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EECS 465: Digital Systems
Lecture Notes # 2
Two-Level Minimization Using Karnaugh Maps
SHANTANU DUTT
Department of Electrical & Computer EngineeringUniversity of Illinois, Chicago
Phone: (312) 355-1314: e-mail: [email protected]: http://www.ece.uic.edu/~dutt
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 is covers h if h=1 g1+ .. + gk =1 (generalized covering).•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.
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AB + BC + ACDA
B
BCACD
f1
Level 2
Level 1
(B + C)(A + D)(C + D) B
C
A
D
CD
f2
3
E.g. f1 = AB + BC + ACD --- 2-level. (SOP)f2 = ( B+C )( A+D )( C+D ) --- 2-level (POS)f3 =AB + AC( B+D ) --- not SOP or POS level 1 I/Ps --- not 2-level
--- 3 level ( can be realized directly by 3 levels of gates)
A
B
A
C
B
D
f3
level 1gates
level 0I/Ps
level 2gates
level 3gate
level 2I/Ps
3
•Defn. A gate in a logic circuit is a level-1 gate if all its inputs areprimary inputs (i.e., literals, A, A, B, B, etc.)
A gate is a level-i gate i > 1 if the highest-level gate whose output feeds the (level-i) gate 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 goal of 2-level minimization is to minimize the number ofliterals (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.
5
E.g. f = ABC + ABC + ABC (non-minimized 9 literals)ABCABCABC
f
(12 gate I/Ps)= Circuit complexity
f = AB + AC (minimized 4 literals)A
B
A
C
f(6 gate I/Ps)= Circuit complexity
6
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)
Multilevel Minimization (Example):
f = x1x2x3 + x2x3x4 + x1x5 + x4x5 (2-level minimized)
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• 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: prop. to # of inputs
Gate Impl.:
x2
x3
x5
x1
x4
f
Multilevelcircuit
Critical path goes through 3 2-i/p gates (6 units of delay) and 2 interconns. (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.
Apply distributive law (factoring)f = x2x3(x1 + x4) + x5(x1 + x4) = (x2x3 + x5)(x1 + x4) -- Multilevel expr.
Cost = 14 (# of gate i/ps in a 2-level impl.)
Defn. Two implicants (or product terms) of a function f are said tobe adjacent (logically) if they have the same literalsexcept in one variable xi which occurs in uncomplemented form(xi) in one implicant and in complemented form (xi) in theother. The 2 implicants are said to differ in xi.
E.g. ABC, ABC (adjacent implicants, differ in B)
ABD, ABC (not adjacent)
NOTE: Adjacent implicants can be combined into one implicantby the combining theorem (ABC + ABC = AC)
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Karnaugh Maps
2-variable TT outputs
A B Z1 Z2
0 0 0 00 1 1 11 0 0 11 1 1 0
Binaryplace-valueordering(Binaryordering)
Physically adjacentbut not logicallyadjacent.
logically adjacent but not physically adjacent AB + AB
BAnother ordering of inputs ( Gray-code ordering)
A B Z1
0 0 00 1 11 1 11 0 0
Z1 = BPhysically as well as logically adjacent.
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n-variable Gray-code ordering
G1 = 0 Base. 1G2 = 0 G1
1 rev(G1)=
0 00 11 11 0
G3 = 0 G2
1 rev(G2)=
A B C 0 0 0 0 0 10 1 1 0 1 01 1 01 1 1 1 0 11 0 0
Gn = 0 Gn-1
1 rev(Gn-1)
Convert to a 2-dimensionalGray-code ordered TT
ABC 00 01 11 10
0
1
x
y •
•
001
110
Throw away variable (A) changing across the 2 squares. We thus obtain BC.
1 1
Defn. Gi = i-bit Gray-code orderingrev(Gi) = reverse order of Gi
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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.
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.
11Example ( 3-var. K-Map)
Function f:AB
C 00 01 11 10
0
1
1 1 1
1 1f = AB + AB + AC + BC
Consensus = BC
A’B’ AB
AC’B’C’
Note: 1) that the consensus theorem is an example of generalized covering discussed earlier.2) AC’ is also a consensus, this time of AB and B’C’. Can’t discard both simultaneously! Need to apply gen. covering (incl. consensus) one at a time.
Gray code ordering
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.
Larger than 2-minterm PIs can also exist:E.g., AB
C 00 01 11 10
0
1
1 1
1 1
AB ABBB
ABC 00 01 11 10
0
1 1 1 1 1
C
• 2-minterm implicants (AB, AB in the 1st example above) can be merged if they are logically adjacent to form a 4-mintermimplicant and so on.
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• 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 theyare 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.
More Examples:AB
C
AB
C
00 01 11 10
0
1
0
11 1
1 1 1
1 1
1 1 1
Symmetric ( logically adjacent)
Not symmetric
2-minterm implicants : Can be “merged” to forma 4-minterm implicant, which will be a PI.
Redrawn
Both PIs(A, BC)areessential.
Thus, f = A + BC is minimized.
00 01 11 10
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4-variable K-Map:
ABCD 00 01 11 10
00
01
11
10
•
•
14
Gray code ordering
Gray code ordering
Example:AB
CD
ABCD
00 01 11 10
00
01
11
10
00
01
11
10
00 01 11 10
1 1 1 1
1 1
1 1 1 1
1 1
1 1
1 1 1
1
AC
f = AC + D Instead of starting with 2-mintermimplicants & growing them, youcan form larger implicants directlyby grouping power of 2 (2, 4, 8,etc.) # of minterms so as to form a rectangle or square ( a convex region).
Convex
Concave
Validgroupings
Invalid grouping(region formed is concave)
f = AB + AC + BCD
D
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Another Example:
A B C D Z0 0 0 0 10 0 0 1 00 0 1 0 10 0 1 1 10 1 0 0 00 1 0 1 10 1 1 0 10 1 1 1 11 0 0 0 11 0 0 1 01 0 1 0 11 0 1 1 11 1 0 0 01 1 0 1 01 1 1 0 11 1 1 1 1
TT:
Binaryorder
ABCD 00 01 11 10
00
01
11
10
1 1
1
1 1 1 1
1 1 1 1
ABD
BD
C
f = C + BD + ABD
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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. 212-214 inKatz text )
• For such combinations, we do not care what the output is, andwe put an ‘x’ in the o/p column for the TT
A B C f0 0 0 1 0 0 1 00 1 0 10 1 1 x1 0 0 01 0 1 11 1 0 x1 1 1 x
ABCD 00 01 11 10
0
1
x x 1 0
0 1 1 0
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.
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by selectively considering the 010cell’s x as 1
• In a K-map, the ‘x’s can be profitably used to form larger implicants (which have fewer literals)
• 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.
Example: f(A,B,C,D) = m(1,3,5,7,9) + d(2,6,12,13)
A B C D0 0 0 0 0 0 0 10 0 1 00 0 1 1
• • • f = CD + AD
or CD + AC If input comb. 2,6 occuro/p is 0.If inputs 2,6 occur o/p = 1.
ABCD 00 01 11 10
00
01
11
10
0 4 12 x 8
1 1 5 1 13 x 9 1
3 1 7 1 15 11
2 x 6 x 14 10
LSBs
MSBs
CD
AC
AD
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• Thus as far as x’s are concerned, we can have the cake and eat it too!
Which to use for lower dynamic power (consumed on a 0 1 transition at the o/p)?
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 of PIs S to cover them. This requires some trial-and error (in fact, this is a hard computational problem).
STEP4: Minimal SOP Expr. for function f:
PI Pi is in set S
f = essential PIs + PiPi
S
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Minimal POS Expr. from a K-Map :
METHOD 1: Obtain minimal SOP expression for the complementfunction f, and complement this SOP expression to get a minimal POS expr. for f (using De-Morgan’s Law)
Example:
F (A,B,C,D) = M(0,2,4,8,10,11,14,15) D(6,12,13)
•
Big M notation
0s of FXs of F
ABCD AB
CD 00 01 11 10
00
01
11
10
00
01
11
10
0 4 12 8
1 5 13 9
3 1 7 1 15 0 11 0
2 0 6 x 14 0 10 0
0 0 x 0
1 1 x 1
Complement 1 1 x 1
0 0 x 00 0 1 1
1 x 1 1AC
D
F =D + AC F = D + AC = D(A +C)
F F
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Example: (Method2: Direct Method)
ABCD 00 01 11 10
00
01
11
10
0 0 x 0
1 1 x 1
1 1 0 0
0 x 0 0
(A + C)
D
(OR terms)
Prime Implicate( Groups of 0s that cannotbe grown any further)
Both D and A + C areEssential Prime Implicates.
F = D( A + C )
( Product of Prime Implicates )
21
METHOD2 : Direct Method :
(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 notcovered by any other PT).
(3) Choose a minimal set T of PTs to cover the 0s not covered bythe set of essential PTs.
(4) The expression for a PT is an OR term obtained by discardingall changing variables, & keeping variables complemented if theyare 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) • qiqi T
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i
5-variable K-map: f(A,B,C,D,E): Juxtapose two 4-var. K-submapsone for the MSB A=0 and the other for A=1:
BCDE
BCDE00 01 11 10
00
01
11
10
00 01 11 10
00
01
11
10
BCDE
BCDE00 01 11 10
00
01
11
10
00 01 11 10
00
01
11
10
A=0 A=1
A=0 A=1
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
16 20 28 24
17 21 29 25
19 23 31 27
18 22 30 26
0 1 4 12 1 8
1 5 13 1 9
3 1 7 1 15 x 11
2 6 14 x 10
16 20 28 x 24
17 21 29 1 25
19 x 23 1 31 x 27
18 22 1 30 1 26
4-var. K-submap 4-var. K-submapAdjacencies of a square: Adjacent squares of the 4-var. K-submapplus the “corresponding” or “mirror” square in the other 4-var.K-submap.
Example:
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EF EF00 01 11 10
00
01
11
10
00 01 11 10
00
01
11
10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
16 20 28 24
17 21 29 25
19 23 31 27
18 22 30 26
6-variable K-map: f(A,B,C,D,E,F)--Juxtapose two 5-var. K-submapsone for the MSB A=0 and other for A=1
B=0 B=1
A=05-var.K-submap
EF EF00 01 11 10
00
01
11
10
00 01 11 10
00
01
11
10
32 36 44 40
33 37 45 41
35 39 47 43
34 38 46 42
48 52 60 56
49 53 61 57
51 55 63 59
50 54 62 58
CD CD
CD CD
A=1
5-var.K-submap
Adjacencies of a square: Adjacent squares of the 5-var.K-submap plus the “corresponding” or “mirror” squarein the other 5-var. K-submap.
B=0 B=1
24
EF EF00 01 11 10
00
01
11
10
00 01 11 10
00
01
11
10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
16 20 28 24
17 21 29 25
19 23 31 27
18 22 30 26
B=0 B=1
A=0
EF EF00 01 11 10
00
01
11
10
00 01 11 10
00
01
11
10
32 36 44 40
33 37 45 41
35 39 47 43
34 38 46 42
48 52 60 56
49 53 61 57
51 55 63 59
50 54 62 58
CD CD
A=1
B=1
6-variable K-map: f(A,B,C,D,E,F,)---Example:
CD CD
B=0
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