Keep the Diagram neat. Thanks. Implement the following boolean function using a) 16:1 mux b) 8:1 mux c) 4:1 mux F(A,B,C,D) =∑(0,1,2,7,8,10,11,13, 15)
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Keep the Diagram neat. Thanks.
Implement the following boolean function using a) 16:1 mux b) 8:1 mux c) 4:1 mux
F(A,B,C,D) =∑(0,1,2,7,8,10,11,13, 15)
Step by step
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- 2. Given a boolean expression below, find the value of the expression if the values of P = 1, Q = 0, R = 1. Y = P.Q + Q.(R+P+Q)2. Look for the dual of the following Boolean expressions:a. xyz + x'y'z'b. xz' + x⋅0 +x'⋅1Implement the following Boolean expressions as they are stated: (a) X = ABC + AB + AC (b) X = AB(C + DE)
- Given a boolean expression consisting of the symbols 0 (false), 1 (true), & (AND), I (OR), and /\ (XOR), and a desired boolean result value result, implement a function to count the number of ways of parenthesizing the expression such that it evaluates to result. The expression should be fully parenthesized (e.g., ( 0) A( 1)) but not extraneously (e.g., ( ( ( 0)) /\ ( 1)) ). EXAMPLE countEval("l/\01011", false) -> 2 countEval("0&0&0&1All0", true)-> 102) Simplify the following Boolean expressions, using three-variable K-maps:F(x,y,z) = x′y′ + yz + x′yz′ PS:Please type instead of paper writingSimplify or reduce the Boolean function below into three literals: F(P,Q,R) = (P + Q’R’)(PQ+R)+PQR’+P’R
- Write the robot program using the VAL+ for the following automation case. First draw the problem; then handwrite the VAL+ code. The robot picks up FIVE blocks of size 100x100x100 mm that are stacked at a fixed position and place them at the corners and center of a square table of size 500 x 500 mm. Select the original position on the table on your own.From the example suggested, this is the output I get if we run the code:n = 4 arcs = [(0, 3), (1, 2), (4, 7), (5, 6), (8, 11), (9, 10), (12, 15), (13, 14)]crossings, nestings = count_crossings_and_nestings(arcs) print(f"Number of crossings: {crossings}")print(f"Number of nestings: {nestings}")However, upon closer inspection of the code, the function is not working as expected because there's an issue with the logic: the condition checks for nestings are repeated.It should be checking for one case where the start of one arc is between the start and end of another arc, and the end of one arc is between the start and end of another arc. The same condition is used twice in the code. And I tried computing nestings and crossings with the smallest examples below: def count_crossings_and_nestings(arcs): n = len(arcs) // 2 crossings = 0 nestings = 0 for i in range(2 * n): for j in range(i + 1, 2 * n): if arcs[i][0] < arcs[j][0] < arcs[i][1] < arcs[j][1]:…We have an Boolean Expression Y= A.B.(A+B') What is the value of Y when A= 1 and B=1.
- Given the following definitions: U={a,b,c, d, e, f, g} A={a, c, e, g} B={a,b,c,d} find A n BLet Q(x, y) denote xy< 0 The domain of x is {1,2,3} and the domain of y is {0, -1, -2, -3). Implement the code that outputs the truth values of the following quantifications: 1. ∀x ∀y Q(x, y) 2. ∀x ∃y Q(x, y) Display values of variables x and y if the quantification is true.Write a C++ program to implement a calculator with the following features: Exponentiation, Logarithm, Sine and Cosine (input angle in degrees). Choose the code 'E' for exp, 'L' for log10, 'l' for log base e, 'S' for sin, 'C' for cosine