CS1022 Computer Programming & Principles Lecture 2 Boolean Algebra

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 pqr  pqr  pqr We can simplify the above expression as follows: pqr  pqr  pqr  (prq  prq)  pqr Comm.& assoc. Laws pr(q q)  pqr Distr. Law pr  pqr Since q q = 1 CS1022

Simplifying Boolean expressions (3) Compare pr  pqr and pqr  pqr  pqr 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 pq pq pq pq r r CS1022

Karnaugh maps (2) pq pq pq pq r 1 r 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, pqr  pqr  pqr has map pq pq pq pq r 1 r CS1022

Karnaugh maps (3) pq pq pq pq r 1 r 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 pq pq pq pq r 1 r 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  pqr  pqr Relabelling columns produces an alternative Karnaugh map and reveals a pair of adjacent minterms pq pq pq pq r 1 r pq pq pq pq r 1 r CS1022

Karnaugh maps (4) pq pq pq pq r 1 r Hence, pqr  pqr  pqr  pqr  pr(q q) pqr  pr pq pq pq pq r 1 r CS1022

Karnaugh maps (5) Let’s simplify pqr  pqr  pqr  pqr  pqr It contains a cluster of 4 (A) and a pair (B) There are no hidden pairs We must try relabelling exhaustively pq pq pq pq r 1 r CS1022

Karnaugh maps (6) Cluster (A) corresponds to pq pq pq pq r 1 r pqr  pqr  pqr  pqr  (p  p)qr  (p  p)qr qr  qr q(r  r) q CS1022

Karnaugh maps (7) Cluster (B) corresponds to So, simplified expression is q  pr pq pq pq pq r 1 r pqr  pqr  pr(q  q) pr 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”: ? p1 q1 p2 q2 p3 q3 ... ... pn qm CS1022

Logic circuits (2) Devices are built of logic gates which perform basic Boolean operations (, , and ) a b a  b OR gate a b a  b AND gate a a NOT gate a b (a  b) NAND gate 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? 4 3 2 1 p q r 5 6 7 CS1022

Logic circuits (4) Example: What is the final output of this circuit? pq pq pqr pqr pqr pqr  pqr 4 3 2 1 p q r 5 6 7 4 3 2 1 p q r 5 6 7 4 3 2 1 p q r 5 6 7 4 3 2 1 p q r 5 6 7 pqr  pqr  pqr CS1022

Logic circuits (5) Solution: pqr  pqr  pqr Gate Inputs Output 1 2 p, q pq 3 pq, r pqr 4 pq, r pqr 5 pq, r pqr 6 pqr, pqr pqr  pqr 7 pqr  pqr, pqr pqr  pqr  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: pqr  pqr  pq(r  r)  pq pqr  pqr  (q  q)pr  pr That is, pqr  pqr  pqr  pq  pr We can further simplify this as p(q  r) pq pq pq pq r 1 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