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Karnaughmap (k-map) It is a simplification tool for managing Boolean algebraic

expressions.

Complex boolean function can be simplified by using

boolean algebra but it lacks specific rules and becomes

tedious sometimes. The map method provides a simple and

straightforward mehtod.

The Karnaugh Map will simplify logic faster and more easily

in most cases.

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Karnaugh map K-map method is a graphical technique for simplifying

Boolean function.

k-map is pictorial arrangement of truth table.Each combinationof variables in a truth table is called as Minterm.

A function of n variables will have 2 minterms. A Booleanfunction is equal to 1 for some minterms and to 0 for others.

The map is a diagram made of squares with each squarerepresenting a minterm.The rows and columns are ordered

according to the principles ofGray code.

Boolean simplification is actually faster than the Karnaughmap for a task involving two or fewer Boolean variables. It isstill quite usable at three variables, but a bit slower. At four

input variables, Boolean algebra becomes tedious.

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Karnaugh map

2 variable map 3 variable map 4 variable map

Note the sequence of numbers across the top of the map. It

is not in binary sequence which would be 00, 01, 10, 11. It

is 00, 01, 11, 10 which is Gray code sequence.

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What is Gray code sequence?

Gray code sequence only changes one bit as we go from one

number to the next in the sequence, unlike binary.

Adjacent cells vary by only one bit because a Gray code

sequence varies by only one bit.

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Karnaugh Maps - Rules of Simplification

The K-Map uses the following rules for simplification of

expressions by grouping together adjascent cells

containing ones.

1. Groups may not include any cell containing a zero

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2. Groups may be horizontal or vertical, but not diagonal.

3. Groups must contain 1, 2, 4, 8, or in general 2n cells.

That is if n = 1, a group will contain two 1's since 21 = 2.

If n = 2, a group will contain four 1's since 22 = 4.

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4. Each group should be as large as possible.

5. Each cell containing a one must be in at least one group.

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6. Groups may overlap.

7. Groups may wrap around the table. The leftmost cell in a

row may be grouped with the rightmost cell and the top

cell in a column may be grouped with the bottom cell.

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8. There should be as few groups as possible, as long as this

does not contradict any of the previous rules.

Summmary:

1. No zeros allowed.

2. No diagonals.

3. Only power of 2 number of cells in each group.

4. Groups should be as large as possible.

5. Every one must be in at least one group.

6. Overlapping allowed.

7. Wrap around allowed.

8. Fewest number of groups possible

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3 variable map

Mapping the four product terms (p-

terms) yields a single group of four,

which is A.

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Mapping the four p-terms above yields a group of four. Visualizethe group of four by rolling up the ends of the map to form a

cylinder, then the cells are adjacent.We normally mark the group of

four as above left. Out of the variables A, B, C, there is a common

variable: C'. C' is a 0 over all four cells. Final result is C'

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Larger 4-variable Karnaugh maps

The above Boolean expression has

seven product terms. The dashed

horizontal group corresponds the

the simplified product term AB. Thevertical 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.

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The four cells above are a group of four because B=0 for the four

cells, and D=0 for the four cells. Thus, these variables (A, B) are not

involved with this group of four. This single group comes out of themap as one product term for the simplified result: Out=B'D'.

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The above 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 to Out=B'

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Boolean function obtained from a k-map may not be unique.

Often times there is more than one minimum cost solution to a

simplification problem. Such is the case illustrated below.

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Both results above have four product terms of three Boolean

variable each. Both are equally valid minimal costsolutions. The

difference in the final solution is due to how the cells are grouped

as shown above. A minimal cost solution is a valid logic design

with the minimum number of gates with the minimum number of

inputs.

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EXMP:

Out=ABCD+ABCD+ABCD

+ABCD+ABCD+ABCD++ABCD+ABCD+ABCD

Ans.=AC+AD+BC+BD

EXMP:////////////////

Out=ABCD+ABCD+ABCD

+ABCD+ABCD+ABCD++ABCD+ABCD+ABCD

Ans.=CD+CD+ABC

Ans.=CD+CD+ABD

EXMP:////////////

Out=ABCD+ABCD+ABCD

+ABCD+ABCD+ABCD+

+ABCD+ABCD+ABCD

Ans.=C +ABD

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EXMP:

F(A,B,C,D)= (1,5,7,9,11,13,15)Ans.=CD+AD+BD = D(A+C+B)

F(A,B,C,D)= (1,3,7,8,10,12,13,15)

Ans.=ABD+BCD+ABC+ABD

F(A,B,C,D)= (1,2,9,10,11,14,15)////////////////

Ans.=BCD+AC+BCD F(A,B,C,D)= (1,5,6,7,11,12,13,15)////////////////

Ans.=ABC+ABC+ACD+ACD

F(A,B,C,D)= (0,1,2,5, 13,15)

Ans.=ABD+ABC+BCD+ABC F(A,B,C,D)= (4,5, 7,11,15)

Ans.=ABC+BCD+ACD

Ans.=ABC+ABD+ACD

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Product-of-sum simplification:

Generally Boolean expressions derived from the maps are

expressed in sum-of-product form. The product terms are ANDterms and sum denotes the ORing of these terms.

But we can obtain product-of-sum form from the map. If we

mark the empty squares with 0s and combine them into groups

of adjacent squares, we obtain the complement of the function,

F. Taking the complement of F we will get expression for F in

product-of-sum form.

F(A,B,C,D)= (0,1,2,5, 8,9,10)

Combining the squares with 1s gives sum-of-product form.F=BD+BC+ACD

A sum-of-products expression can be implemented with NAND

gates.

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If squares marked with 0s are combined, we obtain :

F=AB+CD+BD

(F)=( AB+CD+BD)F=(A+B)(C+D)(B+D)

A product-of- sum expression can be implemented with NOR

gates.

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Dont care conditions:

The 1s and 0s in the map represent the minterms and make the

function equals to 1 or 0.

There are occasions when it does not matter if the function

produces 0 or 1 for a given minterm, we say that we dont care

what the function o/p is to be for this minterm.

Minterms that may produce either 0 or 1 for the function are said

to be dont-care conditions and are marked with X in the map.These dont -care conditions can be used to provide further

simplification of the algebraic expression.

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Another expamle, consider the Boolean function together with

the dont-care minterms:

F(A,B,C)= (0,2,6)

d(A,B,C)= (1,3,5)

By including dont-care minterms the simplified expression is :

F=A+BC (requires 1 AND and 1 OR gate)

If dont-care minterms are not included the simplified expression

would be:

F=AC+BC (requires 2 AND and 1 OR gate)