Fresh features from the #1 AI-enhanced learning platform.Try it free
Fresh features from the #1 AI-enhanced learning platformCrush your year with the magic of personalized studying.Try it free
Question

Show that in a tournament (defined preceding mentioned theorem) it is always possible to rank the contestants so that the person ranked ith beats the person ranked (i +1)st. (Hint: Use the mentioned theorem.)

Solution

Verified
Answered 1 year ago
Answered 1 year ago
Step 1
1 of 2

A tournament is a directed graph G=(V,E)G = (V,E) whose underlying (undirected) graph is KnK_n. We will prove the claim by induction. For n=2n = 2 the graph has only one directed edge (x,y)(x,y). Then we rank xx as 1st and yy as 2nd. Suppose that the claim holds for some nNn \in \mathbb{N} and let us prove it for n+1n+1. Let xVx \in V be any vertex. If we remove xx from GG, we get a tournament GG' with n1n-1 vertices. By the inducitive assumption, the vertices of GG' can be ranked (or renamed) as 11, 22, ..., nn such that ii beats i+1i+1 (that means that GG' contains the directed edge (i,i+1)(i,i+1)).

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Discrete Mathematics and Its Applications 7th Edition by Kenneth Rosen

Discrete Mathematics and Its Applications

7th EditionISBN: 9780073383095Kenneth Rosen
4,283 solutions
Discrete Mathematics 8th Edition by Richard Johnsonbaugh

Discrete Mathematics

8th EditionISBN: 9780321964687 (2 more)Richard Johnsonbaugh
4,246 solutions
Applied Combinatorics 6th Edition by Alan Tucker

Applied Combinatorics

6th EditionISBN: 9780470458389Alan Tucker
844 solutions
Discrete Mathematics and Its Applications 8th Edition by Kenneth Rosen

Discrete Mathematics and Its Applications

8th EditionISBN: 9781259676512Kenneth Rosen
4,397 solutions

More related questions

1/4

1/7