stld notes unit-3 - webnodefiles.harishpola.webnode.com/200000080-848a48583d/unit-3_jwfiles.pdf ·...

Dr. P. Sudhakara Rao, Dean Switching Theory and Logic Design

Vignan Institute of Technology and Science P a g e | 1

UNIT -3 MINIMIZATION OF SWITCHING FUNCTIONS

o Map method o Prime implicants o don’t care combinations o Minimal SOP and POS forms o Tabular Method o Prime – Implicant chart o Simplification rules

Hour No Date Topic/s

04.12.2010 Map method Prime implicants, don’t care combinations, Minimal SOP and POS forms, Tabular Method, Prime – Implicant chart, simplification rules

INTRODUCTION

Minimization of switching functions is to obtain logic circuits with least circuit complexity. This goal is very

difficult since how a minimal function relates to the implementation technology is important. For example, If

we are building a logic circuit that uses discrete logic made of small scale Integration ICs(SSIs) like 7400 series,

in which basic building block are constructed and are available for use. The goal of minimization would be to

reduce the number of ICs and not the logic gates. For example, If we require two 6 and gates and 5 Or gates,

we would require 2 AND ICs(each has 4 AND gates) and one OR IC. (4 gates). On the other hand if the same

logic could be implemented with only 10 nand gates, we require only 3 ICs. Similarly when we design logic on

Programmable device, we may implement the design with certain number of gates and remaining gates may

not be used. Whatever may be the criteria of minimization we would be guided by the following:

PRIME IMPLICANTS : Consider the function

f (x,y,z) = x’y’z + x’yz + xy’z + xyz + xyz’

= z x’ (y’ +y) + zx (y’ + y) + xyz’

= z x’ +zx + xyz’

= z (x’+x) + xyz’

= z +xyz’

= z + xy

The function cannot be reduced further. The terms z and xy have lowest number of litarals. They are the prime

implicants (discussed later) of the function. Original terms are implicants of the function. Also the terms

derived in subsequent steps in the minimisation process are the implicants of the function, since when any

term takes the value 1 the function becomes 1.

Definition: A product term is an implicant of a function f, if it implies f. Any term that appears in the function’s

sum of products form is an implicant of the function.

Example

f (x,y,z)= x’z + xy’ +xy’z, all the terms here are implicants of f. A minterm of a function is its implicant.

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Definition: A prime implicant of a function is a product term that implies the function but does not imply any

other implicant of the function.

Example

As an example consider terms E1, E2, E3 and the function f as shown in table 6.1

E1 E2 E3 f Comments

0 0 1 0 E3 is not an implicant of f

0 1 0 1

1 1 1 1 E1 implies E2 implies f

1 1 0 1 E2 is prime implicant of f

0 0 0 0

0 0 0 0

1 1 1 1

0 0 0 0

Example

Consider a function (x,y,z) = x + yz (For table shown below) . Its canonical form is 3,4,5,6,7 Truth table of the

function f along with some product terms is shown below. From the table, we observe:

If xy’ = 1, then f = 1, i.e., xy’→f, Also xy’ does not imply any other implicant hence, it is prime implicant of f.

Similarly whenever x’yz = 1, yz =1 and also f =1, hence, x’yz → yz→f. Therefore x’yz is not a prime implicant

since it implies yz which in turn implies f. The term x→f, and, x does not imply any other implicant, hence x is a

prime implicant of f. Similarly xy’ →x →f, xy’ is not a prime implicant of f, since it implies x which is another

implicant. It may be seen that prime implicants are the terms that are obtained after eliminating largest

number of variables of the function by combining the minterms.

xyz yz x xy xy’ x’yz f(x.y.z)

000 0 0 0 0 0 0

001 0 0 0 0 0 0

010 0 0 0 0 0 0

011 1 0 0 0 1 1

100 0 1 0 1 0 1

101 0 1 0 1 0 1

110 0 1 1 0 0 1

111 1 1 1 0 0 1

Consider a space of 3 Boolean variables, which is a cube of 8 vertices as shown in Fig. above. There are 3

coordinates x,y and z corresponding to the three variables of the function. Each vertex is the point in the space

and its coordinates are 3-bit binary numbers. The function has value 1 at some points i.e., combinations

(minterms of function). These points are called 1 vertices of the functions. The 1-vertices are marked with the

filled circles. End points of any line in Fig differ only in one variable. In general, n-dimensional space will have

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2n vertices and n dimensional cube as its space. . A subspace is any space with lesser dimension than the

original space. For example in the Fig, A face of the cube (2-dimensional space ) is a sub space or subcube, so is

the line(one-dimensional subcube) and point (0-dimensional subcube). A subcube with function value =1 at all

its point is called 1-subcube. The prime implicant of a function is the largest 1- subcube. An Implicant of the

function is essentially a 1-subcube. In the Fig. above, the face BDHF has all four vertices circles (function f =1) so

it is the prime implicant of f, since no 1-subcube of higher dimension encloses it. Other prime implicants are

subcube GH and sub-cube AB. Both of these are lines not lying entirely in any face having all 1-vertices. In

general in the n-dimensional space, we pick up all largest 1-subcubes such that any common elements do not

form any of these cubes in entirely. In other words pick up largest 1-subcube such that a cube is not covered by

another 1-cube. A 1-cube covered by another 1-cube is only an implicant of the function and not the prime

implicant.

Karnaugh maps are useful in finding minimal expression for functions b y visual inspection of the space of

variables. Subcubes are arranged in way such that 1-vertices of f are neighbors. K-map is the representation of

the n-dimensional space on 2D plane. This also can be looked upon as truth table of the functions arranged in

special way. Consider the 2 variable function f(x,y) = x + y. The truth table of this function has 4 entries. This

table is arranged in 2 –dimensions as shown in Fig.

Two variable maps

The function f(x,y) = x + y is x’y + xy’ + xy (f = 1,2,3). The 2-variable K-map is shown in Fig.6.2, which is nothing

but its truth table shown in different forms. In Fig.6.2(b) it shown as one-dimensional array. Note that, we have

swapped the places for 10 and 11 combinations. This is to make the all physically neighboring combinations as

logical neighbors (differing only in 1-bit position). The end to end cells are neighbors (by rap around). This special

way of coordinate numbering makes us to visualize the largest 1-subcubes . In the present case, we have the

cells with 1 value can be combined to form the prime implicant These are covered showing by covering ellipses

The terms 10 and 11 are covered giving the term (10+11 = xy’ + xy = x(y + y’ )) = x, similarly the terms 01 and 11

are covered by y (x’y + xy = y). These are the prime implicants of the function and both are required in the

minimal expression.

F(x,y) = x+y

xy F

00 0 y

01 1 x 0 1

10 1 0 0 1

11 1 1 1 1

Minimization of Boolean expressions using Karnaugh maps.

Given the following truth table for the majority function.

000 0

001 0

010 0

011 1

100 0

101 1

110 1

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The Boolean algebraic expression is

m = a'bc + ab'c + abc' + abc.

We have seen that the minimization is done as follows.

m = a'bc + abc + ab'c + abc + abc' + abc

= (a' + a)bc + a(b' + b)c + ab(c' + c)

= bc + ac + ab

The abc term was replicated and combined with the other terms. To use a Karnaugh map we draw the

following map which has a position (square) corresponding to each of the 8 possible combinations of the 3

Boolean variables. The upper left position corresponds to the 000 row of the truth table, the lower right

position corresponds to 110. Each square has two coordinates, the vertical coordinate corresponds to the value

of variable a and the horizontal corresponds to the values of b and c.

The 1s are in the same places as they were in the original truth table. The 1 in the first row is at position 011

(a = 0, b = 1, c = 1). The vertical coordinate, variable a, has the value 0. The horizontal coordinates, the

variables b and c, have the values 1 and 1. The minimization is done by drawing circles around sets of adjacent

1s. Adjacency is horizontal, vertical, or both. The circles must always contain 2n 1s where n is an integer.

We have circled two 1s. The fact that the circle spans the two possible values of a (0 and 1) means that

the a term is eliminated from the Boolean expression corresponding to this circle. The bracketing lines shown

above correspond to the positions on the map for which the given variable has the value 1. The bracket

111 1

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delimits the set of squares for which the variable has the value 1. We see that the two circled 1s are at the

intersection of sets b and c, this means that the Boolean expression for this set is bc.

Now we have drawn circles around all the 1s. The left bottom circle is the term ac. Note that the circle spans

the two possible values of b, thus eliminating the b term. Another way to think of it is that the set of squares in

the circle contains the same squares as the set a intersected with the set c. The other circle (lower right)

corresponds to the term ab. Thus the expression reduces to bc + ac + ab as we saw before.

What does adjacency and grouping the 1s together have to do with minimization? Notice that the 1 at position

111 was used by all 3 circles. This 1 corresponds to the abc term that was replicated in the original algebraic

minimization. Adjacency of 2 1s means that the terms corresponding to those 1s differ in one variable only. In

one case that variable is negated and in the other it is not.

For example, in the first map above, the one with only 1 circle. The upper 1 is the term a'bc and the lower

is abc. Obviously they combine to form bc ( a'bc + abc = (a' + a)bc = bc ). That is exactly what we got using the

map. The map is easier than algebraic minimization because we just have to recognize patterns of 1s in the

map instead of using the algebraic manipulations. Adjacency also applies to the edges of the map. Let's try

another 3 variable map.

At first it may seem that we have two sets, one on the left of the map and the other on the right. Actually there

is only 1 set because the left and right are adjacent as are the top and bottom. The expression for all 4 1s is c'.

Notice that the 4 1s span both values of a (0 and 1) and both values of b (0 and 1). Thus, only the c value is left.

The variable c is 0 for all the 1s, thus we have c'. The other way to look at it is that the 1's overlap the

horizontal b line and the short vertical a line, but they all lay outside the horizontal c line, so they correspond

to c'. (The horizontal c line delimits the c set. The c' set consists of all squares outside the c set. Since the circle

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includes all the squares in c', they are defined by c'. Again, notice that both values of a and b are spanned, thus

eliminating those terms.)

Now for 4 Boolean variables. The Karnaugh map is drawn as shown below.

The following corresponds to the Boolean expression

q = a'bc'd + a'bcd + abc'd' + abc'd + abcd + abcd' + ab'cd + ab'cd'

RULE: Minimization is achieved by drawing the smallest possible number of circles, each containing the

largest possible number of 1s.

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Grouping the 1s together results in the following.

The expression for the groupings above is

q = bd + ac + ab

This expression requires 3 2-input and gates and

1 3-input or gate.

We could have accounted for all the 1s in the map as shown below, but that results in a more complex

expression requiring a more complex gate.

The expression for the above is bd + ac + abc'd'. This requires 2 2-input and gates, a 4-input and gate, and a 3

input or gate. Thus, one of the and gates is more complex (has two additional inputs) than required above.

Two inverters are also needed.

PRODUCT OF TERMS SIMPLIFICATION Simplify the following in a) Sum of the products and b) Product of sums F= (A,B,C,D) = Σ(0,1,2,5,8,9,10)

c’d’ c’d cd cd’

a’b’ 1 1 1

a’b 1

ab

ab’ 1 1 1

F = b’d’+a’c’d+b’c’ ..... Ans for a)

c’d’ c’d cd cd’

a’b’ 1 1 0 1

C’

D’

C

DA’

D

B’

A’

C’

D’

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a’b 0 1 0 0

ab 0 0 0 0

ab’ 1 1 0 1

F’ = ∏(3,4,6,7,11,12,13,14,15)…. On grouping ZEROS, we get F’ = bd’ + ab + cd

(F’)’ = F = (bd’ + ab + cd)’ = (b’+d)(a’+b’)(c’+d’) ..... Ans for b)

Don't Cares

Sometimes we do not care whether a 1 or 0 occurs for a certain set of inputs. It may be that

those inputs will never occur so it makes no difference what the output is. For example, we

might have a bcd (binary coded decimal) code which consists of 4 bits to encode the digits 0

(0000) through 9 (1001). The remaining codes (1010 through 1111) are not used. If we had a

truth table for the prime numbers 0 through 9, it would be

The ds in the above stand for "don't care", we don't care whether a 1 or 0 is the value

for that combination of inputs because (in this case) the inputs will never occur.

The circle made entirely of 1s corresponds to the expression a'd and the combined 1 and d circle (actually a

combination of arcs) is b'c. Thus, if the disallowed input 1011 did occur, the output would be 1 but if the

disallowed input 1100 occurs, its output would be 0. The minimized expression is

p = a'd + b'c

Notice that if we had ignored the ds and only made a circle around the 2 1s, the resulting expression would

have been more complex, a'b'c instead of b'c.

5 VARIABLE KARNAUGH MAPS

Let's put our 5-variable Karnaugh Map to use. Design a circuit which has a 5-bit binary input (A,

B, C, D, E), with A being the MSB (Most Significant Bit). It must produce an output logic High for

any prime number detected in the input data. We show the solution above on the older Gray code

(reflection) map for reference. The prime numbers are (1,2,3,5,7,11,13,17, 19,23,29,31). Plot

a 1 in each corresponding cell. Then, proceed with grouping of the cells. Finish by writing the

simplified result. Note that 4-cell group A'B'E consists of two pairs of cell on both sides of the

mirror line. The same is true of the 2-cell group AB'DE. It is a group of 2-cells by being reflected

0000 0

0001 1

0010 1

0011 1

0100 0

0101 1

0110 0

0111 1

1000 0

1001 0

1010 D

1011 D

1100 D

1101 D

1110 D

1111 D

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about the mirror line. When using this version of the K-map look for mirror images in the other

half of the map.

Out = A'B'E + B'C'E + A'C'DE + A'CD'E + ABCE + AB'DE + A'B'C'D

Procedure for Finding Prime Implicants

Find prime implicants by finding all permitted (integer power of 2) maximum sized groups of min-terms.

1. Find essential prime implicants by identifying those prime implicants that contain at least one min-

term not found in any other prime implicant.

Example

Answer: Z’

Two prime implicant {1, 3} {2, 3, 6, 7}. Both are essential Q = Y + X’Z

Relationships

K: Number of literals in product

N: Number of variables

Number of Adjacent

Squares

Number of Literals in a Term in

an variable map

K 2k N=2 N=3 N=4 N=5

0 1 2 3 4 5

1 2 1 2 3 4

2 4 0 1 2 3

3 8 0 1 2

4 16 0 1

5 32 0

Tabular Method:

In order to understand the tabular method of minimisation, it is best you understand the numerical assignment

of Karnaugh map cells and the incompletely specified functions also known as the can't happen conditions

(Don’t care terms). This is because the tabular method is based on these principles. The tabular method which

is also known as the Quine-McCluskey method is particularly useful when minimising functions having a large

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number of variables, e.g. The six-variable functions. Computer programs have been developed employing this

algorithm. The method reduces a function in standard sum of products form to a set of prime implicants from

which as many variables are eliminated as possible. These prime implicants are then examined to see if some

are redundant. The tabular method makes repeated use of the law A + = 1. Note that Binary notation is

used for the function, although decimal notation is also used for the functions. As usual a variable in true form

is denoted by 1, in inverted form by 0, and the absence of a variable by a dash ( - ).

Rules of Tabular Method

Consider a function of three variables f(A, B, C):

Consider the function:

Listing the two minterms shows they can be combined

Now consider the following:

Note that these variables cannot be combined

This is because the FIRST RULE of the Tabular method for two terms to combine, and thus eliminate one

variable, is that they must differ in only one digit position. Bear in mind that when two terms are combined,

one of the combined terms has one digit more at logic 1 than the other combined term. This indicates that the

number of 1's in a term is significant and is referred to as its index.

For example: f(A, B, C, D)

0000 Index 0

0010, 1000 Index 1

1010, 0011, 1001 Index 2

1110, 1011 Index 3

1111 Index 4

The necessary condition for combining two terms is that the indices of the two terms must differ by one logic

variable which must also be the same.

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Example 1:

Consider the function: Z = f(A,B,C) = + C + A + A C

To make things easier, change the function into binary notation with index value and decimal value.

Tabulate the index groups in a colunm and insert the decimal value alongside.

From the first list, we combine terms that differ by 1 digit only from one index group to the next. These terms

from the first list are then separated into groups in the second list. Note that the ticks are just there to show

that one term has been combined with another term. From the second list we can see that the expression is

now reduced to: Z = + + C + A .

From the second list note that the term having an index of 0 can be combined with the terms of index 1. Bear in

mind that the dash indicates a missing variable and must line up in order to get a third list. The final simplified

expression is: Z =

Bear in mind that any un-ticked terms in any list must be included in the final expression (none occurred here

except from the last list). Note that the only prime implicant here is Z = . The tabular method reduces the

function to a set of prime implicants. Note that the above solution can be derived algebraically. Attempt this in

your notes.

Using the Boolean algebra theorem Ax + A = A, we can combine two product terms to generate a reduced

product term. For example, z + yz = z and xy + xyz = xy. The reduced product terms of a function are called

prime implicants (PIs) of the function. Therefore, a Boolean function can be expressed as sum of prime

implicants. Very often, the prime implicants are represented using binary notation, where a missing variable is

represented by `-' . For example, the prime implicant z is represented as 0-1 and xy is represented as 11-. Then

the above function can be expressed as F(x, y, z) = S(0-1, 11 -).

Most of the prime implicant generation algorithms generate more prime implicants than the minimum

requirement. For example, for the above function, minterms 001 and 011 generate the prime implicant 0-1,

minterms 110 and 111 generate the prime implicant 11- , and minterms 011 and 111 generate the prime

implicant -11. But the prime implicant -11 is not required for expressing the function. Therefore, given a set of

minterms and a set of prime implicants for a function, it is required to find a minimal subset of the prime

implicants that covers all the minterms of the function. The problem is indeed an optimization problem.

Example 2:

1

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Consider the function f(A, B, C, D) = (0,1,2,3,5,7,8,10,12,13,15), note that this is in decimal form.

(0000,0001,0010,0011,0101,0111,1000,1010,1100,1101,1111) in binary form.

(0,1,1,2,2,3,1,2,2,3,4) in the index form.

The prime implicants are: + + D + BD + A + AB .

The chart is used to remove redundant prime implicants. A grid is prepared having all the prime implicants

listed at the left and all the minterms of the function along the top. Each minterm covered by a given prime

implicant is marked in the appropriate position.

From the above chart, BD is an essential prime implicant. It is the only prime implicant that covers

the minterm decimal 15 and it also includes 5, 7 and 13. is also an essential prime implicant. It is the only

prime implicant that covers the minterm denoted by decimal 10 and it also includes the terms 0, 2 and 8. The

other minterms of the function are 1, 3 and 12. Minterm 1 is present in and D. Similarly for minterm 3.

We can therefore use either of these prime implicants for these minterms. Minterm 12 is present in A and

AB , so again either can be used. Thus, one minimal solution is: Z = + BD + + A

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Problems

Minimise the function below using the tabular method of simplification:

1. Z = f(A,B,C,D) = + C + A CD + A C + BCD + BC + CD

Using the tabular method of simplification, find all equally minimal solutions for the function below.

2. Z = f(A,B,C,D) = (1,4,5,10,12,14)

-------------------------------------------------------------------------------------------------------------------------------------

1. Consider the function: Z = f(A,B,C,D) = + C + A CD + A C + BCD + BC +

CD

Convert to decimal and binary equivalents:-

Z = f(A,B,C,D) = (0,2,4,5,8,9,12) - decimal equivalent

Z = f(A,B,C,D) = (0000,0010,0100,0101,1000,1001,1100) - binary equivalent

(0, 1, 1, 2, 1, 2, 2) - index values

The simplified answer is: Z = + B + A +

2. Consider Z = f(A,B,C,D) = (1,4,5,10,12,14

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Essential prime Implicant: ACD’ and A’C’

ACD’. It covers Minterm 10 covers 14

A’C’ Covers minterms 0,1,4,5 .The left-out minterm is 12.

This minterm 12 is covered by two terms BC’D’ and ABD’. Hence one of the terms can be dropped

Say BC’D’ is dropped.

Answer is: Z = + AC + AB

Decimal Representation:

The tabulation procedure can be further simplified by using decimal code for minterms. Two minterms can be

combined only if they differ by a power of 2 i.e. 2i.

Consider f(a,b,c,d)=Σ(0,1,8,9).

The difference between minterm 1and 9 is 23 = 8 and is written as 1,9(8)

And the difference between 0 and 8 is 0,8(8)

In the above cases the difference is power of 2 hence can be combined.

Similarly difference beween (4 and 7) is

In this case the difference is not power of 2 and hence cannot be combined.

This condition that the difference between the minterms shall be power of 2 is necessary but not sufficient. If

the index of these terms is same, then they cannot be combined. Also if a term with smaller index has higher

decimal value than the other term cannot be combined.

For example 9 (1001) and 7 (0111) cannot be combined. The index for 9 is 2 and the index for 7 is 3.

X X X X

X X

X

0 4 5 10 14 1

BC’D’

ACD’

ABD’

A’C’

12

X

X X

1(0001) and (1001)

0(0000) and (1000)

4(0100), 7(0111) ( 20+2

1 = 3) i.e. 4,7(3).

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Example:

ƒ(A,B,C,D) = Σ m(4,5,6,8,9,10,13) + Σ d(0,7,15)

Prime Implicant chart

Consider the following chart for function f(v,w,x,y,z) = Σm(13,15,17,18,19,20,21,23,25,27,29,31) +

Σ d(1,2,12,24)

The solution is

A = (17,19,21,23,25,27,29,31) B = (13,15,29,31) C = (24,25) D= (20,21)

E= (18,19) F = (12,13) G= (2,18) H= (1,17)

It may be seen that don’t care terms are not included in the chart shown above. The selection of non-

essential prime implicants is facilitated by initial listing of prime implicants in descending order

(including don’t care terms) to the number of min terms they cover. Draw horizontal lines to separate

the groups. The essential prime implicants are A, B, D cover all the Minterms except 18. This Minterm

can be covered by either E or G. Since both have same number of literals, two minimum expressions

can be found;

f= A + B + D+ E

Column -1 Column-II Column-iii

0√ 0,4(4) √

0,8(8) √

4√

8√

4,5(1) √

4,6(2) √

8,9(1)*

8,10(2)*

4,5,6,7(3)*

5√

6√

9√

10√

5,7(2) √

5,13(8) √

6,7(1) √

9,13(4)*

5,7,13,15(10)*

7√

13√

7,15(8) √

13,15(2) √

15√

(4) Change in Bit position 2.

Therefore 22=4

(4) Change in Bit position 3.

Therefore 23=8

(10) Changes in two Bit positions

1&3. Therefore 21+2

3 = 2 + 8 = 10.

X

X X

X

X X X X X X

13 19 23 17 29 15 18 20 21 25 27 31

A=vz

B =wxz

C =vwx’y’

D = vw’xy’

E = vw’x’y

F =v’wxy’

G = w’x’yz’

H = w’x’y’z

X

X X

X X

X

X

X X

X

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or f = A + B + D+ G

Determination of the set of all irredundant expressions

f(v,w,x,y,z) = Σm(0,1,3,4,7,13,15,19,20,22,23,29,31)

Check-mark the essential prime implicants. Here the essential implicants are A & C. ‘A’ covers the

minterms (13, 15, 29, 31) and implicant ‘C’ covers (3, 7, 19, 23). The Prime implicant ‘B’ covers only

the implicants covered by A&C. Hence ‘B’ can be ignored. Now draw the graph after removing A, B &

C. Now it is necessary to select minimal number of additional prime implicants so as to cover the

entire function.

Let us define a prime implicant function ‘p’. ‘p’ must be equal to one if a column is covered by at least

one prime implicant.

P = (H+I)(G+I)(H+F)(E+F)(D+E) On simplifying we get

= EHI+FEI+DFI+EGH+DFGH

From the above we may conclude that we require at least 3 prime implicants to cover the Minterms in

the reduced graph. i.e. E, H, I or F,E, I or D,F,I, etc. We may drop DFGH as it requires 4 Minterms be

included. Hence we have 4 solutions for ‘F’.

F = A+C + E + H + I

F = A+C + F + E +I

F = A+C + D + F +I

X

X

X

31

I=v’w’x’y’

X

X

X

X

X

X

X

0 7 19 3 23 1 4 13 15 20 22 29

√A=wxz

B =xyz √C =w’yz’

D = vw’xy

E = vw’x’z

F =w’xyz’

G = v’w’x’z

H =v’w’y’z’

X

X

X

X

X

X

X

X

X

X

X

X

X X

0 1 4 20 22

X

I=v’w’x’y’

X

D = vw’xy

E = vw’x’z

F =w’xyz’

G = v’w’x’z

H =v’w’y’z’

X

X

X

X

X

X

X X

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F = A+C + E + G +H

REDUCTION OF CHART:

F(v,w,x,y,z) = Σ (1,3,4,5,6,7,10,11,12,13,14,15,18,19,20, 21,22,23,25,26,27)

A,B,J & K shall be retained in the final solution as they are essential implicants. Hence to derive the

reduced graph, these rows can be deleted. Further the Minterms covered by these prime implicants

can also be deleted (Columns). The Prime implicants A,B,J&K cover

0,3,4,5,6,7,12,13,14,15,20,21,22,23, 25 and 27. Hence the left out columns are 10,11,18,19 and 26.

Now observing the reduced graph, H covers 19 and G covers 19 and 11 and both have same number

of literals. Hence H can be removed. G and I are covering 11 and both have same number of literals.

Hence ‘I’ can be deleted. E and F are both covering 10 and 11 while E is also covering 26. Further both

have same number of literals. Hence F can be removed. C and D are covering 18 and 19. C is also

covering 26 and both have same number of literals. Hence D can be deleted.

18 1 6 11 4 14 3 5 7 10 12 13 15

X

19 20 21 22 25 26 23

X

X X

I=v’yz

X

X

X X X

X

X

√A=w’x

√B =v’x

C = vx’y

D =vw’y

E =wx’y

F =v’wy

G =x’yz

H =w’yz

X X X X

X

X

X

X X

X X

X X

X X

27

X

X X

X

X X

X X X X

X

X

X

X X X

X X X

X X X

X X

√J=v’w’z

√K= vwx’z

18 11 10 19 26

X

X

I=v’yz

X X C = vx’y

D =vw’y

E =wx’y

F =v’wy

G =x’yz

H =w’yz

X

X X

X X

X

X X

X

X

18 11 10 19 26

X X X C = vx’y

E =wx’y

G =x’yz

X X X

X X

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Hence the deleted rows are H, I, F and D. The remaining rows in the reduced graph are C,E and G. Now

further reduced graph is drawn. Considering the reduced graph, C & E cover all the Minterms in the

graph. Hence G can be dropped.

Branching method:

This method is used when we don’t have any essential Prime implicants.

F(w,x,y,z) = Σ (0,1,5,7,8,10,14,15)

00 01 11 10

00 1

01 1 1

11 1 1

10 1 1

0 8 15 5 1 7 10 14

wx yz

X

X

X

A=w’x’y

B =w’y’z

C = w’xz

D =xyz

E =wxy

F =wyz’

G =wx’z

H =x’y’z’

X

X X

X

X

X

X

X

X

X X

X X

8 15 5

1

7 10 14 B =w’y’z

B =w’y’z

C = w’xz

D =xyz

E =wxy

F =wyz’

G =wx’z

H =x’y’z’

X X

X

X

X

X

X

X X

X X

C = w’xz X

X

15 5 7 10 14

X A=w’x’y

D =xyz

E =wxy

F =wyz’

G =wx’z

X

X

X

X

X

X

X

X

X X

X

It may seen that the graph does not have essential

prime implicants. It also does not have dominating

rows and dominated columns. I such case we use

branching method. This chart is called cyclic prime

implicant chart as each Minterm is covered by two

prime implicants and each implicant covers two

Minterms.

i) Consider a column, say column of Minterm ‘0’.

This is covered by A & H. We start by choosing

one of these two prime implicants (Row), say ‘A’.

ii) A covers 0&1. Hence A shall be placed in the

final answer. Further remove the columns ‘0’

and ‘1’. This shown in fig. (A). If we chose to

include ‘H’, then the resultant graph is shown in

fig B.

iii) Follow the proceadure as followed in earlier

problems to find the minimised equation.

iv) We get f= A + C + E + G (As per Fig. A) and

We get f = H + B + E + F (As per Fig. B)

Fig. A

Fig. B

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