From the options its quite evident that we should start from A as our starting vertex.
One of the possible sequences of the prims algorithm is shown here :
|
a |
b |
c |
d |
e |
f |
g |
h |
i |
|
∞ N |
∞ N |
∞ N |
∞ N |
∞ N |
∞ N |
∞ N |
∞ N |
∞ N |
a |
|
4 a |
∞ N |
∞ N |
∞ N |
∞ N |
∞ N |
8 a |
∞ N |
b |
|
|
8 b |
∞ N |
∞ N |
∞ N |
∞ N |
8 a |
∞ N |
c |
|
|
|
7 c |
∞ N |
4 c |
∞ N |
8 a |
2 c |
i |
|
|
|
7 c |
∞ N |
4 c |
6 i |
7 i |
|
f |
|
|
|
7 c |
10 f |
|
2 f |
7 h |
|
g |
|
|
|
7 c |
10 f |
|
|
1 g |
|
h |
|
|
|
7 c |
10 f |
|
|
|
|
d |
|
|
|
|
9 d |
|
|
|
|
e |
|
|
|
|
|
|
|
|
|
As per the order of relaxation we got a choice only between the edges ${a,h}$ and $ {b,c} $ , apart from these we have no choices ever in the entire algorithm.So option a and c are true .
Option d takes non adjacent edges which is true for $Kruskal’s$ algo not for $Prim’s$ and in option b the edge b,h is selected over {a,h} or {b,c} which has higher weight than the latter two.
Correct sequences are $a$ and $c$