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How to solve these kind of questions?

only first l.....just apply algorithmm...a see the valid sequence...

By applying BFS algorithm

first we enter P in Queue and process it now we add neighbors of P that  are Q and R in any order, since all the options have Q as 2nd element first we enqueue Q then R

Queue have |R|Q|

now we process Q and add its neighbors

Queue have |T|R|

process R

queue have |S|M|T|

process T

Queue have |U|N|S|M|

now no new vertex will be added

so PQRTMSNU
by

How to eliminate other options?

How others are wrong?

start from P then according to BFS the nodes which must be viewed should be adjacent of P i.e QR....

but givewn is PQT..False/invalid sequence

you have to maintain tree like structure whenever there are more then one neighbors present.