Lecture Notes 10

Last Time

• Project 1 Assigned
• 4 More BA Rules
• TT POS Simplify w/BA NOR LD
• Karnaugh Maps
• Introduction
• 2,3, & 4 Variables
• Review of Exam 1

This Time

• Boolean Algebra - 4 More Rules
• Factoring
• Karnaugh Maps
• Definitions
• Many Examples
• 5 & 6 Variables

Boolean Algebra

29. A XOR A = 0
30. A XOR A' = 1
31. (A + B)(B' + C)(A + C) = (A + B)(B' + C)
32. AB + B'C + AC = AB + B'C

Derive rule 31 by a truth table proof.

Derive rule 32 by a karnaugh map.

Factoring

• Use rule 14 to help reduce the gate count.
14. AB + AC = A(B + C)

Example

• Y = A' D' + A' C'
• A straightforward implementation requires:
• 3 2-Input NAND's
• 3 Inverters
• After factoring, we have
Y = A' (C'+D')
• This can be implemented more efficiently as shown below
• This requires
• 2 2-Input NAND's
• 2 Inverters

Example

• Y = A' C' + A' D' + AEF + A B'
• A straightforward implementation requires:
• 3 2-Input NAND's
• 1 3-Input NAND
• 1 4-Input NAND (to combine all four products)
• 4 Inverters (for A, B, C, and D)
• This would take 3-4 IC's to implement depending on whether you used one of the spare NAND's to implement the inverter.
• After factoring, we have
Y = A' (C' + D') + A (EF + B')
• This can be implemented as follows
• This requires
• 6 2-Input NAND's
• 1 Inverter
• Assuming you used standard IC's that have 4 2-Input NAND's per an IC, this would take 2 IC's to implement if you used one of the spare NAND's to implement the inverter.

Karnaugh Maps

Today we will discuss Karnaugh maps more formally than last time and discuss a more orderly method for optaining the minimal sum.

Literal
A variable or complement of a variable.
Examples: A, B', C', D
Normal Product Term
A product term in which no variable appears more than once.
Examples: ABC, AB'C', A'B'C'
Counter Examples: AA'BC, AB'B'C', ABCC
n-variable Minterm
A normal product term with n literals. There are 2n such terms.
Example:
 A B 0 0 A'B' 0 1 A'B 1 0 AB' 1 1 AB
Each of the product terms corresponding to a row in the truth table is a minterm.
Minimal Sum
A sum of products (SOP) expression such that no SOP expression for Y has fewer product terms and any SOP expression with the same number of product terms has at least as many literals.
This is what we are trying to produce through the use of Karnaugh maps.
Implicant
A normal product term that implies Y.
Example: For the function Y = AB + ABC + BC, the implicants are AB, ABC, and BC because if any one of those terms are true, then Y is true.
Prime Implicant
An implicant of Y such that if any variable is removed from the implicant, the resulting term does not imply Y.
Example: Y = AB + ABC + BC
Prime Implicants: AB, BC
Not a prime implicant: ABC
ABC is not a prime implicant because the literal A can be removed to give BC and BC still implies Y. Conversely AB is not a prime implicant because you can't remove either A or B and have the remaining term still imply Y.
In truth tables the prime implicants are represented by the largest rectangular groups of ones that can be circled. If a smaller subgroup is circled, the smaller group is an implicant, but not a prime implicant.
PI Theorem
A minimal sum is a sum of prime implicants.
Distinguished 1-Cell
An input combination that is covered by 1 prime implicant. In terms of Karnaugh maps, distinguished 1-cells are 1's that are circled by only 1 prime implicant.
Essential Prime Implicant
A prime implicant that that includes one or more distinguished one cells. Essential prime implicants are important because a minimal sum contains all essential prime implicants.

Karnaugh Map Examples

In the following examples the distinguished 1-cells are marked in the upper left corner of the cell with an asterisk (*). The essential prime implicants are circled in blue, the prime implicants are circled in black, and the non-essential prime implicants included in the minimal sum are shown in red.

Example 1

Prime Implicants: 5
Distinguished 1-Cells: 2
Essential Prime Implicants: 2
Minimal Sums: 1

Example 2

Prime Implicants: 7
Distinguished 1-Cells: 2
Essential Prime Implicants: 2
Minimal Sums: 1

Y = B'D' + AD' + A'C'D + BCD

Example 3

Prime Implicants: 6
Distinguished 1-Cells: 2
Essential Prime Implicants: 2
Minimal Sums: 3

Y = AB'C' + A'CD' + AC'D + BCD

Y = AB'C' + A'CD' + ABD + A'BC

Y = AB'C' + A'CD' + ABD + BCD

Example 4

Prime Implicants: 5
Distinguished 1-Cells: 3
Essential Prime Implicants: 3
Minimal Sums: 1
Y = A'B' + A'C' + ABC + A'D

Example 5

Prime Implicants: 4
Distinguished 1-Cells: 4
Essential Prime Implicants: 4
Minimal Sums: 1

Y = A'C + A'B + BD + CD

Example 6

Prime Implicants: 5
Distinguished 1-Cells: 3
Essential Prime Implicants: 3
Minimal Sums: 1
Y = B'D + BC' + AB

Example 7

Prime Implicants: 8
Distinguished 1-Cells: 0
Essential Prime Implicants: 0
Minimal Sums: 2

Y = A'B'C + A'BD + ABC' + AB'D'

Y = B'CD' + A'CD + BC'D + AC'D'

Example 8

Prime Implicants: 3
Distinguished 1-Cells: 8
Essential Prime Implicants: 3
Minimal Sums: 1

Y = B'C + D + BC'

5-Variable Karnaugh Maps

For these you must circle the prime implicants on each map individually and then the prime implicants on the joint map. The joint essential prime implicants are shown in green.

Prime Implicants: 7
Distinguished 1-Cells: 7
Essential Prime Implicants: 4
Minimal Sums: 2

Y = A'B'C' + BE + ABC' + ACE + A'DE

Y = A'B'C' + BE + ABC' + ACE + CDE

Note that the joint map can help you identify the joint prime implicants.

6-Variable Karnaugh Maps

The prime implicants unique to each map are shown in black.
The prime implicants shared between maps 0 and 1 (A=0) are shown in
aqua.
The prime implicants shared between maps 0 and 2 (B=0) are shown in
violet.
The prime implicants shared between maps 1 and 3 (B=1) are shown in
olive.
The prime implicants shared between maps 2 and 3 (A=1) are shown in
brown.
The prime implicants shared between all 4 maps are shown in
orange.

To find the prime implicants shared among maps it may help to draw out each of the 5 joint maps.

Distinguished 1-Cells: 10
Essential Prime Implicants: 5
Minimal Sums: 2

Essential Prime Implicants
A'EF (Maps 0 & 1)
BCD' (Maps 1 & 3)
B'D'F' (Maps 0 & 2)
ACE' (Maps 2 & 3)
ABDE' (Map 3)

Y = A'EF + BCD' + B'D'F' + ACE' + ABDE' + B'DE'F + A'B'C'F

Y = A'EF + BCD' + B'D'F' + ACE' + ABDE' + B'DE'F + A'B'C'D'