Solution of in - class exercise 1 : Bounding a sequence

Induction step. Let n 2 and suppose that the lemma holds for all insertion permutations of size strictly less than n. Let π = (π(1), . . . , π(n)) be a permutation drawn uniformly at random from Sn. The rst element π(1) will become the root of the tree Tπ. Since the distribution of π(1) is u.a.r. from [n], the root of the tree is chosen uniformly at random, as required by the construction of ~ B[n]. Now let us tackle the two subtrees of the root: Let k 2 [n]. We want to show that, conditioned on π(1) = k, the distribution of the left subtree of the root is the same as ~ B{1,...,k−1}, and the distribution of the right subtree of the root is the same as ~ B{k+1,...,n}. To this end, let π be the sequence of elements in π smaller than k, and let π be the sequence of elements in π larger than k. Note that the insertion sequence will send the keys in π (π) in exactly this order to the left (right) subtree of the root. Since π and π are uniformly random permutations of their respective element sets {1, . . . , k−1} and {k + 1, . . . , n}, we can apply the induction hypothesis to obtain that the left and right subtrees of the root are distributed just as we stated.