- If B represented the set of Boolean values true and false, then the set of Boolean functions B x B → B is... A) finite B) uncountably infinite C) countably infinite - The set of functions from {0,1} to N is... A) finite B) uncountably infinite C) countably infinite - The set of functions from N to {0,1} is... A) finite B) uncountably infinite C) countably infinite

Let F be the set of all functions of the form f : N {0,1}. Also, let G be the set of all functions of the form f: {0, 1}* → {a, b}.Then: O (a) |F| = |G| = |N| O (b) |F| = |P(G)| 미 (c) 1지=1P(N)| □ (d) 지< IgI, but |F # I9| □ (e) |지= 19| %3D %3D Select all possible options that apply. e

2. Let n be a positive integer, and let A be a list of positive integers. We say that the integer n can be factorized by A if there exists a sequence of integers 11, 12,..., ik (with 0 ≤i; < len (A), for j = 1,..., k) such that n is the product of the integers A[i], A[i2],..., A[ik]. Write an efficient algorithm factorizable (n, A) that returns True if n can be factorized by A, and False otherwise. Prove that your algorithm is correct, and bound its running time. Larger scores will be awarded to faster solutions. Example 1: if n = 6 and A [4,2,3], then factorizable (n, A) should return True, L[1] * L[2]. since n == Example 2: if n = 8 and A = L[0] * L[0] * L[0]. [2,5], then factorizable (n, A) should return True, since ,2,3,5,7], then factorizable (n, A) should return Example 3: if n = 13 and A = False since n cannot be factorized by A.

es remaining 8. Consider the function f:NxN-N defined recursively by: 1) Base case: Let meN and define (0,m) = 0 2) Recursive case: For any x,meN, x>0, define f(x,m) = (x-1,m) + (m+m) Prove the following theorem holds using proof by induction: Thereom: For any n,meN, m>0 we have (n.m) I m = n+n Fill in your answer here 9 Help BIU X, x L - ɔE =N E X Format