A prime implicant is essential when it covers at least one 1 that no other prime implicant covers. That 1 is called a distinguished minterm.
Since that 1 has to be covered and only one group can do it, an essential prime implicant appears in every minimal sum of products. That makes them the natural first step:
- List all prime implicants: every group that can't grow without taking in a 0.
- For each 1, count how many prime implicants contain it.
- Any 1 with a count of exactly one marks its group as essential. Circle it.
- Cover the remaining 1s as cheaply as possible from the non-essential ones.
Don't-cares never make a group essential. A cell marked X doesn't need covering, so only real 1s count in step 2.
Some functions have no essential prime implicants at all. Every 1 sits in two or more groups, and you must choose. These are cyclic maps, and they can have more than one minimal answer.
| A\BC | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 1m0 | 1m1 | 1m3 | 0m2 |
| 1 | 0m4 | 0m5 | 1m7 | 0m6 |