In the given network of AND and OR gates, f can be written as:

2008

In the given network of AND and OR gates, f can be written as:

Cascade of alternating AND and OR gates from inputs X0, X1 through Xn producing output f

  1. A.

    X0X1X2Xn + X1X2Xn + X2X3Xn + ⋯ + Xn

  2. B.

    X0X1 + X2X3 + … + Xn-1Xn

  3. C.

    X0 + X1 + X2 + ⋯ + Xn

  4. D.

    X0X1 + X3Xn-1 + X2X3 + X5Xn-1 + ⋯ + Xn-2Xn-1 + Xn

Attempted by 93 students.

Show answer & explanation

Correct answer: A

Concept

A cascade of two-input gates is unfolded into one Boolean expression by reading the chain from the output end backwards and substituting each gate's output into the next. Two laws drive the simplification: the distributive law A·(B + C) = A·B + A·C, and the absorbing/identity behaviour of AND and OR. An alternating AND/OR chain collapses into a sum-of-products (SOP) whose terms are the chain's partial products.

Applying it to this network

Process the gates from the final stage backwards, letting each AND gate form a product and each OR gate accumulate a new term:

  1. The last OR gate outputs f and contributes the standalone term Xn.

  2. The AND gate before it multiplies the carried signal by Xn-1, contributing Xn-1Xn; the next OR re-injects Xn as a separate term again.

  3. Continuing back to the first AND gate (inputs X0, X1), every stage adds one more leading factor, so the accumulated terms are exactly the contiguous suffix products XkXk+1Xn for k = 0, 1, 2, …, n.

Taking the OR of all those suffix products gives:

f = X0X1X2Xn + X1X2Xn + X2X3Xn + ⋯ + Xn

A note on simplification

This is the form the circuit directly produces, and it is the written expression the question asks you to identify. It is worth noting that because every product term ends in Xn while Xn also appears alone, Boolean absorption (A·Xn + Xn = Xn) collapses the whole expression to just Xn. That fully reduced form, Xn, is not one of the listed choices, so the question is answered by the contiguous suffix-product expression above — the written form that the network yields before simplification.

Cross-check

  • Substituting a small chain (for example n = 4) into the gate network and into the expression X0X1X2X3X4 + X1X2X3X4 + X2X3X4 + X3X4 + X4 gives identical truth tables over all 32 input combinations.

  • The plain disjunction X0 + X1 + ⋯ + Xn does not share that truth table (it is 1 whenever any single input is 1, whereas f follows the suffix-product / absorbed-Xn behaviour), so a pure OR of all inputs is a different function.

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