Basic Assembly Language = Close Enough.

[Squeek]

Though I sincerely doubt any of you have a clue about this.

What I'm currently taking classes in is the basics of computer chips and hardware components. As of now, we're doing boolean algebra to learn how to optimize and things like that. Because who wants to use 100 chips when just one will suffice?

Here's the problem. I understood all of chapter one, which was simply converting between bases and doing arithmetic with binary (fun to the max, might I add). Chapter two is giving me problems. These in particular.

Reduce these to the specified literals.

(a) A'C' + ABC + AC in 3 literals.

This one I know. But, I don't know how to do it the CONVENTIONAL way! I know how to do this using methods other than what he wants to see, and if the exams require me to show my work, I won't be getting it right.

The answer is AB + C. You achieve the same output using this correlation and the other one in the code box. However, when I do it the conventional way, I get ABC. Which is obviously wrong. I don't see where I went wrong.

For some background information, note the following.

' = NOT. Inverse. A = 1, A' = 0.

+ = OR. A+B means A OR B.

* = AND. AB (A*B) means A AND B.

Literals are how many "A B C"s you have leftover. AB+C is three literals and two terms (which are separated by the OR).

Hopefully someone out there can explain this to me better than this textbook can.

~Squeek

[soccr743]

First off, how is AB + C right when it doesn't produce the same value for A=0,B=0,C=0

(0' * 0') + (0 * 0 * 0) + (0 * 0) = (0 * 0) + 0
1 + 0 + 0 = 0 + 0
1 = 0???

Not sure if you wrote out the first function right or not... But that is what I come up with...

And as far as simplifying the first function (If it was typed out right):

A'C' + ABC + AC
A'C' + AC(B + 1)
A'C' + AC

Not sure how to get it down more than that...

-----Soccr743-----

EDIT: And I forgot to ask you, but if you come across any really interesting problem or project in your class, can you PM it to me or post it here. I went through a good bit of a college text book called "Digital Logic Circuit Analysis and Design" over the summer, but it would be cool to continue studying it through the year and working on projects that are assigned in college. Thanks

[Squeek]

After a little more reading, I found out that I missed a single prime notation. Sheesh.

Reduce these to the specified literals.

(a) A'C' + ABC + AC' in 3 literals.

Changes things a bit, doesn't it...

A B C | F
0 0 0 | 1
0 0 1 | 0
0 1 0 | 1
0 1 1 | 0
1 0 0 | 1
1 0 1 | 0
1 1 0 | 1
1 1 1 | 1

AB + C

That's how I got it. But I used a Truth Table combined with a Map. This was chapter two. Maps were chapter three. I need to be able to do this without using a Map or anything. Just simple reduction and all.

~Squeek

[soccr743]

Ah, yea I thought so. It seemed like something was a bit messed up.

Works right now.

A'C' + ABC + AC'
C'(A' + A) + ABC
C' + ABC
C + AB

-----Soccr743-----

[Squeek]

Just to update, today's lab was figuring out two things:

1) Make an AND gate using only NOR and NOT gates.

2) Make a function work using only NAND and NOR gates.

#1 was beyond easy. Invert first, NOR second. Simple. #2, we never finished. Gotta work on it now.

F = A + BC + BD

ABCD
0 0 0 0  0
0 0 0 1  0
0 0 1 0  0
0 0 1 1  0
0 1 0 0  0
0 1 0 1  1
0 1 1 0  1
0 1 1 1  1
1 0 0 0  1
1 0 0 1  1
1 0 1 0  1
1 0 1 1  1
1 1 0 0  1
1 1 0 1  1
1 1 1 0  1
1 1 1 1  1

I can probably do it myself once I start working on it.

Edit a day later: done. I'll use % to represent NAND.

A' % B%C % B%D

Take the NAND of BC and BD first, then the NAND of all three and you're done.

Simple as that.

~Squeek