Simplify the following Boolean function f using K-map: m x1 x2 x3 x4 f 0 0000 1 1 0 0 0 1 1 234 10 0 0 1 0 1 0 0 1 1 X 0 1 0 0 0 5 01 0 1 0 6 0 1 1 0 1 7 0 1 1 1 0 8 1 9 1 10 1 11 12 13 0 0 0 X 001 0 1 0 1 10 110 1100 1101 0 14 1110 0 X X 15 1 1 1 10 f=Σ m(0,1,2,6,10) & don't care X=Σ m(3,8,13,14) 3- Write the function in terms of x1, x2, x3 and x4 after K-map: Σ(0, 1, 2, 6, 10) f = X • X₁₂ • X + ×₁ • x₂ 3 X ׸ • × + ×₁ • X₂ •×3 •× 4 Simplified Boolean function of SOP f(x₁ ‚×2 ‚׸‚ ×₁ ) = P₁ + P₁ + P₁₂ + P₁₂ 12 3' 1 2 4 • + X¸ • ×¼ + ×₁ • ×₂ • X•X + X • X•X 1 What are the the NOTS, ANDs, and ORs from the function above NOT & AND' & OR' (1's)

(a) Write the following functions and their docstrings: • between (num1, num2, num3) takes 3 integer arguments and returns True if num2 is between num1 and num3. It is not between them if it is equal to either of the other two. For example, given the inputs 5, 3 and 0, the value returned should be True. Given the inputs -2, 2 and 2, False should be returned. Note that there is no restriction that num1 must be less than num3. • majorityEven (num_list) returns True if more than half of the integers in the num_list are di- visible by 2, with no remainder, otherwise it returns False. The list can be of any size. Recall that zero is divisible by 2 with no remainder. For example, the function should return False for the list [1,2,3,6] (as only two of the four numbers are divisible by 2) and True for the list [0,1,-4] (as two of the three numbers are divisible by 2). (b) Generate at least six test cases for each function you wrote in part (a). You may use white-box and/or black-box test case…

c=3