How can I simplify boolean expressions using K-maps and truth tables?

[Live4eva_2]

Hi...

I'm struggling in grasping how to reduce boolean expressions to K-maps/truth tables.
I drew a diagram in paint illustrating my steps in trying to reduce a function...
could someone please have a look and correct me in my method/point me in the right direction??

much appreciated...

Hope no-one minds but I posted this in the Comp_Sci section as well...because I know that this falls under both discrete maths and comp architecture...So rather check for my post there before wasting time providing help ,when someone else already has...

 

[jhicks]

Hope no-one minds but I posted this in the Comp_Sci section as well...

That's a big no-no.

 

[Live4eva_2]

OK I marked the other thread as solved.
I managed to get a bit further with the K-maps.
Have a look at this .jpg.

I have managed to figure out the ordering of the minterms and the values of the truth table.The function in question is just an arbitrary one I made up so I'm not sure if that's a problem.

In my textbook the K-maps have curly braces indicating where (A,B,C are primed)...I don't quite understand that...

unless say F = A + (B'A') then A would appear primed and unprimed??

 

[chroot]

You're pretty much on the right track. In your first post you indicated the function was f = A + B'C, but in the second, it was f = A XOR B'C. I appears from your K-map that you actually intended the latter.

Also, in your K-map, you have incorrectly ordered the columns. They should read 00, 01, 11, 10. In this case you got lucky, because it doesn't matter.

Now you just need to circle the groups and write out the resulting expression.

- Warren

 

[Live4eva_2]

ah...this is the bit I don't understand.What is the method for writing out the expression once I have grouped everything??Could you explain?

 

[chroot]

Circle the largest groups you can, in powers of two. In other words, look for groups of 8, or 4, or 2, or 1 "ones" all adjacent to each other on the K-map.

Then, look at the row and column headings and figure out the term that represents each group. If you have a lone one that has a circle all by itself, it will result in a term with all of the inputs (A, B, and C in this case) represented. If you have a group of two, however, it'll depend on only two inputs -- one will be redundant.

It's really much easier shown on paper than described here in words. I suggest you see your professor or TA as soon as possible to get some one-on-one help.

- Warren

 

[Live4eva_2]

from the column headings:

binary 100 -------This is my lone minterm. (ABC--->dependent on all inputs)

binary 110 ------
binary 111 ------These constitute my group of 2^1 minterms.(As you say,one input will be redundant.)

So...when all inputs are 1...I get a 1 for output.
Or...when C is 1...I get a 1 for output

Waitaminute,this is where that least significant bit comes in isn't it??Looking at the minterms I would say for the group of 2 minterms that C is redundant.Because of the fact that the truth table shows that if A and B are 1 the ouput is 1.So C is unnecessary??

But all this tells me is that I have ABC and AB appearing in my expression isn't it?