Quine-McCluskey minimization technique - QM using don’t cares, Reduced Prime Implicant table, Map entered variables.

1. Logic Design
PART A
Unit 2:
PRINCIPLES OF COMBINATIONAL LOGIC-2
Quine-McCluskey minimization technique - QM using
don’t cares, Reduced Prime Implicant table, Map
entered variables.

2. Quine-McCluskey method of Simplification
1. It is suitable for large (any) no. of variables.
2. It is suitable for computer solution.
1. It is a tedious and lengthy procedure.

3. 1) Simplify to Minimal form using Quine-McCluskey method.
 )7,6,5,3,1(),,( mcbafY
Minterms a b c
1 0 0 1
3 0 1 1
5 1 0 1
6 1 1 0
7 1 1 1
Next rearrange the Minterms
to form groups such that
each group contains the
same no. of 1’s.
Match minterms of group ‘n’ with those of group n+1
such that they differ in only 1 bit. Put a tick mark across
each matching minterms & ‘—’ in the bit position that
has changed & form a new table.
‘—’ indicates an eliminated variable.

4. Group Minterms a b c
0 1 0 0 1
3 0 1 1
1 5 1 0 1
6 1 1 0
2 7 1 1 1
Group Minterms a b c
0 1,3 0 - 1
1 3,7 - 1 1
0 1,5 - 0 1
1 5,7 1 - 1
1 6,7 1 1 -
Continue as before for
the next new table.
Here 2 minterms that
are grouped should
have a ‘—’ in the same
position.
Group Minterms a b c
0 1,3,5,7 - - 1
The unticked minterms
are 6,7 (ab) & 1,3,5,7
(c). Hence, Y=ab+c

5. 



)15,14,9,8,7,6,3,2,1,0(),,,()3
:)10,8,2,0(),,,()2
mdcbafD
zxPAnsmzyxwfP

6. Minterms a b c d
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
6 0 1 1 0
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1
14 1 1 1 0
15 1 1 1 1
Group Minterms a b c d
0 ‘1’s 0 0 0 0 0
1 0 0 0 1
1 ‘1’s 2 0 0 1 0
8 1 0 0 0
3 0 0 1 1
2 ‘1’s 6 0 1 1 0
9 1 0 0 1
3 ‘1’s 7 0 1 1 1
14 1 1 1 0
4 ‘1’s 15 1 1 1 1

7. Group Min a b c d
0 ‘1’s 0 0 0 0 0
1 0 0 0 1
1 ‘1’s 2 0 0 1 0
8 1 0 0 0
3 0 0 1 1
2 ‘1’s 6 0 1 1 0
9 1 0 0 1
3 ‘1’s 7 0 1 1 1
14 1 1 1 0
4 ‘1’s 15 1 1 1 1
Group Min a b c d
0,1 0 0 0 -
0 0,2 0 0 - 0
0,8 - 0 0 0
1,3 0 0 - 1
1,9 - 0 0 1
1 2,3 0 0 1 -
2,6 0 - 1 0
8,9 1 0 0 -
3,7 0 - 1 1
2 6,7 0 1 1 -
6,14 - 1 1 0
3 7,15 - 1 1 1
14,15 1 1 1 -

8. Group Min a b c d
0 0,1,2,3 0 0 - -
0,1,8,9 - 0 0 -
1 2,3,6,7 0 - 1 -
2 6,7,14,15 - 1 1 -
tablePIthe
formnext,,,,:getuOnce bccacbbaPIs
ba
cb
ca
bc

9. .cabcborbabcbD
c.aorbaeitherconsidercanwecover themto
EPIs.twoby thecoverednotare3and2cells'1'
covered.are6,7,14,15cells'1'thebc,with
&coveredare0,1,8,9cells'1'the,cbwithNow
EPIs.thearebcandbi.e
EPIs.theare)(6,7,14,15and(0,1,8,9)PIs
s.EPI'therepresentmintermscircledThe
it.circle&columnainX''oneonlyforCheck



cc
c

10. PRIME IMPLICANT TABLE:
PIs minterms 0 1 2 3 6 7 8 9 14 15
0,1,2,3 X X X X
0,1,8,9 X X X X
2,3,6,7 X X X X
6,7,14,15 X X X X
ba
cb
ca
bc
cabcborbabcbD  cc

11. table.PIformexpressionMinimalgetThen
.yx,zxz,yz,xPIs:Soln
)14,13,12,11,9,6,5,4,3,1(),,,()4

 mzyxwfR
PRIME IMPLICANT TABLE:
PIs minterms 1 3 4 5 6 9 11 12 13 14
1,3,9,11 X X X X
1,5,9,13 X X X X
4,6,12,14 X X X X
4,12,5,13 X X X X
zx
zy
zx
yx

12. .yxzxxorzyzxxR
.yor xzyeitherconsidercanwecover themto
EPIs.twoby thecoverednotare13and5cells'1'
covered.are4,6,12,14cells'1'the,zwith x
&coveredare1,3,9,11cells'1'the,xwithNow
EPIs.thearezxandxi.e
EPIs.theare)(4,6,12,14and(1,3,9,11)PIs
s.EPI'therepresentmintermscircledThe
it.circle&columnainX''oneonlyforCheck



zz
z
z

13. Quine-McCluskey method of Simplification
Using Don’t Care terms.
  )11,10,9,8()15,13,3,1(),,,()1 dcmdcbafS
Min/dc a b c d
1 0 0 0 1
3 0 0 1 1
13 1 1 0 1
15 1 1 1 1
8* 1 0 0 0
9* 1 0 0 1
10* 1 0 1 0
11* 1 0 1 1
* Represents don’t cares
Group Min/dc a b c d
1 ‘1’s 1 0 0 0 1
8* 1 0 0 0
3 0 0 1 1
2 ‘1’s 9* 1 0 0 1
10* 1 0 1 0
13 1 1 0 1
3 ‘1’s 11* 1 0 1 1
4 ‘1’s 15 1 1 1 1

14. Grp Min/dc a b c d
1,3,9*,11* - 0 - 1
0 8*,9*,10*,
11*
1 0 - -
1 9*,11*,13,
15
1 - - 1
Group Min/dc a b c d
1,3 0 0 - 1
0 1,9* - 0 0 1
8*,9* 1 0 0 -
8*,10* 1 0 - 0
3,11* - 0 1 1
1 9*,11* 1 0 - 1
9*,13 1 - 0 1
10*,11* 1 0 1 -
2 13,15 1 1 - 1
11*,15 1 - 1 1
Exclude because
all 4 are don’t
cares
form.minimalgettotablePItheformNext.,: addbPIs

15. PRIME IMPLICANT TABLE:
PIs minterms 1 3 13 15
1,3,9*,11* X X
9*,11*,13,15 X X
db
ad
Exclude don’t
cares
addbSad.&dbaresEPI' 

16. dbabcfd.ba,bc,:EPIsd.ba,bc,cd,:PIs:Soln
4,15),9,10,12,1(1,3,6,7,8f)3
dacabORdbcabcab,:EPIs.cab,,da,db:PIs:Soln
dc(2,8,13)15),10,12,14,(0,1,4,5,9d)c,b,f(a,2)
.dadccabcba
.dadccabcba:EPIs.da,db,dc,cabc,ba:PIs:Soln
14),10,12,13,(0,2,3,4,8f1)
function,for theformminimalobtain the
henceandPIsofsetobtain themethod,McCluskey-QuineUsing








 

f
f
ASSIGNMENT
method.map-Kusingsolvingbyanswersthecan verifyU

17. Prime Implicant Table Reduction
This involves Column & Row reductions.
Consider this example of a PI table.
PIs Minterms m0 m1 m2 m5 m10 m11 m14 m15
A 0,2 X X
B 0,1 X X
C 1,5 X X
D 14 X
E 10,11,14,15 X X X X
1) COLUMN REDUCTION
Check for a) Equal columns
b) Dominating columns.

18. 1) COLUMN REDUCTION
Check for a) Equal columns (Keep one & delete the others)
b) Dominating columns. (Delete them)
One column dominates another column if in addition
to having ‘X’s in the same rows as that of other
column, it has ‘X’s in other rows also.
a) Columns m10, m11 & m15 are all equal. Therefore we
shall delete any two say, m11 & m15 .
PIs Minterms m0 m1 m2 m5 m10 m11 m14 m15 cost
A 0,2 X X 4
B 0,1 X X 4
C 1,5 X X 4
D 14 X 4
E 10,11,14,15 X X X X 3

19. PIs Minterms m0 m1 m2 m5 m10 m11 m14 m15 cost
A 0,2 X X 4
B 0,1 X X 4
C 1,5 X X 4
D 14 X 4
E 10,11,14,15 X X X X 3
b) Column m0 dominates m2. Therefore we shall delete m0.
Column m1 dominates m5. Therefore we shall delete m1.
Column m14 dominates m10. Therefore we shall delete m14.
Now we have the Reduced PI table.
PIs Minterms m2 m5 m10 cost
A 0,2 X 4
C 1,5 X 4
E 10,11,14,15 X 3
Since the columns
have only one ‘X’ in
them they are the
EPIs. Hence,
ECAf 

20. PIs Minterms m1 m5 m7 m9 m10 m11 cost
A 1,7,9 X X X 3
B 1,7 X X 4
C 5,10 X X 4
D 11 X 3
E 5,10 X X 3
2) ROW REDUCTION
Consider this example of a PI table.
Note this
Check for a) Equal Rows (Keep one & delete the others)
b) Dominated Rows. (Delete them)
One row dominates another row if in addition to
having ‘X’s in the same columns as that of the other
row, it has ‘X’s in other columns also.
a) Rows C & E are all equal. Therefore we shall
delete that row with higher cost. Here delete row C.

21. PIs Minterms m1 m5 m7 m9 m10 m11 cost
A 1,7,9 X X X 3
B 1,7 X X 4
C 5,10 X X 4
D 11 X 3
E 5,10 X X 3
Row A dominates B or B is dominated by A. Hence
delete dominated row if its cost is ≥ cost of dominating
row. Here we delete row B.
Now we have the Reduced PI table.
PIs m1 m5 m7 m9 m10 m11 cost
A X X X 3
D X 3
E X X 3
EDAf
EDAEPIs

,,:

22. Map entered variable method of
simplification (MEV TECHNIQUE)
This technique makes it possible to use smaller
K-maps to handle larger no. of variables.
In a MEV K-map, each cell can contain either a
minterm(1), maxterm(0), don’t care(X) or a single
variable/expression. Eg.
C
X D
A 00 01
00
01
B
C

23. 1) Simplify using a 2 variable K-map. f(a,b,c)=∑m(0,1,4,5,7)
Let us take ‘c’ as the Map entered variable.
Cell no. a b c f entry
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
0 ab=00
1 ab=01
2 ab=10
3 ab=11
(MEV) Compare f&c
f=1
Irrespective
of c
f=0
Irrespective
of c
f=1
Irrespective
of c
f=c
entry
1
0
1
c
Then compare ‘f’ & ‘c’ and fill the entry column.
Next fill the cell nos. of the MEV map which are
decimal equivalent of ab.
Next draw the MEV map and do the grouping.

24. 1 0
1 C
a 0 1
0
1
b
PROCEDURE FOR GROUPING MEVs
1. Group all ‘1’s using don’t cares if any to
your advantage.
2. Next, consider all ‘1’s as d.c s & group all
identical MEV terms using d.c s or ‘1’s.
3. Determine EPIs by reading the map in the
normal way & then AND the MEV variables
with the product term. acby 
b
)(ca

25. 2) Simplify using a 3 variable K-map. f(a,b,c,d)=∑m(2,3,4,5,13,15) + ∑dc(8,9,10,11)
(MEV)
Cell no. a b c d(MEV) f entry
0
abc=000
0 0 0 0 0
0
f=0
irrespective of d0 0 0 1 0
1
0 0 1 0 1
1
f=1
irrespective of d0 0 1 1 1
2
0 1 0 0 1
1
f=1
irrespective of d0 1 0 1 1
3
0 1 1 0 0
0
f=0
irrespective of d0 1 1 1 0
4
1 0 0 0 X
X
f=X
irrespective of d1 0 0 1 X
5
1 0 1 0 X
X
f=X
irrespective of d1 0 1 1 X
6
1 1 0 0 0
d f=d
1 1 0 1 1
7
abc=111
1 1 1 0 0
d f=d
1 1 1 1 1

26. 0 1 0 1
X X d d
a 00 01
0
1
bc
)(da
cb
11 10
cba adcbacby 

27. 3) Simplify using a 3 variable K-map. f(w,x,y,z)=∑m(2,4,5,10,11,14) + ∑dc(7,8,9,12,13,15)
Cell no. w x y z(MEV) f entry
0
wxy=000
0 0 0 0 0
0
f=0
irrespective of z0 0 0 1 0
1
wxy=001
0 0 1 0 1
f =
0 0 1 1 0
2
wxy=010
0 1 0 0 1
1
f=1
irrespective of z0 1 0 1 1
3
wxy=011
0 1 1 0 0
0
f=0
Taking X=00 1 1 1 X
4
wxy=100
1 0 0 0 X
X f=X
1 0 0 1 X
5
wxy=101
1 0 1 0 1
1
f=1
irrespective of z1 0 1 1 1
6
wxy=110
1 1 0 0 X
X f=X
1 1 0 1 X
7
wxy=111
1 1 1 0 1
1
f=1
Taking X=11 1 1 1 X
z z

28. 0 0 1
X 1 1 X
w 00 01
0
1
xy
)(zyx w
11 10
yx
zyxwyxy 
z

29. 4) Simplify using a 2 variable K-map. f(a,b,c)=∑m(0, 4,5,7)
Cell no. a b c f entry
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
0 ab=00
1 ab=01
2 ab=10
3 ab=11
(MEV) Compare f&c
f =
f=0
Irrespective
of c
f=1
Irrespective
of c
f=c
entry
0
1
c
c c
0
1 C
a 0 1
0
1
b
)(cb
ba
c
)(ca
Now here, is not an EPI
because the ‘1’ cell is covered by
PIs of both and c. (MEV & its
complement)
ba
c
accby 

30. 5) Simplify using a 3 variable K-map. f(a,b,c,d)=∑m(0,1,3,5,6,11,13) + ∑dc(4,7)
Cell no. a b c d(MEV) f entry
0
abc=000
0 0 0 0 1
1
f=1
irrespective of d0 0 0 1 1
1
abc=001
0 0 1 0 0
d f=d
0 0 1 1 1
2
abc=010
0 1 0 0 X
1
f=1
Taking X=10 1 0 1 1
3
abc=011
0 1 1 0 1
1
f=1
Taking X=10 1 1 1 X
4
abc=100
1 0 0 0 0
0
f=0
irrespective of d1 0 0 1 0
5
abc=101
1 0 1 0 0
d f=d
1 0 1 1 1
6
abc=110
1 1 0 0 0
d f=d
1 1 0 1 1
7
abc=111
1 1 1 0 0
0
f=0
irrespective of d1 1 1 1 0

31. 1 d 1 1
0 d 0 d
a 00 01
0
1
bc
)(dcb
11 10
ba
dcbcdbcabay 
ca
)(dcb

32. 6) Simplify using a 2 variable K-map. f(a,b,c,d)=∑m(2,3,4,5,13,15) + ∑dc(8,9,10,11)
Cell no. a b c(MEV) d(MEV) f entry
0
ab=00
0 0 0 0 0
c
0 0 0 1 0
0 0 1 0 1
0 0 1 1 1
1
ab=01
0 1 0 0 1
0 1 0 1 1
0 1 1 0 0
0 1 1 1 0
2
ab=10
1 0 0 0 X
X
1 0 0 1 X
1 0 1 0 X
1 0 1 1 X
3
ab=11
1 1 0 0 0
d
1 1 0 1 1
1 1 1 0 0
1 1 1 1 1
c
c
X d
a 0 1
0
1
b
c
)(cb
)(cba
)(da
adcbacbf 