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I know how to simplify a circuit by using the K-map. However, I am still curious whether my result is the simplest. I have heard that when using K-map, try using the biggest and the fewest loops. But, how can I make sure that I have used the one which will be the most straightforward answer?

Besides, when dealing with K-maps, are POS and SOP the algebraically the same? So, I can choose one as I want when simplifying the expression.

Additionally, is there any useful site to learn some digital logic stuff like Karnough map?

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  • \$\begingroup\$ Your question is not an easy one. There's a short answer here at stackexchange. The whole satisfiability question has taken up many papers and minds. Some papers to consider: 1, 2, 3, and 4 for segues into some of the terminology. Matrices, graph theory, and the rank-nullity theorem also play a part. \$\endgroup\$ Sep 12, 2023 at 14:51

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as I noted in https://electronics.stackexchange.com/a/559593/254042, K-maps are based on only one property:

$$AB+A\mathop{\overline{B}}=A$$

This property is extended, for example, to

$$A\mathop{\overline{B}}C+ABC=AC$$

or more complicated cases like:

$$AB\mathop{\overline{C}}\mathop{\overline{D}}+A\mathop{\overline{B}}\mathop{\overline{C}}\mathop{\overline{D}}+AB\mathop{\overline{C}}D+A\mathop{\overline{B}}\mathop{\overline{C}}D=A\mathop{\overline{C}}$$

Consider a Boolean expression represented as a sum-of-products. If we represent each product in a cell, the key is to map the products in a way that adjacent cells have only one variable that is complemented. By this way some rules are defined (see below). You can safely (without errors) apply these rules no matter how complicated the expression is, and you will always get the right answer.

The rules are simple:

  1. select largest groups covering adjacent ones (in a SOP but adjacent zeros in a POS) that must be rectangular of a size which is a power of 2 (2, 4, 8, 16, ...).
  2. the minimum number of groups covering all ones (in a SOP or zeros in a POS) must be selected.

You can apply K-maps on both POS and SOP depending on the number of ones or zeros in the truth table or the number of minterms and maxterms in the expression. In POS you work with zeros and with SOP you work with ones.

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  • \$\begingroup\$ I think that the K-map uses two properties and the one you have mentioned may be just in SOP. It seems that in POS, we use (x+y)(x+z) = x + yz. Besides, I do understand that l must use the largest loop and fewest loop when using K-map. However, l mean that when doing the simplification work, l can't make sure that l have reached the simplest expression. So, is there any method to check that my l have used the fewest loops in K-map? Lastly, does that make sense if a K-map's 1 is more than 0, then l should use SOP. Otherwise, use POS? Thanks \$\endgroup\$
    – mendax1234
    Sep 12, 2023 at 15:22
  • \$\begingroup\$ Only one property, in POS it is the logical negation of the given property and if you apply De Morgan you get (/A+/B)(/A+B)=/A. If you follow the rules you will get the simplest expression. However if your goal is to analyze and validate simplifying expressions then the problem is formulated beyond the k-map operations. concerning the choice of POS or SOP and the number of 1s or 0s filling up less cells will accelerate the operation. \$\endgroup\$ Sep 13, 2023 at 12:39

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