FYBSC IT Digital Electronics Unit II Chapter II Minterm, Maxterm and Karnaugh Maps

Minterm, Maxterm and Karnaugh Maps:
Introduction, minterms and sum of minterm form, maxterm and Product
of maxterm form, Reduction technique using Karnaugh maps –
2/3/4/5/6 variable K-maps, Grouping of variables in K-maps, K-maps
for product of sum form, minimize Boolean expression using K-map
and obtain K-map from Boolean expression, Quine Mc Cluskey
Method.

1. BOOLEAN ALGEBRA
AND LOGIC GATES
&
MINTERM, MAXTERM
AND KARNAUGH
MAPS
UNIT II
DIGITAL
ELECTRONICS
PROF. ARTI GAVAS-PARAB
ANNA LEELA COLLEGE OF COMMERCE AND ECONOMICS, SHOBHA
JAYARAM SHETTY COLLEGE FOR BMS
CHAPTER II

2. UNIT II: CONTENTS
Boolean Algebra and Logic Gates:
Introduction, Logic (AND OR NOT),
Boolean theorems, Boolean Laws, De Morgan’s Theorem,
Perfect Induction, Reduction of Logic expression using Boolean Algebra,
Deriving Boolean expression from given circuit, exclusive OR and Exclusive NOR gates,
Universal Logic gates, Implementation of other gates using universal gates, Input bubbled logic, Assertion level.
Minterm, Maxtermand KarnaughMaps:
Introduction, mintermsand sum of mintermform, maxtermand Product of maxtermform,
Reduction technique using Karnaughmaps –2/3/4/5/6 variable K-maps, Grouping of variables in K-maps, K-maps for product
of sum form,
MminimizeBoolean expression using K-map and obtain K-map from Boolean expression,
Quine Mc CluskeyMethod.
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3. MINTERM, MAXTERMAND KARNAUGHMAPS: INTRODUCTION
MinTerm:-Product term
which contains each of the n
variables in either
complemented or
uncomplimentedform.
MaxTerm:-Sum term which
contains each of the n
variables in either
complemented or
uncomplimentedform.

4. BOOLEAN FUNCTION
A Boolean function is an algebraic form of Boolean expression. A Boolean expression is an expression which
consists of variables, constants (0-false and 1-true) and logical operators which results in true or false.
A Boolean function of n-variables is represented by f(x1, x2, x3….xn).
By using Boolean laws and theorems, we can simplify the Boolean functions of digital circuits.
A brief note of different ways of representing a Boolean function is shown below.
Sum-of-Products (SOP) Form
Product-of-sums (POS) form
Canonical forms
There are two types of canonical forms:
Sum-of-min terms or Canonical SOP
Product-of-max terms or Canonical POS

5. SUM OF MINTERM(SUM OF PRODUCTS (SOP))
The sum-of-products (SOP) form is a method (or form) of simplifying the Boolean expressions of logic gates.
In this SOP form of Boolean function representation, the variables are operated by AND (product) to form a
product term and all these product terms are ORed(summed or added) together to get the final function.
Examples
AB + ABC + CDE
(AB)̅+ ABC + CD E̅
SOP form can be obtained by
Writing an AND term for each input combination, which produces HIGH output.
Writing the input variables if the value is 1, and write the complement of the variable if its value is 0.
OR the AND terms to obtain the output function.

6. SUM OF PRODUCTS (SOP): EXAMPLE
Boolean expression for majority function F = A’BC + AB’C + ABC ‘ + ABC
Truth table:
For a 3-variable (x, y and z) Boolean function, the possible
mintermsare:
x’y’z’, x’y’z, x’yz’, x’yz, xy’z’, xy’z, xyz’ and xyz.
•1 –Minterms= mintermsfor which the function F = 1.
•0 –Minterms= mintermsfor which the function F = 0.

7. PRODUCT OF MAXTERM(PRODUCT OF SUM (POS))
In this POS form, all the variables are ORed, i.e. written as sums to form sum terms.
All these sum terms are ANDed(multiplied) together to get the product-of-sum form. This form is exactly
opposite to the SOP form. So this can also be said as “Dual of SOP form”.
Examples
(A+B) * (A + B + C) * (C +D)
(A+B)̅* (C + D + E̅)
POS form can be obtained by
Writing an OR term for each input combination, which produces LOW output.
Writing the input variables if the value is 0, and write the complement of the variable if its value is 1.
AND the OR terms to obtain the output function.

8. PRODUCT OF SUM (POS): EXAMPLE
Boolean expression for majority function F = (A + B + C) (A + B + C ‘) (A + B’ + C) (A’ + B + C)
Truth table:
For a 3-variable (x, y and z) Boolean function, the possible
maxtermsare:
x + y + z, x + y + z’, x + y’ + z, x + y’ + z’, x’ + y + z, x’ + y + z’,
x’ + y’ + z and x’ + y’ + z’.
•1 –Max terms = max terms for which the function F = 1.
•0 –max terms = max terms for which the function F = 0.

9. CANONICAL FORM (STANDARD SOP FORM)
Any Boolean function that is expressed as a sum of mintermsor as a product of max terms is said to
be in its “canonical form”.
Any Boolean function can be expressed as the sum (OR) of its 1-min terms. The representation of the
equation will be
F(list of variables) = Σ(list of 1-min term indices)
Ex: F (x, y, z) = Σ(3, 5, 6, 7)
The inverse of the function can be expressed as a sum (OR) of its 0-min terms. The representation of
the equation will be
F(list of variables) = Σ(list of 0-min term indices)
Ex: F’ (x, y, z) = Σ(0,1, 2, 4)

10. CANONICAL FORM (STANDARD POS FORM)
In max term, each variable is complimented, if its value is assigned to 1, and each variable is un-
complimented if its value is assigned to 0.
Any Boolean function can be expressed the product (AND) of its 0 –max terms. The representation
of the equation will be
F(list of variables) = Π(list of 0-max term indices)
Ex: F (x, y, z) = Π(0, 1, 2, 4)
The inverse of the function can be expressed as a product (AND) of its 1 –max terms. The
representation of the equation will be
F(list of variables) = Π(list of 1-max term indices)
Ex: F’ (x, y, z) = Π(3, 5, 6, 7)

11. REDUCTION TECHNIQUE USING KARNAUGHMAPS (K-MAPS) RULES
Karnaughmapping is a graphic technique for reducing a Sum-of-Products (SOP) and Product-of-Sums (POS)
expression to its minimum form.
Two, three and four variable k-maps will have 4, 8 and 16 cells respectively.
Each cell of the k-map corresponds to a particular combination of the input variable and between adjacent cells
only one variable is allowed to change (As in Gray Code Sequence).
Use the following steps to reduce an expression using a k-map.
Mark each term of the SOP/POS expression in the correct cell of the k-map. (kind of like the game Battleship)
Circle adjacent cells in groups of 2, 4 or 8 making the circles as large as possible. (NO DIAGONALS!)
Write a term for each circle in a final SOP/POS expression. The variables in a term are the ones that remain constant
across a circle.
The cells of a k-map are continuous left-to-right and top-to-bottom. The wraparound feature can be used to draw
the circles as large as possible.
When a variable does not appear in the original equation, the equation must be plotted so that all combinations
of the missing variable(s) are covered.

12. K-MAPS FOR 2 TO 5 VARIABLES
Karnaughintroduced a method for simplification of Boolean functions in an easy way.
This method is known as Karnaughmap method or K-map method. It is a graphical method, which consists of 2
n
cells for ‘n’ variables. The adjacent cells are differed only in single bit position.
K-Map method is most suitable for minimizing Boolean functions of 2 variables to 5 variables. Now, let us discuss
about the K-Maps for 2 to 5 variables one by one.
would be00, 01, 10, 11. It is00, 01, 11 10, which is Gray code sequence. Gray code sequence only changes one
binary bit as we go from one number to the next in the sequence, unlike binary.

13. 2 VARIABLE K-MAP
The number of cells in 2 variable K-map is four, since the number of variables is two. The following figure shows 2
variable K-Map.
There is only one possibility of grouping 4 adjacent min terms.
The possible combinations of grouping 2 adjacent min terms are {(m
0, m
1), (m
2, m
3), (m
0, m
2) and (m
1, m
3)}.

14. 3 VARIABLE K-MAP
The number of cells in 3 variable K-map is eight, since the number of variables is three. The following figure shows
3 variable K-Map.
There is only one possibility of grouping 8 adjacent min terms.
The possible combinations of grouping 4 adjacent min terms are {(m
0, m
1, m
3, m
2), (m
4, m
5, m
7, m
6), (m
0, m
1, m
4,
m
5), (m
1, m
3, m
5, m
7), (m
3, m
2, m
7, m
6) and (m
2, m
0, m
6, m
4)}.
The possible combinations of grouping 2 adjacent min terms are {(m
0, m
1), (m
1, m
3), (m
3, m
2), (m
2, m
0), (m
4, m
5),
(m
5, m
7), (m
7, m
6), (m
6, m
4), (m
0, m
4), (m
1, m
5), (m
3, m
7) and (m
2, m
6)}.
If x=0, then 3 variable K-map becomes 2 variable K-map.

15. 4 VARIABLE K-MAP
The number of cells in 4 variable K-map is sixteen, since the number of variables is four. The following figure
shows 4 variable K-Map.
There is only one possibility of grouping 16 adjacent min terms.
Let R
1, R
2, R
3and R
4represents the min terms of first row, second row, third row and fourth row respectively.
Similarly, C
1, C
2, C
3and C
4represents the min terms of first column, second column, third column and fourth
column respectively. The possible combinations of grouping 8 adjacent min terms are {(R
1, R
2), (R
2, R
3), (R
3, R
4),
(R
4, R
1), (C
1, C
2), (C
2, C
3), (C
3, C
4), (C
4, C
1)}.
If w=0, then 4 variable K-map becomes 3 variable K-map.

16. 5 VARIABLE K-MAP
The number of cells in 5 variable K-map is thirty-two, since the number of variables is 5. The following figure
shows 5 variable K-Map.
There is only one possibility of grouping 32 adjacent min terms.
There are two possibilities of grouping 16 adjacent min terms.
i.e., grouping of min terms from m
0to m
15and m
16to m
31.
If v=0, then 5 variable K-map becomes 4 variable K-map.

17. MINIMIZATION OF BOOLEAN FUNCTIONS USING K -MAPS
Follow these rules for simplifying K-mapsin order to get standard sum of products form.
Select the respective K-map based on the number of variables present in the Boolean function.
If the Boolean function is given as sum of min terms form, then place the ones at respective min term cells in the
K-map. If the Boolean function is given as sum of products form, then place the ones in all possible cells of K-map
for which the given product terms are valid.
Check for the possibilities of grouping maximum number of adjacent ones. It should be powers of two. Start from
highest power of two and uptoleast power of two. Highest power is equal to the number of variables considered
in K-map and least power is zero.
Each grouping will give either a literal or one product term. It is known as prime implicant. The prime implicant
is said to be essential prime implicant, if atleastsingle ‘1’ is not covered with any other groupings but only that
grouping covers.
Note down all the prime implicantsand essential prime implicants. The simplified Boolean function contains all
essential prime implicantsand only the required prime implicants.

18. MINIMIZATION OF BOOLEAN FUNCTIONS USING K -MAPS
Note 1−If outputs are not defined for some combination of inputs, then those
output values will be represented with don’t care symbol ‘x’. That means, we can
consider them as either ‘0’ or ‘1’.
Note 2−If don’t care terms also present, then place don’t cares ‘x’ in the respective
cells of K-map. Consider only the don’t cares ‘x’ that are helpful for grouping
maximum number of adjacent ones. In those cases, treat the don’t care value as ‘1’.

19. GROUPING OF K-MAP VARIABLES
The square that contains ‘1’ should
be taken in simplifying, at least once.
The square that contains ‘1’ can be
considered as many times as the
grouping is possible with it.
Group shouldn’t include any zeros
(0).
A group should be the as large as
possible.
Groups can be horizontal or vertical.
Grouping of variables in diagonal
manner is not allowed.

20. 2 VARIABLE K-MAP: EXAMPLE
A general representation
of a 2 variable K-map
plot is shown below.
Example
Simplify the given 2-variable
Boolean equation by using
K-map.
F = X Y’ + X’ Y + X’Y’
Here the lower right cell is used in both
groups.
3) After grouping the variables, the next
step is determining the minimized
expression.
By reducing each group, we obtain a
conjunction of the minimized expression
such as by taking out the common terms
from two groups, i.e. X’ and Y’.
4) So the reduced equation will be X’ +Y’.
1) We put 1 at the output terms given
in equation.
In this K-map, we can create 2 groups
by following the rules for grouping,
one is by combining (X’, Y) and (X’, Y’)
terms and the other is by combining
(X, Y’) and (X’, Y’) terms.

21. 2 VARIABLE K-MAP: EXAMPLE
Example 1:
Consider the following map. The
function plotted is:
Z = f(A,B) = A + AB
Z=A

22. 3 VARIABLE K-MAP: EXAMPLE
A typical plot of a 3-variable K-map is
shown below. It can be observed that the
positions of columns 10 and 11 are
interchanged so that there is only change
in one variable across adjacent cells. This
modification will allow in minimizing the
logic.
We put 1 at the output terms
given in equation.
Example
Simplify the given 3-variable Boolean equation by
using k-map.
F = X’ Y Z + X’ Y’ Z + X Y Z’ + X’ Y’ Z’ + X Y Z + X Y’ Z’
And in both the terms, we have ‘Y’ in common. So the group of size
4 is reduced as the conjunction Y. To consume every cell which has
1 in it, we group the rest of cells to form size 2 group, as shown
below.
The 2 size group has no common variables, so they are written
with their variables and its conjugates.
So the reduced equation will be X Z’ + Y’ + X’ Z. In this equation,
no further minimization is possible.

23. 3 VARIABLE K-MAP: EXAMPLE
By using the rules of simplification and ringing
of adjacent cells in order to make as many
variables redundant, the minimisedresult
obtained is
By using the rules of simplification and ringing
of adjacent cells in order to make as many
variables redundant, the minimisedresult
obtained is

24. K-MAP OF 3 VARIABLES (SOP)
Z= ∑A,B,C(1,3,6,7)
Fromredgroup we get product term—
A’C
Fromgreengroup we get product term—
AB
Summing these product termswe get-
Final expression (A’C+AB)

25. K-MAP OF 3 VARIABLES (POS)
Fromredgroup
we findterms
ABC’
Taking complement
of these two
A’B’C
Nowsumup them
(A’ + B’ + C)
Frombrowngroup
we find terms
A’ B’ C’
Taking complement of
these two
A B C
Nowsumup them
(A + B + C)
We will take product of these three terms
:Final expression (A’ + B’ + C) (B’ + C’)
(A + B + C)
Fromgreengroup
we findterms
B C
Taking complement
of these two terms
B’ C’
Now sum up them
(B’+C’)
Example:
F(A,B,C)=π(0,3,6,7)

26. This Boolean expression has seven product terms. They
are mapped top to bottom and left to right on the K-
map above.
For example, the first P-termA’B’CDis thefirst
row,3rd cell, corresponding to map locationA=0, B=0,
C=1, D=1.
The other product terms are placed in a similar manner.
Encircling the largest groups possible, two groups of
four are shown above.
The dashed horizontal group corresponds to the
simplified product termAB. The vertical group
corresponds to Boolean CD. Since there are two
groups, there will be two product terms in the Sum-Of-
Products result ofOut=AB+CD.
Example:
f(A,B,C,D)=
4VARIABLE K-MAP: EXAMPLE

27. 4 VARIABLE K-MAP: EXAMPLE
Fold up the corners of the map here like it is a napkin to make the
four cells physically adjacent.
The four cells here are a group of four because they all have the
Boolean variablesB’andD’in common. In other words,B=0for
the four cells, andD=0for the four cells.
The other variables(A, C)are0in some cases,1in other cases
with respect to the four corner cells.
Thus, these variables(A, C)are not involved with this group of
four. This single group comes out of the map as one product term
for the simplified result:Out=B’D’
Example:
f(A,B,C,D)=

28. 4 VARIABLE K-MAP: EXAMPLE
For the K-map here, roll the top and bottom edges
into a cylinder forming eight adjacent cells.
This group of eight has one Boolean variable in
common:B=0.
Therefore, the one group of eight is covered by one
p-term:B’.
The original eight-term Boolean expression
simplifies toOut=B’
Example:
f(A,B,C,D)=

29. MISSING-TERMS IN 4 VARIABLE K MAPS
The Boolean expression here has nine p-terms, three
of which have three Booleans instead of four. The
difference is that while four Boolean variable product
terms cover one cell, the three Boolean p-terms
cover a pair of cells each.
The six product terms of four Boolean variables map
in the usual manner above as single cells. The three
Boolean variable terms (three each) map as cell pairs,
which is shown above.
Note that we are mapping p-terms into the K-map,
not pulling them out at this point.
For the simplification, we form two groups of eight.
Cells in the corners are shared with both groups. This
is fine. In fact, this leads to a better solution than
forming a group of eight and a group of four without
sharing any cells. Final Solution isOut=B’+D’
Example:
f(A,B,C,D)=

30. 4 VARIABLE K-MAP: EXAMPLE
Here we map the un-simplified Boolean
expression to the Karnaughmap.
Three of the cells form into groups of two
cells.
A fourth cell cannot be combined with
anything, which often happens in “real world”
problems. In this case, the Boolean p-
termABCDis unchanged in the simplification
process.
Result:
Out= B’C’D’+A’B’D’+ABCD
Example:
f(A,B,C,D)=

31. 4 VARIABLE K-MAP: EXAMPLE
Often times there is more than one
minimum cost solution to a simplification
problem. Such is the case illustrated below.
Both results above have four product terms
of three Boolean variable each. Both are
equally validminimal costsolutions.
The difference in the final solution is due to
how the cells are grouped as shown here.
A minimal cost solution is a valid logic
design with the minimum number of gates
with the minimum number of inputs.
Example:
f(A,B,C,D)=

32. 4 VARIABLE K-MAP: EXAMPLE
Below we map the un-simplified
Boolean equation as usual and form a
group of four as a first simplification
step. It may not be obvious how to
pick up the remaining cells.
Pick up three more cells in a group of
four, center above. There are still two
cells remaining. the minimal cost
method to pick up those is to group
them with neighboring cells as groups
of four as at above right.
On a cautionary note, do not attempt
to form groups of three.
Groupings must be powers of 2, that
is, 1, 2, 4, 8 ...
Example:
f(A,B,C,D)=

33. K-MAP FOR 4 VARIABLES (SOP)
F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15)
Fromredgroup we get product term—
QS
Fromgreengroup we get product
term—
Q’S’
Summingthese product termswe get-
Final expression (QS+Q’S’)

34. K-MAP FOR 4 VARIABLES (POS)
F(A,B,C,D)=π(3,5,7,8,10,11,12,13)
Fromgreengroup we findterms C’DB
Taking their complement and summing them (C+D’+B’)
Fromredgroup we find terms CDA’
Taking their complement and summing them (C’+D’+A)
Frombluegroup we findterms AC’D’
Taking their complement and summing them (A’+C+D)
Frombrowngroup we findterms AB’C
Taking their complement and summing them (A’+B+C’)
Finally we express these as product –
(C+D’+B’).(C’+D’+A).(A’+C+D).(A’+B+C’)

35. K-MAP FOR 4 VARIABLES (DON’T CARES)
Don’t cares in a Karnaughmap, or truth table, may be either1s or0s, as long as we don’t care what the output is
for an input condition we never expect to see. We plot these cells with an asterisk, *, among the normal1s
and0s.
When forming groups of cells, treat the don’t care cell as either a1or a0, or ignore the don’t cares.
This is helpful if it allows us to form a larger group than would otherwise be possible without the don’t cares.
There is no requirement to group all or any of the don’t cares.
Only use them in a group if it simplifies the logic.

36. QUINE-MCCLUSKEYTABULAR METHOD
Quine-McClukeytabular method is a tabular method based on the concept of prime implicants.
The prime implicantis a product or sum term, which can’t be further reduced by combining with any other
product or sum terms of the given Boolean function.
Quine-McClukeytabular method is a tabular method based on the concept of prime implicants. We know that
prime implicantis a product or sum term, which can’t be further reduced by combining with any other product
or sum terms of the given Boolean function.

37. QUINE-MCCLUSKEYTABULAR METHOD: STEPS FOR SIMPLIFYING BOOLEAN FUNCTIONS
Step 1−Arrange the given min terms in anascending orderand make the groups based on the number of ones
present in their binary representations. So, there will beat most ‘n+1’ groupsif there are ‘n’ Boolean variables in a
Boolean function or ‘n’ bits in the binary equivalent of min terms.
Step 2−Compare the min terms present insuccessive groups. If there is a change in only one-bit position, then take
the pair of those two min terms. Place this symbol ‘_’ in the differed bit position and keep the remaining bits as it is.
Step 3−Repeat step2 with newly formed terms till we get allprime implicants.
Step 4−Formulate theprime implicanttable. It consists of set of rows and columns. Prime implicantscan be placed
in row wise and min terms can be placed in column wise. Place ‘1’ in the cells corresponding to the min terms that are
covered in each prime implicant.
Step 5−Find the essential prime implicantsby observing each column. If the min term is covered only by one prime
implicant, then it isessential prime implicant. Those essential prime implicantswill be part of the simplified Boolean
function.
Step 6−Reduce the prime implicanttable by removing the row of each essential prime implicantand the columns
corresponding to the min terms that are covered in that essential prime implicant. Repeat step 5 for Reduced prime
implicanttable. Stop this process when all min terms of given Boolean function are over.

38. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 1
simplify the following Boolean function, f(W,X,Y,Z)=∑m(2,6,8,9,10,11,14,15) using Quine-McClukeytabular method.
Group Name Min termsW X Y Z
GA1
2 0 0 1 0
8 1 0 0 0
GA2
6 0 1 1 0
9 1 0 0 1
10 1 0 1 0
GA3
11 1 0 1 1
14 1 1 1 0
GA4 15 1 1 1 1
The given Boolean function is
insum of min termsform.
It is having 4 variables W, X, Y & Z.
The given min terms are 2, 6, 8,
9, 10, 11, 14 and 15.
The ascending order of these
min terms based on the
number of ones present in
their binary equivalent is 2, 8, 6,
9, 10, 11, 14 and 15.
The following table shows
thesemin terms and their
equivalent
binaryrepresentations.
Min
Term
Binary
2 0010
6 0110
8 1000
9 1001
10 1010
11 1011
14 1110
15 1111

39. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 2
The given min terms are arranged into 4 groups based on the number of ones present in their binary equivalents.
The following table shows the possiblemerging of min termsfrom adjacent groups.
Group Name Min terms W X Y Z
GB1
2,6 0 - 1 0
2,10 - 0 1 0
8,9 1 0 0 -
8,10 1 0 - 0
GB2
6,14 - 1 1 0
9,11 1 0 - 1
10,11 1 0 1 -
10,14 1 - 1 0
GB3
11,15 1 - 1 1
14,15 1 1 1 -

40. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 3
The min terms, which are differed in only one-bit position from adjacent groups are merged. That differed bit is
represented with this symbol, ‘-‘. In this case, there are three groups and each group contains combinations of two
min terms. The following table shows the possiblemerging of min term pairsfrom adjacent groups.
Group Name Min terms W X Y Z
GB1
2,6,10,14 - - 1 0
2,10,6,14 - - 1 0
8,9,10,11 1 0 - -
8,10,9,11 1 0 - -
GB2
10,11,14,151 - 1 -
10,14,11,151 - 1 -

41. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 4
The successive groups of min term pairs, which are differed in only one-bit position are merged. That differed bit
is represented with this symbol, ‘-‘. In this case, there are two groups and each group contains combinations of
four min terms. Here, these combinations of 4 min terms are available in two rows. So, we can remove the
repeated rows. The reduced table after removing the redundant rows is shown below.
Group Name Min terms W X Y Z
GC1 2,6,10,14 - - 1 0
8,9,10,11 1 0 - -
GC2 10,11,14,151 - 1 -

42. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 5
Further merging of the combinations of min terms from adjacent groups is not possible, since they are differed in
more than one-bit position. There are three rows in the above table. So, each row will give one prime implicant.
Therefore, theprime implicantsare YZ’, WX’ & WY.
Theprime implicanttableis shown below.
Min terms / Prime
Implicants
2 6 8 9 10 11 14 15
YZ’ 1 1 1 1
WX’ 1 1 1 1
WY 1 1 1 1

43. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 6
The prime implicantsare placed in row wise and min terms are placed in column wise. 1s are placed in the
common cells of prime implicantrows and the corresponding min term columns.
The min terms 2 and 6 are covered only by one prime implicantYZ’. So, it is anessential prime implicant.
This will be part of simplified Boolean function. Now, remove this prime implicantrow and the corresponding min
term columns. The reduced prime implicanttable is shown below.
Min terms / Prime
Implicants
8 9 11 15
WX’ 1 1 1
WY 1 1

44. QUINE-MCCLUSKEYTABULAR METHOD: EXAMPLE: STEP 7
The min terms 8 and 9 are covered only by one prime implicantWX’. So, it is anessential prime implicant.
This will be part of simplified Boolean function. Now, remove this prime implicantrow and the corresponding min
term columns. The reduced prime implicanttable is shown below.
Min terms / Prime Implicants15
WY 1
The min term 15 is covered only by one prime implicantWY. So, it is anessential prime implicant. This will be
part of simplified Boolean function.
In this example problem, we got three prime implicantsand all the three are essential.
Therefore, thesimplified Boolean functionis
F(W,X,Y,Z)= YZ’ + WX’ + WY.

45. THANK YOU!