GATE CS 2010 Q34 — Algorithm Design: Dynamic Programming

To find the relation between $X$ and $Y$, we can analyze the structure of the optimal subsequence of $a_0, a_1, \dots, a_{n-1}$.

Let $S$ be any subsequence of $a_1, a_2, \dots, a_{n-1}$ with elements $s_0, s_1, \dots, s_{k-1}$. By definition, the weight of $S$ is:

The maximum possible weight of such a subsequence is $Y = \max_{S} W(S)$.

Now, let $T$ be a subsequence of the entire sequence $a_0, a_1, \dots, a_{n-1}$. There are two mutually exclusive and exhaustive cases for $T$:

1. Case 1: $a_0$ is not included in $T$.
   In this case, $T$ is entirely a subsequence of $a_1, a_2, \dots, a_{n-1}$.
   The maximum possible weight of $T$ in this case is simply $Y$.


2. Case 2: $a_0$ is included in $T$.
   Since $a_0$ is the first element of the original sequence, it must be the first element of $T$. Thus, $T$ can be written as $a_0$ followed by some subsequence $S = s_0, s_1, \dots, s_{k-1}$ of $a_1, a_2, \dots, a_{n-1}$.
   The weight of $T$ is: $$W(T) = a_0 + \frac{s_0}{2} + \frac{s_1}{2^2} + \dots + \frac{s_{k-1}}{2^k} = a_0 + \frac{1}{2} \left( s_0 + \frac{s_1}{2} + \dots + \frac{s_{k-1}}{2^{k-1}} \right) = a_0 + \frac{W(S)}{2}$$ To maximize $W(T)$, we must maximize $W(S)$, which yields $Y$.
   Therefore, the maximum possible weight of $T$ in this case is $a_0 + \frac{Y}{2}$.

Since the overall maximum weight $X$ must be the maximum over these two cases, we get:

Thus, the correct option is B.