Intro to Logic Circuit Design (University of Utah) - Karnaugh Map (Discussion 3)
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Overview
A University of Utah logic-design discussion applies Karnaugh maps to three minimization problems: POS from maxterms, SOP from five-variable minterms, and SOP after converting four-variable maxterms to minterms. The solutions identify essential prime implicants through uniquely covered cells, then prove minimum nonessential covers by testing group-count and group-size combinations, including a six-zero POS cover using two 4-cell groups and one 2-cell group.
Key takeaways
- A prime implicant is essential when it covers at least one target cell that no other prime implicant covers; selecting these groups first reduces the remaining cover problem.
- A good Karnaugh-map starting cell is covered by only one prime implicant, while a cell with multiple candidate groups can lead to a suboptimal early choice.
- Karnaugh-map edges wrap around, making corner cells adjacent and allowing groups to cross the map boundary.
- To justify a minimum cover, compare candidate group counts and sizes systematically, ruling out smaller combinations because they cannot cover all remaining cells.
- For POS minimization, group zeros; for SOP minimization, group ones. A function given in maxterms can be converted for SOP by listing the missing indices as minterms.
- The demonstrated nonessential covers are two 4-cell groups plus one 2-cell group for six remaining POS zeros, and one 4-cell plus one 2-cell group for four remaining SOP ones.
Chapters
- For the first POS problem, the function is given in maxterms, so the map is minimized by grouping its zeros.
- A useful starting cell is covered by exactly one prime implicant; a cell covered by multiple candidates is not a reliable starting point, even if one candidate is larger.
- The corner cells of a Karnaugh map are adjacent through wraparound, so groups crossing map edges must be considered.
- Two essential prime implicants are identified because each covers at least one zero that no other prime implicant covers.
- After the two essential groups are selected, six zero cells remain uncovered and must be covered by nonessential prime implicants.
- The demonstrated minimum cover uses two 4-cell groups and one 2-cell group.
- The minimum is justified by testing candidate covers in order: one group cannot cover six zeros, two 4-cell groups leave zeros uncovered, and smaller group combinations also fail.
- Other covers may use different groups, but a cover with the same group sizes has the same stated cost.
- The second problem requests SOP directly from minterms, so the listed minterms can be placed on the five-variable Karnaugh map without conversion.
- Cells with few adjacent 1s are used as starting points; three groups are identified as essential because each uniquely covers at least one 1.
- Four 1-cells remain uncovered after selecting the essential groups, and the minimum nonessential cover is one 4-cell group plus one 2-cell group.
- The cover is checked against alternatives: one group cannot cover all four cells, and two 4-cell groups cannot cover a cell that is only covered by a 2-cell group.
- The final problem asks for SOP although the function is given as maxterms, so the listed maxterm indices are complemented to obtain the minterm indices.
- For a four-variable function with indices from 0 through 15, the absent indices are minterms 3, 6, 11, and 13.
- Those four minterms are placed on a four-variable Karnaugh map, where isolated cells and available adjacencies determine the essential groups.
- The key method is to convert maxterms to the complementary minterm set before grouping 1s for SOP minimization.
Transcript excerpt
0:12 All right. Good afternoon. I hope you all are doing well. Today we have our discussion three and we are going to go over K-map questions today. So, this is the first question. We are asked to find a minimum cost in POS format of this function that is given. Let's give it 5 minutes and then we're going to start with it. All right. Let's solve with this together. We're going to solve it in two ways. One is from finding the good starting point cell and the other is through prime implicants implicants which are
1:00 prime that that direction. Okay. So, which cell is a good starting point? Here. Um So, let's see. So, is this a good starting point? This cell. So, this cell is covered by two groups, by two prime by two um implicants. By two prime implicants. Okay. The one of them is this and the other one is this one. Okay. So, there are two prime implicants that cover this. So, this won't be a good starting point even though that the size of this one is four and the size of
1:50 this one is two. Okay. So, we cannot make the argument now that which one is better to go with at this point. Okay? We might go with the 2 by 2 one. Sorry, by 1 by 2 one, the green one. And finally, we come up with a better design. That's why we don't want to start with this. Okay? So, this is a starting point. It's not matter of how large the other groups are. As long as there are two prime implicant that cover one cell, we don't want to start with them. Okay? So, keep that in mind. All right. So, is this a good starting point?
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