The Square of a Directed Graph and At least one vertex doubles its
An undirected graph always has a symmetric adjacency matrix this is not always the case with a directed graph. I have introduced the terms input graph as the
23.1-5 The square of a directed graph G = (V E) is the graph G2 = (V
Given an adjacency matrix we can check in constant time whether a given edge exists. To discover whether there is an edge (u w) 2G2
1. 22.1-2 Give an adjacency-list representation for a complete binary
22.1-5 The square of a directed graph G = (V E) is the graph G2 = (V
International Journal of Recent Technology and Engineering (IJRTE)
30 nov. 2019 According to the square of oriented graph conjecture (SOGC) ... study
Lecture 15: Breadth/Depth-First Search (1997) Steven Skiena
The square of a directed graph G = (VE) is the graph. G2 = (V
Graph Theory
10 août 2007 A squared graph takes the original graph and adds an arc (ac) for each pair of arcs of the form. (ab
CSci 231 Homework 10 Solutions
[CLRS 22.1-5] Give and analyse an algorithm for computing the square of a directed graph G given in (a) adjacency-list representation and (b)
Effective Resistance Preserving Directed Graph Symmetrization
11 janv. 2019 The resulting undirected Laplacian can be thought of as a symmetrization of a directed. Laplacian where a metric on graphs square root of ...
arXiv:1912.04524v3 [cs.CC] 11 Mar 2022
11 mars 2022 properties of random walks on general directed graphs (with polynomial ... regular digraphs the derandomized square produces a graph ˜.
Derandomized Squaring of Graphs
We introduce a “derandomized” analogue of graph squaring. This op- eration increases the connectivity of In directed graphs it equals the square root.
[PDF] 231-5 The square of a directed graph G = (VE) is
Given an adjacency matrix we can check in constant time whether a given edge exists To discover whether there is an edge (u w) 2G2 for each possible
[PDF] The Square of a Directed Graph and At least one vertex doubles its
A directed graph is a simple graph (no loops or multiple edges) with each edge assigned a direction Given vertices u and v this direction can be any of
[PDF] The Square of A Directed Graph
30 nov 2019 · Abstract: The square of an oriented graph is an oriented graph such that if and only if for some both and exist According to the square
The Square of a Directed Graph Request PDF - ResearchGate
The square of an oriented graph is an oriented graph such that if and only if for some both and exist According to the square of oriented graph
[PDF] Directed Graphs (digraphs) - csPrinceton
Directed Graphs digraph search Every square matrix is a weighted digraph Identical to undirected version (substitute Digraph for Graph)
[PDF] Directed Graphs - csPrinceton
Directed graphs (digraphs) Every square matrix is a weighted digraph Identical to undirected version (substitute Digraph for Graph)
[PDF] Derandomized Squaring of Graphs
A K-regular directed graph X on N vertices with ?(X) ? ? will be called an (NK?)-graph We define g(X)=1 ? ?(X) to be the spectral gap of X The “best
[PDF] Graph Theory
directed edges of the directed graph We use notation Note: vertex matrix uniquely determines connectivity of the graph Now square :
[PDF] Graph Theory - University of Chicago Math
10 août 2007 · A squared graph takes the original graph and adds an arc (ac) for each pair of arcs of the form (ab bc) An oriented graph is a directed
[PDF] Properties of Adjacency Matrix of a Graph and Its Construction
In the adjacency matrix of a directed graph aij equals For any given square symmetric and binary matrix A of order n there exists a graph
What is the square of a directed graph?
The square of a directed graph G = (V, E) is the graph G2 = (V, E2) such that (u,v) ? E2 if and only if G contains a path with at most two edges between u and v. Describe efficient algorithms for computing G2 from G for both the adjacency-list and adjacency-matrix representations of G.How do you find the square of a graph?
The square of a graph is obtained from by adding new edges between every two vertices having distance two in . A block graph is one in which every block is a clique.- The square of an adjacency matrix A^2=(s_{ij}) has the property that s_{ij} represents the number of walks of length two from vertex i to vertex j. With this information, the motivating question behind this paper was to determine what conditions on a matrix S are needed to have S=A(G)^2 for some graph G.
The Square of a Directed Graph
andAt least one vertex doubles it's out-degree
Jon Perry
October 2013
Abstract
I begin with a brief examination of the concepts directed graphs use, and develop a definition of the squaring process. Next I demonstrate that this process always at least doubles the out-degree of at least one vertex, thus proving a conjecture at:Directed Graphs
A directed graph is a simple graph (no loops or multiple edges) with each edge assigned a direction. Given vertices u and v, this direction can be any of u->v, v->u or both, i.e. u<->v. This is similar to a one-way traffic system. The edge is now called an arc. The out-degree is the number of arcs that are leaving a given vertex u and the in- degree is the number of arcs coming into u. To square a directed graph, we consider vertices u, v and w. If u->v and v->w then we add a new arc (assuming it is not already there) of (u,w). Note that if u<->v then we create 2 loops, namely (u,u) and (v,v). If this is the case, the resulting directed graph is no longer simple. An adjacency matrix for a graph consists of a table with each (labelled) vertex forming the co-ordinates. An entry (i,j) is 1 if i->j and 0 otherwise. An undirected graph always has a symmetric adjacency matrix, this is not always the case with a directed graph. I have introduced the terms input graph as the initial directed graph which we are about to square, and output graph as the resultant graph of squaring the input graph.Squaring a Directed Graph
To begin with we examine an input graph and develop it's adjacency matrix. If we label the vertices 1 to 6 (top three are 1, 2 and 3, bottom three from left to right are 4, 5 and 6), we get the following adjacency matrix:123456
1000000
2100000
3000001
4000000
5111100
6000010
The main diagonal is all 0's as this is part of the criteria for the input graph - no loops.The square of this graph is:
This graph has the following adjacency matrix:
123456
1000000
2100000
3000011
4000000
5111101
6111110
The arc (5,1) is not doubled up because it already exists. Generating the adjacency matrix for the output graph There is a simple rule for generating the adjacency matrix for the square of a given input graph. Let ri be the contents of the i-th row of the input adjacency matrix. Let rij be the contents of the i-th row and j-th column of the input adjacency matrix. Let Ri be the contents of the i-th row of the output adjacency matrix. Let Rij be the contents of the i-th row and j-th column of the output adjacency matrix.Then we have:
Ri = ri + rj [ if rij = 1 ], with each Rij being limited to 1.Proof of the conjecture
If a row consists of entirely zeroes, the vertex in question has an out-degree of 0. Double of 0 is 0, and so we have trivially proved the conjecture if this is the case. This means that the input graph must have every vertex with an out-degree of 1 or more. Consider an input graph with at least one vertex with out-degree exactly 1. We have two situations to consider. If we call our vertex u and its arc-neighbour v, then either u->v->w where w is not u, or u->v->u. In the first case we now have u->w and because the out-degree of u is 1, this is a new arc and so the new out-degree of u is 2. In the second case we now have u->u, a loop, and the out-degree of u is also now 2. So we can conclude that the input graph must have every vertex with out-degree at least 2, and we consider an input graph with at least one vertex with out-degree exactly 2. The argument that follows will prove that we need an input graph with every vertex of at least out-degree 3, and more specifically at least one vertex of exactly out- degree 3. And we can use the same argument to extend 3 to 4 to 5 and so on, thus proving the conjecture. If a vertex u has out-degree 2, then we may assume it points to vertices v and w. In the best case, v and w are not connected, so u has at least 3 new arcs, to vertices x, y and z, where v and w both point to x. If v points to w (or vice versa), it must at least point to x say, and w at worst can point to x and y, giving u two new arcs (u,x) and (u,y). So the out-degree must be at least 3. We can now see that u points to k vertices, and even in the worst case of each vertex pointing to as many 'in common' vertices as possible, u will always double its out- degree. QED.quotesdbs_dbs12.pdfusesText_18[PDF] squid proxy server centos 7
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