# Why do we need to relax all the edges at most V-1 times

*To understand this example, it is recommended to have a brief idea on Bellman-Ford single source shortest path algorithm which can be found **here*

In Bellman-Ford algorithm, to find out the shortest path, we need to *relax* all the edges of the graph. This process is repeated at most **(V-1)** times, where **V** is the number of vertices in the graph.

The number of iterations needed to find out the shortest path from **source** to all other vertices depends on the order that we select to *relax* the edges.

Let’s take a look at an example:

Here, the **source** vertex is 1. We will find out the shortest distance between the **source** and all the other vertices. We can clearly see that, to reach **vertex 4**, in the worst case, it’ll take **(V-1)** edges. Now depending on the order in which the edges are discovered, it might take **(V-1)** times to discover **vertex 4**. Didn’t get it? Let’s use Bellman-Ford algorithm to find out the shortest path here:

We’re going to use this sequence:

+--------+--------+--------+--------+ | Serial | 1 | 2 | 3 | +--------+--------+--------+--------+ | Edge | 3->4 | 2->3 | 1->2 | +--------+--------+--------+--------+

For our first iteration:

**d[3]**+**cost[3][4]**=**infinity**. It won’t change anything.**d[2]**+**cost[2][3]**=**infinity**. It won’t change anything.**d[1]**+**cost[1][2]**=**2**<**d[2]**.**d[2]**=**2**.**parent[2]**=**1**.

We can see that our *relaxation* process only changed **d[2]**. Our graph will look like:

Second iteration:

**d[3]**+**cost[3][4]**=**infinity**. It won’t change anything.**d[2]**+**cost[2][3]**=**5**<**d[3]**.**d[3]**=**5**.**parent[3]**=**2.**- It won’t be changed.

This time the *relaxation* process changed **d[3]**. Our graph will look like:

Third iteration:

**d[3]**+**cost[3][4]**=**7**<**d[4]**.**d[4]**=**7**.**parent[4]**=**3**.- It won’t be changed.
- It won’t be changed.

Our third iteration finally found out the shortest path to **4** from **1**. Our graph will look like:

So, it took **3** iterations to find out the shortest path. After this one, no matter how many times we *relax* the edges, the values in **d[]** will remain the same. Now, if we considered another sequence:

+--------+--------+--------+--------+ | Serial | 1 | 2 | 3 | +--------+--------+--------+--------+ | Edge | 1->2 | 2->3 | 3->4 | +--------+--------+--------+--------+

We’d get:

**d[1]**+**cost[1][2]**=**2**<**d[2]**.**d[2]**=**2**.**d[2]**+**cost[2][3]**=**5**<**d[3]**.**d[3]**=**5**.**d[3]**+**cost[3][4]**=**7**<**d[4]**.**d[4]**=**5**.

Our very first iteration has found the shortest path from **source** to all the other nodes. Another sequence **1->2**, **3->4**, **2->3** is possible, which will give us shortest path after **2** iterations. We can come to the decision that, no matter how we arrange the sequence, it won’t take more than **3** iterations to find out shortest path from the **source** in this example.

We can conclude that, for the best case, it’ll take **1** iteration to find out the shortest path from **source**. For the worst case, it’ll take **(V-1)** iterations, which is why we repeat the process of *relaxation* **(V-1)** times.