Solve K-Map Homework: Practice Using K-Maps

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The discussion revolves around solving K-map homework problems, specifically minimizing the expression F = a \bar b + b \bar c d + cd + \bar a c d + a \bar b \bar c d. Participants express concerns about the accuracy of the K-map and the use of don't care conditions. Corrections are suggested, emphasizing the need for simpler algebraic steps to leverage the K-map's efficiency. Ultimately, a revised expression is proposed, leading to the conclusion that the final minimized form can be expressed as F = \bar d [ a + \bar b + c ]. The conversation highlights the importance of careful analysis and simplification in K-map problems.
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Homework Statement



1. Draw the K-Map for ##F = a \bar b + b \bar c d + cd + \bar a c d + a \bar b \bar c d## and minimize the expression.

2. Find a simplified expression for the K-map:

Screen Shot 2015-02-15 at 2.20.14 PM.png


Homework Equations

The Attempt at a Solution



My work for each question is shown in the image below. I hope that I have done everything properly:

IMG_0527.jpg


If someone could verify my work it would be much appreciated.

Thank you.
 
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Why do you have don't cares? either you are not using them correctly, or there is information you are not giving us.
Also I am fairly certain your kmap is not correct.
 
donpacino said:
Why do you have don't cares? either you are not using them correctly, or there is information you are not giving us.
Also I am fairly certain your kmap is not correct.

The don't cares are given as stated, and the questions are given exactly as I've mentioned in the first post.

What leads you to believe my K-Map for the first problem is incorrect?
 
Zondrina said:
The don't cares are given as stated, and the questions are given exactly as I've mentioned in the first post.

What leads you to believe my K-Map for the first problem is incorrect?
ohhh my god, I thought the Kmap show above was the Kmap from the first problem. disregard what I said before...For problem #1 I would check your work again
 
for number 2, you can get it simpler. Why are you doing so much algebra. The beauty of the K map allows you to create the logic functions simply.
I only did one algebra step.
 
donpacino said:
ohhh my god, I thought the Kmap show above was the Kmap from the first problem. disregard what I said before...For problem #1 I would check your work again

No problem, perhaps it was a little confusing.

I assume the K-Map for the first problem is wrong because it should look like this:

e e 1 e
e 1 1 e
e 1 1 e
1 1 1 e

Where I used e to denote an empty spot on the map.

For the second question, I can't see how to get it any simpler. I thought the loops I used were as large as possible.
 
Zondrina said:
No problem, perhaps it was a little confusing.

I assume the K-Map for the first problem is wrong because it should look like this:

e e 1 e
e 1 1 e
e 1 1 e
1 1 1 e

Where I used e to denote an empty spot on the map.

For the second question, I can't see how to get it any simpler. I thought the loops I used were as large as possible.
your correction to #1 is still not correct. look at your AB' term.

for the second question I have two comments. You don't need all that algebra. It defeats the purpose of using a kmap. it should be 1, maybe two steps.
That being said, you have a redundant term in your final answer
 
Oh whoops, I missed a one on the bottom right corner there, it should be:

e e 1 e
e 1 1 e
e 1 1 e
1 1 1 1

Then taking the biggest loops I get ##F = a \bar b + bd + \bar a c d## for the first problem.

Upon looking at the second problem with a different vision now, I believe It should be:

$$F = c \bar d + a \bar b + a + \bar a \bar b \bar c \bar d = a + c \bar d + \bar a \bar b \bar c \bar d$$.
 
#1.) Nice job!

#2.)that is not correct, i know that because the expression a does not work in this case.
look at your final answer for #2 in your original work. It is very close to being correct. You can get to that point simply by looking at the loops.
 
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  • #10
That's weird, thought I had it that time. I see the problem now though, I have an extra loop along the bottom row that's been confusing me.

So removing that extra loop, I have three loops. The one along the right column, the one along the bottom half of the left column, and the one surrounding the four corners. Reading the loops off I get:

$$F = c \bar d + a \bar c \bar d + \bar b \bar d = \bar d [ c + a \bar c + \bar b] = \bar d [ a + \bar b + c ]$$
 
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  • #11
thats what I got.
 
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