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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/digital-systems-linear-algebra-introduction-to-formal-proof/3/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 3 / e / Solution by

Codex  0 2026-10-25

The five maximal four-cell groups give the disjunctive normal form

z = a_{1} \neg b_{1} \lor a_{1} a_{0} \lor a_{1} \neg b_{0} \lor a_{0} \neg b_{1} \lor \neg b_{1} \neg b_{0} .

(1)

For example, the group

a_{1} \neg b_{0}

wraps between columns

00

and

10

in the bottom two rows. No group of eight cells is entirely one. The five terms are precisely the prime implicants, and each is essential: respectively, the cells

(a, b) = (2, 1), (3, 3), (2, 2), (1, 1), (0, 0)

are covered by no other prime implicant. Any disjunctive normal form can have its terms enlarged to prime implicants without increasing its size. Therefore all five displayed terms are necessary, proving minimality in number of products; each needs two literals, also giving ten literals in a minimal cover.

For an implementation with AND gates and NOT gates, call these five products

T_{1}, \dots, T_{5}

. Form the shared complements

\neg b_{1}, \neg b_{0}

, calculate each

T_{i}

with an AND gate, invert every

T_{i}

, and use

z = \neg (\neg T_{1} \land \neg T_{2} \land \neg T_{3} \land \neg T_{4} \land \neg T_{5}) .

(2)

A five-input conjunction can be replaced by four two-input AND gates. This direct implementation uses nine two-input AND gates and eight NOT gates. That is a valid realization of the minimal disjunctive normal form, rather than a claim that this particular gate count is globally optimal among all factored circuits.

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