CENG 342 Digital Systems Tabular Minimization Larry Pyeatt - - PowerPoint PPT Presentation

CENG 342 – Digital Systems

Tabular Minimization Larry Pyeatt

SDSM&T

Tabular Minimization

Karnaugh Maps are good for up to six input variables, but cannot be extended beyond that. Karnaugh Maps are not easily implemented in a computer program.

The tabular method1 overcomes both of these limitations.

1Also known as the Quine-McCluskey method, because it was developed by Willard V. Quine and extended by Edward

• J. McCluskey between 1950 and 1956.

From Truth Table to Minterms

A B C D f 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 d 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 d 1 1 1 1 1 F(A, B, C, D) =

• m(4, 8, 10, 11, 12, 15) + d(9, 14)

Recall that d stands for don’t-care. The output does not matter in those cases.

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Group the minterms by how many bits are set to 1.

Number

• f 1s

Term Binary 1 m4 0100 m8 1000 2 (m9) 1001 m10 1010 m12 1100 3 m11 1011 (m14) 1110 4 m15 1111 Size 2

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Compare each term in group 1 to each term in group 2. Combine terms m12 and m4. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000 2 (m9) 1001 m10 1010 m12 1100

• 3

m11 1011 (m14) 1110 4 m15 1111 Size 2 m(4, 12)

• 100

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m8 and m9. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010 m12 1100

• 3

m11 1011 (m14) 1110 4 m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100-

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m8 and m10. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011 (m14) 1110 4 m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m8 and m12. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011 (m14) 1110 4 m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Compare each term in group 2 to each term in group 3. Combine terms m9 and m11. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110 4 m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00 m(9, 11) 10-1

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m10 and m11. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110 4 m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00 m(9, 11) 10-1 m(10, 11) 101-

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m10 and m14. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00 m(9, 11) 10-1 m(10, 11) 101- m(10, 14) 1-10

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m12 and m14. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111 Size 2 m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00 m(9, 11) 10-1 m(10, 11) 101- m(10, 14) 1-10 m(12, 14) 11-0

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Compare each term in group 3 to each term in group 4. Combine terms m11 and m15. (Only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00 m(9, 11) 10-1 m(10, 11) 101- m(10, 14) 1-10 m(12, 14) 11-0 m(11, 15) 1-11

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m14 and m15. (Only one bit differs.) All groups in this column have been combined. Start a new column.

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100- m(8, 10) 10-0 m(8, 12) 1-00 m(9, 11) 10-1 m(10, 11) 101- m(10, 14) 1-10 m(12, 14) 11-0 m(11, 15) 1-11 m(14, 15) 111- Size 4

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Compare each term in group 1 to each term in group 2. Combine terms m(8, 9) and m(10, 11). (Dashes match and only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0 m(8, 12) 1-00 m(9, 11) 10-1 m(10, 11) 101-

• m(10, 14)

1-10 m(12, 14) 11-0 m(11, 15) 1-11 m(14, 15) 111- Size 4 m(8, 9, 10, 11) 10--

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m(8, 10) and m(9, 11). (Dashes match and only one bit differs) The term m(8, 9, 10, 11) is already listed, so don’t add it again, but do put a checkmark by the terms m(8, 10) and m(9, 11)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0

• m(8, 12)

1-00 m(9, 11) 10-1

• m(10, 11)

101-

• m(10, 14)

1-10 m(12, 14) 11-0 m(11, 15) 1-11 m(14, 15) 111- Size 4 m(8, 9, 10, 11) 10--

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m(8, 10) and m(12, 14). (Dashes match and only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0

• m(8, 12)

1-00 m(9, 11) 10-1

• m(10, 11)

101-

• m(10, 14)

1-10 m(12, 14) 11-0

• m(11, 15)

1-11 m(14, 15) 111- Size 4 m(8, 9, 10, 11) 10-- m(8, 10, 12, 14) 1--0

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m(8, 12) and m(10, 14). (Dashes match and only one bit differs) The term m(8, 10, 12, 14) is already listed, so don’t add it again, but do put a checkmark by the terms m(8, 12) and m(10, 14)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0

• m(8, 12)

1-00

• m(9, 11)

10-1

• m(10, 11)

101-

• m(10, 14)

1-10

• m(12, 14)

11-0

• m(11, 15)

1-11 m(14, 15) 111- Size 4 m(8, 9, 10, 11) 10-- m(8, 10, 12, 14) 1--0

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Compare each term in group 2 to each term in group 3. Combine terms m(10, 14) and m(11, 15). (Dashes match and only one bit differs)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0

• m(8, 12)

1-00

• m(9, 11)

10-1

• m(10, 11)

101-

• m(10, 14)

1-10

• m(12, 14)

11-0

• m(11, 15)

1-11

• m(14, 15)

111- Size 4 m(8, 9, 10, 11) 10-- m(8, 10, 12, 14) 1--0 m(10, 11, 14, 15) 1-1-

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Combine terms m(10, 11) and m(14, 15). (Dashes match and only one bit differs.) The term m(10, 11, 14, 15) is already listed, so don’t add it again, but do put a checkmark by the terms m(10, 11) and m(14, 15)

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0

• m(8, 12)

1-00

• m(9, 11)

10-1

• m(10, 11)

101-

• m(10, 14)

1-10

• m(12, 14)

11-0

• m(11, 15)

1-11

• m(14, 15)

111-

• Size 4

m(8, 9, 10, 11) 10-- m(8, 10, 12, 14) 1--0 m(10, 11, 14, 15) 1-1-

Phase 1: Finding Prime Implicants

F(A, B, C, D) = (

• m(4, 8, 10, 11, 12, 15)) + d(9, 14)

Cannot combine anything from group 1 of the size 4 column with anything in group 2 of the size 4 column, so we are done. Everything without a checkmark is a prime implicant.

Number

• f 1s

Term Binary 1 m4 0100

• m8

1000

• 2

(m9) 1001

• m10

1010

• m12

1100

• 3

m11 1011

• (m14)

1110

• 4

m15 1111

• Size 2

m(4, 12)

• 100

m(8, 9) 100-

• m(8, 10)

10-0

• m(8, 12)

1-00

• m(9, 11)

10-1

• m(10, 11)

101-

• m(10, 14)

1-10

• m(12, 14)

11-0

• m(11, 15)

1-11

• m(14, 15)

111-

• Size 4

m(8, 9, 10, 11) 10-- m(8, 10, 12, 14) 1--0 m(10, 11, 14, 15) 1-1-

Phase 2: Finding Essential Prime Implicants

Build a table with one column for each minterm in the original function. Do not include columns for the don’t care terms. Minterm Term Pattern Expr m4 m8 m10 m11 m12 m15

Phase 2: Finding Essential Prime Implicants

Add a row for the first prime implicant, m(4, 12). Put an X in the m4 column, and an X in the m12 column. Minterm Term Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100

B¯ C¯ D X X

Phase 2: Finding Essential Prime Implicants

Add a row for m(8, 9, 10, 11). Put an X in the m8, m10, and m11 columns. The m9 term is a don’t care, so there is no column for it. Minterm Term Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100

B¯ C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X

Phase 2: Finding Essential Prime Implicants

Add a row for m(8, 9, 12, 14). Put an X in the m8 and m12 columns. The m9 and m14 terms are don’t care, so there is no column for them. Minterm Term Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100

B¯ C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0 A¯ D X X

Phase 2: Finding Essential Prime Implicants

Add a row for m(10, 11, 14, 15). Put an X in the m10, m11, and m15 columns. The m14 term is a don’t care, so there is no column for it. Minterm Term Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100

B¯ C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0 A¯ D X X m(10, 11, 14, 15) 1-1- AC X X X

Phase 2: Finding Essential Prime Implicants

Any column with only one X indicates an essential prime implicant. Those terms must be part of the final result. Minterm Term Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100

B¯ C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0 A¯ D X X m(10, 11, 14, 15) 1-1- AC X X X

Phase 2: Finding Essential Prime Implicants

There is only one X in the m4 column, on row m(4, 12), so m(4, 12) is essential. Mark columns m4 and m12, and mark row m(4, 12). Minterm

• Term

Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100
• B¯

C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0 A¯ D X X m(10, 11, 14, 15) 1-1- AC X X X

Phase 2: Finding Essential Prime Implicants

There is only one X in the m15 column, on row m(10, 11, 14, 15), so m(10, 11, 14, 15) is

• essential. Mark columns m10, m11, and m15, and mark row m(10, 11, 14, 15).

Minterm

• Term

Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100
• B¯

C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0 A¯ D X X m(10, 11, 14, 15) 1-1-

• AC

X X X

Phase 2: Finding Essential Prime Implicants

Choose the minimal set of terms necessary to cover the remaining columns. In this case either of the remaining terms would cover the column, and they are both the same size, so there are two equivalent minimal equations. Minterm

• Term

Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100
• B¯

C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0 A¯ D X X m(10, 11, 14, 15) 1-1-

• AC

X X X

Phase 2: Finding Essential Prime Implicants

Solution 1: We have m(4, 12), m(10, 11, 14, 15), and m(8, 9, 11, 10): f = B¯ C¯ D + AC + A¯ B Minterm

• Term

Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100
• B¯

C¯ D X X m(8, 9, 10, 11) 10--

• A¯

B X X X m(8, 9, 12, 14) 1--0 A¯ D X X m(10, 11, 14, 15) 1-1-

• AC

X X X

Phase 2: Finding Essential Prime Implicants

Solution 2: We have m(4, 12), m(10, 11, 14, 15), and m(8, 9, 12, 14): f = B¯ C¯ D + AC + A¯ D Minterm

• Term

Pattern Expr m4 m8 m10 m11 m12 m15 m(4, 12)

• 100
• B¯

C¯ D X X m(8, 9, 10, 11) 10-- A¯ B X X X m(8, 9, 12, 14) 1--0

• A¯

D X X m(10, 11, 14, 15) 1-1-

• AC

X X X