Minimization of Boolean Functions Last Updated : 19 May, 2025 Comments Improve Suggest changes Like Article Like Report Boolean functions are used to represent logical expressions in terms of sum of minterms or product of maxterms. Number of these literals (minterms or maxterms) increases as the complexity of the digital circuit increases. This can lead to large and inefficient circuits. By minimizing Boolean functions, we can reduce the number of logic gates, simplify circuit design, and improve performance in terms of speed, cost, and power consumption.For example, let the Boolean function:F2 = x’y’z + x’yz + xy’This function can be further minimized byGrouping first two terms: x’y’z + x’yz = x’(y’z + yz)Simplify inside: y’z + yz = z, so it becomes x’zAdd the remaining term: x’z + xy’The minimized Boolean function will be: F2 = xy' + x’zIn the diagram below we can see the implementation of the Boolean function:Minimization of Boolean FunctionInstead of building a circuit with 3 big parts (one for each term), the minimized version needs only 2. Read more about Representation of Boolean FunctionsVarious methods and techniques, such as Karnaugh maps, Quine-McCluskey algorithm, and the use of Boolean algebra, help achieve this simplification.Main Methods for Minimizing Boolean ExpressionsThe two main methods for minimizing Boolean expressions are:1. Boolean AlgebraBoolean algebra involves using a set of rules and laws (like distributive, associative, and complement laws) to simplify Boolean expressions. This method focuses on applying algebraic manipulations to reduce the complexity of the expression by eliminating redundant terms.Law/RuleExpressionIdentity LawA ⋅ 1 = A, A + 0 = ANull LawA ⋅ 0 = 0, A + 1 = 1Idempotent LawA ⋅ A = A, A + A = AComplement LawA ⋅ A′ = 0, A + A' = 1Domination LawA ⋅ 0 = 0, A + 1 = 1Double Negation Law(A′)′ = ADistributive LawA ⋅ (B + C) = A ⋅ B + A ⋅ CDe Morgan’s Law(A ⋅ B)′ = A′ + B', (A + B)′ = A′ ⋅ B′Absorption LawA ⋅ (A + B) = A, A + (A ⋅ B) = AComplementation LawA ⋅ A′ = 0, A + A′ = 1Consensus TheoremAB + A'C + BC = AB + A'CRead more about Boolean Algebraic TheoremsExample: Simplify the Boolean function F = AB + (AC)′ + AB′C(AB + C). Solution: F = AB + (AC)′ + AB′C(AB + C) = AB + A′ + C′+ AB′C.AB + AB′C.C = AB + A′ + C′ + 0 + AB′C (B.B′ = 0 and C.C = C) = ABC + ABC′ + A′ + C′ + AB′C (AB = AB(C + C′) = ABC + ABC′) = AC(B + B′) + C′(AB + 1) + A′ = AC + C′+A′ (B + B′ = 1 and AB + 1 = 1) = AC + (AC)′ = 1 2. K-MapThe Karnaugh Map is a graphical technique used to simplify Boolean expressions by grouping adjacent cells containing 1s (minterms). This visual method makes it easier to identify patterns and minimize the expression by combining terms that can be grouped together. It is especially useful for functions with 4 or fewer variables.Questions Based on Minimization of Boolean FunctionsExample: Minimize the following Boolean function using algebraic manipulation F = ABC'D' + ABC'D + AB'C'D + ABCD + AB'CD + ABCD' AB'CD' Solution: 1. Using Boolean LawsF = ABC'(D' + D) + AB'C'D + ACD(B + B') ACD'(B + B')= ABC' + AB'C'D + ACD + ACD'= ABC' + AB'C'D + AC(D + D') = ABC' + AB'C'D + AC = A(BC' + C) + AB'C'D = A(B + C) + AB'C'D = AB + AC + AB'C'D = AB + AC + AC'D = AB + AC + AD 2. Using K-Mapk-mapThe above figure highlights the prime implicants in green, red and blue. The green one spans the whole third row, which gives us – ABThe red one spans 4 squares, which gives us – ADThe blue one spans 4 squares, which gives us – ACSo, the minimized Boolean expression is - AB + AC + ADGATE CS Corner Questions:1. GATE CS 2012, Question 302. GATE CS 2007, Question 323. GATE CS 2014 Set-3, Question 174. GATE CS 2005, Question 185. GATE CS 2004, Question 176. GATE CS 2003, Question 457. 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