a simple graph with n vertices and k components can have at most (n k)(n k+1)/2 edges
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Graph Theory
Consider vertices u and v from different components of G Consider a shortest u-v-path P in G Then P has an edge e = xy with exactly one vertex x in |
This is called a complete graph.
The maximum number of edges in the complete graph containing 5 vertices is given by K5: which is C(5, 2) edges = “5 choose 2” edges = 10 edges.
Since 12 > 10, it is not possible to have a simple graph with more than 10 edges.
How many edges can a simple graph have with n vertices?
The maximum number of edges in an undirected graph with n vertices is n(n-1)/2.
What are the components of a graph?
The components of any graph partition its vertices into disjoint sets, and are the induced subgraphs of those sets.
A graph that is itself connected has exactly one component, consisting of the whole graph.
Components are sometimes called connected components.
What are the edges and vertices of a graph?
A graph is a slightly abstract represention of some objects that are related to each other in some way, and of these relationships.
The objects are drawn as dots called vertices; a line (or curve) connects any two vertices representing objects that are related or adjacent; such a line is called an edge.
Durée : 19:37
Postée : 9 oct. 2020Autres questions
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Number Theory and Graph Theory Chapter 6
Theorem 4. A simple graph with n vertices and k components can have at most have (n?k)(n? k+1)/2 edges. Proof. Let X be a graph with k components. |
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A 1. Define Graph. Ans: A graph G = (VE) consists of
Prove that a simple graph with n vertices and k components can have at most. ( n - k )( n - k + 1 ). 2 edges. Let the number of vertices of the ith |
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CPS 102: Discrete Mathematics for Computer Science Homework VI
2. (15 points) Show that isomorphism of simple graphs is an equivalent with n vertices and k connected components has at most. (n-k)(n-k+1). 2 edges. |
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Tight query complexity bounds for learning graph partitions
We prove that for n ? k ? 2 learning the components of an n-vertex hidden graph with k components requires at least (k -. 1)n-( k. 2) membership queries. |
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Chapter 5 Connectivity
Similarly a graph is k-edge connected if it has at least two vertices and (ii) for any i < n |
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Tight query complexity bounds for learning graph partitions
17 juin 2022 We prove that for n ? k ? 2 learning the components of an n-vertex hidden graph with k components requires at least (k ? 1)n ? (k. 2. ) ... |
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CS 171 Lecture Outline
14 avr. 2010 has at least n ? k connected components. Induction Step: We want to prove that a graph G |
| On the sizes of $ k $-edge-maximal $ r $-uniform hypergraphs |
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On some estimates for Erd os-Renyi random graph
2 avr. 2019 with n vertices where the probability of an edge is equal to p. ... k=0. Bn |
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A variant of the Erdos-Renyi random graph process
28 juin 2018 of edges (union of a complete graph on n ? k + 1 vertices and k ? 1 isolated ... have weight 1 or (ii) have i.i.d. random weights. |
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Connected and Disconnected graphs Unit V Connected - SGRRITS
A graph G is said t be disconnected if and only if its vertex V can be portioned into two would be at least one edge whose one end vertex would be in V1 and the other in must be belong to the same component and hence must have a path Prove that a simple graph with n vertices must be connected if it has more than |
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CPS 102: Discrete Mathematics for Computer Science Homework VI
(15 points) Show that isomorphism of simple graphs is an equivalent relation Each connect component with ni vertices can have at most C(ni,2) edges |
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Maximum number of edges in connected graphs with a - CORE
number of vertices and a given domination number can have However where 2 s d s n, then the number of edges of G is at most [(n - d)(n - d + 2)/2J the above paragraph we see that each component Ci must consist of a single cycle, and |
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Solutions to InClass Problems Week 5, Mon
4 oct 2005 · Theorem 1 1 In every graph, there are an even number of vertices of odd degree Proof vertex in this component can have degree at most n − 1 □ together with the edge, would be a simple cycle that traversed the edge |
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Maximum number of edges in connected graphs with a given
number of vertices and a given domination number can have However where 2 s d s n, then the number of edges of G is at most [(n - d)(n - d + 2)/2J the above paragraph we see that each component Ci must consist of a single cycle, and |
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Math 108 Homework 1 Solutions - Tufts Math
Suppose that G is a simple graph on 10 vertices that is not connected Prove that G has of a connected component, there are no edges in G between vertices in A and vertices in Moreover, equality can occur, if G is the union of the is some integer k between 1 and n − 1 such that two or more vertices have degree k |
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MATH 2113 - Assignment 7 Solutions
Therefore, in a graph with 5 vertices, no vertex could have degree 5 11 1 21 - Here have an edge to at most n − 1 vertices of the graph since it cannot be con- nected to cycles and each component must have at most k vertices Therefore |
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Lecture 21 - Outline - UMD CS
10 juil 2017 · We will prove this claim by doing induction on m Base Case: m = 0 A graph with n vertices and no edges has n connected components as each One could also have proved the above claim directly by observing that a |
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HOMEWORK 2 (1) Let G be a simple graph where the vertices
of a vertex in G? How many vertices have each degree? How many edges we will start out by assuming that the degree of each of the vertices of G is greater separate connected components of G Thus the vertex u lies in H, a connected Solution: Clearly, Nn, the simple graph with n vertices and no edges, and Kn |
