What is wrong with the following "proof"? In addition to finding a counterexample, you should explain what is fundamentally wrong with this approach, and why it demonstrates the danger of build-up error. False Claim: If every vertex in an undirected graph has degree at least 1, then the graph is con- nected. Proof? We use induction on the number of vertices n ≥ 1. Base case: There is only one graph with a single vertex and it has degree 0. Therefore, the base case is vacuously true, since the if-part is false. Inductive hypothesis: Assume the claim is true for some n ≥ 1. Inductive step: We prove the claim is also true for n + 1. Consider an undirected graph on n vertices in which every vertex has degree at least 1. By the inductive hypothesis, this graph is connected. Now add one more vertex x to obtain a graph on (n+1) vertices, as shown below. n-vertex graph All that remains is to check that there is a path from x to every other vertex z. Since x has degree at least 1, there is an edge from x to some other vertex; call it y. Thus, we can obtain a path from x to z by adjoining the edge {x,y} to the path from y to z. This proves the claim for n + 1.

What is wrong with the following "proof"? In addition to finding a counterexample, you should explain what is fundamentally wrong with this approach, and why it demonstrates the danger of build-up error. False Claim: If every vertex in an undirected graph has degree at least 1, then the graph is con- nected. Proof? We use induction on the number of vertices n ≥ 1. Base case: There is only one graph with a single vertex and it has degree 0. Therefore, the base case is vacuously true, since the if-part is false. Inductive hypothesis: Assume the claim is true for some n ≥ 1. Inductive step: We prove the claim is also true for n + 1. Consider an undirected graph on n vertices in which every vertex has degree at least 1. By the inductive hypothesis, this graph is connected. Now add one more vertex x to obtain a graph on (n+1) vertices, as shown below. n-vertex graph All that remains is to check that there is a path from x to every other vertex z. Since x has degree at least 1, there is an edge from x to some other vertex; call it y. Thus, we can obtain a path from x to z by adjoining the edge {x,y} to the path from y to z. This proves the claim for n + 1.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

What is wrong with the following "proof"? In addition to finding a counterexample, you should
explain what is fundamentally wrong with this approach, and why it demonstrates the danger of
build-up error.
False Claim: If every vertex in an undirected graph has degree at least 1, then the graph is con-
nected.
Proof? We use induction on the number of vertices n ≥ 1.
Base case: There is only one graph with a single vertex and it has degree 0. Therefore, the base
case is vacuously true, since the if-part is false.
Inductive hypothesis: Assume the claim is true for some n ≥ 1.
Inductive step: We prove the claim is also true for n + 1. Consider an undirected graph on n vertices
in which every vertex has degree at least 1. By the inductive hypothesis, this graph is connected.
Now add one more vertex x to obtain a graph on (n+1) vertices, as shown below.
n-vertex graph
All that remains is to check that there is a path from x to every other vertex z. Since x has degree
at least 1, there is an edge from x to some other vertex; call it y. Thus, we can obtain a path from x
to z by adjoining the edge {x,y} to the path from y to z. This proves the claim for n + 1.

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