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Task:
Construct a graph of the following relationship:
Define its properties.
Solution:
Construct graph G( X ) with the set of vertices , and two vertices X i and X j are connected by an edge if and only if . Since the relation is symmetric, graph G( X ) is undirected.
Let’s construct the adjacency matrix (vertices) A:
X1 | X 2 | X 3 | X 4 | X 5 | X 6 | |
X1 | 1 | 1 | 1 | 1 | 1 | 1 |
X 2 | 1 | 1 | 1 | 1 | 1 | 0 |
X 3 | 1 | 1 | 1 | 1 | 0 | 0 |
X 4 | 1 | 1 | 1 | 0 | 0 | 0 |
X 5 | 1 | 0 | 0 | 0 | 0 | 0 |
X 6 | 1 | 0 | 0 | 0 | 0 | 0 |
Here Aij element is the number of edges going from vertex X i to vertex X j . As our graph is undirected, the adjacency matrix is symmetric.
Let’s construct the incidence matrix (ribs) R:
g1 | g2 | g3 | g4 | g5 | g6 | g7 | g8 | g9 | g10 | g11 | g12 | |
x1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
x2 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
x3 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
x4 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
x5 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
x6 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Here, the element Rij is 1 if the top vertex of the X i incident to edge g j and 0 otherwise.
Let’s construct the distance matrix D:
x1 | x2 | x3 | x4 | x5 | x6 | |
x1 | 0 | 1 | 1 | 1 | 1 | 1 |
x2 | 1 | 0 | 1 | 1 | 1 | 2 |
x3 | 1 | 1 | 0 | 1 | 2 | 2 |
x4 | 1 | 1 | 1 | 0 | 2 | 2 |
x5 | 0 | 0 | 0 | 1 | 0 | 1 |
x6 | 0 | 0 | 0 | 0 | 1 | 0 |
Here, the element Dij is the length the shortest path from vertex X i to vertex X j .
Since our graph is undirected, the distance matrix is symmetric.
We find the distance vector d, each of which is defined as the component
(the maximum distance from the vertex X i to every other vertex).
Distance vector d = (1, 2, 2, 2, 2, 2). Center is the peak vertex X1 , since it corresponds to the smallest distance (1 = d1 < d j , j = 2,…,6) .
Peripheral tops: X 2 , X 3 , X 4 , X 5 , X 6 , since it corresponds to the maximum remoteness (d j = 2, j = 2,…,6) .
Graph radius G( X ) is the center distance, r(G) = 1 .
Graph diameter G( X ) is the removal of the peripheral vertex, Diam(G) = 2 .
Let’s find the number for the internal and external stability graph. The largest set of internal stability for our graph has the form S = {X 4 , X 5 , X 6 } (the addition of any other peaks will receive adjacent vertices). Accordingly, graph G( X ) number is equal to the internal stability card(T ) = 1.
The smallest set for external stability for our graph is T X1 (as any other vertex (not owned to T) is connected to the apex of X1 from T). The number for the external sustainability graph G( X ) is equal to card(T ) 1.