Question

1. Modify BFS to only use a single bit of information to represent a vertex’s label. Based on your pseudocode, write an invariant proof to show that the modified algorithm still reaches every reachable vertex.

Answer 1

Breadth First Search (BFS)

BFS could be a traversing algorithm where you ought to begin navigating from a chosen hub (source or beginning node) and navigate the chart layerwise hence exploring the neighbor hubs (hubs which are straightforwardly associated to source node). You must at that point move towards the next-level neighbour nodes.

As the title BFS recommends, you're required to navigate the chart breadthwise as follows:

1. First move on a level plane and visit all the hubs of the current layer

2. Move to the another layer

for example;

s 2 3 5 CS Scanned with CamScanner

The navigating will begin from the source node and thrust s in queue. s will be checked as 'visited'.

To begin with iteration

1 .s will be popped from the queue

2 .Neighbors of s i.e. 1 and 2 will be traversed

3. 1 and 2, which have not been navigated prior, are navigated. They will be:

a. Pushed within the queue

b. 1 and 2 will be stamped as gone to

Second iteration

1. 1 is popped from the queue

2. Neighbors of 1 i.e. s and 3 are traversed

3. s is disregarded since it is checked as 'visited'

4. 3, which has not been navigated prior, is navigated. It is:

a. Pushed within the queue

b. Stamped as gone by

Third iteration

1. 2 is popped from the queue

2. Neighbors of 2 i.e. s, 3, and 4 are traversed

3. 3 and s are disregarded since they are stamped as 'visited' 4. 4, which has not been navigated prior, is navigated. It is:

a. Pushed within the queue

b. Stamped as gone to

Fourth iteration

1. 3 is popped from the queue

2. Neighbors of 3 i.e. 1, 2, and 5 are traversed

3. 1 and 2 are overlooked since they are stamped as 'visited'

4. 5, which has not been navigated prior, is navigated. It is:

a. Pushed within the queue

b. Checked as gone by

Fifth iteration

1. 4 will be popped from the queue

2. Neighbors of 4 i.e. 2 is traversed

3. 2 is overlooked since it is as of now stamped as 'visited'

6th iteration

1. 5 is popped from the queue

2. Neighbors of 5 i.e. 3 is traversed

3. 3 is disregarded since it is as of now stamped as 'visited'

The queue is purge and it comes out of the circle. All the nodes have been navigated by utilizing BFS.