Minimization via KMaps

1
Minimization via K-Maps
A
Row A B C F(A,B,C) 0 0 0 0
0 1 0 0 1 0 2 0 1 0
1 3 0 1 1 0 4 1 0 0
0 5 1 0 1 0 6 1 1 0
1 7 1 1 1 0
1
0
BC
0
0
00
01
0
0
11
0
0
1
1
10
F(A,B,C) ? m(2,6) ABC
ABC BC(AA)
BC
Boolean adjacency can be used to minimize
functions!
2
Groupings on K-Maps
A
1
0
BC
Grouping can be read DIRECTLY as BC by looking
at what is COMMON within the circled group.
0
0
00
01
0
0
11
0
0
1
1
10
F(A,B,C) BC
3
Example Groupings on 3-Variable K-Maps
A
1
0
F(A,B,C) AB
BC
1
0
00
01
1
0
1
0
11
BC
0
0
1
1
00
0
0
10
01
0
0
11
0
0
F(A,B,C) C
A
1
0
BC
1
10
1
1
1
00
Remember that top, bottom of map are adjacent!!
01
1
1
11
0
0
F(A,B,C) B
0
0
10
4
Multiple Groupings
A
1
0
BC
Want to cover all 1s with largest possible
groupings.
1
0
00
01
1
1
11
0
0
F(A,B,C) BC AB
0
0
10
A
1
0
BC
Groupings of only a single 1 are ok if larger
groupings cannot be found.
0
1
00
01
0
0
11
F(A,B,C) ABC AB
1
0
1
0
10
5
Illegal Groupings
A
1
0
BC
1
0
00
Illegal Grouping! Minterms are not boolean
adjacent!
01
0
1
11
0
0
ABC , ABC will NOT reduce to a single
product term ABC ABC B(ACAC)
0
0
10
Valid groupings will always be a power of 2.
(will cover 1, 2, 4, 8, etc minterms).
6
Groupings on four Variable Maps
AB
01
00
11
10
CD
0
0
00
0
0
01
0
0
0
0
11
1
0
1
0
1
1
10
0
1
F(A,B,C,D) ABC ACD BCD ABCD
7
Other Groupings
AB
01
00
11
10
CD
1
0
00
0
1
01
1
0
0
1
11
1
0
0
1
1
0
10
0
1
F (A,B,C,D) B
8
More than one way to group..
AB
01
00
11
10
CD
1
1
00
1
1
F (A,B,C,D) BD CD CD
01
1
0
0
1
11
1
0
0
1
1
1
10
1
1
AB
Want LARGEST groupings that can cover 1s.
01
00
11
10
CD
1
1
00
1
1
01
1
0
0
1
F (A,B,C,D) B D
11
1
0
0
1
1
1
10
1
1
9
Four Corner Grouping on 4-Variable Map
AB
01
00
11
10
CD
1
0
00
0
1
01
0
0
0
0
4 Corner grouping is valid on four variable map
11
0
0
0
0
1
0
10
0
1
F (A,B,C,D) BD
10
Some Definitions
A
Implicant Any single 1 or any group of 1s is
called an implicant of F. Any possible grouping
of 1s is an implicant.
1
0
ABC
BC
0
1
00
01
0
0
BC
11
1
1
1
1
10
B
Prime Implicant A covering that cannot be
combined with some other covering to eliminate a
variable.
A
AC
1
0
BC
0
1
00
01
0
0
11
1
1
B
1
1
10
11
Minimum SOPs
The minimum SOP expression consists of some (but
not necessarily all) of the prime implicants of a
function.
If a SOP expression contains a term which is NOT
a prime implicant, then it CANNOT be minimum.
12
Prime Implicants
AB
01
00
11
10
CD
EACH of the coverings is a PRIME IMPLICANT.
0
1
00
1
0
01
1
1
1
0
11
BC , ACD , ABD
1
0
0
0
0
0
10
0
0
Minimum SOP will have some or all of these prime
implicants. The included prime implicants must
cover all of the ONEs.
F(A,B,C,D) BC ABD (minimum
of PIs) BC ABD
ACD (valid, but not minimum)
? ABD ACD (both PIs, but all 1s not
included!)
13
Non-Essential vs. Essential Prime Implicants
AB
01
00
11
10
CD
EACH of the coverings is a PRIME IMPLICANT.
0
1
00
1
0
01
1
1
1
0
11
BC , ACD , ABD
1
0
0
0
0
0
10
0
0
F(A,B,C,D) BC ABD (minimum
of PIs)
Prime Implicant ACD is a NON-ESSENTIAL prime
implicant because its 1s are covered by other
PIs . A PI is ESSENTIAL if it covers a MINTERM
that cannot be covered by any other PI.
14
An example with more than one solution
AB
EACH of the coverings is a PRIME IMPLICANT.
01
00
11
10
CD
AC
1
1
00
0
0
01
1
1
0
0
11
0
1
1
1
ACD
0
0
10
0
0
ABD
BCD
Recall that a covering is a Prime Implicant if it
cannot be combined with another covering to
eliminate a variable.
15
Two Solutions
AB
EACH solution is equally valid.
01
00
11
10
CD
F(A,B,C,D) AC ACD ABD
1
1
00
0
0
01
1
1
0
0
Essential PIs
11
0
1
1
1
Non-Essential PIs
0
0
10
0
0
AB
01
00
11
10
CD
1
1
00
0
0
01
1
1
0
0
F(A,B,C,D) AC ACD BCD
11
0
1
1
1
0
0
10
0
0
16
Minimal Solution
A minimal SOP will consist of prime implicants. A
minimal SOP equation will have all of the
essential prime implicants on the map. By
definition, these cover a minterm that may not be
covered by some other prime implicant. The
minimal SOP equation may or may not include
non-essential prime implicants. It will include
non-essential prime implicants if there are 1s
remaining that have not been covered by an
essential prime implicant.
17
Dont Cares
Row A B C D F(A,B,C,D) 0 0 0 0
0 0 1 0 0 0 1 0 2 0 0
1 0 1 3 0 0 1 1 1 4 0
1 0 0 0 5 0 1 0 1 0 6
0 1 1 0 1 7 0 1 1 1 0 8
1 0 0 0 0 9 1 0 0 1 010
1 0 1 0 x11 1 0 1 1 x12
1 1 0 0 x13 1 1 0 1 x14
1 1 1 0 x15 1 1 1 1 x
Recall that Dont Cares are labeled as Xs in
truth table. Can treat Xs as either 0s or 1s

A
F(ABCD)Recognize BCD numbers 2,3,6
B
C
D
Non BCD numbers are dont cares because will
never be applied as inputs.
18
Dont Cares treated as 0s or 1s
AB
01
00
11
10
CD
Treat Xs as 1s when can get a larger grouping.
All Xs do not have to be covered.
0
0
00
X
0
01
0
0
X
0
11
1
0
X
X
1
1
10
X
X
F(A,B,C,D) CD BC
19
Minimizing 0s
Grouping 0s produces an equation for F.
1
0
BC
1
1
00
01
0
0
F(A,B,C) C
11
0
0
1
10
1
F(A,B,C) C
20
Minimize 0s, then Complement to get POS
AB
01
00
11
10
CD
F(A,B,C,D) C BD Take inverse of both
sidesF(A,B,C,D) (C BD)
C (BD) C (BD)
0
0
00
X
0
01
0
0
X
0
11
1
0
X
X
1
1
10
X
X
Minimizing zeros, then applying inverse to both
sides is a way to get to minimum POS form!!!!!
21
What do you need to know?

• How to minimize functions using 2,3,4 variable
  Kmaps.
• Group 1s to get to minimal SOP form
• Group 0s then take complement to get to minimal
  POS form.
• Definitions of implicant, prime implicant,
  non-essential prime implicant, essential
  prime-implicant.
• Be able to recognize these on a K-map.
• How to treat Xs on a K-map.