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I hope this is the right place to ask this. I have the following problems: 1. Using a K-Map technique perform the following: Simplify the following function: f = (A,B,C,D) = ∑ m (0,1,2,3,6,7,8,9,13,15) Show all the "prime implicants" and "essential prime implicants"

and Find a minimum SOP expression for: f(w,x,y,z) = ∑ m (1,3,5,9,11,14)+d(4,6,7,12) Show all the "prime implicants" and "essential prime implicants" This is what I came up with: enter image description here

But as you can see, I don't have any essential prime implicants whatsoever. So I'm worried that I'm doing something horribly wrong. Could someone please help me out?

Updated progress: enter image description here

enter image description here

Fixed up the blue circles a bit to make it more clear.

user215717
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2 Answers2

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I'm familiar with K-maps, though not as familiar with the terminology, so I've had to resort to Wikipedia. First of all, you're allowed to overlap your groupings. Since you haven't 2 of your implicants from part a are not prime.

You have a square in the upper right corner for a prime implicant of A'C. Similarly, you have another square in the upper/lower left for another prime implicant of B'C'. The other 2 prime implicants you have for that part appear to be correct. Of these 4 prime implicants, it appears 3 are essential.

According to Wikipedia, an implicant is prime if it cannot be covered by an implicant with fewer literals. A'BC is not a prime implicant since it can be covered by A'C. Also, from Wikipedia, a prime implicant is essential if covers an output not covered by any combination of other prime implicants. No other prime implicants cover either output of ABD, so it is essential. AB'C'D' is covered only by B'C', so it too is essential. But every output covered by A'B' is covered by other prime implicants; therefore it is not essential.

Mike
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  • Thank you for your help, I now see that I can make 2 more squares in part A that will cover 4 of the ones. Could you please clarify how you determine what the essential ones are and which are not? Are the 3 essential ones due to the 3 squares of 4 ones? (Hope that makes sense) – user215717 Feb 18 '15 at 04:55
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    @user215717 A'B' is not essential. The upper left is covered by B'C' and the upper right is covered by A'C. All outputs covered by A'B' are covered by these 2 prime implicants. – Mike Feb 18 '15 at 05:22
  • Ah so the other 3 that are not essential are: ABD, A'BC, AB'C'. Does that seem right? – user215717 Feb 18 '15 at 05:27
  • @user215717 I'll update the answer. – Mike Feb 18 '15 at 05:32
  • Ok, I updated it to what I have now. Including my attempt at part B. – user215717 Feb 18 '15 at 05:48
  • @user215717 I see you've updated the groupings, but you haven't rewritten the prime implicants. And z does not appear to be an implicant. wxz should not be covered. – Mike Feb 18 '15 at 06:05
  • I'm not sure what you mean. I don't have wxz, I have wxz'. – user215717 Feb 18 '15 at 06:25
  • @user215717 What I mean is that the grouping for z contains 2 zeroes. – Mike Feb 18 '15 at 06:31
  • Sorry, it's still not getting through to me. What are the zeros? Is it the x's? – user215717 Feb 18 '15 at 06:37
  • @user215717 The squares you have left blank. Where your Boolean function is supposed to be false. – Mike Feb 18 '15 at 06:43
  • But I don't think I covered any of the blank ones. I just went through my calculation again and it doesn't seem to contain any of the blank ones. Sorry if this is something super simple and I'm just being blind. – user215717 Feb 18 '15 at 07:01
  • @user215717 Third row, center columns. Been talking about part b of course. – Mike Feb 18 '15 at 07:07
  • Ooooooh here's why there is a confusion right now: I'm not meaning to include them in my grouping! I realize the blue lines leak into them a bit, but that was just my way of trying to not cover up the 1's and x's. Here, let me try to touch it up and I'll update the diagram. – user215717 Feb 18 '15 at 07:11
  • @user215717 It's not the drawing. You listed z as an essential prime implicant. z covers the 2 middle columns. – Mike Feb 18 '15 at 07:16
  • I'm starting to see what you mean. But how do I allow for my logic function to exclude those 2? If I do the calculation for the group of 6, I end of with just Z. – user215717 Feb 18 '15 at 07:22
  • @user215717 You can't do a grouping of 6. The groupings can't have a dimension of size 3. You'll have to use two 2x2 groupings instead. – Mike Feb 18 '15 at 07:27
  • Ah, didn't realize that since it was divisible by 2. Ok, I've updated it with prime implicants: w'z, xz', wx'z – user215717 Feb 18 '15 at 07:33
  • @user215717 The first 2 are good. The third is not prime. And there's a fourth you had drawn in originally. – Mike Feb 18 '15 at 07:39
  • What if I change the third one to x'z? – user215717 Feb 18 '15 at 07:43
  • @user215717 Looks good. And the fourth prime implicant? – Mike Feb 18 '15 at 07:46
  • I have wx’z listed in prime implicant, just not as essential because essential is covered in 3. – user215717 Feb 18 '15 at 07:50
  • @user215717 Again, going by Wikipedia's definitions wx'z is not prime since it can be covered by an implicant with fewer literals: x'z. – Mike Feb 18 '15 at 07:53
  • Can't it be prime but just not essential prime? Sorry, I've probably been staring at this for too long. But thank you for all your help! I'll take another look at it tomorrow. – user215717 Feb 18 '15 at 08:00
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Computer the prime implicants (PI) and essential prime implicants (EPI) in the following Boolean function. F(W, X, Y, Z) = п M(1, 3, 5, 7, 8, 9, 11, 12) + d(2, 10) Computer a minimal SOP expression for F and determine whether it is unique.

Sachin Singh
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