This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.
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You're on a roll — d) i) Deduce the Boolean expression for sum and carry, and then draw the logic circuit diagram. From the truth table: | A | B | Sum | Carry | |---|---|-----|-------| | 0 | 0 | 0 | 0 | | 0 | 1 | 1 | 0 | | 1 | 0 | 1 | 0 | | 1 | 1 | 0 | 1 | Step 1: Deduce the Boolean expression for Sum. The Sum output is 1 when A is 0 and B is 1, OR when A is 1 and B is 0. This corresponds to the XOR operation. Sum = A B Alternatively, in Sum of Products (SOP) form: Sum = AB + AB Step 2: Deduce the Boolean expression for Carry. The Carry output is 1 only when A is 1 AND B is 1. This corresponds to the AND operation. Carry = A · B Step 3: Draw the logic circuit diagram. The circuit consists of one XOR gate for the Sum output and one AND gate for the Carry output, with inputs A and B. ` A ---+--- XOR --- Sum | / | / B ---+-- | +--- AND --- Carry ` Note: Due to text-based limitations, the circuit diagram is represented using ASCII art. In a graphical representation, an XOR gate symbol and an AND gate symbol would be used. d) ii) Using Boolean identities, prove that XNOR gate is the complement of a XOR gate. Step 1: Write the Boolean expression for an XOR gate. The XOR operation (A B) is defined as: A B = AB + AB Step 2: Find the complement of the XOR expression. We need to prove that A B = A B. Let's take the complement of the XOR expression: A B = AB + AB Step 3: Apply De Morgan's Theorem (X+Y = XY). AB + AB = (AB) · (AB) Step 4: Apply De Morgan's Theorem again (XY = X+Y). = (A + B) · (A + B) Step 5: Simplify using the double complement rule (X = X). = (A + B) · (A + B) Step 6: Expand the expression (distributive law). = AA + AB + BA + BB Step 7: Apply Boolean identities (XX = 0 and X · 0 = 0). = 0 + AB + AB + 0 Step 8: Simplify the expression. = AB + AB Step 9: Recognize the resulting expression as the XNOR gate. The expression AB + AB is the definition of the XNOR operation (A B). Therefore, we have shown that: A B = A B This proves that the XNOR gate is the complement of an XOR gate. Got more? Send 'em!