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_{1},x

_{2},x

_{3}= \(\displaystyle \sum\)m(1,2,3,5)

I need to use

**Karnaugh Maps**. How do I do this with a Karnaugh map, I understand how to get the rows and columns with a K map. But i don;t understand after that.

- Thread starter shamieh
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I need to use

- Jan 30, 2012

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How do you measure cost?Find the minimum cost sum of products and product of sums forms for the function : $f$(x_{1},x_{2},x_{3}= \(\displaystyle \sum\)m(1,2,3,5)

Could you explain more precisely what you understand and what you don't? You know how to get the rows and columns of what? Can you post (or at least describe) the Karnaugh map for this function?I need to useKarnaugh Maps. How do I do this with a Karnaugh map, I understand how to get the rows and columns with a K map. But i don;t understand after that.

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I think it would be best to leave this topic, so that when you get your other question answered, you may want to revisit this topic and show what you learned by completing this question here for the benefit of our readers.Please delete this post. I have created another, with a more details to question on a particular K-map problem. Please void this post.

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finally figured out how to use K-maps!

Can anyone check my work? Want to make sure I am doing it the right way.

Find the minimum cost Sum of Product form for the function $f$(x

I got x

Thank you for your time,

Sham

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Since this is the same question originally posted in this topic, I have merged the two topics to avoid redundancy.finally figured out how to use K-maps!

Can anyone check my work? Want to make sure I am doing it the right way.

Find the minimum cost Sum of Product form for the function $f$(x_{1},x_{2,}x_{3}) = \(\displaystyle \sum\)m(1,2,3,5)

I got x_{3}x_{4}! + x_{1}!x_{2}! Is this correct?

Thank you for your time,

Sham

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- Jan 26, 2012

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Where did the $x_4$ come from? I get $\bar{x}_1 x_2+\bar{x}_2 x_3$.finally figured out how to use K-maps!

Can anyone check my work? Want to make sure I am doing it the right way.

Find the minimum cost Sum of Product form for the function $f$(x_{1},x_{2,}x_{3}) = \(\displaystyle \sum\)m(1,2,3,5)

I got x_{3}x_{4}! + x_{1}!x_{2}! Is this correct?

Thank you for your time,

Sham

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- #8

Wait a minute, I'm an idiot. Was writing down the wrong numbers and was looking at another problem, let me re-work the problem.Where did the $x_4$ come from? I get $\bar{x}_1 x_2+\bar{x}_2 x_3$.

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by Truth table I put all 1's in the spot of the numbers. drew a karnaugh map and got

my

- 1 1 1 1 <---I grouped these 4 1's horizontally..Ialso grouped the 2 1s vertically.
- 1 1 0 1 <-- I grouped the 4 1's from the 1st and second row.
- 0 1 1 0 <---- I grouped 2 1s horizontally
- 1 0 1 1 <--- grouped 2 1s horizontally.

- Jan 30, 2012

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(Yay, I learned to generate images with transparent background!)

The K-map is correct. Now you need to write the SOP and POS forms. Note that for POS you look at 0's in the K-map and the convention for writing maxterms is the opposite from that for minterms: 0 for a variable means no negation; 1 means negation.

by Truth table I put all 1's in the spot of the numbers. drew a karnaugh map and got

myKMAP

- 1 1 1 1 <---I grouped these 4 1's horizontally..Ialso grouped the 2 1s vertically.
- 1 1 0 1 <-- I grouped the 4 1's from the 1st and second row.
- 0 1 1 0 <---- I grouped 2 1s horizontally
- 1 0 1 1 <--- grouped 2 1s horizontally.

Here is a nice online K-map generator. However, it only shows minimal SOP forms.

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Wow I love you. Can I ask you a question. Suppose I am grouping and I had a pair of **1 & d** next to each other.. I noticed on this particular problem my teacher gave us, he didn't group that pair, and thus; left out a term. My question is, if there is a grouping to be made of 1's. Why would you *not* make that grouping? Because that d can just be a 1 right? So why not group that? Wouldn't it minimize your cost? Isn't that the whole point! To find all groupings? Otherwise aren't you missing something? Example Below. We are dealing with Sum of Products by the way just FYI.. **NOTE: The x's are my don't cares or ****"d"s**

Last edited:

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- Jan 26, 2012

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It isn't always advantageous to make a 'd' a '1'. Clumping on K-maps works like this: they always have to be contiguous rectangles (they can wrap around, both horizontally and vertically), where the height and width are powers of 2. In your case, the 'd' in the $ABCD$ location allows you to form the more efficient term $BC$. However, the 'd' in the $ \bar{A} \bar{B}C \bar{D}$ position does not allow you to form a contiguous rectangle of width a power of 2 and height a power of 2 that includes any ones you can't already get from a different grouping. That is, the '1' in the $\bar{A} \bar{B} \bar{C} \bar{D}$ position you've already got from $\bar{B} \bar{C} \bar{D}$, and the '1' in the $\bar{A}BC \bar{D}$ position you've already gotten from the $BC$ term.Wow I love you. Can I ask you a question. Suppose I am grouping and I had a pair of1 & dnext to each other.. I noticed on this particular problem my teacher gave us, he didn't group that pair, and thus; left out a term. My question is, if there is a grouping to be made of 1's. Why would younotmake that grouping?

Because that d can just be a 1 right? So why not group that? Wouldn't it minimize your cost? Isn't that the whole point! To find all groupings? Otherwise aren't you missing something? Example Below. We are dealing with Sum of Products by the way just FYI..NOTE: The x's are my don't cares or"d"s

View attachment 1402

I'll let the master, Evgeny.Makarov, answer that one, as I have no experience forming PoS from K-maps.Also, another question..When we are solving using a K-map for theProduct of Sumsdo we evaluate our K-Map the exact same way except group the 0s? Or do we evaluate it differently?

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- Jan 30, 2012

- 2,502

Leaving outSuppose I am grouping and I had a pair of1 & dnext to each other.. I noticed on this particular problem my teacher gave us, he didn't group that pair, and thus; left out a term.

Yes, you group zeros, but the rest is the other way around. With a SOP, 0 means a negation, so 0000 produces W!X!Y!Z!. With a POS, 1 means a negation, and besides terms are sums (it's a Product Of Sums). So, with a POS, 0000 produces W + X + Y + Z.Apparently you group all the zeros, but when you are evaluating, if you have 00 00 for example..and you have WXYZ for example... Instead ofWXYZ(which you would put in a SOP form Kmap) you just readW!X!Y!Z!in aform kmap.POS

Basically, in constructing a POS, everything is dual compared to a SOP: instead of 1's in a truth table and a K-map, you look at 0's; instead of products, terms are sums, and instead of negations, 0's correspond to simple variables.