Minterm and Maxterm Relationship

In Boolean algebra, every logic function can be expressed in two canonical forms: the Sum of Products (SOP) using minterms, or the Product of Sums (POS) using maxterms. Understanding the relationship between minterms and maxterms is fundamental for digital circuit design and is a recurring topic in GATE examinations.

Core Concept Explanation

A minterm is a product (AND) term in which every variable appears exactly once, either in complemented or uncomplemented form, such that the term evaluates to 1 for exactly one combination of input values. For n variables there are exactly 2^n minterms, each corresponding to one row of the truth table. The minterm for row i is denoted mi.

A maxterm is a sum (OR) term in which every variable appears exactly once, such that the term evaluates to 0 for exactly one input combination. The maxterm Mi evaluates to 0 at row i and is 1 everywhere else. The two forms are fundamentally dual to each other under the principle of duality in Boolean algebra.

The critical relationship is: the complement of minterm mi equals maxterm Mi, and vice versa. That is, mi' = Mi and Mi' = mi. This follows directly because mi = 1 only at row i, so mi' = 0 only at row i, which is exactly the definition of Mi. This complement duality is the bridge between SOP and POS canonical forms.

A Boolean function expressed as a canonical SOP (sum of minterms) lists all rows where output is 1. Its complement function lists all rows where output is 0, giving the canonical POS (product of maxterms) of the original function. So if F = sum of minterms {m1, m3}, then F' = sum of minterms {m0, m2}, and F = product of maxterms {M0, M2}.

Mathematical Expression

For a function of n variables, if the set of minterm indices where F = 1 is denoted S, and the remaining indices form the set S' (where F = 0), then the two canonical forms are related as follows. The canonical SOP form is F = sum(mi for i in S), and the canonical POS form is F = product(Mi for i in S'). Together these give the conversion rule: every minterm index missing from the SOP appears as a maxterm index in the POS, and every maxterm index missing from the POS appears as a minterm index in the SOP.

Mathematically, for a minterm mi with index i written in binary as b(n-1)...b1 b0, the variable xk appears uncomplemented if bk = 1 and complemented if bk = 0. For the corresponding maxterm Mi, the rule is exactly reversed: xk appears uncomplemented if bk = 0 and complemented if bk = 1. This inversion is what makes mi' = Mi precise and algebraically consistent.

Practical Understanding

In circuit design, SOP and POS forms directly map to two-level AND-OR and OR-AND gate networks respectively. The minterm-maxterm relationship tells designers that converting between these two networks requires only identifying the complementary set of indices. If a truth table has more 1s than 0s, the POS form (using fewer maxterms) gives a more compact expression and vice versa for SOP.

This relationship also underlies the K-map method: grouping 1s gives a simplified SOP, while grouping 0s gives a simplified POS. Both approaches simplify the same function through the same duality principle. GATE problems frequently test whether students can switch between the two canonical forms given only one of them.

Given:
F(A,B,C) = sum of minterms (1, 3, 5, 7)  [3-variable function]

Why this formula applies:
The complement set of minterm indices gives maxterm indices for the same function.
Total minterms for 3 variables = 2^3 = 8, indices 0 to 7.

Formula:
F = product of Mi for all i NOT in the minterm list
Missing minterm indices = {0, 2, 4, 6}

Substitution:
M0 = (A+B+C), M2 = (A+B'+C), M4 = (A'+B+C), M6 = (A'+B'+C)

Calculation:
F = M0 . M2 . M4 . M6
  = (A+B+C)(A+B'+C)(A'+B+C)(A'+B'+C)

Final Answer:
F(A,B,C) = product of maxterms (0,2,4,6)
The SOP had 4 minterms; the POS also has 4 maxterms (equal split for this function).

Exam Tip: In GATE, if F = sum of minterms(S), then F expressed as product of maxterms uses ALL indices NOT in S. Do not complement the variable labels — only swap the index set. A very common mistake is to also complement the variable encoding inside each maxterm unnecessarily.

Minterm to Maxterm Conversion Mechanism

• Minterms correspond to rows where F = 1; maxterms correspond to rows where F = 0.
• The complement of minterm mi is maxterm Mi: mi' = Mi and Mi' = mi for the same index i.
• To convert SOP to POS: collect all minterm indices where F = 0 and write the product of those maxterms.
• Variable encoding in maxterm Mi is the bitwise complement of the variable encoding in minterm mi.
• For n variables, the total number of minterms in SOP plus maxterms in POS always equals 2^n.

Quick Revision

• Minterm mi = 1 only at row i; Maxterm Mi = 0 only at row i.
• Key relationship: mi' = Mi and Mi' = mi for same index i.
• SOP uses minterms (indices where F=1); POS uses maxterms (indices where F=0).
• Conversion rule: SOP minterm index set and POS maxterm index set are complements of each other within {0, ..., 2^n - 1}.
• Exam trap: Do not change variable complements when converting — only swap the index sets.
• For n variables: (number of minterms in SOP) + (number of maxterms in POS) = 2^n.
• Variable encoding: in mi, bit=1 means uncomplemented; in Mi, bit=1 means complemented — exact reversal.