Karnaugh Maps:
A Karnaugh map is a planar area subdivided into 2n equal cells each representing a point 
for functions of n variables. Each variable x is used to split the area into two equal halves in a different way, i.e., one for x and other for x’. The cells corresponding to the arguments for which the function has the value 1 contains 1.
Example1: When the number of variables n = 1, the karnaugh map is like as shown in fig:
2. When the number of variables n =2, the karnaugh map is like as shown in fig:
3. When the number of variables n =3, the karnaugh map is like as shown in fig:
Simplification of Boolean Functions using K-Map:
Boolean Functions can be simplified with k-map. It is based on the principle of combining terms in adjacent cells. Two cells are said to be adjacent if they differ in only one variable. In adjacent cells, one of the variables is the same, whereas the other variable appears in the un-complemented form in one and in the complemented form in the other cell.
Minimization of SOP Form:
The following algorithm can be used by which minimized expression can be obtained:
- Identify the ones which cannot be combined with any other ones encircle them.
- Identify the ones that can be combined in groups of two in only one way and encircle them as groups.
- Identify the ones that can be combined with three other ones, to make a group of four adjacent ones, in only way and encircle them as groups.
- Identify the ones that can be combined with seven other ones, to make a group of eight adjacent ones, in only way and encircle them as groups.
- After identifying the essential groups of 2, 4 and 8 ones, if there still remain some ones which have not been encircled then these are to be combined with each other or with other already encircled ones, i.e., we should connect the leftover ones in largest possible groups and in as few grouping as possible.
Example1: Minimize the following Boolean Expression using k-map:
       f(A, B) = A’ B+BA
Solution: First of all draw the 2-variables k-map and insert 1’s in the corresponding cells as shown in fig:
The required minimized Boolean Expression is f=B.
Example2: Minimize the following Boolean Expression using k-map:
       AB + A’ B+BA’
Solution: Draw the two-variable k-map and insert 1’s in the corresponding cells as shown in fig:
The required minimized Boolean Expression is f=A+B.
Example3: Minimize the following Boolean Expression using k-map:
       f(A, B, C) = AB’ C+A’ BC+AB+A’ B’ C
Solution: Draw the 3-variable k-map and insert 1’s in the corresponding cells as shown in fig:
The required minimized Boolean Expression is f=AB+C’
Minimisation of Boolean Functions not in Minterms/Max-terms:
One way to minimize such functions is to convert them into standard forms i.e., SOP or POS, then make the k-map and obtain the minimized function.
Another way is to prepare the k-map using the following algorithm directly
- Enter ones for min-terms and zeros for max-terms.
- Enter a pair of one’s/zeros for each of the terms with one variable less than the total number of variables.
- Enter four adjacent one’s/zeros for terms with two variables less than the total number of variables.
- Repeat for other terms in a similar way.
Example: Minimize the four variable logic function
       f (A, B, C, D) = A B C’D + A’ BCD+A’ B’ C’+A’ B’ D’+AC’+AB’ C+B’
Solution: The k-map is obtained by following way
(a)Enter 1 in the cell with A=1, B=1, C=0, D=1 corresponding to the min-term A B C’D
(b) Enter 1 in the cell with A=0, B=1, C=1, D=1 corresponding to the min-term A’ BCD
(c) Enter 1’s in the two cells with A=0, B=0, C=0 corresponding to the term A’ B’ C’
(d) Enter 1’s in the two cells with A=0, B=0, D=0 corresponding to the term A’ B’ D’
(e) Enter 1’s in the two cells with A=1, B=0, C=1 corresponding to the term AB’ C
(f) Enter 1’s in the four cells with A=1,C=0 corresponding to the term AC’
(g) Enter 1’s in the eight cells with B=0 corresponding to the term B’
The minimized expression is B’+ AC’+A’ CD.