Solve Directed Graphs Problem 4: Construct Bipartite Graph H

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In summary, we can prove that any tournament on n vertices contains a subgraph which is a transitive tournament on log2n + 1 vertices by induction on n.
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Problem 4. Let k be an integer and let D be a directed graph with the property that
deg+(v) = k = deg-(v) for every v IN V (D). Prove that there exist vertex disjoint directed
cycles C1,...Ck so that SUm of V(Ci) = V (D). (HInt: construct a bipartite graph H from D
so that each vertex in D splits into two vertices in H.)

No idea how to even start...

Def: A transitive tournament is a tournament with no directed cycles. Equivalently, it is
a tournament so that the vertices can be ordered v1; v2,...vn so that (vi, vj) is an edge
whenever i < j.

Problem 6. Let T be a tournament on n vertices. Prove that T contains a subgraph which is a transitive tournament on log2n + 1 vertices.

I have no idea how to do it, i made a ton of graphs with up to 8 vertices and found subgraphs that work, but can't find any way to prove this.


So confused
 
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about this problem.Solution: Let T be a tournament on n vertices. We can prove that T contains a subgraph which is a transitive tournament on log2n + 1 vertices by induction on n. Base Case: When n = 1, the only possible tournament is the empty graph, which has no edges and is thus a transitive tournament on log2n + 1 = 1 vertices. Inductive Step: Assume that the statement is true for some n ≥ 1 and consider a tournament T on n+1 vertices. Since T has n+1 vertices, we can partition the vertices into two sets V1 and V2 of size n1 = ⌊(n+1)/2⌋ and n2 = ⌈(n+1)/2⌉ respectively. By the induction hypothesis, T[V1] contains a subgraph which is a transitive tournament on log2n1 + 1 vertices and T[V2] contains a subgraph which is a transitive tournament on log2n2 + 1 vertices. Now, let T' be the subgraph of T consisting of the edge (u,v) where u ∈ V1 and v ∈ V2. Note that T' has n1 + n2 − 1 = n edges, so it is a tournament on n vertices. Furthermore, any two vertices in T' can be ordered in a transitive order, since the edges in T' are all of the form (u,v) where u ∈ V1 and v ∈ V2. Thus, T' is a transitive tournament on log2n + 1 vertices. Therefore, T contains a subgraph which is a transitive tournament on log2n + 1 vertices, as desired.
 

FAQ: Solve Directed Graphs Problem 4: Construct Bipartite Graph H

1. What is a directed graph?

A directed graph is a type of graph where the edges/connections between nodes have a specific direction. This means that there is a specific flow or relationship between the nodes, rather than just a connection.

2. What is a bipartite graph?

A bipartite graph is a type of graph where the nodes can be divided into two distinct groups or sets, and all edges connect nodes from different sets. In other words, there are no edges between nodes within the same set.

3. How do you construct a bipartite graph?

To construct a bipartite graph, you first need to identify the two distinct sets of nodes. Then, you can draw the graph by connecting nodes from different sets with edges. It is important to ensure that there are no edges between nodes within the same set.

4. What is the purpose of constructing a bipartite graph?

Bipartite graphs are useful for representing relationships between two different types of entities or objects. They can also be used to solve various problems related to matching, scheduling, and allocation.

5. How can I solve Directed Graphs Problem 4: Construct Bipartite Graph H?

To solve this problem, you will need to follow the steps for constructing a bipartite graph. Start by identifying the two distinct sets of nodes and then draw the graph by connecting nodes from different sets. Make sure there are no edges between nodes within the same set. You can also use a computer program or software to help you construct the graph.

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