Hello. Please answer the attached Data Structures question correctly and follow all directions. Create a proof for this problem. *If you answer correctly, I will provide a thumbs up. Thank you. 3. Prove that the bound \( \frac{2}{2^d} \) in Lemma 4.1 is the best possible bound by showing that, if \( x = 2^{w-d-1} \) and \( y = 3x \), then \( \Pr\{ \text{hash}(x) = \text{hash}(y) \} = \frac{2}{2^2} \). (Hint: look at the binary representations of \( zx \) and \( z3x \) and use the fact that \( z3x = zx + 2zx \).)**Lemma 4.1.** Let \( x \) and \( y \) be any two values in \( \{0, \ldots, 2^w - 1 \} \) with \( x \neq y \). Then \( \Pr\{ \text{hash}(x) = \text{hash}(y) \} \leq \frac{2}{2^d} \).With **Lemma 4.1**, the performance of remove(\( x \)) and find(\( x \)) are easy to analyze:

Given Assume one has access to a d-bit hash function called h. Let's act as though one possesses it. One of the main goals of this study is to demonstrate that Prhash(x) = hash(y) = 2/22 is true in the case where x = 2ω-d-1 and y = 3x. In order to succeed in this quest, one must investigate the binary representations of the variables zx and z3x. It is well known that z3x = zx + 2zx.Let's examine the first d bits of each integer now that we've taken care of everything else. They understand that the value zx's initial d bits will consist entirely of zeros. If the first d bits of 2zx are all zeros, the first d bits of z3x will either be all ones or all zeros, depending on the state of the first d bits of 2zx. If the first d bits of 2zx are all ones, they will all be ones (if the first d bits of 2zx are all 1s). They can therefore draw the conclusion that the probability of h(x) = h(y) is equal to 2/2d. The answer proves that the preceding query was addressed and that the bound 2/2d specified in Lemma 4.1 is the tightest bound that can be discovered. This is because there is a 2/2d chance that h(x) = h(y) in the case where x = 2w-d-2 and y = 3x are both true. This explains why things are the way they are right now.Conclusion This is due to the fact that the first d bits in the case of zx will only contain zeroes. This provides justification. If the first d bits of 2zx are all zeros, the first d bits of z3x will either be all ones or all zeros, depending on the state of the first d bits of 2zx. If the first d bits of 2zx are all ones, they will all be ones (if the first d bits of 2zx are all 1s). Therefore, 2/2d represents the likelihood that h(x) and h(y) are equivalents. This is the best limitation that can be obtained because a probability higher than 2/2d is not attainable. This makes it possible to show that the bound 2/2dd given in Lemma 4.1 is the best bound that can be imagined.