Question

fundamentals of algorithmics.pdf (page 200 of 530) Q Search Protiom 8.24. For heapsort, what are the best and the worst initial arrangements of the elements to be sorted, as far as the execution time of the algorithm is concerned? Justify your answer Problem 5.25. Prove that the binomial tree B defined in Section 5.8 contains 2 nodes, of which (i) are at depth k, 0 s k si Problem 5.26. Prove that a binomial heap con Tlg nl binomial trees, the largest of which contains 2 n items comprises at most items. Problem 5.27. Consider the algorithm for inserting a new item into a binomial heap H given in Section 5.8. A simpler method would be to create a binomial tree B0 as in step (i) of the algorithm, make it into a binomial heap, and merge this new heap with H. Why did we prefer the more complicated algorithm? Problem 5.28. Using the accounting trick described in Section 5.8, what is the amortized cost of deleting the largest item from a binomial heap? Problem 5.29. (Fibonacci trees) It is convenient to define the Fibonacci tree F-i to consist of a single node. Then the i-th Fibonacci tree Fit z0, is defined recursively to consist of a root node with i children, where the j-th child, 1 s j s i, is in turn the root of a Fibonacci tree Fj-2. Figure 5.23 shows Fo to Fs. Prove that the Fibonacci tree Fii 2 0, has fii nodes, where fi is the k-th member of the Fibonacci sequence; see Section 1.6.4

Need help with 5.25. I have attached the definition of the i-th binomial tree.

Answer 1

Proof The proof is by induction on k.For each property, the basis is the binomial tree B_0. Verifying that each property holds for B_0 is trivial. For the inductive step, we assume that the lemma holds B_k - 1. 1. Binomial tree B_k consist of two copies of B_k - 1, so B_k has 2^k - 1 + 2^k - 1 = 2^k nodes. 2. Because of the way in which the two copies of B_k - 1 are linked to form B_k, the maximum depth of anode in B_k is one greater than the maximum depth in B_k - 1. BY the inductive hypothesis, this maximum depth is (k - 1) + 1 = k. Let D(k, i) be the number of nodes at depth I of binomial tree B_k. Since B_k once at depth I + 1. In other words, the number of nodes at depth I in B_k is the number of nodes at depth I in B_k - 1 plus the number of nodes at