q-m technique lect 3

7/31/2019 Q-M Technique Lect 3

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Introduction

Determining the Prime Implicants

What are these Prime Implicants?

Selecting a Minimum Set

Another Example

Rules for Table Reduction

Incompletely Specified Functions

Review


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Quine-McCluskey Methodis a step-by-step

procedure which is guaranteed to produce asimplified standard form for given function.

Advantages:

can be applied to problems of many variables

suitable for machine computation

BUT tedious and prone-to-errors for humans.

Quine-McCluskey Tabulation Method

Introduction


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Introduction

Relies on repeated applications on terms using UnifyingTheorem:

axb + ax'b = ab

where a and b stand for product terms;Example: wxyz + wx'yz = wyz

Using binary representations(for the product terms), abovetheorem correspond to:

a1b + a0b = a-b

Examples: 111 + 101 = 1-1 (abc + ab'c = ac)

101 + 100 = 10- (ab'c + ab'c' = ab')

11-10 + 11-00 = 11--0 (abde' + abd'e' = abe')


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Introduction

Two steps procedure:

Find all prime implicants

Smallest set of prime implicants to coverthe minterms of function


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Consider a 4-variable function.

F(A,B,C,D) = m(0,1,4,5,9,11,14,15)

Procedure:

Arrange the minterms in groups according to the number of1s (see Column 1)

Combine terms from adjacent groups which differ by only 1bit. (see Column 2)

Repeat step 2 with a new set of terms (see Column 3); untilno further combinations are possible.


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The prime implicants (PIs) are those withouta tickmark, namely: {-001, 10-1, 1-11, 111-, 0-0-}

Column 1

0 0000

1 0001

4 0100

5 0101

9 1001

11 1011

14 1110

15 1111

Column 2 Column 3

P

P

0, 1 000-P

P

0, 4 0-00P

P

1, 5 0-01

P

P

1, 9 -001

P

P

4, 5 010-P

P 9,11 10-1P

P

P

P11,15 1-11

14,15 111-

P

P

0,1,4,5 0-0-P

P

or {B'C'D, AB'D, ACD, ABC, A'C'}


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A prime implicant (PI) is a product termobtained by combining the maximum possible

number of minterms from adjacent squares inthe map.

Largest groupings of minterms in K-mapcorrespond to prime implicants.


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What are these Prime Implicants? For comparison, the equivalent K-map with the various

groupings of minterms are as follows:

Column 1

0 0000

1 0001

4 0100

5 0101

9 1001

11 1011

14 1110

15 1111

Column 2 Column 3

P

P

0, 1 000-P

P

0, 4 0-00P

P

1, 5 0-01

P

P

1, 9 -001

P

P

4, 5 010-P

P 9,11 10-1P

P

P

P 11,15 1-1114,15 111-

P

P

0,1,4,5 0-0-P

P

1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1 1

1

1

1

1


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For comparison, the equivalent K-map with the variousgroupings of minterms are as follows:

Note that some of these groupings (primeimplicants) are redundant.

So we need to find the minimum set whichwould cover all minterms step 2 of thetabulation technique. 1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1 1

1

1

1

1


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Next step is to select a minimum set of prime implicants torepresent the function the covering problem.

Prepare a table for prime-implicant vs minterms a primeimplicant chart(PI Chart):

0 1 4 5 9 11 14 15

1,9 -001 X X

9,11 10-1 X X

11,15 1-11 X X

14,15 111- X X

0,1,4,5 0-0- X X X X


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Choose essential prime implicants (EPIs) those withunique minterm in it (i.e. single minterm in the column).

0 1 4 5 9 11 14 15

1,9 -001 X X

9,11 10-1 X X

11,15 1-11 X X

14,15 111- X X0,1,4,5 0-0- X X X X

Unique minterms: 0,4,5,14

Therefore, EPIs are: A'C', ABC.

P PP P

1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1 1

1

1

1

1


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Next, select one or more prime implicants which wouldcover all the remaining minterms. (namely, AB'D)

9 11

1,9 -001 X

9,11 10-1 X X

11,15 1-11 X

Hence, minimum set of prime implicants to

cover all the minterms is { A'C', ABC, AB'D }

F = A'C' + ABC + AB'D

1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1 1

1

1

1

1


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Another Example

Try tabulation technique with function:

F(A,B,C,D) = m(2,4,6,8,9,10,12,13,15)

Column 1 Column 2 Column 3

2 0010 2,6 0-10 PI2 8,9,12,13 1-0- PI1

4 0100 2,10 -010 PI3

8 1000 4,6 01-0 PI4

6 0110 4,12 -100 PI5

9 1001

8,9 100-

10 1010 8,10 10-0 PI6

12 1100 8,12 1-00

13 1101 9,13 1-01

15 1111 12,13 110-

13,15 11-1 PI7


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Another Example

Prime implicants: 1-0-, 0-10, -010, 01-0, -100, 10-0, 11-1

or AC', A'CD', B'CD', A'BD', BC'D', AB'D', ABD

2 4 6 8 9 10 12 13 15

8,9,12,13 1-0- X X X X

2,6 0-10 X X

2,10 -010 X X

4,6 01-0 X X4,12 -100 X X

8,10 10-0 X X

13,15 11-1 X X

P P

EPIs: 1-0-, 11-1 or AC', ABD


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Another Example

Reduced PI chart

2 4 6 8 9 10 12 13 15

8,9,12,13 1-0- X X X X

2,6 0-10 X X

2,10 -010 X X

4,6 01-0 X X

4,12 -100 X X

8,10 10-0 X X

13,15 11-1 X X

P P

2 4 6 10

2,6 0-10 X X

2,10 -010 X X

4,6 01-0 X X

4,12 -100 X

8,10 10-0 X

EPIs: AC', ABD

F = AC' + ABD +B'CD' + A'BD'


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Another Example

Questions:

How would you obtain simplified product-of-sums

(POS) expressions with the tabulation technique?

How would you handle dont care conditions inan incompletely specified function?


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After obtaining the EPIs, the rules for PI chart

reduction are: Rule 1: A row that is covered byanother row may be

eliminated from the chart. When identical rows arepresent, all but one of the rows may be eliminated.

Rule 2: A column that coversanother column may beeliminated. All but one column from a set of identicalcolumns may be eliminated.


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Example: Given this reduced PI chart:

5 10 11 13

1,5,9,13 --01 X X

5,7,13,15 -1-1 X X

8,9,10,11 10-- X X9,11,13,15 1--1 X X

10,11,14,15 1-1- X X

Apply rule 1: 2nd and 5th rows can be eliminated.

Apply rule 2: 3rd and 4th columns can be eliminated.

5 10

1,5,9,13 --01 X

8,9,10,11 10-- X


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Incompletely Specified Functions

Functions with dont-cares.

Step 1: Finding prime implicants (no change).

Step 2: PI chartomit the dont-cares.

Example:

F(A,B,C,D,E) = m(2,3,7,10,12,15,27)+ d(5,18,19,21,23)


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Incompletely Specified FunctionsF(A,B,C,D,E) = m(2,3,7,10,12,15,27) + d(5,18,19,21,23)

Column 1 Column 2 Column 3

2 00010 2,3 0001- 2,3,18,19 -001- PI1

3 00011 2,10 0-010 PI4 3,7,19,23 -0-11 PI2

5 00101 2,18 -0010 5,7,21,23 -01-1 PI3

10 01010 3,7 00-11

12 01100 PI7 3,19 -0011

18 10010 5,7 001-1 7 00111 5,21 -0101

19 10011 18,19 1001-

21 10101 7,15 0-111 PI5

15 01111 7,23 -0111 23 10111 19,23 10-11 27 11011 19,27 1-011 PI6

21,23 101-1

EPIs: PI4, PI5, PI6 and

PI7.

2 3 7 10 12 15 27PI1 X X

PI2 X X

PI3 X

PI4 X X

PI5 X X

PI6 X

PI7 X

PI chart

3

PI1 X

PI2 X

Reduced

PI chart:F = PI1 + PI4 + PI5 + PI6 + PI7

or

PI2 + PI4 + PI5 + PI6 + PI7


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Review Quine-McCluskeyTechnique

Relies on unifying theorem:axb + axb = ab

Procedure:

(i) Find all prime implicants

arrange minterms according to the number of 1s.

exhaustively combine pairs of terms from adjacent groupswhich differ by 1 bit, to produce new terms.

tick those terms which have been selected.

repeat with new set of terms until none possible.

terms which are unticked are the prime implicants.


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Review Quine-McCluskey Technique

(ii) select smallest set to cover function prepare prime-implicants chart.

select essential prime implicants for which one or more of itsminterms are unique (only once in the column).

obtain a new reduced PI chart for remaining prime-implicants

and the remaining minterms. select one or more of remaining prime implicants which will

cover all the remaining minterms.


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Review Quine-McCluskeyTechnique

Example:F(A,B,C,D) = m(0,1,4,5,9,11,14,15)

Column 1

0 0000

1 0001

4 0100

5 0101

9 1001

11 1011

14 1110

15 1111

Column 2 Column 3

P

P

0, 1 000-P

P

0, 4 0-00P

P

1, 5 0-01

P

P

1, 9 -001

P

P

4, 5 010-P

P 9,11 10-1P

P

P

P11,15 1-11

14,15 111-

P

P

0,1,4,5 0-0-P

P

Review Quine McCluskey


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Review

Quine-McCluskeyTechnique

The prime implicants (those without a tick mark) are:{ B'C'D, AB'D, ACD, ABC, A'C' }

0 1 4 5 9 11 14 15

1,9 -001 X X

9,11 10-1 X X11,15 1-11 X X

14,15 111- X X

0,1,4,5 0-0- X X X X

9 11

1,9 -001 X

9,11 10-1 X X

11,15 1-11 X

Essential prime implicants : ABC, A'C'

F = ABC + A'C' + AB'D


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Petricks method

Consider the PI table of

f(ABCD) = Sm(4,5,7,12,14,15)

PI terms minterms

4 5 7 12 14 15

ABC a X X

BCD d X X

ABD c X X

ABD d X X

BCD e X X

ABC f X X


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There are no essential prime implicants in this PI table. Also itis not possible to apply column and row reduction. Now weapply Petricks method. We formulate a p-expression for the PI

table. The first column of the table indicates that either primeimplicant a or b must be selected to cover minterm 4.Corresponding to this we construct a term (a+b).Similarly,prime implicant a or c must be selected to cover minterm5.Accordingly a term (a+c) is constructed. In the same way the

other similar terms are constructed, which are as followsPrime implicants

p = (a+b)(a+c)(c+e)(b+d)(d+f)(e+f)

Petricks method


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Petricks method

p = (a+b)(a+c)(c+e)(b+d)(d+f)(e+f)---- (1)

This expression can be manipulated using Boolean algebraic methods. It is tobe expressed in SOP form

p=(a+bc)(d+bf)(e+cf)------------- (2)

since (x+y)(x+z)= x.x+y.x+x.z+y.z= {x+x(y+z)} + yz

= x+yz

Simplify equn 2 we have

P= ade+adcf+bcde+bcdf+abcf+bcfe+bcf+ abfe

= ade+adcf+bcde+bcf(1+d+a+e)+abfe= ade+adcf+bcde+bcf+ abfe ----------------- (3)

From equn 3, we obtain the condition that p must be equal to 1 only when asufficient subset of the prime implicants is selected to cover the PI table.


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From equn (3), we can write five possible minimalexpressions corresponding to each term as follows

f1(ABCD)= a + d + e =ABC +ABD + BCD ------ (4)f2(ABCD)= a + d + c + f =ABC+ABD +ABD + ABC ------ (5)

f3(ABCD)= b + c + d + e =BCD+ABD+ABD+BCD ------ (6)

f4(ABCD)= b + c + f = BCD+ABD+ABC ------ (7)

f5(ABCD)= a + b + f + e =ABC+BCD+BCD+ABC ------ (8)

Thus we see that the Petricks method not only provides minimalexpression, it also gives all the possible minimal expressions. Now, wecan select one of these expression, Logically all are equivalent, but mayneed different number of components for realization resulting in differentcosts.

Petricks method


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From equn (4) 3+3+3+3 =12

From equn (5) 3+3+3+3+4 =16

From equn (6) 3+3+3+3+4 =16

From equn (7) 3+3+3+3 =12

From equn (8) 3+3+3+3+4 =16

So equn 4and equn 7 gives the optimalsolution.

Petricks method

E l 2


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Example 2

PI terms MINTERMS

0 4 10 13 15 16 22 23 26ABDE a X X

ACD b X X

ABC c X X

ABD d X XBCDE e X X

ACE f X X

BCDE g X X

ABDE h XABCD i X X

ABCE j X


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In the PI table, the row i dominates

over rows h and j, therefore, h & j rows are eliminated.

Now, columns 22 and 23 have single X, therefore,row is secondary EPI, i.e. ABCD term will be a partof minimal expression. Now we use Petricks

method.

Then expression for p is given byp = (a+g)(a+b)(d+e)(b+c)(c+d)(f+g)(e+f)

Example 2

E l 2


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p = (a+g)(a+b)(d+e)(b+c)(c+d)(f+g)(e+f)

=(a+bg)(d+ce)(b+c)(f+ge)

=(ad+bdg+ace+bceg)(bf+cf+beg+ceg)

=(abdf+acdf+abdeg+acdeg+bdfg+bcdfg+bdeg+bcdeg+abcef+acef+abceg+a

ceg+bcefg+bcefg+bceg+bceg

=abdf+acdf+(abdeg+bdeg+bcdeg)+(acdeg+aceg)+

(bdfg+bcdfg)+(abcef+acef))+abceg+(bcefg+bceg)

=abdf+acdf+bdeg+aceg+bdfg+acef+(abceg+bceg)=abdf+acdf+bdeg+aceg+bdfg+acef+bceg

There are seven minimal functions corresponding to the above p expression

Example 2


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f1(ABCDE) =a+b+d+f =ABDE+ACD+ABD+ACE

COMPONENT = 4+3+3+3+4 =17

f2(ABCDE)=a+c+d+f =ABDE+ABC+ABD+ACE

COMPONENT = 4+3+3+3+4 =17f3(ABCDE)=b+d+e+g =ACD+ABD+BCDE+BCDE

COMPONENT = 3+3+4+4+4 =18

f4(ABCDE)=a+c+e+g =ABDE+ABC+BCDE+BCDE

COMPONENT = 4+3+4+4+4 =19

f5(ABCDE)=b+d+f+g =ACD+ABD+ACE+BCDECOMPONENT = 3+3+3+4+4 =17

f6(ABCDE)= a+c+e+f =ABDE+ABC+BCDE+ ACE

COMPONENT = 3+3+4+4+4 =18

f6(ABCDE)= b+ c+e+g =ACD+ ABC+BCDE+BCDE

COMPONENT = 3+3+4+4+4 =18The term ABCD corresponding to i in table will be included in each of the

above functions. This will add 5 to the component cost of each function. Thecost is minimum for the functions f1,f2, and f5. Any one of these can beselected.

q-m technique lect 3

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