1. EECS 465: Digital Systems Lecture Notes # 2
Two-Level Minimization Using Karnaugh Maps
SHANTANU DUTT
Department of Electrical & Computer Engineering
University of Illinois, Chicago
Phone: (312) :
URL:

2. Concepts in 2-Level Minimization
Defn: An implicant g of a function is a product term (e.g., g=xyz) s.t. when g=1  f=1
Defn: A minterm is an implicant whose product term representation consists of all n variables of an n-var. function (each in either compl.or uncompl. form)
Defn: Let g, and h be 2 implicants of a function f. g is said to cover h if h=1  g=1 (e.g., g = xz, h = xyz)
Defn: Let G = {g1, .., gk} be a set of implicants of func f, and let h be another implicant of f. Then G covers h if h=1  g gk =1 (generalized covering).

3. 3
Defn: An SOP or POS expression is also called a 2-level expression
Defn: A 2-level AND-OR (OR-AND) gate realization of an SOP (POS) expression is one in which all product terms (OR terms in POS) in the SOP (POS) expression are implementedby multiple input AND (OR) gates & the ORing (ANDing) of the product (OR) terms is realized by a multiple input OR (AND) gate.
AB + BC + ACD A B C D f1
Level 2
Level 1

4. 4
An OR-AND 2-level circuit implementing a POS expression
Level 2
Level 1
(B + C)(A + D)(C + D) B C A D f2

5. 3
A multi-level circuit implementing a multi-level expression
E.g.
f1 = AB + BC + ACD level. (SOP)
f2 = ( B+C )( A+D )( C+D ) level (POS)
f3 =AB + AC( B+D ) not SOP or POS
level 1 I/Ps not 2-level
level ( can be realized
directly by 3 levels of gates) A B A f3 C B
level 2
I/Ps D
level 2
gates
level 3
gate
level 0
I/Ps
level 1
gates

6. 6
Defn.
A gate in a logic circuit is a level-1 gate if all its inputs are
primary inputs (i.e., literals, A, A, B, B, etc.)
A gate is a level-i gate g, i > 1, if the highest-level gate whose
output feeds g is a level-(i-1) gate.
NOTE:
If a given expr. is not 2-level, we can convert it to a
2-level one using the distributive law.
The traditional goal of 2-level minimization is to minimize the number of literals (a literal is a var. in complemented or uncompl. form) in a 2-level logic expr.
This approx. reduces the total # of inputs over all level-1 gates (AND gates in an SOP expr. , OR gates in a POS expr.) in the circuit in a 2-level gate realization.
Cost of a circuit = total # of gate inputs across all its gates (approx. = (number of MOS transistors)/2; exact for nand/nor/and/or/not gates)
Cost of a 2-level expr. f = cost of a 2-level circuit implementing f
A better 2-level minimization goal is to minimize the number of literals plus number of product terms as this corresponds exactly to total # of gate inputs across all its gates in a 2-level implementation

7. E.g. f = ABC + ABC + ABC (non-minimized 9 literals) A B (12 gate I/Ps) 7
E.g. f = ABC + ABC + ABC (non-minimized 9 literals) A B
(12 gate I/Ps)
= Circuit complexity C A f B C A B C
f = AB + AC (minimized 4 literals) A f B
(6 gate I/Ps)
= Circuit complexity A C
Note: Minimizing # of literals + # of product terms  minimizing total # of gate inputs in a 2-level impl.  approx. min. total # of gate i/ps of a multi-level impl. (reqd. when gates of certain sizes reqd. by the minimized expression is not available)

8. Multilevel Minimization Example 8
Multilevel Minimization Example
f = x1x2x3 + x2x3x4 + x1x5 + x4x (2-level minimized)
Applying the distributive law (factoring)
f = x2x3(x1 + x4) + x5(x1 + x4)
= (x2x3 + x5)(x1 + x4) -- Multilevel expr. See gate impl. below
Cost = 14 (# of gate i/ps in a 2-level impl.)
Cost = 8 (# of gate i/ps).
Can increase delay though due to more levels of gates & more interconnects that the critical path has to go through
Simple gate delay model: delay prop. to # of inputs
Critical path goes through 3 2-i/p gates (6 units of delay) and 2 interconn. delays (not counting i/p and o/p interconnects). The critical path in a 2-level impl. of the above SOP expr. will go through a total of 7 gate units of delay but only 1 interconnect delay.
An interconnect’s delay is generally much more than a single gate i/p delay
So mostly, and esp. for large ckts, cost decrease via multi-level min. can cause delay increase. x2 x3 x5 x1 x4 f
Multilevel
circuit

9. Two implicants (or product terms) of a function f are said to 9
Defn.
Two implicants (or product terms) of a function f are said to
be logically adjacent if they have the same literals
except in one variable xi which occurs in uncomplemented form
(xi) in one implicant and in complemented form (xi) in the
other. The 2 implicants are said to differ in xi.
E.g.
ABC, ABC (adjacent implicants, differ in B)
ABD, ABC (not adjacent)
Adjacent implicants can be combined into one implicant
by the combining theorem (ABC + ABC = AC) C 1
A’B’ AC AB
ABC
AB’C
A’B’C’
A’B’C
AC’ A

10. Another ordering of inputs ( Gray-code ordering) A B Z1 0 0 0 0 1 1 10
Karnaugh Maps
2-variable TT outputs A B Z1 Z2
Binary
place-value
ordering
(Binary
ordering)
Physically adjacent
but not logically
adjacent.
logically adjacent but not physically adjacent
AB + AB B
Another ordering of inputs ( Gray-code ordering)
A B Z1
Z1 = B
Physically as well as
logically adjacent.

11. n-variable Gray-code ordering11
n-variable Gray-code ordering
Defn.
Gi = i-bit Gray-code ordering
rev(Gi) = reverse order of Gi
Gn = 0 Gn-1
1 rev(Gn-1)
G1 = Base. 1
G2 = 0 G1
1 rev(G1)
0 0
0 1
1 1
1 0
A B C =
G3 = 0 G2
1 rev(G2) =
Convert to a 2-dimensional
Gray-code ordered TT AB x
110 C y 1
Throw away variable (A)
changing across the 2 squares.
We thus obtain BC. 1 1
001

12. 12 Example ( 3-var. K-Map) Consensus = BC Function f:
Gray code ordering
Consensus = BC AB C 1
A’B’
AC’
B’C’
Function f:
f = AB + AB + AC + BC
Note: 1) The consensus theorem is an example of generalized covering discussed earlier—the consensus is covered by the sum of the 2 implicants whose consensus it is.
2) AC’ is also a consensus, this time of AB and B’C’. Can’t discard both A’C’ and B’C’ simultaneously! Need to apply gen. covering (incl. consensus) one at a time.
Defn. A Prime Implicant (PI) of a function f is an implicant of f that
is not covered by any other implicant of f.
Defn. An Essential PI is a PI that covers/includes at least one minterm
that is not covered by any other PI.
In the above K-map,
four PIs AB, BC, AC, AB formed. Not all these PIs are needed in the expression. Only a minimum set that covers all minterms is needed.
In the above example, AB & AB are essential PIs. These will need to be present in any SOP expr. of the function. To form a minimal set, the PI BC can be used. Thus f = AB + AB + BC is a minimized expression.

13. Larger than 2-minterm PIs can also exist: E.g.,13
Larger than 2-minterm PIs can also exist: AB AB
E.g., C C 1 1 AB AB B C B
2-minterm implicants (AB, AB in the 1st example above) can
be merged if they are logically adjacent to form a 4-minterm
implicant and so on.

14. When an implicant can not be “grown” any further, then it is a PI.14
When an implicant can not be “grown” any further, then it is a PI.
Defn. In a K-map, 2 implicants can be logically adjacent if they
are symmetric,
i.e., if : (1) They are disjoint (no common minterms).
(2) They cover the same # of minterms.
(3) Each minterm in one implicant is adjacent to a
unique minterm in the other implicant. AB
More Examples:
Both PIs
(A, BC)
are
essential. C AB
Redrawn 1 C 1
Not symmetric
Thus, f = A + BC
is minimized.
Symmetric ( logically adjacent)
2-minterm implicants : Can be “merged” to form
a 4-minterm implicant, which will be a PI.

15. 15 4-variable K-Map: Gray code ordering AB CD Gray code ordering 00 01 11 10 CD
Gray code ordering

16. Instead of starting with 2-minterm implicants & growing them, you 16
Example:
f = AC + D
Instead of starting with 2-minterm
implicants & growing them, you
can form larger implicants directly
by grouping power of 2 (2, 4, 8,
etc.) # of minterms so as to form
a rectangle or square ( a convex
region). AB CD 00 01 AC 11 10
Convex D
Concave AB CD
Invalid grouping
(region formed is concave)
f = AB + AC + BCD 00 01 11 10
Valid
groupings 1

17. 17 Another Example: AB CD BD ABD C f = C + BD + ABD TT: 00 01 11 10
A B C D Z 00 01
ABD 1 C 11 10
Binary
order
f = C + BD + ABD

18. Many times certain input combinations are invalid for a function 18
Don’t Cares in K-Maps
Many times certain input combinations are invalid for a function
(E.g. BCD to 7-segment display functions; see pp in
Katz text )
For such combinations, we do not care what the output is, and
we put an ‘x’ in the o/p column for the TT AB CD
A B C f x x x 1
x x
f = BC + AB
by considering all
x’s as 0s
f = B
The x’s can, however, be profitably
used to make larger PIs & thus
further simplify the function.
by selectively
considering the 010
cell’s x as 1

19. In a K-map, the ‘x’s can be profitably used to form larger
implicants (which have fewer literals) 19
However, when we need to form a minimal PI cover, we need to
worry only about covering the minterms ( the 1s), and not the Xs.
• Thus as far as x’s are concerned, we can have the cake and eat it too!
Example:
f(A,B,C,D) =
m(1,3,5,7,9) +
d(2,6,12,13) AB CD 00 01 11 10 x x
2 x x
LSBs
MSBs AC AD
A B C D
f = CD + AD
or CD + AC
If input comb. 2,6 occur
o/p is 0.
If inputs 2,6 occur o/p = 1.
Which to use for lower dynamic power (consumed on a 0  1 transition at the o/p)?

20. STEPS IN K-MAP MINIMIZATION (sop EXPR.)20
STEPS IN K-MAP MINIMIZATION (sop EXPR.)
STEP1: Form all PIs by including the Xs with the 1s to form larger
PIs. (Do not form PIs covering only Xs !)
STEP2: Identify all essential PIs by visiting each minterm (1s) &
checking if it is covered by only one PI. If so, then that
PI is an essential PI.
STEP3: Identify all 1s not covered by essential PIs. Select a minimal- cost set S of PIs to cover them. This requires some trial-and error (in fact, this is a hard computational problem).
STEP4: Minimal SOP Expr. for function f:
f =
essential PIs + Pi Pi S
PI Pi is in set S

21. Minimal POS Expr. from a K-Map : 21
METHOD 1: Obtain minimal SOP expression for the complement
function f, and complement this SOP expression to get a minimal
POS expr. for f (using De-Morgan’s Law)
Big M notation
Example:
F (A,B,C,D) =
M(0,2,4,8,10,11,14,15)
D(6,12,13)
Xs of F
0s of F AB F F AB CD CD D x x 00 01 11 10 x x x 00 01 11 10
Complement x AC
F =D + AC F = D + AC = D(A +C)

22. Example: (Method2: Direct Method)22
Example: (Method2: Direct Method) AB
(OR terms) CD x x x 00 01 11 10
An implicate of f is an OR term which when 0  f = 0.
An implicate g covers implicate h if h = 0  g = 0
Prime implicate is an implicate not covered by any other implicate. (groups of 0s that cannot be grown any further).
A prime implicate that covers at least one 0 [or maxterm] not covered by any other prime implicate is an essential prime implicate.
Both D and A + C are essential prime implicates. D
(A + C)
F = D( A + C )
( Product of Prime Implicates )

23. METHOD2 : Direct Method : 23
(1) Obtain all prime implicates (PTs) (largest groups of 0s & Xs
of size 2 , i
0, forming a convex region, that can not be “grown”
any further).
(2) Identify all essential PTs (those that cover at least one 0 not
covered by any other PT).
(3) Choose a minimal set T of PTs to cover the 0s not covered by
the set of essential PTs.
(4) The expression for a PT is an OR term obtained by discarding
all changing variables, & keeping variables complemented if they
are constant at 1 or uncomplemented if they are constant at 0 &
taking the OR (sum) of these literals.
(5) Minimal POS Expr.:
f =
(essential PTs) qi qi T

24. 5-variable K-map: f(A,B,C,D,E): Juxtapose two 4-var. K-submaps 24
5-variable K-map: f(A,B,C,D,E): Juxtapose two 4-var. K-submaps
one for the MSB A=0 and the other for A=1: BC BC DE DE 00 01 11 10 00 01 11 10
A=0
A=1
4-var. K-submap
4-var. K-submap
Adjacencies of a square: Adjacent squares of the 4-var. K-submap
plus the “corresponding” or “mirror” square in the other 4-var.
K-submap.
Example: BC BC DE DE 00 01 11 10 00 01 11 10
x 24
19 x x 27
x 11
x 10
A=0
A=1

25. 6-variable K-map: f(A,B,C,D,E,F)--Juxtapose two 5-var. K-submaps 25
6-variable K-map: f(A,B,C,D,E,F)--Juxtapose two 5-var. K-submaps
one for the MSB A=0 and other for A=1 CD CD EF EF 00 01 11 10 00 01 11 10
5-var.
K-submap
A=0
B=0
B=1 CD CD EF EF 00 01 11 10 00 01 11 10
5-var.
K-submap
A=1
B=1
B=0
Adjacencies of a square: Adjacent squares of the 5-var.
K-submap plus the “corresponding” or “mirror” square
in the other 5-var. K-submap.

26. 6-variable K-map: f(A,B,C,D,E,F,)---Example:26
6-variable K-map: f(A,B,C,D,E,F,)---Example: CD CD EF EF 00 01 11 10 00 01 11 10
A=0
B=0
B=1 CD CD EF EF 00 01 11 10 00 01 11 10
A=1
B=1
B=0

27. f = (C’+D’)(B’+C’)(C+B)(A’+C+D)
f = A’BC’ + BC’D + B’CD’ AB 00 01 11 10 CD 1 1 1 1 1
f = (C’+D’)(B’+C’)(C+B)(A’+C+D)

28. f = (C’+D’)(A’+B)(A’+C’)(C+B)(A’+D)
f = A’BC’ + BC’D + A’CD’ AB 00 01 11 10 CD 1 1 1 1 1
f = (C’+D’)(A’+B)(A’+C’)(C+B)(A’+D)

29. f = (C’+D’)(A’+B)(C+B)
f = BC’ + A’CD’ + BCD’ AB 00 01 11 10 CD 1 1 1 1 1 1 1
f = (C’+D’)(A’+B)(C+B)