Unit 1(stld)

Minimization of
Switching Functions
Unit-II

Topics:
1. Karnaugh Map Method
2. Prime Implicants
3. Don’t Care Combinations
4. Minimal SOP and POS forms
5. Quine Mc Cluskey Tabular Method
6. Prime Implicant Chart
7. Simplification Rules

Simplifying Switching Functions
 SOP and POS expressions
==> 2-level circuits
Minimum SOP/POS expression:
 Minimize the number of literals
 Minimum number of terms
How?
 Algebraically: I.e. using the axioms and theorems of
Boolean algebra.
 Karnaugh Map
 McCluskey Method

Simplifying Switching Functions:
K-Map
Simplifying Theorem:
XY + X’Y = Y
Definition: Logical Adjacency
Two terms are logically adjacent if they differ in only one literal: the
literal is complemented in one term and non-complemented in the other.
Two Logically adjacent terms can be combined into one term consisting
of only the common literals

1. Karnaugh-Map (K-Map)
2-dimensional representation of a truth table.
Logically adjacent terms are physically adjacent in the map.
2-Variable Functions: F(X,Y) = XY + X’Y
X Y
0 0
0 1
1 0
1 1
F(X,Y)
m0
m1
m2
m3

2-Variable K-Map
X
Y
0 1
0
1
X
Y
0 1
0
1
00
01
10
11
X’Y’
X’Y
XY’
XY
Note:Note:
• Logically adjacent cells are
physically adjacent in the k-map
• Each cells has two adjacent cells
X
Y
0 1
0
1
m0 m2
m1 m3

Function Minimization Using K-Maps
1. Each square (minterm) in a k-map of 2 variables has 2 logically
adjacent squares, each square in a 3-variable k-map has 3
adjacent squares, etc.
2. Combine only the minterms for which the function is 1.
3. When combining terms on a k-map, group adjacent squares in
groups of powers of 2 (I.e. 2, 4, 8, etc.). Grouping two squares
eliminates one variables, grouping 4 squares eliminates 2
variables, etc.
 Can't combine a group of 3 minterms

Function Minimization Using K-Maps
4. Group as many squares together as possible; the larger the group is
the fewer the number of literals in the resulting product term
5. Select as few groups as possible to cover all the minterms of the
functions. A minterm is covered if it is included in at least one
group. Each minterm may be covered as many times as it is
needed; however, it must be covered at least once.
6. In combining squares on the map, always begin with those squares
for which there are the fewest number of adjacent squares (the
“loneliest" squares on the map).

2. Prime Implicants
Definitions:
 Implicant: a product term that could be used to cover one or more
minterms
 Prime Implicant: A product term obtained by combining the
maximum number of adjacent squares in the map.
 Essential Prime Implicant: A prime implicant that covers at least
one minterm that is not covered by any other prime implicant.
 All essential prime implicants must be included in the final
minimal expression.

Definitions (Cont.)
 Cover of function: is a set of prime implicants for which
each minterm of the function is covered by at least one
prime implicant.
 All essential prime implicants must be included in the
cover of a function.

3. Don’t care conditions
 In practice there are some applications where the function is not
specified for certain combinations of the variables.
 In most applications we simply don’t care what value is assumed by
the function for the unspecified min terms.
 It is customary to call the unspecified min terms of a function don’t-
care conditions.
 These don’t-care conditions can be used on a map to provide
further simplification of the Boolean Expression.

4. Algorithm for Deriving the
Minimal SOP
1. Circle all prime implicants on the k-map.
2. 2. Identify and select all essential prime implicants.
3. Select a minimum subset of the remaining prime implicants to cover
those minterms not covered by the essential prime implicants.

00 01
00
01
11
10
11 10
Four Variable K-map

ABC
DE
000 001
00
01
11
10
011 010 100 101 111 110
m16
m17
m19
m18
m20
m21
m23
m22
m28
m29
m31
m30
m24
m25
m27
m26
ABC
DE
00
01
11
10
m0
m1
m3
m2
m4
m5
m7
m6
m12
m13
m15
m14
m8
m9
m11
m10
Five Variable K-map

More Examples
Find min. SOP and POS expression for each of the
following functions
1. F(A,B,C,D) = Σ m(2,3,5,7,10,11,13,14,15)
2. G(W,X,Y,Z)= Π M(1,3,4,5,7)
3. H(A,B,C,D) = Σm(1,3,4,7,11)+d(5,12,13,14,15)
4. F2(A,B,C,D)= Σ m(1,2,7,12,15)+d(5,9,10,11,13)
5. F3(A,B,C,D,E) = Σ
m(0,1,2,4,5,6,13,15,16,18,22,24,26,29)

TabulationTabulation
MethodMethod

5. Tabulation Method
 Map method is a trial-and-error procedure
 Tabulation method performs thorough search
 It starts with SOM and consists of 2 steps:
 PIs generation
 group minterms by number of 1s
 compare minterms & find pairs that differ in 1 variable
 generate subcubes
 repeat the above 3 steps to generate subcubes until no more
subcubes can be generated
 Minimal cover generation
 find EPIs through a selection table
 find minimal cover through the POS of PIs

 K-map representation:
 PIs generation:
 0-subcubes
Example: simplify w’y’z’ + wz + xyz + w’y
00 01 11 10
00 1 0 1 1
01 1 0 1 1
11 0 1 1 0
10 0 1 1 0
yz
wx

 1-subcubes
 2-subcubes
Example: simplify w’y’z’ + wz + xyz + w’y (cont.)

 Minimal cover generation:
 EPIs selection
 PI list: w’z’, w’y, yz, wz
 EPI list: w’z’, wz
 POS: (P2 + P3)(P2 + P3) = P2 + P3
 Minimal cover expressions:
 F1 = w’z’ + wz + w’y
 F2 = w’z’ + wz + yz
Example: simplify w’y’z’ + wz + xyz + w’y (cont.)

 K-map representation:
 PIs generation:
 0-subcubes, 1-subcubes
Another example
00 01 11 10
00 0 0 0 1
01 0 0 1 1
11 0 1 1 0
10 1 1 0 0
yz
wx

 Minimal cover generation:
 EPIs selection
 PI list: w’yz’, x’y’z, w’xy, wx’z, xyz, wyz
 EPI list: w’yz’, x’y’z
 POS: (P3 + P5)(P4 + P6)(P5 + P6) =
(P3 + P5)(P4P5 + P5P6 + P4P6 + P6) =
P3P4P5 + P4P5 + P3P6 + P5P6
 Minimal cover expressions:
 F1 = w’yz’ + x’y’z + wx’z + xyz
 F2 = w’yz’ + x’y’z + w’xy + wyz
 F3 = w’yz’ + x’y’z + xyz + wyz
Another example (cont.)

EXAMPLE
f on
= {m0, m1, m2, m3, m5, m8, m10, m11, m13, m15} = ∑ (0, 1, 2, 3, 5, 8, 10, 11, 13,
15)
Minterm Cube
0 0 0 0 0 
1 0 0 0 1 
2 0 0 1 0 
8 1 0 0 0 
3 0 0 1 1 
5 0 1 0 1 
10 1 0 1 0 
11 1 0 1 1 
13 1 1 0 1 
15 1 1 1 1 
Minterm Cube
0,1 0 0 0 - 
0,2 0 0 - 0 
0,8 - 0 0 0 
1,3 0 0 - 1 
1,5 0 - 0 1 PI=D
2,3 0 0 1 - 
2,10 - 0 1 0 
8,10 1 0 - 0 
3,11 - 0 1 1 
5,13 - 1 0 1 PI=E
10,11 1 0 1 - 
11,15 1 - 1 1 PI=F
13,15 1 1 - 1 PI=G
Minterm Cube
0,1,2,3 0 0 - - PI=A
0,8,2,10 - 0 - 0 PI=C
2,3,10,11 - 0 1 - PI=B
f on
= {A,B,C,D,E,F,G} = {00--, -01-, -0-0, 0-01, -101, 1-11, 11-1}

6. Construct Cover Table
 PIs Along Vertical Axis (in order of # of literals)
 Minterms Along Horizontal Axis
0 1 2 3 5 8 10 11 13 15
A x x x x
B x x x x
C x x x x
D x x
E x x
F x x
G x x

Finding the Minimum Cover
 Extract All Essential Prime Implicants, EPI
 EPIs are the PI for which a Single x Appears in a Column
0 1 2 3 5 8 10 11 13 15
A x x x x
B x x x x
C x x x x
D x x
E x x
F x x
G x x
• C is an EPI .
• Row C and Columns 0, 2, 8, and 10 can be Eliminated
Giving Reduced Cover Table
• Examine Reduced Table for New EPIs

Reduced Table
0 1 2 3 5 8 10 11 13 15
A x x x x
B x x x x
C x x x x
D x x
E x x
F x x
G x x
1 3 5 11 13 15
A x x
B x x
D x x
E x x
F x x
G x x
Essential row
Distinguished Column
•The Row of an EPI is an
Essential row
•The Column of the Single x in
the Essential Row is a
Distinguished Column

The Reduced Cover Table
 Initially, Columns 0, 2, 8 and 10 Removed
1 3 5 11 13 15
A x x
B x x
D x x
E x x
F x x
G x x
• No EPIs are Present
• No Row Dominance Exists
• No Column Dominance Exists
• This is Cyclic Cover Table
• Must Solve Exactly OR Use a Heuristic

Unit 1(stld)

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Unit 1(stld)