Introduction

The tabular method which is also known as the Quine-McCluskey method is particularly useful when minimising functions having a large number of variables, e.g. The six-variable functions. Computer programs have been developed employing this algorithm. The method reduces a function in standard sum of products form to a set of prime implicants from which as many variables are eliminated as possible. These prime implicants are then examined to see if some are redundant.

The tabular method makes repeated use of the law A + = 1. Note that Binary notation is used for the function, although decimal notation is also used for the functions. As usual a variable in true form is denoted by 1, in inverted form by 0, and the abscence of a variable by a dash ( - ).

Consider the function:

Listing the two minterms shows they can be combined

Now consider the following:

Note that these variables cannot be combined

This is because the

Bear in mind that when two terms are combined, one of the combined terms has one digit more at logic 1 than the other combined term. This indicates that the number of 1's in a term is significant and is referred to as its index.

For example: f(A, B, C, D)

0000...................Index 0

0010, 1000.............Index 1

1010,
0011, 1001.......Index 2

1110, 1011.............Index 3

1111...................Index 4

The necessary condition for combining two terms is that the indices of the two terms must differ by one logic variable which must also be the same.

Consider the function: Z = f(A,B,C) = + C + A + AC

To make things easier, change the function into binary notation with index value and decimal value.

Tabulate the index groups in a colunm and insert the decimal value alongside.

From the first list, we combine terms that differ by 1 digit only from one index group to the next. These terms from the first list are then seperated into groups in the second list. Note that the ticks are just there to show that one term has been combined with another term. From the second list we can see that the expression is now reduced to: Z = + + C + A

From the second list note that the term having an index of 0 can be combined
with the terms of index 1. Bear in mind that the dash indicates a missing
variable and **must** line up in order to get a third list. The final
simplified expression is: Z =

Bear in mind that any unticked terms in any list must be included in the final expression (none occured here except from the last list). Note that the only prime implicant here is Z = .

The tabular method reduces the function to a set of prime implicants.

Note that the above solution can be derived algebracially. Attempt this in your notes.

**Example 2:**

Consider the function f(A, B, C, D) = (0,1,2,3,5,7,8,10,12,13,15), note that this is in decimal form.

(0000,0001,0010,0011,0101,0111,1000,1010,1100,1101,1111) in binary form.

(0,1,1,2,2,3,1,2,2,3,4) in the index form.

The prime
implicants are: +
+ D + BD + A + AB

The chart is used to remove redundant prime implicants. A grid is prepared
having all the prime implicants listed at the left and all the minterms of the
function along the top. Each minterm covered by a given prime implicant is
marked in the appropriate position.

From the above chart, BD is an essential prime implicant. It
is the only prime implicant that covers the minterm
decimal 15 and it also includes 5, 7 and 13. is also an essential
prime implicant. It is the only prime implicant that covers the minterm denoted
by decimal 10 and it also includes the terms 0, 2 and 8. The other minterms of
the function are 1, 3 and 12. Minterm 1 is present in and D. Similarly for
minterm 3. We can therefore use either of these prime implicants for these
minterms. Minterm 12 is present in A and AB, so again either can
be used.

Thus, one minimal solution is: Z = + BD + + A

- Minimise the function below using the tabular method of simplification:

Z = f(A,B,C,D) = + C + ACD + AC + BCD + BC + CD - Using the tabular method of simplification, find all equally minimal
solutions for the function below.

Z = f(A,B,C,D) = (1,4,5,10,12,14)

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