I was wondering how you got that. Did you draw the map from the original statement, and do a truth table from that, then everything from there by assuming no errors were made in making the K-Map or truth table?

i thought it was two different questions, since your function values didn't match the statement you originally posted.

omg forgot something so small, the truth table! LOL I actually came up with the k-map by looking at the equation and then giving them the value that would make it true

AC' - 1001,1100
B'D - 0011, 1001
A'CD - 0111, 0011
ABCD - 1111

Thats how I got the sum of minterms but it's clearly wrong since I'm missing 4 more minterns

 

omg forgot something so small, the truth table! LOL I actually came up with the k-map by looking at the equation and then giving them the value that would make it true

AC' - 1001,1100
B'D - 0011, 1001
A'CD - 0111, 0011
ABCD - 1111

Thats how I got the sum of minterms but it's clearly wrong since I'm missing 4 more minterns

If you have a function of four variables, then a four-variable term corresponds to one minterm. If it only has three of the four variables, then the fourth is a don't care and so it maps to two minterms. So a two-variable term would map to how many minterms?

 

a two-variable term would have 2^2 = 4 minterms?

 

See if this is correct (DOUBLE CHECK YOUR EQUATION GIVEN IN BOOK/WORKSHEET!), do a screen cap and draw circles around each term.

[table="head";]AB\CD|00|01|11|10
00|0|1|1|0
01|0|0|1|0
11|1|1|1|0
10|1|1|1|0
[/table]


This is a cheat sheet for making a truth table from a K-Map:
[table="head";]AB\CD|00|01|11|10
00|0|1|3|2
01|4|5|7|6
11|C|D|F|E
10|8|9|B|A
[/table]
Hex: A=10, B=11, C=12, D=13, E=14, F=15 if they don't read that way to you already.

 

a two-variable term would have 2^2 = 4 minterms?

Yes, and a 1 variable miniterm covers 8, see cheat sheet above.

 

But I end up with AC' + CD + A'B'D, which differs from the solution/your answer
CD+AC'+B'D

 

You forgot to wrap around A'B'D + AB'D = B'D

 

But I end up with AC' + CD + A'B'D, which differs from the solution/your answer
CD+AC'+B'D

Your answer is not the minimal solution, but it is valid.

Now, essentially, do the same thing, where zeros match zeros (rather than ones matching ones). Circle and proceed the same way.

You'll then have PoS.

 

But I end up with AC' + CD + A'B'D, which differs from the solution/your answer
CD+AC'+B'D

Now that you have the maps for both of these and see that they are the same, try to figure out how you can use Boolean manipulation in order to reduce your answer to their answer. It's straightforward, but not obvious because all three terms play a part.

 

Highlight the parts after my question marks in post 17 to see the huge mess your incorrectly derived function was, those ARE the minimum terms, as far as I can get!

 

You forgot to wrap around A'B'D + AB'D = B'D

Was looking at the k-map the longest time to try and find out where that wrap around is and alittle confused. I thought I got everything already.

Now that you have the maps for both of these and see that they are the same, try to figure out how you can use Boolean manipulation in order to reduce your answer to their answer. It's straightforward, but not obvious because all three terms play a part.

Alright, I gave boolean manipulation a go but got stuck. Here's my attempt at it

AC' + CD + A'B'D
= AC' + CD + A'B'CD + A'B'C'D
= AC' + CD(1+A'B') +A'B'C'D
= AC' + CD + A'B'C'D
= AC'(BD+B'D') + CD + A'B'C'D
= ABC'D + AB'C'D + CD + A'B'C'D
= AC'D ( B+B') + CD + A'B'C'D
= AC'D + CD + A'B'C'D

at this point, it felt like I made the equation more complicated than simplified

 

Was looking at the k-map the longest time to try and find out where that wrap around is and alittle confused. I thought I got everything already.

Let's consider the most extreme example, the corners.

AB\CD|00|01|11|10
00|1|0|0|1
01|0|0|0|0
11|0|0|0|0
10|1|0|0|1

What is important to note is that there is nothing magical about having the first column be 00. All that is important is that, as we move from one column to the next, that only one of the variables changes. But as we go from the extreme right edge, with CD=10, and "wrap around" to the extreme left edge with CD=00, the only thing that changed was C. From a grouping standpoint, we can group 1's that are in these two columns together as long as they are on the same row. The same thing applies to the rows.

To see this better, we can just reorder the rows and columns as long as we hold to the rule that only one variable can change at a time.

AB\CD|01|00|10|11
01|0|0|0|0
00|0|1|1|0
10|0|1|1|0
11|0|0|0|0

Do you see that these two tables represent exactly the same truth table?

In both cases, we can use

Y=B'D'

Alright, I gave boolean manipulation a go but got stuck. Here's my attempt at it

AC' + CD + A'B'D
= AC' + CD + A'B'CD + A'B'C'D
= AC' + CD(1+A'B') +A'B'C'D
= AC' + CD + A'B'C'D
= AC'(BD+B'D') + CD + A'B'C'D
= ABC'D + AB'C'D + CD + A'B'C'D
= AC'D ( B+B') + CD + A'B'C'D
= AC'D + CD + A'B'C'D

at this point, it felt like I made the equation more complicated than simplified

You usually have to make things more complicated before you can simplify them.

The make sure we know what we are trying to do, first.

AC' + CD + A'B'D

We are trying to make A' go away from the final term.

But if we can make it go away that means that it doesn't matter. So why doesn't it matter?

For it to matter, there must be a situation in which it and it alone makes the difference between the result being True and the result being False. For that to be the case, we know that we are talking about those cases when B'D is True but that both of the first terms are False. For B'D to be True, that means that D must be True. For the second term, CD, to be False if D is True means that C must be False. But if C is False then C' is True and the only way that the first term can then be False is if A is False, which means that A' is True.

This means that changing A won't make a difference if B'D is True. If A' is True, then A'B'D is True and the overall result is True. But if A' is False, then both A and D are True which guarantees that one of the first two terms will be True regardless of what value C has and, again, the overall result is True.

So let's see if we can get there algebraically.

AC' + CD + A'B'D
= AC' + CD + A'B'D(C+C')
= AC' + CD + A'B'CD + A'B'C'D
= AC' + A'B'C'D + [CD + A'B'CD]
= AC' + A'B'C'D + CD(1 + A'B')
= AC' + A'B'C'D + CD
= C'(A + A'B'D) + CD
= C'[A(1+B'D) + A'B'D] + CD
= C'[A + AB'D + A'B'D] + CD
= C'[A + (A+A')B'D] + CD
= C'[A + B'D] + CD
= AC' + B'C'D + CD
= AC' + [B'C' + C]D
= AC' + [B'C' + (B'+1)C]D
= AC' + [B'C' + B'C + C]D
= AC' + [B'(C'+C) + C]D
= AC' + [B' + C]D
= AC' + B'D + CD

 

Let's consider the most extreme example, the corners.

AB\CD|00|01|11|10
00|1|0|0|1
01|0|0|0|0
11|0|0|0|0
10|1|0|0|1

What is important to note is that there is nothing magical about having the first column be 00. All that is important is that, as we move from one column to the next, that only one of the variables changes. But as we go from the extreme right edge, with CD=10, and "wrap around" to the extreme left edge with CD=00, the only thing that changed was C. From a grouping standpoint, we can group 1's that are in these two columns together as long as they are on the same row. The same thing applies to the rows.

To see this better, we can just reorder the rows and columns as long as we hold to the rule that only one variable can change at a time.

AB\CD|01|00|10|11
01|0|0|0|0
00|0|1|1|0
10|0|1|1|0
11|0|0|0|0

Do you see that these two tables represent exactly the same truth table?

In both cases, we can use

Y=B'D'



You usually have to make things more complicated before you can simplify them.

The make sure we know what we are trying to do, first.

AC' + CD + A'B'D

We are trying to make A' go away from the final term.

But if we can make it go away that means that it doesn't matter. So why doesn't it matter?

For it to matter, there must be a situation in which it and it alone makes the difference between the result being True and the result being False. For that to be the case, we know that we are talking about those cases when B'D is True but that both of the first terms are False. For B'D to be True, that means that D must be True. For the second term, CD, to be False if D is True means that C must be False. But if C is False then C' is True and the only way that the first term can then be False is if A is False, which means that A' is True.

This means that changing A won't make a difference if B'D is True. If A' is True, then A'B'D is True and the overall result is True. But if A' is False, then both A and D are True which guarantees that one of the first two terms will be True regardless of what value C has and, again, the overall result is True.

So let's see if we can get there algebraically.

AC' + CD + A'B'D
= AC' + CD + A'B'D(C+C')
= AC' + CD + A'B'CD + A'B'C'D
= AC' + A'B'C'D + [CD + A'B'CD]
= AC' + A'B'C'D + CD(1 + A'B')
= AC' + A'B'C'D + CD
= C'(A + A'B'D) + CD
= C'[A(1+B'D) + A'B'D] + CD
= C'[A + AB'D + A'B'D] + CD
= C'[A + (A+A')B'D] + CD
= C'[A + B'D] + CD
= AC' + B'C'D + CD
= AC' + [B'C' + C]D
= AC' + [B'C' + (B'+1)C]D
= AC' + [B'C' + B'C + C]D
= AC' + [B'(C'+C) + C]D
= AC' + [B' + C]D
= AC' + B'D + CD

Thanks for helping me out, I'm learning alot from you and others that replied in this thread Alot of tips and clarification you gave in your reply, I'll keep in mind that you gotta complicate things in order to simplify them to simplest form. Thanks again

 

Here is the wrap around that would give you B'D without having to go through all that algebra.

 

Thanks Mr.Chip , didn't really thought about that one because after I did the ones in the black I didn't want to overlap so I just left it as it is.

 

Thanks Mr.Chip , didn't really thought about that one because after I did the ones in the black I didn't want to overlap so I just left it as it is.

But do you see if you use the wraparound, you remove a redundant variable in a miniterm.

That is important, and a feature some design software will sort out for you because it is important.