Download

Lattice and Boolean Algebra






Advertisement
/ 24 []
Download Presentation
Comments
barb
From:
|  
(1124) |   (0) |   (0)
Views: 96 | Added:
Rate Presentation: 0 0
Description:
Lattice and Boolean Algebra. Algebra. An algebraic system is defined by the tuple A,o 1 , …, o k ; R 1 , …, R m ; c 1 , … c k , where, A is a non-empty set, o i is a function A p i A, p i is a positive integer, R j is a relation on A, and c i is an element of A. Lattice.
Lattice and Boolean Algebra

An Image/Link below is provided (as is) to

Download Policy: Content on the Website is provided to you AS IS for your information and personal use only and may not be sold or licensed nor shared on other sites. SlideServe reserves the right to change this policy at anytime. While downloading, If for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.











- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -




Lattice and boolean algebraSlide 1

Lattice andBoolean Algebra

AlgebraSlide 2

Algebra

  • An algebraic system is defined by the tuple A,o1, …, ok; R1, …, Rm; c1, … ck, where, A is a non-empty set, oi is a function Api A, pi is a positive integer, Rj is a relation on A, and ci is an element of A.

Lattice and Boolean Algebra

LatticeSlide 3

Lattice

  • The lattice is an algebraic system A, , , given a,b,c in A, the following axioms are satisfied:

    • Idempotent laws: a  a = a, a  a = a;

    • Commutative laws: a  b = b  a, a  b = b  a

    • Associative laws: a  (b  c) = (a  b)  c, a  (b  c) = (a  b)  c

    • Absorption laws: a  (a  b) = a, a  (a  b) = a

Lattice and Boolean Algebra

Lattice exampleSlide 4

Lattice - Example

  • Let A={1,2,3,6}.

  • Let a  b be the least common multiple

  • Let a  b be the greatest common divisor

  • Then, the algebraic system A, ,  satisfies the axioms of the lattice.

Lattice and Boolean Algebra

Distributive latticeSlide 5

Distributive Lattice

  • The lattice A, ,  satisfying the following axiom is a distributive lattice

    5. Distributive laws: a  (b  c) = (a  b)  (a  c), a  (b  c) = (a  b)  (a  c)

Lattice and Boolean Algebra

ExamplesSlide 6

Examples

distributive

non-distributive

Lattice and Boolean Algebra

Complemented latticeSlide 7

Complemented Lattice

  • Let a lattice A, ,  have a maximum element 1 and a minimum element 0. For any element a in A, if there exists an element xa such that a  xa = 1 and a  xa = 0, then the lattice is a complemented lattice.

  • Find complements in the previous example

Lattice and Boolean Algebra

Boolean algebraSlide 8

Boolean Algebra

  • Let B be a set with at least two elements 0 and 1. Let two binary operations  and , and a unary operation  are defined on B. The algebraic system B, ,  , , 0,1 is a Boolean algebra, if the following postulates are satisfied:

    • Idempotent laws: a  a = a, a  a = a;

    • Commutative laws: a  b = b  a, a  b = b  a

    • Associative laws: a  (b  c) = (a  b)  c, a  (b  c) = (a  b)  c

    • Absorption laws: a  (a  b) = a, a  (a  b) = a

    • Distributive laws: a  (b  c) = (a  b)  (a  c), a  (b  c) = (a  b)  (a  c)

Lattice and Boolean Algebra

Boolean algebra1Slide 9

Boolean Algebra

  • Involution:

  • Complements: a  a = 1, a  a = 0;

  • Identities: a  0 = a, a  1 = a;

  • a  1 = 1, a  0 = 0;

  • De Morgan’s laws:

Lattice and Boolean Algebra

Huntington s postulatesSlide 10

Huntington’s Postulates

  • To verify whether a given algebra is a Boolean algebra we only need to check 4 postulates:

    • Identities

    • Commutative laws

    • Distributive laws

    • Complements

Lattice and Boolean Algebra

ExampleSlide 11

Example

  • prove the idempotent laws given Huntington’s postulates:

    a = a  0

    = a  aa

    = (a  a)  (a  a)

    = (a  a)  1

    = a  a

Lattice and Boolean Algebra

Models of boolean algebraSlide 12

Models of Boolean Algebra

  • Boolean Algebra over {0,1}

    B={0,1}. B, ,  , , 0,1

  • Boolean Algebra over Boolean Vectors

    Bn = {(a1, a2, … , an) | ai  {0,1}}

    Let a=(a1, a2, … , an) and b = (b1, b2, … , bn)  Bn

    define

    a  b = (a1  b1, a2  b2, … , an  bn)

    a  b = (a1  b1, a2  b2, … , an  bn)

    a=(a1, a2, … , an)

    then Bn, ,  , , 0,1 is a Boolean algebra, where, 0 = (0,0, …, 0) and 1 = (1,1, …, 1)

  • Boolean Algebra over Power Set

Lattice and Boolean Algebra

Examples1Slide 13

Examples

P({a,b,c})

{n  | n|30}

B3

Lattice and Boolean Algebra

Isomorphic boolean algebraSlide 14

Isomorphic Boolean Algebra

  • Two Boolean algebras A, ,  , , 0A,1A and B, ,  , , 0B,1B are isomorphic iff there is a mapping f:AB, such that

  • for arbitrary a,b  A, f(ab) = f(a)f(b), f(a  b) = f(a)  f(b), and f(a) = f(a)

  • f(0A ) = 0B and f(1A ) = 1B

    An arbitrary finite Boolean algebra is isomorphic to the Boolean algebra Bn, ,  , , 0,1

    Question: define the mappings for the previous slide.

Lattice and Boolean Algebra

De morgan s theoremSlide 15

De Morgan’s Theorem

  • De Morgan’s Laws hold

  • These equations can be generalized

Lattice and Boolean Algebra

DefinitionSlide 16

Definition

  • Let Bn, ,  , , 0,1 be a Boolean algebra. The variable that takes arbitrary values in the set B is a Boolean variable. The expression that is obtained from the Boolean variables and constants by combining with the operators ,  , and parenthesis is a Boolean expression. If a mapping f:Bn B is represented by a Boolean expression, then f is a Boolean function. However, not all mappings f:Bn B are Boolean functions.

Lattice and Boolean Algebra

TheoremSlide 17

Theorem

  • Let F(x1, x2, …, xn) be a Boolean expression. Then the complement of the complement of the Boolean expression F(x1, x2, …, xn) is obtained from F as follows

    • Add parenthesis according to the order of operations

    • Interchange  with 

    • Interchange xi with xi

    • Interchange 0 with 1

      Example

Lattice and Boolean Algebra

Principle of dualitySlide 18

Principle of Duality

  • In the axioms of Boolean algebra, in an equation that contains , , 0, or 1, if we interchange  with  , and/or 0 with 1, then the other equation holds.

Lattice and Boolean Algebra

Dual boolean expressionsSlide 19

Dual Boolean Expressions

  • Let A be a Boolean expression. The dual AD is defined recursively as follows:

    • 0D = 1

    • 1D = 0

    • if xi is a variable, then xiD = xi

    • if A, B, and C are Boolean expressions, and A = B  C, then AD= BD  CD

    • if A, B, and C are Boolean expressions, and A = B  C, then AD= BD  CD

    • if A and B are Boolean expressions, and A = B, then

Lattice and Boolean Algebra

Examples2Slide 20

Examples

  • Given xy  yz = xy  yz  xz

    the dual (x  y)(y  z) = (x  y)(y  z)(x  z)

  • Consider the Boolean algebra B={0,1,a,a} check if f is a Boolean function.

    f(x) = xf(0)  xf(1)

    f(x) = x  a  x  1

    f(a) = a  a  a  1 = a

Lattice and Boolean Algebra

Logic functionsSlide 21

Logic Functions

  • Let B = {0,1}. A mapping Bn B is always represented by a Boolean expression–a two-valued logic function.

    f  g = h  f(x1,x2,…,xn)  g(x1,x2,…,xn) = h(x1,x2,…,xn)

    f = g  f(x1,x2,…,xn) = g(x1,x2,…,xn)

Example

Lattice and Boolean Algebra

Logical expressionsSlide 22

Logical Expressions

  • Constants 0 and 1 are logical expressions

  • Variables x1,x2,…,xn are logical expressions

  • If E is a logical expression, then E is one

  • If E1 and E2 are logical expressions, then (E1 E2) and (E1 E2) are also logical expressions

  • The logical expressions are obtained by finite application of 1 - 4

Lattice and Boolean Algebra

Evaluation of logical expressionsSlide 23

Evaluation of logical Expressions

  • An assignment mapping :{xi} {0,1} (i = 1, … , n)

  • The valuation mapping |F| of a logical expression is obtained:

    • |0| = 0 and |1| = 1

    • If xi is a variable, then | xi | = (xi)

    • If F is a logical expression, then |F| = 1 |F| = 0

    • If F and G are logical expressions, then |F  G| = 1 (|F| = 1 or |G| = 1)

    • If F and G are logical expressions, then |F  G| = 1 (|F| = 1 and |G| = 1)

      Example: F:x  y  z

      (x) = 0, (y) = 0, (z) = 1

Lattice and Boolean Algebra

Equivalence of logic expressionsSlide 24

Equivalence of Logic Expressions

  • Let F and G be logical expressions. If |F| = |G| hold for every assignment , then F and G are equivalent ==> F  G

  • Logical expressions can be classified into 22n equivalence classes by the equivalence relation ()

Lattice and Boolean Algebra


Copyright © 2014 SlideServe. All rights reserved | Powered By DigitalOfficePro