Proving Matching Size In Bipartite Graphs

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.

Save Time On Research and Writing
Hire a Pro to Write You a 100% Plagiarism-Free Paper.
Get My Paper
  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.

Save Time On Research and Writing
Hire a Pro to Write You a 100% Plagiarism-Free Paper.
Get My Paper
  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.

Calculate your order
Pages (275 words)
Standard price: $0.00
Client Reviews
4.9
Sitejabber
4.6
Trustpilot
4.8
Our Guarantees
100% Confidentiality
Information about customers is confidential and never disclosed to third parties.
Original Writing
We complete all papers from scratch. You can get a plagiarism report.
Timely Delivery
No missed deadlines – 97% of assignments are completed in time.
Money Back
If you're confident that a writer didn't follow your order details, ask for a refund.

Calculate the price of your order

You will get a personal manager and a discount.
We'll send you the first draft for approval by at
Total price:
$0.00
Power up Your Academic Success with the
Team of Professionals. We’ve Got Your Back.
Power up Your Study Success with Experts We’ve Got Your Back.