I am working on the Kmap for the Segment C on the 7 Segment Display.

The list of numbers that light up the 7 segment displays segment C are as follows:

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The Kmap I came up with, without grouping yet, is as follows:

enter image description here

Then, when I start grouping, I came up with multiple variations. Some utilize the don't care conditions, and some don't. I'd like to know which one(s) are correct. It seems to me that they are all correct because I am following the grouping rules. If there isn't a single correct answer, then how do I know if the one I have is an acceptable answer?

Here is the list I came up with:

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Finally, looking on the web, I encountered a site where they also have their version of the "solution". However, in one of their grouping, they have grouped 12 1s. One of the rules for grouping was that you can have only 2^n number of elements in a group. So, with that said, I don't know how their solution is correct. Here is the link that I am referring to: other segment c kmap solution

Can somebody please clarify my confusions? Thanks!

  • \$\begingroup\$ Just a reference to consider further: 7-seg kmaps. \$\endgroup\$
    – jonk
    Commented Oct 18, 2018 at 1:25
  • 1
    \$\begingroup\$ In the link, they made a mistake (or perhaps a confusing simplification) when drawing it. If you look at the formula below it, there are two terms that both correspond to eight of those twelve: the ~C and D. \$\endgroup\$
    – Hearth
    Commented Oct 18, 2018 at 2:16

1 Answer 1


All of your solutions are correct. You have covered all the 1 cases in valid terms, and not covered any 0 cases.

However, of them, the last one is closest to optimal. It uses the fewest terms, and its terms are the simplest. It's possible to optimize it slightly by "expanding" one of the terms.

Specifically, the red term can be expanded to the left, changing it from b⋅c to simply b. This makes the formula b + c̅ + d.

  • \$\begingroup\$ In addition to using the fewest terms, the most optimal solution will generally make each of the groupings as large as possible (which you imply in reference to "simplest", but did not explicitly state why). \$\endgroup\$
    – mith
    Commented Oct 18, 2018 at 12:44

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