# Quine mccluskey method

In this article we will cover Quine Mccluskey method with examples. I will show you the simplification of Boolean expression by Quine Mccluskey method. This process is lengthy Read this full article in desktop view, I am sure you will get a clear concept.

Quine Mccluskey method is also known as tabular method or the method of prime implicants.

Generally, we use K Map ( Karnaugh Map ) to simplify a Boolean function. K Map can easily simplify Boolean functions with less variable. The Quine Mccluskey method can simplify greater number of variables than K Map. Quine Mccluskey method provides a symmetrical approach that are easily programmed into a computer for digital simplification.

For accurate use of Quine Mccluskey method the expression needs to be in SOP form.

Quine Mccluskey method is easy but make sure you have a clear concept about –

1. Min term ( SOP ) and max term ( POS )
2. Decimal to binary conversion.
3. BCD to expression conversion.

Question

### Y = Σ m = ( 2, 4, 5, 9, 12, 13 )

Convert these decimals to Binary Coded Decimal

 Decimal Binary A B C D 2 0 0 1 0 4 0 1 0 0 5 0 1 0 1 9 1 0 0 1 12 1 1 0 0 13 1 1 0 1

Now make a group of decimals that contains logic 1 only once. As shown in table below –

 Min Term A B C D m2 0 0 1 0 m4 0 1 0 0

Again make a group of decimals that contains logic 1 twice. As shown in table below –

 Min Term A B C D m5 0 1 0 1 m9 1 0 0 1 m12 1 1 0 0

Again make a group of decimals that contains logic 1 thrice. As shown in table below –

 Min Term A B C D m13 1 1 0 1

This process is done until all the decimal numbers given in question are covered. So now we have covered all the decimals given in question.

The combined table of above three groups is shown below. You directly need to write the combined table. I separated it to explain.

 Group MinTerm A B C D G 1 m2 0 0 1 0 m4 0 1 0 0 G 2 m5 0 1 0 1 m9 1 0 0 1 m12 1 1 0 0 G 3 m13 1 1 0 1

#### Now we compare group 1 and 2 then 2 and 3.

First to compare group 1 with 2. Compare m2 with m5, m9, m12. A new table will be made and comparison will only be noted when there is difference in only one bit. The comparison of

0, 0 will give 0

1, 1 will give 1

0,1 will give X

1, 0 will give X

X, X will give X

During comparison similar digits input will give similar output and different digits input will give X.

When m2 and m5 is compared, more than 1 bit B, C & D are different so output will not be noted.

Again when m2 and m9 is compared, more than 1 bit A, C & D are different so output will not be noted.

Now when m2 and m12 is compared, again more than 1 bit A, B & C are different so the output will not be noted.

Now we compare m4 with m5, m9, m12.

When m4 is compared with m5. Only one bit D is  different. So the output will be noted as shown in table below.

 MinTerm A B C D m4 – m5 0 1 0 X

Again we will compare m4 with m9. More than one bit A,B & D are different. so the output will not be noted.

Lastly, we will compare m4 with m12. Only one bit A is different. So the output will be noted as shown in table below.

 MinTerm A B C D m4 – m12 X 1 0 0

Comparison between group 1 and 2 is now complete. Now we will compare group 2 and 3. That means m5, m9, m12 will be compared with m13.

When m5 and m13 is compared. Only bit A is different so comparison will be noted as shown in figure below.

 MinTerm A B C D m5 – m13 X 1 0 1

When m9 and m13 is compared. Only bit B is different so comparison will be noted as shown in figure below.

 MinTerm A B C D m9 – m13 1 X 0 1

When m12 and m13 is compared. Only bit D is different so comparison will be noted as shown in figure below.

 MinTerm A B C D m12 – m13 1 1 0 X

The combined table of above 5 noted comparisons is shown below. You directly need to write the combined table. I separated it to explain.

 Group MinTerm A B C D 4 m4 – m5 0 1 0 X m4 – m12 X 1 0 0 5 m5 – m13 X 1 0 1 m9 – m13 1 X 0 1 m12 – m13 1 1 0 X

We will again compare group 4 and 5 as we have done above. For other questions these process of making and comparing groups goes on until no group is formed.

First we will compare ( m4 – m5 ) with ( m5 – m13 ), ( m9 – m13 ) & ( m12 – m13 ).

There is more than 1 bit difference when ( m4 – m5 ) is compared with ( m5 – m13 ), ( m9 – m13 ). So comparison will not be noted.

There is only one bit difference when ( m4 – m5 ) is compared with ( m12 – m13 ). So comparison will be noted as shown in table below.

 MinTerm A B C D m4 – m5 – m12 – m13 X 1 0 X

Again we will compare ( m4 – m12 ) with ( m5 – m13 ), ( m9 – m13 ) & ( m12 – m13 ).

There is more than 1 bit difference when ( m4 – m12 ) is compared with ( m9 – m13 ), ( m12 – m13 ). So comparison will not be noted.

There is only one bit difference in ( m4 – m12 ) & ( m5 – m13 ). So again comparison will be noted as shown in table below.

 MinTerm A B C D m4 – m12 – m5 – m13 X 1 0 X

The combined table of above two noted comparison is shown in table below. With their expression ( E ) in SOP form.

 MinTerm A B C D E m4 – m5 -m12 – m13 X 1 0 X BC’ m4 – m12 –m5 – m13 X 1 0 X BC’

We get BC’ two times in the output.

The output reduced expression can be written as –

Y = BC’

## Prime and essential prime implicant

The prime implicant and essential prime implicant is shown in the table below.

The last table of comparison contains m4, m5, m12 & m13. So cross for essential prime implicant is made under 4, 5, 12 & 13.

 Group Min Term 2 4 5 9 12 13 BC’ X X X X PrimeImplicant Essential PrimeImplicant

The advantage of Quine Mccluskey method over K Map is – this method can solve a large number of inputs where as K Map gets too long or near impossible.

Author

Akash Sharma

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