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In this lecture, we delve into the principles of Boolean algebra, focusing on methods for simplifying Boolean expressions, including the use of Karnaugh maps. We discuss techniques for transforming complex expressions into simpler forms, utilizing disjunctive normal form (DNF) and the importance of logical equivalences. The lecture also covers the design of logic circuits using basic Boolean operations, with examples illustrating how simplifications can optimize circuit designs. Gain a solid understanding of these essential concepts in computer programming and digital logic design.
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CS1022Computer Programming & Principles Lecture 9.2 Boolean Algebra (2)
Plan of lecture • Simplifying Boolean expressions • Karnaugh maps • Logic circuits CS1022
Simplifying Boolean expressions (1) • Given a Boolean expression, can we find a “simpler” equivalent expression? • “Simpler” = “fewer symbols” • Technique applied to disjunctive normal form (DNF) • Even though it might be more complex than original • Process: original expr. DNF simplified expr. • Technique restricted to at most 3 Boolean variables • Extension to more variables is possible CS1022
Simplifying Boolean expressions (2) • To simplify, let’s suppress “” • Just like “” in algebra can be dropped (e.g., 2y = 2 y) • For instance, the expression in disjunctive normal form (p q r) (p q r) (p q r) Will be represented as pqr pqr pqr • We can simplify the above expression as follows: CS1022
Simplifying Boolean expressions (3) • Compare pr pqr and pqr pqr pqr • They are equivalent – all changes done via a “law” • It is much simpler than original expression • Very important: simplification can be automated • Each step in the process is an operation • Sequence of steps terminates at some point • Notice: • First step puts together two minterms which differed by one symbol • Second step reduced two terms to one CS1022
Karnaugh maps (1) • Simplification done via a Karnaugh map • “Device” invented in the 1950s to help circuit design • Uses a visual display to indicate which pairs of minterms can be merged as a simpler expression • For Boolean expressions with 3 variables (p, q and r) • Karnaugh map is a table with 2 rows and 4 columns • Columns labelled with all possible disjunctions of p and q and their negations • Rows labelled with r and r CS1022
Karnaugh maps (2) • As we move from column to column precisely one change occurs in the column label • Cells correspond to 8 different minterms stemming from 3 Boolean variables • For a given Boolean expression, we write “1” in each cell whose row/column expression appears • For instance, pqr pqr pqr has map CS1022
Karnaugh maps (3) • Required simplification suggested by “clusters” of 1s • Shaded in green here • There is only one such cluster in the example • Corresponds to terms we combined using the laws CS1022
Karnaugh maps (3) • Relabelling columns preserves adjacency • Some clusters may be “hidden” • NB relabelling must be consistent with requirement that adjacent columns differ by one symbol only • Example: Karnaugh map of pqr pqr pqr • Relabelling columns produces an alternative Karnaugh map and reveals a pair of adjacent minterms CS1022
Karnaugh maps (4) • Hence, CS1022
Karnaugh maps (5) Let’s simplify pqr pqr pqrpqr pqr • Karnaugh map: • It contains a cluster of 4 (A) and a pair (B) • There are no hidden pairs • We must try relabelling exhaustively CS1022
Karnaugh maps (6) Cluster (A) corresponds to CS1022
Karnaugh maps (7) Cluster (B) corresponds to So, simplified expression is q pr CS1022
Logic circuits (1) • One of the main application of Boolean algebra is the design of binary devices • They accept a (finite) number of inputs • They produce a (finite) number of outputs • Binary: only two possible values for each input & output • Circuits as “black boxes”: ? q1 p1 q2 p2 q3 p3 ... ... qm pn CS1022
Logic circuits (2) • Devices are built of logic gates which perform basic Boolean operations (, , and ) OR gate ANDgate NOT gate a a a a a b a (a b) a b b b b NANDgate CS1022
Logic circuits (3) • Interconnecting gates produces a logical circuit • A logical circuit evaluates a Boolean expression • Example: What is the final output of this circuit? p 4 3 1 5 q 6 7 2 r CS1022
Logic circuits (4) • Example: What is the final output of this circuit? pqr pqr pqr pqr pqr pq pq pqr pqr pqr p p p p 3 5 5 4 3 1 4 1 3 3 1 5 4 1 5 4 q q q q 6 7 6 7 6 6 7 7 2 2 2 2 r r r r CS1022
Logic circuits (5) • Solution: pqr pqr pqr CS1022
Logic circuits (6) • Logic circuits can be simplified if we allow AND and OR gates to have more than 2 inputs • More dramatic simplifications w/ Karnaugh maps • 2 clusters (one hidden) • We can simplify things: • pqrpqr pq(r r) pq • pqr pqr (q q)pr pr • That is, pqr pqr pqr pq pr • We can further simplify this as p(q r) CS1022
Logic circuits (7) • With p(q r), we simplify previous circuit to obtain p q r CS1022
Further reading • R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd. 2002. (Chapter 9) • Wikipedia’s entry on Boolean algebra • Wikibooks entry on Boolean algebra CS1022