I was trying to solve the attached question from the book Digital Design: With an Introduction to Verilog HDL, 5e however the values of f1 and f2 for the entry (1,1,0) are missing.


If we use sum of product or product of sum form the function we get would be valid for all the values however it would also give a result for (1,1,0).

Will that value be the missing value from the table?

I tried both POS and SOP and the result for 1,1,0 is 0 in both the cases so is it correct to say the original truth table must have had 0 for the following entry or do we not talk about it since it was not one of the readings which we had.

(Note: I have no idea about K Maps just yet so please refrain from using those to explain unless that's the only way.)

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    \$\begingroup\$ I really encourage learning how to use karnaugh maps, they make things very easy. \$\endgroup\$
    – Hearth
    Jan 24 at 15:29
  • \$\begingroup\$ Yeah i started watching their tutorials they'll be covered later in course, this exercise wasn't supposed to be done using K Maps so I wanted a way to interpret the results with Boolean Algebra @Hearth \$\endgroup\$ Jan 24 at 15:31

K-maps really are the easiest.

Outputs f1 and f2 are X for 110, so they can be 1 or 0 in answer. So in boolean algebra, 110 can be added or excluded as required.

$$f_1 = \bar a \bar b \bar c + \bar a b \bar c + \bar a b c + a \bar b \bar c + abc$$

You have \$ab \bar c\$ as a don't care. So looking at the inputs of your don't care you have a & b at 1 and c at 0. You look for other common terms in your function. There are three \$\bar c\$, so there may be an advantage to add it.

$$f_1 = \bar a \bar b \bar c + \bar a b \bar c + \bar a b c + a \bar b \bar c + abc + ab \bar c$$

Extract common factor \$\bar c\$.

$$f_1 = \bar c (\bar a \bar b + \bar a b + a \bar b + ab) + \bar a b c + abc$$ $$f_1 = \bar c (\bar a (\bar b + b) + a (\bar b + b)) + \bar a b c + abc$$ $$f_1 = \bar c (\bar a + a) + \bar a b c + abc$$ $$f_1 = \bar c + \bar a b c + abc$$

So 3 terms and a don't care became 1 term.

I excluded \$\bar a b c + abc\$ from the reduction, but a similar process is required for them. Go back to origional function and see if they can be reduced.

Partial K-map:

enter image description here

I have solved for half of f1. You have to finish it, but this should get you started.

  • \$\begingroup\$ Thanks for the answer, so would it be appropriate to say a TLDR of the answer is it doesn't really matter as anyways the term is going to disappear when you use reductions? So it can be both 0 or 1 whichever one wants? \$\endgroup\$ Jan 24 at 15:33
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    \$\begingroup\$ Yes. A 1 or 0, but it cannot be both for one function. 110 is 1 for f1. It can be 1 or 0 for f2. Work through it and edit your answer putting in your answers. Someone will confirm it, but we don't do the work for you. A nudge in general direction is all I've done. \$\endgroup\$ Jan 24 at 15:57
  • \$\begingroup\$ Got it thanks :) \$\endgroup\$ Jan 24 at 16:10

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