Question

Let XX be a 3-variable Boolean function that produces output as '1' when at least two of the input variables are '1'. Which of the following statement(s) is/are CORRECT, where a,b,c,d,ea, b, c, d, e are Boolean variables?

Choose every correct option
Official answer and solution
Official answer
B and D
From
Official GATE 2025 answer key

Solution

Write X(x,y,z)=xy+yz+zxX(x, y, z) = xy + yz + zx (the 3-input majority function) and simplify each expression.

The function X(x,y,z)=xy+yz+zxX(x, y, z) = xy + yz + zx is the 3-variable majority function. Note that for any u,v,wu, v, w:

  • X(u,v,v)=uv+v+uv=vX(u, v, v) = uv + v + uv = v.
  • If a=ba = b, then X(a,b,w)=aX(a, b, w) = a; if a≠ba \ne b (one is 00 and the other is 11), then X(a,b,w)=wX(a, b, w) = w.

Checking each option:

  • (A) False: Majority is not associative. For (a,b,c,d,e)=(1,1,0,0,0)(a, b, c, d, e) = (1, 1, 0, 0, 0), LHS=X(1,1,X(0,0,0))=X(1,1,0)=1\text{LHS} = X(1, 1, X(0,0,0)) = X(1,1,0) = 1, whereas RHS=X(X(1,1,0),0,0)=X(1,0,0)=0\text{RHS} = X(X(1,1,0), 0, 0) = X(1,0,0) = 0.
  • (B) True: In X(a,b,X(a,b,c))X(a, b, X(a, b, c)), if a=ba = b then both sides equal aa; if a≠ba \ne b then X(a,b,w)=wX(a, b, w) = w, so LHS=X(a,b,c)\text{LHS} = X(a, b, c).
  • (C) False: For (a,b,c,d)=(0,1,1,1)(a, b, c, d) = (0, 1, 1, 1), LHS=X(0,1,X(0,1,1))=X(0,1,1)=1\text{LHS} = X(0, 1, X(0, 1, 1)) = X(0, 1, 1) = 1, whereas RHS=X(0,1,0)⋅X(1,1,1)=0⋅1=0\text{RHS} = X(0, 1, 0) \cdot X(1, 1, 1) = 0 \cdot 1 = 0.
  • (D) True: Since X(a,c,c)=cX(a, c, c) = c and XX is symmetric in its three arguments, RHS=X(a,X(a,b,c),c)=X(a,c,X(a,b,c))=X(a,b,c)\text{RHS} = X(a, X(a, b, c), c) = X(a, c, X(a, b, c)) = X(a, b, c) by property (B).
Final answer

Options (B), (D)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question
24 of 65
Section
Computer Science and Information Technology

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