1. Karnaugh Maps

2. 2018/9/13
Karnaugh Maps
Karnaugh maps (K-maps) are graphical representations of boolean functions.
One map cell corresponds to a row in the truth table.
Also, one map cell corresponds to a minterm or a maxterm in the boolean expression
Multiple-cell areas of the map correspond to standard terms.
Boolean Algebra

3. Definitions
Minterm: A product term in which all the variables appear exactly once, either complemented or uncomplemented.
Ex: ABC, A’B’C, ABC’ etc.
Maxterm: A sum term in which all the variables appear exactly once, either complemented or uncomplemented.
Ex : A+B+C, A’+B’+C, A+B+C’ etc.

4. Truth Table notation for Minterms and Maxterms
2018/9/13
Truth Table notation for Minterms and Maxterms
Minterms and Maxterms are easy to denote using a truth table.
Example: Assume 3 variables x,y,z (order is fixed) x y z
Minterm
Maxterm
x’y’z’ = m0
x+y+z = M0 1
x’y’z = m1
x+y+z’ = M1
x’yz’ = m2
x+y’+z = M2
x’yz = m3
x+y’+z’= M3
xy’z’ = m4
x’+y+z = M4
xy’z = m5
x’+y+z’ = M5
xyz’ = m6
x’+y’+z = M6
xyz = m7
x’+y’+z’ = M7
Boolean Algebra

5. 2018/9/13
Shorthand: ∑ and ∏
f1(a,b,c) = ∑ m(1,2,4,6), where ∑ indicates that this is a sum-of- products form, and m(1,2,4,6) indicates that the minterms to be included are m1, m2, m4, and m6.
f1(a,b,c) = ∏ M(0,3,5,7), where ∏ indicates that this is a product-of- sums form, and M(0,3,5,7) indicates that the maxterms to be included are M0, M3, M5, and M7.
Since mj = Mj’ for any j, ∑ m(1,2,4,6) = ∏ M(0,3,5,7) = f1(a,b,c)
Boolean Algebra

6. Minimization as SOP using K-map
2018/9/13
Minimization as SOP using K-map
Minimization of POS using K-map
Enter 1s in the K-map for each product term in the function
Group adjacent K-map cells containing 1s to obtain a product with fewer variables. Group size must be in power of 2 (2, 4, 8, …)
Enter 0’s in the K-map for each sum term in the function
Group adjacent K-map cells containing 0’s to obtain a sum with fewer variables. Group size must be in power of 2 (2, 4, 8, …)
Handle “boundary wrap” for K-maps of 3 or more variables.
Boolean Algebra

7. SOP 2 variable K-maps There are 4 cells (2↑2) in the 2-variable k-map.
The following table shows the positions of all the possible outputs of 2-variable Boolean function on a K-map.

8. 3 variable K-maps For a 3-variable Boolean function, there is a possibility of 8 output min terms. The general representation of all the min terms using 3-variables is shown below.

9. A typical plot of a 3-variable K-map is shown below.

10. 4 variable K-maps There are 16 possible min terms in case of a 4-variable Boolean function. The general representation of minterms using 4 variables is shown below.

11. Four-variable Map Simplification
2018/9/13
Four-variable Map Simplification
One square represents a minterm of 4 literals.
A rectangle of 2 adjacent squares(pair) represents a product term of 3 literals as a pair removes one variable which change its state.
A rectangle of 4 squares(quad) represents a product term of 2 literals as a quad removes two variables which change its state.
A rectangle of 8 squares(Octet) represents a product term of 1 literal as it removes 3 variables that are changing their states.
Boolean Algebra

12. POSSIBILITIES WHILE GROUPING
OVERLAPPING GROUPS

13. Example Simplify the given 2-variable Boolean equation by using K-map.
2 variable k-map
Simplify the given 2-variable Boolean equation by using K-map.
F = X Y’ + X’ Y + X’Y’
First, let’s construct the truth table for the given equation,
                                                      
We put 1 at the output terms given in equation.
Reduced equation will be X’ +Y’.

14. 3 variable k-map Given function, F = Σ (1, 2, 3, 4, 5, 6)
First we need to look for an octet i.e. 8 adjacent 1′s. There is none, so we should now look for a quad i.e. 4 adjacent 1′s. Again, there is none, so we will look for pairs.
There are 3 pairs circled in red.
(1,3) – A’C (Since B is the changing variable between these two cells, it is eliminated)
(2,6) – BC’ (Since A is the changing variable, it is eliminated)
(4, 5) – AB’ (Since C is the changing variable, it is eliminated)
Thus, F = A’C + BC’ + AB’

15. 4 varible map

16. POS
4 Variable map

17. Expected type of Questions
Obtain a simplified form for a Boolean expression:
F(U,V,W,Z)=∏(0,1,3,5,6,7,10,14,15)
2. Reduce the following Boolean expression using K-Map:
F(A,B,C,D)=∑(0,1,2,4,5,6,8,10)
3. Reduce the following Boolean expression using K-Map:
F(P,Q,R,S)= ∑(1,2,3,4,5,6,7,8,10)
4. Obtain a simplified form for a Boolean expression using K-Map:
F(U,V,W,Z)=∏(5,6,7,8,9,12,13,14,15)