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Question:

Consider a directed graph ( G = (V, E) ) where ( V ) is the set of vertices and ( E ) is the set of directed edges. Let ( n = |V| ) be the number of vertices in the graph. For each vertex ( v \in V ), there is an associated positive integer ( d(v) ) representing the outdegree of the vertex, i.e., the number of outgoing edges from ( v ).

 

Prove that the sum of the outdegrees of all vertices in the graph is equal to the total number of edges in the graph.

 

Let ( A ) be the adjacency matrix of the graph, where ( A[i][j] = 1 ) if there is a directed edge from vertex ( i ) to vertex ( j ), and ( A[i][j] = 0 ) otherwise. Define a new matrix ( B ) as follows: ( B[i][j] = d(i) \times A[i][j] ), where ( d(i) ) is the outdegree of vertex ( i ). Show that the sum of all entries in matrix ( B ) is equal to the total number of edges in the graph.

 

Consider a path in the graph, which is a sequence of distinct vertices ( v_1, v_2, \ldots, v_k ) such that there is a directed edge from ( v_i ) to ( v_{i+1} ) for ( 1 \leq i < k ). Prove that the sum of the outdegrees of the vertices along this path is equal to the total number of edges in the path.

 

Let ( G' ) be the transpose of the original graph ( G ). That is, ( G' = (V, E') ) where ( E' ) consists of all the edges in ( G ) with their directions reversed. Show that the sum of the indegrees of all vertices in ( G' ) is equal to the total number of edges in the graph ( G ).

 

Solution:

 

1. Proof of Outdegrees Sum:

 

Let ( d(v) ) be the outdegree of vertex ( v ). The sum of the outdegrees is given by:

 

[ \sum_{v \in V} d(v) ]

 

Consider each edge in the graph, it contributes 1 to the outdegree of its starting vertex. Therefore, the sum of outdegrees is precisely the total number of edges in the graph.

 

[ \sum_{v \in V} d(v) = \text{Number of Edges in } G ]

 

2. Matrix ( B ) Sum:

 

The sum of all entries in matrix ( B ) is given by:

 

[ \sum_{i \in V} \sum_{j \in V} d(i) \times A[i][j] ]

 

Consider each term in the double sum. If ( A[i][j] = 1 ), then ( d(i) ) contributes to the sum. This is precisely the outdegree of vertex ( i ). Thus, the double sum counts the sum of outdegrees, which by part 1, is equal to the total number of edges in the graph.

 

[ \sum_{i \in V} \sum_{j \in V} d(i) \times A[i][j] = \text{Number of Edges in } G ]

 

3. Sum of Outdegrees on a Path:

 

Consider a path ( P: v_1, v_2, \ldots, v_k ). The sum of outdegrees along this path is:

 

[ \sum_{i=1}^{k-1} d(v_i) ]

 

Each term in the sum represents the outdegree of the respective vertex on the path. Again, by part 1, this sum is equal to the total number of edges in the path.

 

[ \sum_{i=1}^{k-1} d(v_i) = \text{Number of Edges in } P ]

 

4. Transpose Graph Indegrees Sum:

 

Let ( G' ) be the transpose of ( G ). The sum of indegrees in ( G' ) is given by:

 

[ \sum_{v \in V} \text{indegree}_{G'}(v) ]

 

Consider an edge in ( G ) from ( u ) to ( v ). This edge becomes an edge from ( v ) to ( u ) in ( G' ), contributing to the indegree of ( v ). Thus, the sum of indegrees in ( G' ) is equal to the total number of edges in ( G ).

 

[ \sum_{v \in V} \text{indegree}_{G'}(v) = \text{Number of Edges in } G ]

 

Conclusion:

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