# Sol2 - AMS 301 Spring 2011 Homework Set 2 Solution Notes 4 Supplement II A graph with n vertices can have at most n = n(n2 1 vertices this would be

This preview shows pages 1–2. Sign up to view the full content.

AMS 301 Spring, 2011 Homework Set # 2 — Solution Notes # 4, Supplement II: A graph with n vertices can have at most ( n 2 ) = n ( n - 1) 2 vertices; this would be the case of a complete graph, K n , in which all n vertices have degree ( n - 1). Thus, if we have 3 vertices, there are at most 3 edges, if we have 4 vertices, there are at most 6 edges, etc. If we have n = 10, there are at most ( 10 2 ) = 45 edges, so it is impossible to have a 10-vertex graph with 50 edges. (NOTE: We are assuming here that it is a simple graph, not a multigraph; in a multigraph, even if there is only ONE vertex, it is possible to have 50 (or 50,000) edges — just make them all self-loops!) If we have n = 11, there are at most ( 11 2 ) = 55 edges (and K 11 has exactly 55 edges). Thus, with 11 vertices it is possible to have 50 edges (e.g., take K 11 and delete any 5 edges), and it is not possible to have 50 edges if there are fewer than 11 vertices. #6, 1.3: The number of edges in

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 05/28/2011 for the course AMS 301 taught by Professor Estie during the Spring '11 term at University of Florida.

### Page1 / 2

Sol2 - AMS 301 Spring 2011 Homework Set 2 Solution Notes 4 Supplement II A graph with n vertices can have at most n = n(n2 1 vertices this would be

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online