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In Tutorial we prove the claim that every directed acyclic graph (DAG) has a vertex v out-degree(v)=0. with In the lecture on paths, cycles and trees (week 8) we proved a similar claim for undirected graphs, namely that every undirected graph has a vertex v with either degree(v)=0 OR degree(v)=1. Why is the proof for DAGS able to be more specific than for undirected graphs? (Why can we guarantee a out-degree O vertex for DAGS but not necessarily a degree-O vertex for undirected graphs?) Select the option that best describes why. a) Because one is directed and one is undirected. Ob) Both proofs construct maximal simple paths ugu₁... Uk. In the case when k > 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex u but undirected graphs are not because {uk, Uk-1} is an edge in the undirected graph, but (Uk, Uk-1) is not an edge in the DAG. Oc) Both proofs construct maximal simple paths U₁・・・ Uk. In the case when k > 0, directed acyclic graphs are able to guarantee there is no edge to a previous vertex u; in the simple path for 0 ≤ j≤ k − 2, but undirected graphs are not. d) Both proofs construct maximal simple paths uu₁ ..Uk. In the case when k = 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex Un but undirected graphs are not.

In Tutorial we prove the claim that every directed acyclic graph (DAG) has a vertex v out-degree(v)=0. with In the lecture on paths, cycles and trees (week 8) we proved a similar claim for undirected graphs, namely that every undirected graph has a vertex v with either degree(v)=0 OR degree(v)=1. Why is the proof for DAGS able to be more specific than for undirected graphs? (Why can we guarantee a out-degree O vertex for DAGS but not necessarily a degree-O vertex for undirected graphs?) Select the option that best describes why. a) Because one is directed and one is undirected. Ob) Both proofs construct maximal simple paths ugu₁... Uk. In the case when k > 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex u but undirected graphs are not because {uk, Uk-1} is an edge in the undirected graph, but (Uk, Uk-1) is not an edge in the DAG. Oc) Both proofs construct maximal simple paths U₁・・・ Uk. In the case when k > 0, directed acyclic graphs are able to guarantee there is no edge to a previous vertex u; in the simple path for 0 ≤ j≤ k − 2, but undirected graphs are not. d) Both proofs construct maximal simple paths uu₁ ..Uk. In the case when k = 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex Un but undirected graphs are not.

Database System Concepts
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
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In Tutorial we prove the claim that every directed acyclic graph (DAG) has a vertex v out-degree(v)=0. with In the lecture on paths, cycles and trees (week 8) we proved a similar claim for undirected graphs, namely that every undirected graph has a vertex v with either degree(v)=0 OR degree(v)=1. Why is the proof for DAGS able to be more specific than for undirected graphs? (Why can we guarantee a out-degree 0 vertex for DAGS but not necessarily a degree-O vertex for undirected graphs?) Select the option that best describes why. O a) Because one is directed and one is undirected. Ob) Both proofs construct maximal simple paths uu₁... Uk. In the case when k > 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex Ubut undirected graphs are not because {Uk, Uk-1} is an edge in the undirected graph, but (uk, Uk-1) is not an edge in the DAG. O c) Both proofs construct maximal simple paths U₁... Uk. In the case when k > 0, directed acyclic graphs are able to guarantee there is no edge to a previous vertex u, in the simple path for 0 ≤ ≤ k − 2, but undirected graphs are not. d) Both proofs construct maximal simple paths uu₁... Uk. In the case when k = 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex Un but undirected graphs are not.
Transcribed Image Text:In Tutorial we prove the claim that every directed acyclic graph (DAG) has a vertex v out-degree(v)=0. with In the lecture on paths, cycles and trees (week 8) we proved a similar claim for undirected graphs, namely that every undirected graph has a vertex v with either degree(v)=0 OR degree(v)=1. Why is the proof for DAGS able to be more specific than for undirected graphs? (Why can we guarantee a out-degree 0 vertex for DAGS but not necessarily a degree-O vertex for undirected graphs?) Select the option that best describes why. O a) Because one is directed and one is undirected. Ob) Both proofs construct maximal simple paths uu₁... Uk. In the case when k > 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex Ubut undirected graphs are not because {Uk, Uk-1} is an edge in the undirected graph, but (uk, Uk-1) is not an edge in the DAG. O c) Both proofs construct maximal simple paths U₁... Uk. In the case when k > 0, directed acyclic graphs are able to guarantee there is no edge to a previous vertex u, in the simple path for 0 ≤ ≤ k − 2, but undirected graphs are not. d) Both proofs construct maximal simple paths uu₁... Uk. In the case when k = 0, directed acyclic graphs are able to guarantee an out-degree 0 vertex Un but undirected graphs are not.
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