Graph theory
Strong connected components
dfi means depth first index
Question
dfi(v) = dfi(z)+ n . Then show there exists a v-z path by induction on n

To prove that there exists a \( v \)-\( z \) path using induction on \( n \) where \( \text{dfi}(v) = \text{dfi}(z) + n \), let's proceed as follows:

### **Base Case ( \( n = 1 \) ):**

If \( n = 1 \), then \( \text{dfi}(v) = \text{dfi}(z) + 1 \).

- In this case, \( v \) is the immediate successor of \( z \) in the Depth-First Search (DFS) ordering.
- By definition of DFS, there must be an edge from \( z \) to \( v \), implying that there is a direct path from \( z \) to \( v \).

### **Inductive Hypothesis:**

Assume that for some \( k \geq 1 \), if \( \text{dfi}(v) = \text{dfi}(z) + k \), then there exists a path from \( z \) to \( v \).

### **Inductive Step:**

Now, consider \( \text{dfi}(v) = \text{dfi}(z) + (k+1) \).

- Let \( u \) be a vertex such that \( \text{dfi}(u) = \text{dfi}(z) + k \).
- By the inductive hypothesis, there exists a path from \( z \) to \( u \).
- Since \( \text{dfi}(v) = \text{dfi}(u) + 1 \), \( v \) is the immediate successor of \( u \) in the DFS ordering.
- This implies that there is a direct edge

Learn how to prove the existence of a path from vertex v to vertex z in strongly connected components using Depth-First Search (DFS) and induction. This problem involves analyzing Depth-First Index (dfi) relationships and applying mathematical induction to establish the path's existence.

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