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See in the image i got a 1(min term) at the bottom right corner which I can't assign it to any groups . What should i do ? Should i leave this 1 as it is and do other simplification or i should count it as single group ?

Schl....r
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2 Answers2

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You create groups with the size of 2^N elements, this includes the size 1 .(1=2^0). So you create indeed a single group with only that single element at AB'CD'.

Progman
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  • I think my procedure of making groups are ok? I can make a group with 1 and 9 by binding rule , should i do that? Or the way i drew its ok? – Schl....r Nov 30 '20 at 20:47
  • @Schl....r You can create such a group for `1` and `9`, but this wouldn't give you any new information. You already have collected all the `1`s in your grid. But you could use the group `1`+`9` *instead* of `13`+`9`, which would also get you all `1`s collected. – Progman Nov 30 '20 at 20:54
  • @Schl....r Yes, you can collect all the `1`s with both combinations. – Progman Nov 30 '20 at 20:59
  • I will get two different answers for two combination , so if i build a machine with such result shouldn't i worry about my outlput as its showing two different results?? – Schl....r Nov 30 '20 at 21:02
  • @Schl....r You are not getting different results, you get the same results. You can use any combination of groups, as long as they collect all the `1`s and not collect any `0`s. You can even create 6 single groups for the 6 `1`s you have and it would still get you the same result of the function. – Progman Nov 30 '20 at 21:08
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    Yeah i know that but i got confused while working with "dont". Suppose : (1,4,10,15)+d(3,5,13,14) can you tell me the answer ? – Schl....r Nov 30 '20 at 21:11
  • @Schl....r Note sure what you are asking. `(1,4,10,15)` is not a valid group in your k-map, neither is `(3,5,13,14)`. – Progman Nov 30 '20 at 21:20
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It would be: AB'CD' as those are the constant variables.

The larger the group, the less variables (ABCD) you will need to represent it.

It is considered an essential prime implicant (meaning that it is required to be represented as it can only be grouped in a single way) - therefore it has it's own grouping of 1 (in this case). Groupings work for powers of 2, so a grouping of a single element is the same as 2^0. You can have groupings of: 1,2,4,8, etc. This will be more relevant once you work on 5-variable or 6-variable K-Maps

A P
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