Universal sink in a directed graph

A universal sink is a vertex which has no edge emanating from it, and all other vertices have an edge towards the sink. Determine whether a universal sink exists in a directed graph

Example:

Input:

👁 420047153

Output: Sink found at vertex 2

Input:

👁 420047152

Output: No Sink

Adjacency Matrix Elimination Approach - O(V^2) Time and O(V^2) Space

Instead of examining each vertex one by one, we can efficiently eliminate non-sink vertices by using the adjacency matrix.

We begin with the first vertex (i = 0) and the first column (j = 0) and traverse the matrix following simple rules: if adj[i][j] == 1, it means vertex i cannot be a sink, so we increment i; if adj[i][j] == 0, it means vertex j cannot be a sink, so we increment j.

This process continues until either i or j exceeds the total number of vertices, leaving only one candidate vertex. Finally, we verify this candidate by checking that its row contains all 0s and its column contains all 1s except for the diagonal element to confirm whether it is indeed a universal sink.

Example: Consider the graph

Step 1: Create adjacency matrix

Step 2: Eliminate non-sink vertices

Start with i = 0 and j = 0.

Rules:

• If adj[i][j] equals 1, vertex i cannot be a sink since it has an outgoing edge to j, so increment i.
• If adj[i][j] equals 0, vertex j cannot be a sink since it is missing an incoming edge from i, so increment j.

Iteration:

• i=0, j=0, adj[0][0]=0, increment j to j=1
• i=0, j=1, adj[0][1]=1, increment i to i=1
• i=1, j=1, adj[1][1]=0, increment j to j=2
• i=1, j=2, adj[1][2]=0, increment j to j=3
• i=1, j=3, adj[1][3]=0, increment j to j=4
• i=1, j=4, adj[1][4]=0, increment j to j=5
• i=1, j=5, adj[1][5]=0, increment j to j=6

j exceeds number of vertices - stop scanning. Candidate sink = vertex 1 (0-based index), which is vertex 2 in 1-based indexing.

Step 3: Verify candidate sink

• Row of v2 - all 0s
• Column of v2 - all 1s except diagonal
• v2 is the universal sink.

Sink found at vertex 2

There exist better approaches to solve this problem. Please refer The Celebrity Problem for more efficient approaches.