Determine whether a universal sink exists in a directed graph
Determine whether a universal sink exists 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.
Input : v1 -> v2 (implies vertex 1 is connected to vertex 2) v3 -> v2 v4 -> v2 v5 -> v2 v6 -> v2 Output : Sink found at vertex 2 Input : v1 -> v6 v2 -> v3 v2 -> v4 v4 -> v3 v5 -> v3 Output : No Sink
We try to eliminate n – 1 non-sink vertices in O(n) time and check the remaining vertex for the sink property.
To eliminate vertices, we check whether a particular index (A[i][j]) in the adjacency matrix is a 1 or a 0. If it is a 0, it means that the vertex corresponding to index j cannot be a sink. If the index is a 1, it means the vertex corresponding to i cannot be a sink. We keep increasing i and j in this fashion until either i or j exceeds the number of vertices.
Using this method allows us to carry out the universal sink test for only one vertex instead of all n vertices. Suppose we are left with only vertex i.
We now check for whether row i has only 0s and whether row j as only 1s except for A[i][i], which will be 0.
v1 -> v2 v3 -> v2 v4 -> v2 v5 -> v2 v6 -> v2 We can visualize the adjacency matrix for the above as follows: 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0
We observe that vertex 2 does not have any emanating edge, and that every other vertex has an edge in vertex 2. At A (A[i][j]), we encounter a 0, so we increment j and next
look at A. Here we encounter a 1. So we have to increment i by 1. A is 0, so we keep increasing j. We notice that A, A.. etc are all 0, so j will exceed the
number of vertices (6 in this example). We now check row i and column i for the sink property. Row i must be completely 0, and column i must be completely 1 except for the index A[i][i]
v1 -> v6 v2 -> v3 v2 -> v4 v4 -> v3 v5 -> v3 We can visualize the adjacency matrix for the above as follows: 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
In this example, we observer that in row 1, every element is 0 except for the last column. So we will increment j until we reach the 1. When we reach 1, we increment i as long as
the value of A[i][j] is 0. If i exceeds the number of vertices, it is not possible to have a sink, and in this case, i will exceed the number of vertices.
input set 1: Sink found at vertex 6 input set 2: No Sink
This program eliminates non-sink vertices in O(n) complexity and checks for the sink property in O(n) complexity.
You may also try The Celebrity Problem, which is an application of this concept
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