Following questions have been asked in GATE CS 2011 exam.

**1) An undirected graph G(V, E) contains n ( n > 2 ) nodes named v1 , v2 ,….vn. Two nodes vi , vj are connected if and only if 0 < |i – j| <= 2. Each edge (vi, vj ) is assigned a weight i + j. A sample graph with n = 4 is shown below.**

**What will be the cost of the minimum spanning tree (MST) of such a graph with n nodes?
**(A) 1/12(11n^2 – 5n)

(B) n^2 – n + 1

(C) 6n – 11

(D) 2n + 1

Answer: (B)

Minimum spanning tree for 2 nodes would be

(v1) _ (v2)

Total weight 3

Minimum spanning tree for 3 nodes would be

(v1) _ (v2) | (v3)

Total weight= 3 + 4 = 7

Minimum spanning tree for 4 nodes would be

(v1) _ (v2) _ (v4) | (v3)

Total weight= 3 + 4 + 6 = 13

Minimum spanning tree for 5 nodes would be

(v1) _ (v2) _ (v4) | (v3) | (v5)

Total weight= 3 + 4 + 6 + 8 = 21

Minimum spanning tree for 6 nodes would be

(v1) _ (v2) _ (v4) _ (v6) | (v3) | (v5)

Total weight= 3 + 4 + 6 + 8 + 10 = 31

We can observe from above examples that when we add kth node, the weight of spanning tree increases by 2k-2. Let T(n) be the weight of minimum spanning tree. T(n) can be written as

T(n) = T(n-1) + (2n-2) for n > 2

T(1) = 0, T(2) = 0 and T(2) = 3

The recurrence can be written as sum of series (2n – 2) + (2n-4) + (2n-6) + (2n-8) + …. 3 and solution of this recurrence is n^2 – n + 1.

2) The length of the path from v5 to v6 in the MST of previous question with n = 10 is

(A) 11

(B) 25

(C) 31

(D) 41

Answer: (C)

Any MST which has more than 5 nodes will have the same distance between v5 and v6 as the basic structure of all MSTs (with more than 5 nodes) would be following.

(v1) _ (v2) _ (v4) _ (v6) _ . . (more even numbered nodes) | (v3) | (v5) | . . (more odd numbered nodes)

Distance between v5 and v6 = 3 + 4 + 6 + 8 + 10 = 31

**3) Consider two binary operators ‘↑ ‘ and ‘↓’ with the precedence of operator ↓ being lower than that of the ↑ operator. Operator ↑ is right associative while operator ↓ is left associative. Which one of the following represents the parse tree for expression (7 ↓ 3 ↑ 4 ↑ 3 ↓ 2)?**

Answer: (B)

Let us consider the given expression (7 ↓ 3 ↑ 4 ↑ 3 ↓ 2).

Since the precedence of ↑ is higher, the sub-expression ([3 ↑ 4 ↑ 3) will be evaluated first. In this sub-expression, 4 ↑ 3 would be evaluated first because ↑ is right to left associative. So the expression is evaluated as ((7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2). Also, note that among the two ↓ operators, first one is evaluated before the second one because the associativity of ↓ is left to right.

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