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Suppose I face a situation where I can neither form octect nor quad nor pair with an element and it remains single. What should I do with it ? Should I ignore it?

HostileFork says dont trust SE
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Neer
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  • If the element is a "don't care" value, it will always join a different group (unless it is the only element, in which case the whole expression resolves to a "don't care"). If the value is concrete, then it can live on it's own as a group of size 1, and it equals the expression that makes it up. For example, x/y : { FF / FT }. The T term is clearly on it's own, and the original expression is (x and not(y)) and is perfectly fine. – SecurityMatt Apr 11 '13 at 06:20
  • And what should I do with it? I have got the question to simplify the expression using k-map. And what do you mean by dont care value? – Neer Apr 11 '13 at 06:22
  • Don't care term: https://en.wikipedia.org/wiki/Don't-care_term – SecurityMatt Apr 11 '13 at 06:30
  • You don't need to do anything with it. It is a term in its own right. The karnaugh map comprised of the terms xy=F, x!y=F, !xy=F, !x!y=T resolves to !x!y. – SecurityMatt Apr 11 '13 at 06:30

1 Answers1

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An isolated element like this (as long as it doesn't represent a "don't care" variable), is an essential prime implicant, and therefore must be included

Simon MᶜKenzie
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  • What is a don't care case? – Neer Apr 11 '13 at 06:24
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    It's an input condition you never expect to see, and therefore you don't mind if it's a `1` or a `0`. You can use these to aid grouping. See here for more details: http://www.allaboutcircuits.com/vol_4/chpt_8/10.html – Simon MᶜKenzie Apr 11 '13 at 06:27
  • So if its not not a don't care situation would just write the expression without simplifying it. Right? – Neer Apr 11 '13 at 06:29
  • Yes, if it's not a "don't care", the element will remain the same and cannot be simplified. – Simon MᶜKenzie Apr 11 '13 at 06:31