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Lattice and Boolean Algebra






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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

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Slide 1

Lattice andBoolean Algebra

Slide 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

Slide 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

Slide 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

Slide 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

Slide 6

Examples

distributive

non-distributive

Lattice and Boolean Algebra

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 13

Examples

P({a,b,c})

{n  | n|30}

B3

Lattice and Boolean Algebra

Slide 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

Slide 15

De Morgan’s Theorem

  • De Morgan’s Laws hold

  • These equations can be generalized

Lattice and Boolean Algebra

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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


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