Tabular Method

If a prime implicant covers all the minterms of another prime implicant , then we say row dominates row (row dominance) and prime implicant is redundant

If the prime implicants of minterm cover all prime implicants another minterm , then we say column dominates column (column dominance) and minterm is redundant (opposite of row dominance)

1. List all minterms in a table and group them based on the number of 1s in the binary representation of their minterm number
   • Don't cares can be treated as a minterm
2. Combine implicants of one group with an implicant in their successive group that differ by only 1 bit location
   • Create a list, replacing the differing bit with a don't care.
3. Repeat until further combinations cannot be made. Any un-combined minterms are prime implicants.
4. List prime implicants and the minterms they cover, identify any essential prime implicants using inspection and either column or row dominance

Consider the function . Find the minimum-cost realization for this function.

solution
We have:

• 0: 0000
• 4: 0100
• 8: 1000
• 10: 1010
• 11: 1011
• 12: 1100
• 13: 1101
• 15: 1111
  (see Number Systems#Binary)

Now we group the minterms by the number of s in their binary representations:

Now we combine adjacent minterms. For example, we can combine 0000 with 0100 to get 0x00. In essence, this is what a K-map is doing. We can show this like so:

Finally, we combine once more:

Any implicant that was not checked off is a prime implicant. We will label these from 1 to 6:

Now we list all the prime implicants and the minterms they cover (exclude any don't cares from list 1):

Now we remove because it is the only prime implicant which covers and . As a result, any columns covers will also be removed, and is an EPI:

Note that while covers , covers and . This is called row dominance. Since and have the same number of literals, we keep and remove . The same goes for and . We arrive at a new table:

Now, by inspection, we need to to chose and for a full cover, while is redundant. The final cover is .

Therefore, the minimum-cost implementation is:

Consider the following table without any EPIs:

Find all necessary prime implicants.

solution
Note that column and column are covered by the same prime implicants. Since has more checkmarks than , dominates . Recall when a column dominates another, we remove the dominating column.

Same goes for and .

Note that is dominated by and is dominated by .

Observe that and each cover at least 1 minterm exclusively. So these implicants must be part of the cover.

Finally, dominates both and , so it is part of the cover.

Therefore, our final cover is .