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Example 1:
•
•
•
•
•
•
•
•
LHS
xy+xy+x y = x+y
x(y+y)+ x y
using distributive law
x •1+ x y
using y+y = 1 theorem
x(1+ y ) +x y (1+y) =1 theorem
x+ x y + x y using distributive law
x + y(x+ x)
x + y• 1 = x +y RHS
•Example 2:
•(x+y)(x+z) = xz+xy using distributive law
•x x + x z + x y + y z using P5 postulate
•x z + x y +(x+x) yz using distributive law
•xz+ x y + xyz+xyz
using distributive law
•xz(1+y) +x y(1+z)
using T2 theorem
•xz•1 +xy • 1 = xz + xy RHS
•PRINCIPLE OF DUALITY
• One can transform the given expression by interchanging the operation (+) and
(•) as well as the identity elements 0 and 1 . Then the expression will be referred
as dual of each other. This is known as the principle of duality.
• Example x + x = 1 then the dual expression is
•
x•x =0
•BOOLEAN FORMULAS AND FUNCTIONS
•Boolean expressions or formulas are constructed by using Boolean constants and
variables with the Boolean operations like (+) , (•) and ‘not’
• Example: (x + y) z
• f(x,y.z) = (x +y) z or f = (x +y) z
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xyz
000
001
010
011
100
101
110
111
x y
0
0
0
0
0
0
1
1
xz
xyz
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
0
f
0
0
0
0
1
0
1
1
Truth table for the above Boolean expression is
•Example 2:
•
xyz
Write a truth table for following function
f=xyz+xy+xz
x y
xz
xyz
f
000
001
010
011
100
101
110
111
0
0
0
0
0
0
1
1
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
1
1
•NORMAL FORMULAS
• Boolean expression can be represented by
following structures
•1. Sum of products ( SOP)
or disjunctive normal form
•2. Product of sum (POS)
or Conjunctive form
•In SOP normal form is a Boolean formula that is written as a single product
term or as a sum ( also called disjunctive) of product terms is said to be in the
sum of product form or disjunctive normal form.
•Example:
•f(w,x,y,z) = x + w y + w y z
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•In the POS normal form is a Boolean formula which is written as a single
sum term or as a product of sum ( also called conjunctive) terms is said to be
in product of sums form or conjunctive normal form.
•Example:
• f(w,x,y,z) = z (x + y) (w + y + z)
•CANONICAL FORMULAS
• A procedure which will be used to write Boolean expressions form truth table
is known as canonical formula. The canonical formulas are of two types
•1. Minterm canonical formulas
•2. Maxterm canonical formulas
•1. MINTERM CANONICAL FORMULAS
•Minterms are product of terms which represents the functional values of the
variables appear either in complemented or un complemented form.
•Ex: f(x,y,z) = x y z + x y z + x y z
• The Boolean expression whichis represented above is also known as SOP or
disjunctive formula
•The
truth table is
x y z
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
f
0
1
0
1
1
0
0
0
•m- NOTATION
• To simplify the writing of a minterm in canonical formula for a function is
performed using the symbol mi. Where i stands for the row number for which the
function evaluates to 1.
•The m-notation for 3- variable an function Boolean function
• f(x,y,z) = x y z + x y z + x y z is written as
• f(x,y,z) = m1+ m3 + m4 or
• f(x,y,z) = m(1,3,4)
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•A three variable m- notation truth variable
Decimal designator of Minterm
xyz
row
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0
1
2
3
4
5
6
7
x
x
x
x
x
x
x
x
y
y
y
y
y
y
y
y
z
z
z
z
z
z
z
z
m-notation
m0
m1
m2
m3
m4
m5
m6
m7
•2. MAXTERM CANONICAL FORM
• Maxterm are sum terms where the variable appear once either in
complement or un-complement
forms and these terms corresponds to a functional value representing 0.
Ex. f(x,y,z) = ( x+ y+ z ) ( x+ y+z ) ( x + y + z )
= M( 0, 2, 5)
= M0, M2, M5
M-NOTATION
•A maxterm in a canonical form can be represented as Mi. Where i stands for
row number for which the the function evaluates to 0. A product
of
maxterms are represented as
M.
•MANIPULATIONS OF BOOLEAN FORMULAS
•A Boolean function is described by different formulas. i.e.by applying postulate
and theorems of Boolean algebra. Even it is possible to translate the Boolean
expressions in different forms.
•1 EQUATION COMPLEMENTATION
•For every Boolean function ‘f’ there is associated a complementary function
‘f’ in which
•f(x1,x2,….xn) = 1 if f(x1,x2,….xn) = 0 and
•f(x1,x2,….xn) = 0 if f(x1,x2,….xn) =1
•for all combinations values of x1,x2,….xn.
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•Example 1: f= x y + x y
•Example 2: f = w x z+ w ( x + y z)
•2. EXPANSION ABOUT VARIABLE
• There are occasions when it is desirable to single out a variable and
rewrite a Boolean formula f(x1,x2,…xi,…..xn) so that
•
xi g1 + xi g2
or
•
(xi + h1) (xi + h2)
•Where g1, g2, h1, and h2 are expression not containing the variable xi. This
theorem is known as Shannon expansion theorem.
•Shannon expansion theorem
•Theorem 1: f(x1, x2, …..xi, ….xn) can be represented in SOP form as
•= xi  f(x1,x2,…,1….xn) + xi f(x1,x2,…0,...xn)
•Theorem 2 : f(x1, x2, …..xi, ….xn) can be represented in POS form as
•= xi + f(x1,x2,…,0….xn)  xi+ f(x1,x2,…1,...xn)
•Example : f(w,x,y,z) = w x + (w x + y ) z
•3. EQUATION SIMPLIFICATION
• It is a direct method using postulates and theorem of a Boolean algebra
enable the manipulations of a formula into equivalent forms.
•Example: ( x+ x y) ( x + y) + y z
•4. THE REDUCTION THEOREM
• For the purpose of obtaining simple Boolean formula two additional
theorems are useful,
•These theorems are known as Shannon’s reduction theorems.
•theorem 3:
• xi• f( x1, x2, …..xi…..xn) = xi• f( x1, x2, …..1…..xn)
• xi+ f( x1, x2, …..xi…..xn) = xi+ f( x1, x2, …..0…..xn)
•Where f( x1, x2, …..k,…..xn) for k= 0,1.
•Similarly
•xi• f( x1, x2, …..xi…..xn) = xi• f( x1, x2, …..0…..xn)
•xi+ f( x1, x2, …..xi…..xn) = xi• f( x1, x2, …..1…..xn)
• Where f( x1, x2, …..k,…..xn) for k= 0,1.
•Example : f(w,x,y,z)= x+ x y + w x ( w+ z) ( y+ w z )
•5. MINTERM CANONICAL FORMULA
• 1. Apply DeMorgan’s law a sufficient no. of times until all the
NOT operations appear only with the single variables.
•2. Apply distributive of AND over OR (•) over (+)
i.e. x•(y+z) = xy+xz in order to manipulate the formula
into disjunctive normal formula.
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•3.
Remove duplicate literals and turns by idempotent law well as
any term that are identically zero (x•x = 0)
•4. If any product in the disjunctive normal formula does not
contain all the variables of the Boolean function then these
missing variables are introduced by ANDing the terms with
logic 1 in the form of xi+xi where xi is the missing variable being
introduced. This process is repeated for each missing variables in
each of the product turns of the disjunctive normal form
•5.
Apply distributive of ( •) over (+) again so that each
variable appears exactly once in each term.
6. Remove duplicate term if any.
Example : ( x + y) + ( y + x z )( x + y)
•6. MAXTERM CANONICAL FORMULA
•1. Apply DeMorgan’s law until all the NOT operations appear with single
variables.
•2. Apply distributive law of (+) over (•)
•
i.e. (x+yz) = (x+y)(x+z) and bring the expressions into its conjugate normal
form (POS).
•3. The missing variables are introduced into the sum terms by OR ing logic
0’s in the form xi •xi = 0 where xi is missing variable.
•4. Distributive law of (+) over (•) is again applied.
•5. Duplicate literals are deleted.
•Ex:
f(x,y,z) = ( x + y ) + ( y+ x z )( x + y )
•7. COMPEMENTS OF CANONICAL FORMULAS
• Even by taking complements minterm expression may result different
expressions.
• Ex : f(x,y,z) = m(0, 2, 4, 6) its complement expression is
•
f(x,y,z) = m(1, 3, 5,7)
•Similarly
• A Maxterm canonical expression may be represented in completed for as
follows
•Ex : f(x,y,z) =  M (1, 2, 4, 7) its complement expression is
•
f(x,y,z) =  M (0 3, 5,6)
•GATES AND COMBINATIONAL NETWORKS
• Gates are electronics circuits whose terminal characteristics correspond to the
various Boolean operations. Where as logic networks is a network of
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interconnections of gates and flip – flops. The gates are operated with two
possible steady state voltage signal values appear at the terminals of the
circuits at any time.
•An electronic circuit in which the output signal is a logic-1 if and only if all
its inputs are at logic-1 is called AND gate.
•An electronic circuit in which the output signal is a logic-1 if and only if at
least one of its input signal is at logic-1 is called OR gate.
3. An electronic circuit in which the output signal is always opposite logic with
input signal is called as NOT gate or inverter gate.
•COMBINATIONAL NETWORKS
•1. COMBINATIONAL NETWORK
• The inter connections 0f gates result in a gate network. If the network has
the property that its outputs at any time are determined strictly by the inputs
at that time then the network is said to be a combinational network. Ex.
Adders. Multiplexers. etc
• Let us consider an set of n signals at any time is called input state or input
vector of the network. While a set of resulting signals appearing at the m output
terminals is called the output state or output vector. The network can be expressed
as
•z1, z2,……zm as Boolean function then
•Zi = fi(x1,x2,….xn) for i= 1,2,…m.
•2. SEQUENTIAL NETWORKS
• A second type of logic network is the sequential networks. Sequential network
have memory property, so that the the outputs from these networks are
dependent not only upon the current inputs but upon previous input as well.
Feed back path are used in the sequential circuits. Ex. Counters. Shift registers
etc.
•ANALYSIS PROCEDURE
• Analysis procedure for a combinational network is as follows
•1. Each gate output that is only a function of the input variables is labeled.
•2. Boolean algebraic expression for the outputs of each of these gates are
then written.
•3. Next these gates outputs that are a function of just inputs variables and
previously labeled gate outputs.
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•4. Then each of the previously defined labels is replaced by the already
written Boolean expression and this process is continued until the output of
the network is labeled and till final expression is obtained.
•f(w,x,y,z) = w•(y+z) + wxy
•
= w•G1 + G2
•f(w,x,y,z) = G2 + G3
•SYNTHESIS PROCEDURE
•The synthesis procedure or logic design will be obtained by having
specification of the desired behavior of a gate network. The truth table will
provide total information of network specifications. If the variables and their
compliments are used as inputs then such logic is said to be DOUBLE –RAIL
logic. If the variables and their complements are not available, then not-gates
are required then it is called SINGLE – RAIL logic.
•The gate input signals will pass through input to out is called as levels of
logic.
•Example1: f(w,x,y,z) = w x +x (y+z)
•In the above logic circuit their exists 3- logic levels G1, G2 & G3 with double rail logic.
•Example 2: f(w,x,y,z) = w x +x y + x z
•In the above logic circuit there exists two level logic levels with double -rail.
•INCOMPLETE BOOLEAN FUNCTION WITH DONOT CARE
CONDITIONS
•In this type of Boolean functions having n-variables so that it may have 2n
combinations of subsets and if all values are not specified such functions are
called incompletely specified functions.
•Ex. Let us consider a three variable function with following truth table
•We can describe the incomplete function in the SOP form as
•f(x,y,z) = m(0,1,7) +dc(3,5)
•Also we can represent the function in the pos form as
•f(x,y,z) = M(2,4,6) +dc(3,5)
•The odd parity generation result output expression
•f(w,x,y,z) = m(0,3,5,6,9) +dc(10,11,12,13,14,15)
•f(w,x,y,z)=w x y z + x y z + x y z + x y z + w z
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•ADDITIONAL BOOLEAN OPERATIONS AND GATES
•The NAND and NOR gates are referred as universal gates or universal building
blocks.
•1. NAND FUNCTION
• nand(x1, x2, ….xn) = (x1,x2,….. xn)
Ay applying DeMorgan’s law we get
x1, x2,…..xn
= x1+x2+….+xn
•NAND- GATE REALIZATIONS
•Any Boolean functions can be realized by nand gates alone
•1. Apply DeMorgan’s law to the expression to obtain in unary operation i.e.
appear only with single variables. Draw the logic diagram using and and-or
gates.
•2. Replace each and –gate symbol by nand-gate and each or-gate symbol by
nand-gates.
•3. Check for bubbles occurring on all lines between two gate symbols. so that
two gates will be there.
•4. Whenever an input variable enters a gate symbol at a bubble then
Complement the variable of the gate. If the output has a bubble then insert
an output not-gate symbol.
•5. Replace all not-gates by a nand-gate equivalent if desired.
• Example: f(w,x,y,z) = w z + w z ( x + y)
•2. NOR FUNCTION
•nor(x1, x2, ….xn) = (x1+x2+…..+xn)
Ay applying DeMorgan’s law we get
x1+x2+….+xn =
x1, x2,…..xn
NOR-GATE REALIZATIONS
To realize an Boolean function we must follow the procedure listed below
•1. Apply DeMorgan’s law to the expression to obtain expression in unary
operations i.e. appear only with single variables. Draw the logic diagram using
and and-or gates.
•2. Replace each or –gate symbol by nor-gate and each and-gate symbol by norgates symbol.
•3. Check for bubbles occurring on all lines between two gate symbols. so that
two gates will be there.
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•4. Whenever an input variable enters a gate symbol at a bubble then
Complement the variable of the gate. If the output has a bubble then insert
an output not-gate symbol.
•5. Replace all not-gates by a nand-gate equivalent if desired.
•Example: f(w,x,y,z) = w z + w z ( x + y)
•EX-OR FUNCTION
•A EX-OR gate will have two input and a output. The EX-OR gate will be
logic 1 if and only if both inputs are not at same logic state.
• The Boolean output expression of EX-OR gate is
• f1 = x y +x y
•f2 = ( x + y )( x + y )
•f3 = x y + x y
•f4 = (x + y ) (x + y)
•EX-NOR FUNCTION
•A EX-NOR gate will have two input and a output. The EX-NOR gate will be
logic 1 if and only if both inputs are at same logic state.
• The Boolean output expression of EX-NOR gate is
• f1 = x y + x y
•f2 = ( x + y )( x + y )
•f3 = x y + x y
•f4 = ( x + y ) (x + y)
•The truth table of Ex-OR gate
x
0
0
1
1
y
0
1
0
1
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f =xy+xy
0
1
1
0
10
x
0
0
1
1
y
0
1
0
1
f =xy+x y
1
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0
1
•The truth table of Ex-NOR gate
Example 1:Using Boolean identities prove the following
xy+xy+xy =x+y
Example 2:Using perfect induction method prove the following identities.
f(x, y,z) = x z + x y = ( x + y)(x + z)
Example 3: Express the following in min term canonical formulas and construct the
truth table.
f(x,y,z) = m(0,2,4,5,7)
Example 4: Express the following in maxterm canonical formulas and construct the
truth table.
f(x,y,z) =  M(2,4,7)
Example 5:Express the following by a min -term canonical formulas without
constructing truth table
f(x,y,z) = x y + x z + z
Ans f = m(0,1,2,3,4,6)
Example 6: Minimize the following Boolean output expression using Boolean
algebra
f(A, B, C) = ( A +(A +B))(B B C ) ( C ABC )
CHAPTER 2: SIMPLICATION OF BOOLEAN EXPRESSIONS
SIMPLIFICATION OF BOOLEAN ALGEBRA :
The given Boolean output expression has to be simplified in terms its
equation nor the capacity. The following are the method used
1. Graphical method- Karnaugh method (K-map method)
2. Tabular method – Quine McCluskey method.
FORMULATION OF THE SIMPLIFICATION PROBLEM : The merit of the
network can be evaluated referring to the following
1. The cost of network.
2. The reliability of the network.
3. Propagation delay.
4. Fabrication density of the network.
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•
CRITERIA OF MINIMALITY : Minimal overall response time- minimum
levels, minimum terms to represent the Boolean function either in SOP
or POS, Single output network is to be considered.
•
SIMPLIFICATION OF PROBLEM :Using Boolean algebra X+X=1,
X.X= 0, X +X=X and X*X= X one can simplify the given Boolean
output expression without altering basic function . Only the no. Literal
sum or product terms are minimized.
PRIME IMPLICANTS AND IRRENDUNDANT DISJUCTIVE EXPRESSION :
IMPLIES: Consider two complete function of n -variables
f1 and f2 . The function f1 implies the function f2 if there is no assignment
of values to the n-variables that makes f1 equal to 1 and f2 equals to 0.
Ex. f1(x,y,z) = xy +yz and f2(x,y,z) = xy + yz+ x z
x y z
f1
f2
0
0
0
0
1
1
1
1
0
0
0
1
0
0
1
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
Similarly in the conjunctive normal formula if f3 implies f4 when if there is no
assignment of values to the n-variable makes f4 equal to 0 and f3 equals to 1.
Ex.f3(x,y,z) = (x+y)(y+z)(x+z) and
f4(x,y,z) = (x+y)(y+z)
SUSUMES: It is a comparison between two product terms or two sum terms is
also possible. A term t1 is said to be subsume of term t2 if all the literal of the
term t2 are also literal of the term t1.
Ex. x y z is subsume of the product x z
similarly ( x + y+ z ) is subsume of term (x + z)
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IMPLICANTS : A product term said to be implicant of a complete function if
the product term implies the function, which describes the functional values of
1.
The term x is equals to 0 for the four 3-tuples of (x,y,z) = (0,0,0), (0,0,1), (0,1,0)
and (0,1,1).
Hence the x is said to be implicant and other implicant that represent value
1 is y z , where it represent value 1 for two 3-tuples(x,y,z) = ( 0,0,1) and (
1,0,1).
Truth table
x y z
0
0
0
0
1
1
1
1
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
f
1
1
1
1
0
1
0
0
PRIME IMPLICANT: If the implicant does not subsumes any other
implicant with fewer literals of the same function. In other words if we remove
prime implicant term from the expression the remaining product terms no longer
implies the function
Ex. x and y z are prime implicants
Theorem 1: When he cost assigned by some criterion, for a minimal
Boolean formula is such that decreasing the number of literals in the
disjunctive normal formula does not increase the cost of the formula, there is
at lest one minimal disjunctive normal formula that corresponds to a sum of
the prime implicants.
IRRENDUNDANT DISJUNCTIVE NORMAL FORMULAS :An
irredundant disjunctive normal formula describing a complete function is
defined as an expression in SOP form such that (1) every product term in
the expression is a prime implicant and (2) no product term may be
eliminated from the expression without changing the function described
by the expression.
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IMPLICATE: A sum term is said to be implicate of a complete Boolean
function if the function implies the sum term. Which describes the function
value of 0.
Ex. A sum term x+y has the value 0 for two- 3 tuples (x,y,z) = (1,0,0)
and (1,1,0) The sum term x +y is implied by the function hence it is referred
as implicate of the function.
PRIME IMPLICATE: If the implicate does not subsumes any other
implicate with fewer literals of the same function. In other words if we remove
prime implicate term from the expression the remaining sum terms no longer
implies the function
Ex. x and (y + z) are prime implicates
x y z
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
f
0
0
0
0
1
0
1
1
IRRENDUNDANT CONJUNCTIVE NORMAL FORMULAS :An irredundant
conjunctive normal formula describing a complete function is defined as an
expression in POS form such that (1) every sum term in the expression is a
prime implicate and (2) no sum term may be eliminated from the
expression without changing the function described by the expression.
Note : It should be noted that prime implicates are dual concept of prime
implicants and irredundant conjunctive normal formulas are dual concept of
irredundant disjunctive normal formulas. Prime implicant f is exactly complement f
will be prime implicate and vice versa.
KARNAUGH MAPS ( K- MAP)
A method for graphically
determining implicants and implicates
of a Boolean function was developed by Veitch and modified by
Karnaugh . The method involves a diagrammatic representation of a Boolean
algebra. This graphic representation is called map.
It is seen that the truth table can be used to represent complete function of nvariables. Since each variable can have value of 0 or 1. The truth table has 2n
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rows. Each rows of the truth table consist of two parts (1) an n-tuple which
corresponds to an assignment to the n-variables and (2) a functional value.
A
Karnaugh
map (K-map) is a
such
that each of the
n-tuples
uniquely locates a cell on the map. The
tuples are
placed as entries in the
associated cell.
geometrical
configuration of 2n cells
corresponds to a row of a truth table
functional values assigned to the ncells, i.e. 0 or 1 are placed in the
An significant about the construction of K-map is the arrangement of
the cells. Two cells are physically adjacent within the configuration if and
only if their respective n-tuples differ in exactly by one element. So that the
Boolean law x+x=1 cab be applied to adjacent cells. Ex. Two 3- tuples (0,1,1)
and (0,1,0) are physically a djacent since these tuples vary by one element.
One variable : One variable needs a map of 21= 2 cells map as shown below
x
0
1
f(x)
f(0)
f(1)
TWO VARIABLE : Two variable needs a map of 22 = 4 cells
x y f(x,y)
0 0
f(0,0)
0 1
f(0,1)
1 0
f(1,0)
1 1
f(1,1)
THREE VARIABLE : Three variable needs a map of 23 = 8 cells. The
arrangement of cells are as follows
x y z
f(x,y,z)
0 0 0
f(0,0,0)
0 0 1
f(0,0,1)
0 1 0
f(0,1,0)
0 1 1
f(0,1,1)
1 0 0
f(1,0,0)
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1 0 1
1 1 0
1 1 1
f(1,0,1)
f(1,1,0)
f(1,1,1)
FOUR VARIABLE : Four variable needs a map of 24 = 16 cells. The arrangement
of cells are as follows
w x y z f(w,x,y,z)
w x y z
f(w,x,y,z)
0 0 0 0 f(0,0,0,0)
1 0 1 0
f(1,0,1,0)
0 0 0 1 f(0,0,0,1)
1 0 1 1
f(1,0,1,1)
0 0 1 0 f(0,0,1,0)
1 1 0 0
f(1,1,0,0)
0 0 1 1 f(0,0,1,1)
1 1 0 1
f(1,1,0,1)
0 1 0 0 f(0,1,0,0)
1 1 1 0
f(1,1,10)
0 1 0 1 f(0,1,0,1)
1 1 1 1
f(1,1,1,1)
0 1 1 0 f(0,1,1,0)
0 1 1 1 f(0,1,1,1)
1 0 0 0 f(1,0,0,0)
1 0 0 1 f(1,0,0,1)
Four variable K-map.
0000 0001
0011
0010
0100 0101
0111
1010
1100 1101
1111
1110
1000 1001
1011
1010
Ex. Obtain the minterm canonical formula of the three variable problem given below
f(x, y,z) = x y z+ x y z + x y z + x y z
f(x,y,z) = m(0,2,4,5)
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00
01
11
11
1
0
0
1
1
1
0
0
Ex. Express the minterm canonical formula of the four variable K-map given below
yz
00
01
11
10
1
1
0
1
1
1
0
0
0
0
0
0
1
0
0
1
f(w,x,y,z) = w x y z + w x y z + w x y z + w x y z + w x y z + w x y z
f(w,x,y,z) =  m(0, 1, 2, 4, 5,
Ex. Obtain the max term canonical formula
(POS) of the three variable problem stated above
f(x,y,z) = ( x + y +z)( x + y +z)(x + y +z)
(x + y +z)
f(x,y,z) = M(1,3,6,7)
Ex Obtain the max term canonical formula
(POS) of the four variable problem stated above
f(w,x,y,z) = (w + x + y + z) (w + x + y + z) (w + x + y + z)
(w + x + y + z) (w + x + y + z) (w + x + y + z)
(w + x + y + z) (w + x + y + z) (w + x + y + z)
f(w,x,y,z) = M(3,6,7,9,11,12,13,14,15)
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PRODUCT AND SUM TERM REPRESENTATION OF
K –MAP
1.The importance of K-map lies in the fact that it is possible to determine the
implicants and implicates of a function from the pattern of 0’s and 1’s appearing
in the map. The cell of a K-map has entry of 1’s is refereed as 1-cell and that of
0,s is referred as 0-cell.
2. The construction of an n-variable map is such that any set of 1-cells or 0cells which form a 2ax2b rectangular grouping describing a product or sum term with
n-a-b variables , where a and b are non-negative no.s
3. The rectangular grouping of these dimensions referred as Subcubes. The
subcubes must be the power of 2 i.e. 2 a+b equals to 1,2,4,8 etc.
4. For three variable and four variable K-map it must be remembered that the
edges are also adjacent cells or subcubes hence they will be grouped together.
5. Given an n-variable map with a pair of adjacent 1-cells or 0-cellscan result n-1
variable.Where as if a group of four adjacent subcubes are formed than it can
result n-2 variables. Finally if we have eight adjacent cells are grouped may
result n-3 variable product or sum term.
Typical pair of subcubes
wxz
1
1
1
1
1
1
1
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Typical group of four adjacent subcubes
1
1
1
1
1
1
1
1
1
1
1
1
Typical group of four adjacent subcubes.
1
1
1
1
1
1
1
1
Typical group of eight adjacent subcubes.
1 1 1 1
1 1 1 1
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1 1
1 1
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1 1
1 1
1 1 1 1
1 1 1 1
1
1
1
1
1
1
1
1
Typical map subcubes describing sum terms
0
0
0
0
0
0
0
0
0
0
0
0
0
0
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USING K-MAP TO OBTAIN MINIMAL EXPRESSION
COMPLETE BOOLEAN FUNCTIONS :
FOR
How to obtain a minimal expression of SOP or POS of given function is
discussed.
PRIME IMPLICANTS and K-MAPS :
CONCEPT OF ESSENTIAL PRIME IMPLICANT
00
01
11
10
0
0
0
1
0
0
1
1
f(x,y,z)= xy+ yz
ALGORITHM TO FIND ALL PRIME IMPLICANTS
A General procedure is listed below
1. For an n-variable map make 2n entries of 1’s. or 0’s.
2. Assign I = n , so that find out biggest rectangular group with
dimension 2ax2b = 2 n-1.
3. If bigger rectangular group is not possible I = I-1 form the subcubes
which consist of all the previously obtained subcube repeat the step till
all 1-cell or 0’s are covered.
4. Remaining is essential prime implicants
1. Essential prime implicants
2. Minimal sums
3. Minimal products
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MINIMAL EXPRESSIONS OF INCOMPLETE BOOLEAN FUNCTIONS
1. Minimal sums
2. Minimal products.
EXAMPLE TO ILLUSTRATE HOW TO OBTAIN ESSENTIAL PRIMES
1. f(x,y,z) = m(0,1,5,7)
Ans f(x,y,z) = xz + x y
2. f(w,x,y,z) = m(1,2,3,5,6,7,8,13)
Ans. f(w,x,y,z) = w z +w y+xyz+w x y z
MINIMAL SUMS
f(w,x,y,z)=m(0,1,2,3,5,7,11,15)
MINIMAL PRODUCTS
F(w,x,y,z)=m(1,3,4,5,6,7,11,14,15)
MINIMAL EXPRESSIONS OF INCOMPLETE BOOLEAN FUNCTIONS
f(W,X,Y,Z)=m(0,1,3,7,8,12) +dc(5,10,13,14)
QUINE – McCLUSKEY METHOD
Using K-maps for simplification of Boolean expressions with more than six
variables becomes a tedious and difficult task. Therefore a tabular method illustrate
below can be used for the purpose.
ALGORITHM FOR GENERATING PRIME IMPLICANTS
The algorithm procedure is listed below
1.Express each minterm of the function in its binary representation.
2. List the minterms by increasing index.
3. Separate the sets of minterms of equal index with lines.
4. Let i = 0.
5. Compare each term of index I with each term of index I+1. For each pair of
terms that can combine which has only one bit position difference.
6. Increase I by 1 and repeat step 5 . The increase of I continued until all terms are
compared. The new list containing all implicants of the function that have one less
variable than those implicants in the generating list.
7. Each section of the new list formed has terms of equal index. Steps 4,5, and 6 are
repeated on this list to form another list. Recall that two terms combine only if
they have their dashes in the same relative positions and if they differ in exactly
one bit position.
8. The process terminates when no new list is formed .
9. All terms without check marks are prime implicants.
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Example: Find all the prime implicants of the function
f(w,x,y,z) = m(0,2,3,4,8,10,12,13,14)
Step 1: Represent each minter in its 1-0 notation
no.
minterm
1-0 notation
index
0
2
3
4
8
10
12
13
14
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
wxyz
0000
0010
0011
0100
1000
1010
1100
1101
1110
0
1
2
1
1
2
2
3
3
Step 2: List the minterm in increasing order of their index.
No.
wxyz
index
0
2
4
8
3
10
12
13
14
0000
0010
0100
1000
0011
1010
1100
1101
1110
Index 0
Index 1
Index 2
Index 3
Wxyz
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0.2
0,4
0,8
2,3
2,10
4,12
8,10
8,12
10,14
12,13
12,14
00–0
0–00
- 000
001–
- 010
- 100
10–0
1–00
1–10
110–
11-0
wxyz
(0, 2, 8, 10)
(0, 4, 8,12 )
__ 0 __ 0
__ __ 0 0(index 0)
(8,10,12,14)
1__ __ 0 (index 1)
F(w,x,y,z)=x z + y z +w z+w x y +w x z
PETRICK’S METHOD OF DETERMINING IRREDUNDANT EXPRESSIONS
FIND THE PRIME IMPLICANTS AND IRREDUNDANT EXPRESSION
F(W,X,Y,Z)= M(0,1,2,5,7,8,9,10,13,15)
A=X Y , B= X Z C= Y Z D= X Z
P = (A+B)(A+C) (B)(C+D)(D)(A+B)(A+C)(B)(C+D)(D)
P = (A +C)(BD) = ABD +BCD
F1(W,X,Y,Z)= ABD =X Y +X Z +X Z
F2(W,X,Y,Z) = BCD = X Z + Y Z +X Z
DECIMAL METHOD FOR OBTAINING PRIME IMPLICANTS
The prime implicants can be obtained for decimal number represented minterms.In
this procedure binary number are not used to find out prime implicants
f(w, x,y,z) =m(0,5,6,7,9,10,13,14,15)
fsop= xy +xz+xyz+wyz+w x y z
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MAP ENTERED VARIABLE(MEV)
It is graphical approach using k-map to have a variable of order n. Where in we are
using a K-map of n-1 variable while map is entered with ouput function and
variable.
f(w,x,y.z) = m(2,3,4,5,13,15) +dc(8,9,10,11)
Ans.fsop= w z +x y + w x y
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