# GATE | GATE-CS-2014-(Set-1) | Question 65

• Difficulty Level : Hard
• Last Updated : 28 Jun, 2021

Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}. There is an edge between (a, b) and (c, d) if |a − c| <= 1 and |b − d| <= 1.
The number of edges in this graph is __________.

(A) 500
(B) 502
(C) 506
(D) 510

Explanation:

```Given:
The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}.
There is an edge between (a, b) and (c, d) if |a − c| <= 1 and
|b − d| <= 1.

There can be total 12*12 possible vertices. The vertices are (1, 1),
(1, 2) ....(1, 12) (2, 1), (2, 2), ....

The number of edges in this graph?
Number of edges is equal to number of pairs of vertices that satisfy
above conditions. For example, vertex pair {(1, 1), (1, 2)} satisfy
above condition.

For (1, 1), there can be an edge to (1, 2), (2, 1), (2, 2). Note that
there can be self-loop as mentioned in the question.
Same is count for (12, 12), (1, 12) and (12, 1)

For (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3),
(1, 3)
Same is count for (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11)

For (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1),
(2, 3), (3, 1), (3, 2), (3, 3)
Same is count for remaining vertices.

For all pairs (i, j) there can total 8 vertices connected to them if
i and j are not in {1, 12}

There are total 100 vertices without a 1 or 12.  So total 800 edges.

For vertices with 1, total edges = (Edges where 1 is first part) +
(Edges where 1 is second part and not first part)

=  (3 + 5*10 + 3) + (5*10) edges

Same is count for vertices with 12

Total number of edges:
= 800 + [(3 + 5*10 + 3) + 5*10]  + [(3 + 5*10 + 3) + 5*10]
= 800 + 106 + 106
= 1012

Since graph is undirected, two edges from v1 to v2 and v2 to v1
should be counted as one.

So total number of undirected edges = 1012/2 = 506. ```
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