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Chapter 6: Problem 18

Prove that a tournament with no cycles is transitive.

Short Answer

Expert verified
A tournament without cycles is transitive because any sequence of directed edges would form a cycle unless the third edge completes the transitive property.

Step by step solution

01

Define a Tournament

A tournament is a directed graph where there is exactly one directed edge between every pair of vertices. Therefore, for any two vertices \(u\) and \(v\), either \((u, v)\) or \((v, u)\) exists, but not both.
02

Define a Cyclic and Transitive Tournament

A tournament is called transitive if for any three vertices \(u, v,\) and \(w\), if \((u, v)\) and \((v, w)\) exist, then \((u, w)\) must also exist. A cycle means there is a sequence \(u_1, u_2, ..., u_k\) such that \((u_i, u_{i+1})\) exists for all \(i\), and \((u_k, u_1)\) also exists.
03

Assume No Cycle Exists

Assume the tournament graph has no cycles. This means there is no sequence of edges forming a loop such that you can return to the starting vertex through directed edges.
04

Establish Transitivity

To prove that the tournament is transitive, we must show that for any three distinct vertices \(u, v,\) and \(w\), if \((u, v)\) and \((v, w)\) exist, then \((u, w)\) exists. If \((u, w)\) did not exist, there would exist a \(w, u\) edge, which combined with \((u, v)\) and \((v, w)\) would form a cycle \((u, v, w, u)\). This contradicts the assumption of no cycles, therefore \((u, w)\) must exist, establishing transitivity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transitive Tournament
Imagine a tournament as a special type of directed graph where each pair of vertices is connected by exactly one directed edge. This ensures that between any two nodes, there's a clear-cut winner and loser. Transitivity comes into play when you have three nodes: if the graph shows a winner between node A and B, and B and C, then there should logically be a winner between A and C as well. This is what makes the tournament transitive.
  • If A beats B, and B beats C, then A must beat C for transitivity.
  • Transitive tournaments resemble actual competitive tournaments where clear victors emerge.
  • Without cycles, there's always a clear hierarchy or ranking among the competitors.
This clear hierarchical structure ensures that no cycle or loophole contradicts who is deemed the winner.
Directed Graph
A directed graph, often abbreviated as digraph, contains vertices connected by edges that have a direction. You can think of it as a one-way street, where each edge has a head and a tail, showing direction from one node to another.
  • In the context of tournaments, this means each pair of competitors has a clear winner.
  • This directional nature helps in determining straightforward relationships among vertices.
  • By analyzing directed edges, it becomes easier to see patterns of dominance or hierarchy.
All tournaments are directed graphs by nature. This means one competitor beats another, never ending in a tie or going both ways. It's crucial to establish direction to uphold the rules without ambiguity.
Cyclic Tournament
Cyclic tournaments occur when there's a cycle among the vertices. This means you can start at one vertex, follow the directed edges, and end up back at the original vertex without retracing any step in reverse.
  • Cyclic behavior implies that there's no clear winner circle.
  • Every participant in a cycle can outdo another, creating a closed loop.
  • Breaking the cycle is essential to knowing who truly outranks another without contradiction.
In tournaments without cycles, no set of participants can return to the start via directed edges without creating contradicting paths.
Removing cycles simplifies the tournament, allowing for a clear resolution and ranking of competitors.

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Most popular questions from this chapter

Draw a minimal depth decision tree that represents sorting four elements from a linearly ordered set. Only five levels are required.

Represent by a digraph the partial order divides defined on the integers 0 through \(11 .\)

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Prove that a connected undirected graph is orientable if and only if each edge is contained in a cycle.

Show that 1,2,2,3,4 is graphical but that 1,3,3,3 is not. Prove the theorem of Havel-Hakimi that for \(n \geq 1,\) the sequence \(d_{1}, d_{2}, \ldots, d_{n}\) is graphical if and only if \(d_{1}, d_{2} \ldots \ldots d_{n-d_{n}}, d_{n-d_{n}+1}-1, \ldots, d_{n-1}-1\) is graphical.
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