CS1022 Computer Programming &

Principles

Lecture 9.1Boolean Algebra (1)

Plan of lecture• Introduction• Boolean operations• Boolean algebra vs. logic• Boolean expressions• Laws of Boolean algebra• Boolean functions• Disjunctive normal form

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IntroductionBoolean algebra• Is a logical calculus– Structures and rules applied to logical symbols– Just like ordinary algebra is applied to numbers

• Uses a two-valued set {0, 1}– Operations: conjunction, disjunction and negation

• Is related with– Propositional logic and– Algebra of sets

• Underpins logical circuits (electronic components)

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Boolean operations (1)• Boolean algebra uses set B 0, 1 of values• Operations: – Disjunction (),– Conjunction () and – Negation ()

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0 10 0 01 0 1

0 10 0 11 1 1

1 00 1

Boolean operations (2)• If p and q are Boolean variables– “Variables” because they vary over set B 0, 1 – That is, p, q B

• We can build the following tables:

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p q p q p q0 0 0 00 1 1 01 0 1 01 1 1 1

p p0 11 0

Boolean algebra vs. logicBoolean algebra is similar to propositional logic• Boolean variables p, q,... like propositions P, Q,...• Values B 0, 1 like F (false) and T (true)Operators are similar: • p q similar to P or Q• p q similar to P and Q• p similar to not PTables also called “truth tables” (as in logic)

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• As in logic, we can build more complex constructs from basic variables and operators , and – Complex constructs are called Boolean expressions

• A recursive definition for Boolean expressions:– All Boolean variables (p, q, r, s, etc.) are Boolean exprns.– If α and β are Boolean expressions then• α is a Boolean expression (and so is β)• (α β) is a Boolean expression• (α β) is a Boolean expression

• Example: ((p q) (r s))

Boolean expressions

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• Two Boolean expressions are equivalent if they have the same truth table– The final outcome of the expressions are the same, for

any value the variables have• We can prove mechanically if any two expressions

are equivalent or not– “Mechanic” means “a machine (computer) can do it”– No need to be creative, clever, etc.– An algorithm explains how to prove equivalence

Equivalence of expressions

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• Some equivalences are very useful– They simplify expressions, making life easier– They allow us to manipulate expressions

• Some equivalences are “laws of Boolean algebra”– We have, in addition to truth-tables, means to operate

on expressions– Because equivalence is guaranteed we won’t need to

worry about changing the meaning

Laws of Boolean algebra (1)

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Laws of Boolean algebra (2)

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Commutative Laws

p q q pp q q p

Associativity Laws

p (q r) (p q) rp (q r) (p q) r

Distributive Laws

p (q r) (p q) (p r)p (q r) (p q) (p r)

Idempotent Laws

p p pp p p

Absorption Laws

p (p q) pp (p q) p

De Morgan’s Laws

(p q) p q(p q) p q

• Do they look familiar?– Logics, sets & Boolean algebra

are all related• Summary of correspondence:

Laws of Boolean algebra (3)

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Logical operation

Set operation

Boolean operation

not or

and

• Very important:– Laws shown with Boolean variables (p, q, r, etc.)– However, each Boolean variable stand for an expression– Wherever you see p, q, r, etc., you should think of “place

holders” for any complex expression– A more “correct” way to represent e.g. idempotent law:

Laws of Boolean algebra (4)

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• What do we mean by “place holders”?– The laws are “templates” which can be used in many

different situations• For example:

((p q) (q s)) ((p q) (q s))((p q) (q s)) ((p q) (q s))

((p q) (q s))((p q) (q s))

Laws of Boolean algebra (5)

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• We should be able to prove all laws– We know how to build truth tables (lectures 2.1 & 2.2)– Could you work out an algorithm to do this?– There is help at hand...

Laws of Boolean algebra (6)

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• Using the laws of Boolean algebra, show that– (p q) (p q) is equivalent to p

• Solution:(p q) (p q) (p q) (p q) De Morgan’s

(p q) (p q)since p p p (q q) Distr. Law p 0 since (q q) 0 p from def. of

Laws of Boolean algebra (7)

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• A Boolean function of n variables p1, p2, ..., pn is– A function f : Bn B – Such that f(p1, p2, ..., pn) is a Boolean expression

• Any Boolean expression can be represented in an equivalent standard format– The so-called disjunctive normal form

• Format to make it easier to “process” expressions– Simpler, fewer operators, etc.

Boolean functions (1)

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• Example: function minterm m(p, q, r)– It has precisely one “1” in the final column

of truth table–m(p, q, r) 1 if p 0, q 1 and r 1

•We can define m(p, q, r) p q r• The expression p q r is the product

representation of the minterm m

Boolean functions (2)

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p q r m0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 01 1 0 01 1 1 0

• Any minterm m(p1, p2, ..., pr) can be represented as a conjunction of the variables pi or their negations

• Here’s how:– Look at row where m takes value 1– If pi 1, then pi is in the product representation of m

– If pi 0, then pi is in the product representation of m–We thus have m(p, q, r) p q r

Boolean functions (3)

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p q r m0 1 1 1

• We can express any Boolean function uniquely as a disjunction (a sequence of “”) of minterms– This is the disjunctive normal form

• The idea is simple:– Each minterm is a conjunction (sequence of “”) which, if

all holds, then we have output (result) “1” – The disjunction of minterms lists all cases which, if at

least one holds, then we have output (result) “1”– In other words: we list all cases when it should be “1” and

say “at least one of these is 1”

Disjunctive normal form (1)

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• Find disjunctive normal form for (p q) (q r)• Let f (p q) (q r). Its truth table is as below• The minterms are (rows with f 1)– p q r– p q r– p q r– p q r

• Disjunctive normal form is(p q r) (p q r) (p q r) (p q r)

Disjunctive normal form (2)

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p q r f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

p q r f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

p q r f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

p q r f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

p q r f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

p q r f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

• Any Boolean function can be represented with Boolean operators , , • This is a complete set of operators– “Complete” = “we can represent anything we need”

• It is not minimal, though:– De Morgan’s (p q) p q, so (p q) (p q)–We can define “” in terms of ,

•Minimal set of Boolean operators: , or , – Actually, any two operators suffice

•Minimality comes at a price: expressions become very complex and convoluted

Minimal set of operators

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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

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