tabular method

124 FOUNDATION OF SWITCHING THEORY AND LOGIC DESIGN

3.3 MINIMIZATION USING QUINE-McCLUSKEY (TABULAR) METHODThe K-map method is suitable for simplification of Boolean functions up to 5 or 6

variables. As the number of variables increases beyond this, the visualization of adjacentsquares is difficult as the geometry is more involved.

The ‘Quine-McCluskey’ or ‘Tabular’ method is employed in such cases. This is a system-atic step by step procedure for minimizing a Boolean expression in standard form.

Procedure for Finding the Minimal Expression1. Arrange all minterms in groups, such that all terms in the same group have same

number of 1’s in their binary representation. Start with the least number of 1’s andcontinue with grouping of increasing number of 1’s, the number of 1’s in each termis called the index of that term i.e., all the minterms of same index are placed ina same group. The lowest value of index is zero. Separate each group by a thickline. This constitutes the I stage.

2. Compare every term of the lowest index (say i) group with each term in thesuccessive group of index (say, i + 1). If two minterms differ in only one variable,that variable should be removed and a dash (–) is placed at the position, thus a newterm with one less literal is formed. If such a situation occurs, a check mark (✔) is

F(A, B, C, D, E) = Σ(0, 1, 4, 5, 16, 17, 21, 25, 29)

23. Simplify the following function in (a) s–o-p

and (b) p–o–s

F(A, B, C, D) = ∏(3, 4, 6, 7, 11, 12, 13, 14, 15)

24. Simplify the Boolean function using tabular method.

F(A, B, C, D, E) = Σ(0, 1, 4, 5, 16, 17, 21, 25, 29, 30)

25. Simplify the Boolean function using tabular method

F(A, B, C, D, E, F) = Σ(6, 9, 13, 18, 19, 27, 29, 41, 45, 57, 61, 63)

22. Simplify the Boolean function by tabular method

BOOLEAN FUNCTION MINIMIZATION TECHNIQUES 125

placed next to both minterms. After all pairs of terms with indices i and (i + 1) havebeen considered, a thick line is drawn under the last terms.When the above process has been repeated for all the groups of I stage, one stageof elimination have been completed. This constitutes the II stage.

3. The III stage of elimination should be repeated of the newly formed groups ofsecond stage. In this stage, two terms can be compared only when they have dashesin same positions.The process continues to next higher stages until no further comparisons arepossible. (i.e., no further elimination of literals).

4. All terms which remain unchecked (No ✔ sign) during the process are consideredto be prime implicants (PIs). Thus, a set of all PIs of the function is obtained.

5. From the set of all prime implicates, a set of essential prime implicants (EPIs) mustbe determined by preparing prime implicant chart as follow.(a) The PIs should be represented in rows and each minterm of the function in a column.(b) Crosses should be placed in each row corresponding to minterms that makes

the PIs.(c) A complete PIs chart should be inspected for columns containing only a single

cross. PIs that cover minterms with a single cross in their column are called EPIs.6. The minterms which are not covered by the EPIs are taken into consideration and

a minimum cover is obtained form the remaining PIs.Now to clarify the above procedure, lets do an example step by step.Example 1. Simplify the given function using tabular method.

F (A, B, C, D) = ∑ (0, 2, 3, 6, 7, 8, 10, 12, 13)Solution. 1. The minterms of the function are represened in binary form. The binary

represented are grouped into a number of sections interms of the number of 1’s index asshown in Table of Fig. 3.15.

Minterms Binary No. Minterms Index BinaryABCD of 1's Group ABCD

m0 0 0 0 0 0 m0 0 0 0 0 0 ✔

m2 0 0 1 0 1 m2 0 0 1 0 ✔

m3 0 0 1 1 2 m81 1 0 0 0 ✔

m6 0 1 1 0 2 m3 0 0 1 1 ✔

m7 0 1 1 1 3 m6 20 1 1 0 ✔

m8 1 0 0 0 1 m10 1 0 1 0 ✔

m10 1 0 1 0 2 m12 1 1 0 0 ✔

m12 1 1 0 0 2 m7 0 1 1 1 ✔

m13 1 1 0 1 3 m133 1 1 0 1 ✔

Fig. 3.15

2. Compare each binary term with every term in the adjacent next higher category.If they differ only by one position put a check mark and copy the term into the nextcolumn with (–) in the place where the variable is unmatched, which is shown innext Table of Fig. 3.16.


Minterm Binary

Group A B C D

0, 2 0 0 – 0 ✔

0, 8 – 0 0 0 ✔

2, 3 0 0 1 – ✔

2, 6 0 – 1 0 ✔

2, 10 – 0 1 0 ✔

8, 10 1 0 – 0 ✔

8, 12 1 – 0 0 PI

3, 7 0 – 1 1 ✔

6, 7 0 1 1 – ✔

12, 13 1 1 0 – PI

Fig. 3.16

Minterm Binary

Group A B C D

0, 2, 8, 10 – 0 – 0 PI

0, 8, 2, 10 – 0 – 0 PI eliminated

2, 3, 6, 7 0 – 1 – PI

2, 6, 3, 7 0 – 1 – PI eliminated.

Fig. 3.17

3. Apply same process to the resultant column of Table of Fig. 3.16 and continue untilno further elimination of literals. This is shown in Table of Fig. 3.17 above.

4. All terms which remain unchecked are the PIs. However note that the mintermscombination (0, 2) and (8, 10) form the same combination (0, 2, 8, 10) as thecombination (0, 8) and (2. 10). The order in which these combinations are placeddoes not prove any effect. Moreover, as we know that x + x = x, thus, we caneliminate one of these combinations.

The same occur with combination (2, 3) and (6, 7).

5. Now we prepare a PI chart to determine EPIs as follows shown in Table of Fig. 3.18.

Minterms

Prime Implicants 0 2 3 6 7 8 10 12 13

(8, 12) × ×

(12, 13) * × ×

(0, 2, 8, 10) * × × × ×

(2, 3, 6, 7) * × × × ×

✔ ✔ ✔ ✔ ✔ ✔

Fig. 3.18


(a) All the PIs are represented in rows and each minterm of the function in acolumn.

(b) Crosses are placed in each row to show the composition of minterms thatmake PIs.

(c) The column that contains just a single cross, the PI corresponding to the rowin which the cross appear is essential. Prime implicant. A tick mark is partagainst each column which has only one cross mark. A star (*) mark is placedagainst each. EPI.

6. All the minterms have been covered by EPIs.

Finally, the sum of all the EPIs gives the function in its minimal SOP form

EPIs. Binary representation Variable Representation

A B C D

12, 13 1 1 0 – ABC'

0, 2, 8, 10 – 0 – 0 B'D'

2, 3, 6, 7 0 – 1 – A'C

Therefore, F = ABC' + B'D' + A'C.

If don't care conditions are given, they are also used to find the prime implicating, butit is not compulsory to include them in the final simplified expression.

Example 2. Simplify the given function using tabular method.

F(A, B, C, D) = ∑ (0, 2, 3, 6,7)

d (5, 8, 10, 11, 15)

Solution. 1. Step 1 is shown in Table of Fig. 3.19. The don’t care minterms are alsoincluded.

Minterms Binary No. Minterms Index BinaryA B C D of 1’s Group ABCD

m0 0 0 0 0 0 m0 0 0 0 0 0 ✔

m2 0 0 1 0 1 m2 0 0 1 0 ✔

m3 0 0 1 1 2 m8 1 1 0 0 0 ✔

m5 0 1 0 1 2 m3 0 0 1 1 ✔

m6 0 1 1 0 2 m5 2 0 1 0 1 ✔

m7 0 1 1 1 3 m6 0 1 1 0 ✔

m8 1 0 0 0 1 m10 1 0 1 0✔

m10 1 0 1 0 2 m7 3 0 1 1 1 ✔

m11 1 0 1 1 3 m11 1 0 1 1 ✔

m15 1 1 1 1 4 m15 4 1 1 1 1 ✔

Fig. 3.19

2. Step 2 is shown in Table of Fig. 3.20.

3. Step 3 is shown in Table of Fig. 3.21.


Minterm Binary

Group A B C D

0, 2 0 0 – 0 ✔

0, 8 – 0 0 0 ✔

2, 3 0 0 1 – ✔

2, 6 0 – 1 0 ✔

2, 10 – 0 1 0 ✔

8, 10 1 0 – 0 ✔

3, 7 0 – 1 1 ✔

3, 11 – 0 1 1 ✔

5, 7 0 1 – 1 PI

6, 7 0 1 1 – ✔

10, 11 1 0 1 – ✔

7, 15 – 1 1 1 ✔

11, 15 1 – 1 1 ✔

Fig. 3.20

Minterm Binary

Group A B C D

0, 2, 8, 10 – 0 – 0 PI

0, 8, 2, 10 – 0 – 0 PI Eliminated

2, 3, 6, 7 0 – 1 – PI

2, 3 10, 11 – 0 1 – PI

2, 6, 3, 7 0 – 1 – PI Eliminated

2, 10, 3, 11 – 0 1 – PI Eliminated

3, 7, 11, 15 – – 1 1 PI

3, 11, 7, 15 – – 1 1 PI Eliminated

Fig. 3.21

4. All the terms which remain unchecked are PIs. Moreover, one of two same com-binations is eliminated.

5. Step 5 is to prepare a PI chart to determine EPIs as shown in Table of Fig. 3.22.

Note, however, that don’t care minterms will not be listed as column headings inthe chart as they do not have to be covered by the minimal (simplified) expression.


Prime Implicants Minterms

0 2 3 6 7

(5, 7) ×

(0, 2, 8, 10) * × ×

(2, 3, 6, 7) * × × × ×

(2, 3, 10, 11) × ×

(3, 7, 11, 15) × ×

✔ ✔

Fig. 3.22


Therefore F (A, B, C, D) = B'D' + A'C

Example 3. Simplify the given function using tabular method:

F (A, B, C, D, E, F, G) = Σ (20, 28, 38, 39, 52, 60, 102, 103, 127)

Solution. Step 1 is shown in Table of Fig. 3.23.

Minterms Binary No. Minterms Index BinaryABCDEFG of 1’s Group ABCDEFG

m20 0 0 1 0 1 0 0 2 m20 2 0 0 1 0 1 0 0 ✔

m28 0 0 1 1 1 0 0 3 m28 0 0 1 1 1 0 0 ✔

m38 0 1 0 0 1 1 0 3 m38 3 0 1 0 0 1 1 0 ✔

m39 0 1 0 0 1 1 1 4 m52 0 1 1 0 1 0 0 ✔

m52 0 1 1 0 1 0 0 3 m39 4 0 1 0 0 1 1 1 ✔

m60 0 1 1 1 1 0 0 4 m60 0 1 1 1 1 0 0 ✔

m102 1 1 0 0 1 1 0 4 m102 1 1 0 0 1 1 0 ✔

m103 1 1 0 0 1 1 1 5 m103 5 1 1 0 0 1 1 1 ✔

m127 11 1 1 1 1 1 7 m127 7 1 1 1 1 1 1 1 PI

Fig. 3.23

2. Step 2 is shown in Table (Fig. 3.24).

3. Step 3 is shown in Table (Fig. 3.25).

Minterms Binary

Group A B C D E F G20, 28 0 0 1 – 1 0 0 ✔

20, 52 0 – 1 0 1 0 0 ✔

28, 60 0 – 1 1 1 0 0 ✔

38, 39 0 1 0 0 1 1 – ✔

38, 102 – 1 0 0 1 1 0 ✔

52, 60 0 1 1 – 1 0 0 ✔

39, 103 – 1 0 0 1 1 1 ✔

102, 103 1 1 0 0 1 1 – ✔

Fig. 3.24


Mintesms Binary

Group A B C D E F G

20, 28, 52, 60 0 – 1 – 1 0 0 PI

20, 52, 28, 60 0 – 1 – 1 0 0 PI Eliminated

38, 39 102, 103 – 1 0 0 1 1 – PI

38, 102, 39, 103 – 1 0 0 1 1 – PI Eliminated

Fig. 3.25

4. All the terms which remain unchecked are PIs. Moreover one of two same com-binations is eliminated.

5. PI chart to determine EPIs is shown in Table Fig. 3.26.

Prime Implicants Minterms

20 28 38 39 52 60 102 103 127

127 * ×

(20, 28, 52, 60) * × × × ×

(38, 39, 102, 103) * × × × ×

✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔

Fig. 3.26


Therefore, F (A, B, C, D, E, F, G) = ABCDEFG + A'CEF'G' + BC'D'EF

3.4 EXERCISE1. Using Boolean algebra simplify each of the following logic expressions as much as

possible:

(a) Z = A(A + AB) (A + ABC) (A + ABCD)

(b) C = �� ′ + ′ ′

2. Draw the simplest possible logic diagram that implements the output of the logicdiagram given below.

AB

C

H

3. Write the logic expression and simplify it as much as possible and draw a logicdiagram that implements the simplified expression.


A

BC

X

D

4. Obtain the simplified expression in s-of-p for the following Boolean functions:

(a) �� + ′ ′ ′ + ′ ′

(b) ABD + A′C′D′ + A′B + ACD + AB′D′

(c) � � � �� ′ + ′ ′ + ′ + ′�

(d) � �� = Σ � � �

(e) � �� Σ

5. Use a K-map to simplify each of the following logic expressions as much as possible:

(i) � � �� ′ ′

(ii) � � � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′(iii) � � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′(iv) � � � � � � � �� ′ ′ ′ ′ ′ ′

6. Simplify the following logic expressions using K-maps and tabular method.

(a) F(A, B, C) = A′C + B′C+AB′C′(b) G(A, B, C, D) = B′CD + CD′ + A′B′C′D + A′B′C

7. Simplify the Boolean function F(ABCDE) = Σ(0, 1, 4, 5, 16, 17, 21, 25, 29)

8. Simplify the following Boolean expressions using K-maps and Tabular method.

(i) BDE + B′C′D + CDE + ABCE + ABC + BCDE

(ii) ABCE + ABCD + BDE + BCD + CDE + BDE

(iii) F(ABCDEF) = Σ(6, 9, 13, 18, 19, 27, 29, 41, 45, 57, 61)

9. Draw Karnaugh maps for the following expressions:

F = � �� ′ ′ ′ ′ ′ ′� � � �� ′ ′ ′ ′ ′ ′� � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′

� � � � � � �� ′ ′ ′ ′ ′ ′ ′� �

10. Simplify the following logic expressions using karnaugh maps. Draw logic diagramsfor them using only (a) NAND, (b) NOR gates, assuming inputs A, B, C, and D onlyare available.

� � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′� � � � � � � � � � �

+ ′ ′ ′ ′ ′� ��


� � � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′� � � � � � � � � � �

� � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′� � � � � � � � � � �

+ ′ ′ ′ ′ ′ ′ ′ ′� ��

� � �� ′ ′ ′ ′ ′ ′ ′ ′ ′11. The institute’s pool room has four pool tables lined up in a row. Although each table

is far enough from the walls of the room, students have found that the tables aretoo close together for best play. The experts are willing to wait until they canreserve enough adjacent tables so that one game can proceed unencombered bynearby tables. A light board visible outside the pool room shows vacant tables. Themanager has developed a digital circuit that will show an additional light wheneverthe experts’ desired conditions arise. Give a logic equation for the assertion of thenew light signal. Simplify the equation using a K-Map.

12. Simlify the Boolean functions using tabular method and verify result with K-map.

(a) � �� = Σ � � � � � � � � ��

(b) � �� = Σ � � ��

(c) � �� = Σ � � � ��

(d) � �� = Σ � ��

(e) � � � �� = Σ � ��

13. Simplify the Boolean function F using the don’t care conditions d, in (I) SOP and(II) POS:

� � � � � � � � � � �′ ′ ′ ′ � = ′ ′ ′ ′� � � � ��

� � �� ′ ′ + ′ ′ + + ′ ′ + � � � � � � �� = ′ ′ + ′ +�

� � �� ′ ′ ′ ′ ′ � = ′ ′ ′ ′ ′��

14. Use a Karnaugh map to simplify each of the following logic expressions as muchas possible.

(a) � � � � � � �� ′ ′ ′ ′ ′ ′ ′ ′ ′

Solution. F = ABD + A′B′D + B′C′

(b) � � � � � � �� ′ ′ ′ ′ ′

Solution. � � �� !�� ′ ′ ′ ′ ′ ′

(c) � � � � � � � � � � � � � ′ ′ ′ ′ ′ ′ ′ ′

Solution. � � � � � � � � � � � � � ′ ′ ′ ′ ′ ′ ′ ′

(d) � � � � � � �� ′ ′ ′

Solution. � � � � �′

15. Use a Karnaugh map to simplify each of the following logic expressions as muchas possible.

(a) � � �� ′ ′ ′ ′ ′�


(b) " � � � � � � � � � � � � � � � �+ ′ ′ + ′ + ′

16. Using Boolean Algebra simplify

(a) �� (b) � � ��

(c) �� + (d) ��

(e) ��

17. Use a karnaugh map to simplify each function to a minimum sum-of-products form:

(a) � � �� (b) � � � ��

(c) � � ��

18. A B C F1 A B C F2

0 0 0 1 0 0 0 0

0 0 1 0 0 0 1 1

0 1 0 0 0 1 0 1

0 1 1 0 0 1 1 1

1 0 0 0 1 0 0 0

1 0 1 1 1 0 1 1

1 1 0 0 1 1 0 1

1 1 1 1 1 1 1 0

Transfer the input-output specifications for F1 and F2 given above to 3 variableKarnaugh maps.

19. Using a Karnagh map simplify the following equations

(a) � � ��

(b) � � ��

(c) � � � ��

(d) � � ��

20. Simplify the following using Boolean Algebra

(a) � � � � � �= +� � �

(b) � � � � �= + +� � �

(c) � � � � � � � � � �= + + +� � � � �

(d) � � � � � � � � �= + + + +� ��

21. Consider the function

� � � � � � � � � � � � � �= = + +� � � � � � � � ��

(a) Draw a schematic diagram for a circuit which would implement this function.

tabular method

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