GATE | GATE CS 2010 | Question 65

The weight of a sequence a0, a1, …, an-1 of real numbers is defined as a0+a1/2+…+ aa-1/2n-1. A subsequence of a sequence is obtained by deleting some elements from the sequence, keeping the order of the remaining elements the same. Let X denote the maximum possible weight of a subsequence of a0, a1, …,an-1 and Y the maximum possible weight of a subsequence of a1, a2, …,an-1. Then X is equal to
(A) max(Y, a0+Y)
(B) max(Y, a0+Y/2)
(C) max(Y, a0+2Y)
(D) a0+Y/2


Answer: (B)

Explanation: Using concepts of Dynamic Programming, to find the maximum possible weight of a subsequence of X, we will have two alternatives:
1. Do not include a0 in the subsequence: then the maximum possible weight will be equal to maximum possible weight of a subsequence of {a1, a2,….an} which is represented by Y
2. Include a0: then maximum possible weight will be equal to a0 + (each number picked in Y will get divided by 2) a0 + Y/2. Here you can note that Y will itself pick optimal subsequence to maximize the weight.

Final answer will be Max(Case1, Case2) i.e. Max(Y, a0 + Y/2). Hence B).

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