26.1.2.2.2. Case 2: k-sized Subset with Exactly k Neighbours

So, my case 2 is talking about a possibility where in my graph G, I have a subset A of k nodes which has exactly k neighbours in the subset V. So, again for demonstration here I am taking the case of k = 3.

In this scenario, we are examining a subset A of size k from a bipartite graph G. This subset has exactly k neighbours in another part of the graph, denoted as V. For example, if k equals 3, that means we look at a group of three nodes in A, which precisely connects to three nodes in V.

Imagine three friends who have exactly three favorite restaurants. Each friend chooses exactly one restaurant to go to. In this context, the friends are the nodes in subset A, and the restaurants are the nodes in subset V.

Even though this condition is true, as part of that condition you have a subset A of k nodes in V which has exactly k neighbours in V. This does not violate the condition (|N(A)| ≥ |A|).

This scenario does not violate Hall's condition since having exactly k neighbours for k nodes still satisfies the requirement that the number of neighbours is greater than or equal to the number of nodes. In other words, the equality condition still upholds the theorem's necessary conditions.

Consider a class of three students, where each student has exactly one study partner. Each study partner corresponds to a different subject. This is similar to how our subset A of three students ties exactly to three subjects, fulfilling the overall requirements.

Since |A| = k, I can use my inductive hypothesis and since the number of neighbours of S is as large as the number of nodes in S, a complete matching is there. So, call that complete matching as M.

With the assumption that every subset of size k has as many neighbours as nodes, we can apply our inductive hypothesis. This allows us to infer that a complete matching exists for the nodes involved. The complete matching M connects these nodes perfectly, meaning every node in subset A can be matched to a different node in subset V.

Think about organizing a little league sports team. If you have exactly three players (k) and exactly three positions (neighbours), you can match each player to a unique position, ensuring a complete assignment and functioning team.

Now this is a complete matching from S to T not from V to V.

Here, the current matching only considers a smaller part of the original graph, connecting S (part involving nodes) to T (their respective neighbours). This doesn't yet consider the entire set of nodes in V; we're reducing our problem to apply our hypothesis effectively.

This is akin to completing only one set of tasks for a project (like answering questions for a survey) rather than the entire project scope, with the idea that successfully completing the smaller task means we can address the larger one later.

Now if I take the union of the matching M from the subset S to the subset T and the matching M’ which is a complete matching from V ’ to V ’, that will ensure a complete matching from V to V.

Once we form two matchings—one from our smaller set of nodes S to their neighbours T, and the other for remaining unmatched nodes—we combine both to establish a comprehensive matching across the entire graph. This final step ensures that all nodes in V are paired, thus fulfilling the conditions of Hall's theorem.

Imagine completing your team assignments (S to T) first while ensuring you cover all positions. Then, you find additional roles for any remaining individuals. Together, all the elements and roles are filled seamlessly, ensuring no one is left out.

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