph H the edges incident with X did not change so the same set of edges is a matching of size |M| = |X| also in the original graph G. Conversely if there is a matching M of size |X| in G then this matching has to touch every vertex in X. Thus the edges of M are still edges in H, matching every vertex in X to some vertex in Y. There might be some vertices in Y that are not matched by M, but the construction of H made sure that the graph induced by these vertices is a clique on an even number of vertices, enabling us to complete the matching.

Also, assuming that G satisfies Hall’s condition for subsets S ⊂ X and let T ⊂ V (H). If Y ⊂ T then there are at most |X| vertices left - each a connected component. But by our assumption it is clear that |Y | ≥ |X| so that we are okay. Assume therefore that Y 6⊂ T, since Y forms a clique in H there is one connected component B of H T containing all the vertices Y T. Let S = X V (B). By construction N(S) ⊂ T so that by assumption |S| ≤ |T|.

The connected components of H T are exactly B and the separate vertices of S. If |V (B)| is even we are therefore done. If not write the vertices of H as a disjoint union V (H) = S tT tV (B), since the total number of vertices is even either |S| or |T| is even and the other is odd. In particular we have  and we are done again.

Now by assuming the condition of Hall’s marriage theorem, namely that |N(S)| ≥ |S| for every S that is contained either in X or in Y. By the above two paragraphs we have a matching of size |X| and another matching of size |Y | in the bipartite G. This means that |X| = |Y | and hence both of these matchings are perfect hence proof.

References

D´?az, G., & Grammatikopoulos, A., & Kaporis, L., & Kirousis, X., & Perez, D., & Sotiropoulos (2008).  5-regular graphs are 3-colorable with positive probability. In Algorithms - ESA 2008 (Brodal and Leonardi, eds.), pp. 215–225. LNCS 3669, Springer.

Janson, T., & Luczak, B., & Rucin´ski, A (2011). Random Graphs: Wiley, New York.

Krza¸ka La., & Pagnani B., & Weight. M (2010). Threshold values, stability analysis and high-q asymptotic for the coloring problem on random graphs, Phys. Rev. E 70, 04678.

M´ezard M., & Zecchina, R (2008). Random K-Satisfiability: From an Analytic Solution to a new efficient algorithm, Phys. Rev. E 66, 056126.

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