# Simplifying a Four Variable Truth Table Equation

Discussion in 'Math' started by Freak80MC, Feb 26, 2015.

1. ### Freak80MC Thread Starter New Member

Feb 26, 2015
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(In the picture, an "i" at the end of a letter is used to show it is to be inverted. I don't know how to type characters with lines over them in Google Sheets so I just did this)

So I am trying to create a simplified equation using the sum of products method for a circuit that has 4 inputs and 1 ouput, where the output turns off when 2 or more of the 4 inputs are on. Only one problem, I am having trouble simplifying it as I either can't get very far or when I do successfully simplify it by a lot, the equation doesn't actually work as intended. So I was just wondering if you all could help me out to simplify it, or if it is even possible to simplify all that much.

Thanks in advance for anybody that can help me with this!

2. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
It can be simplified to 4 product terms.

Show us what you come up with during your simplification.

By the way, negation is usually written as a ', as 'i' could be mistaken for a variable. Also, you needn't specify the AND between the variables - like multiplication, putting two variable next to each other implied the AND. So, A*Bi*C in your notation would be AB'C.

3. ### MikeML AAC Fanatic!

Oct 2, 2009
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Are you trying to reduce the equation so that you can implement it using just NAND gates, or just NOR gates, or a mixture of AND, OR and INV functions?

4. ### Freak80MC Thread Starter New Member

Feb 26, 2015
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I am just really trying to simplify it to use the least amount of operations. So whether that can be done with just NAND gates, just NOR gates, or a combination of gates doesn't matter to me.

5. ### Freak80MC Thread Starter New Member

Feb 26, 2015
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Thanks for telling me that last bit! And the most I could simplify it myself was to get it to A'B'(C+D')+A'BC'D'+AB'C'D' even though when I hooked it up into Logisim it didn't work right. It just really sucks that the whole thing about simplifying equations was just sort of skimmed over in my high school digital electronics class.

6. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
Well, one way to check is to undue your proposed minimized expression and see if that matches the expression derived from your truth table.

Let's take a look at the expression you came up with:
You have A'B'(C+D')+A'BC'D'+AB'C'D'.
Distributing the A'B' in the first term, we get:
A'B'C+A'B'D'+A'BC'D'+AB'C'D'.

Expanding A'B'C, we get:
A'B'CD+A'B'CD'+A'B'D'+A'BC'D'+AB'C'D'.

Now expanding A'B'D', we get:
A'B'CD+A'B'CD'+A'B'CD'+A'B'C'D'+A'BC'D'+AB'C'D'.

And we can group like terms:
A'B'CD+A'B'CD'+A'B'C'D'+A'BC'D'+AB'C'D'

This should match the initial expression, but it doesn't - which means there's a mistake somewhere.

Try posting your minimization process, that way, we can see where you are going astray.

Have you learned about Karnaugh maps (K-maps)?

The eBook on this site does a decent job at explaining minimization - you may want to check it out.

Last edited: Feb 27, 2015
7. ### WBahn Moderator

Mar 31, 2012
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4,917
I'll emphasize what tshuck just said -- we really need to see your work in order to provide must assistance.

8. ### Freak80MC Thread Starter New Member

Feb 26, 2015
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0
No, I have not learned about K-maps yet. And here is my simplification process:

A'B'C'D'+A'B'C'D+A'B'CD'+A'BC'D'+AB'C'D'
A'B'C'(D'+D)+A'B'CD'+A'BC'D'+AB'C'D'
A'B'C'1+A'B'CD'+A'BC'D'+AB'C'D'
A'B'C'+A'B'CD'+A'BC'D'+AB'C'D'
A'B'(C'+CD')+A'BC'D'+AB'C'D'
A'B'(C+D')+A'BC'D'+AB'C'D'

And this is about as far as I got. I know it can be simplified more as I have messed around in logisim and gotten a significantly smaller circuit than the one that that last equation shows, but I want to know how to work it out myself from the equations and from simplifying them.

9. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
First thing to note is that what you have above is not sum-of-products (SOP), which is what you said you were looking for.

The approach you started off with is the route you want to go.

A'B'C'D'+A'B'C'D+A'B'CD'+A'BC'D'+AB'C'D'
A'B'C'(D'+D)+A'B'CD'+A'BC'D'+AB'C'D'
A'B'C'+A'B'CD'+A'BC'D'+AB'C'D'

Now are there any other pairs of terms that are the same except for one literal (which is straight in one term and complemented in the other).

Also keep in mind that you can reuse terms since A=A+A. This means that even after you have simplified a pair of terms, for instance (A'B'C'D' and A'B'C'D), you can always add one or both of them back in if it can be combined with another term, such as, say, A'BC'D'.

10. ### Freak80MC Thread Starter New Member

Feb 26, 2015
6
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Wait, how is that not sum of products? I took the products of the rows which equaled one (to get stuff like A'B'C'D' and AB'C'D') and then added them all together. Isn't that what you are supposed to do for sum of products?

11. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
This: A'B'C'D'+A'B'C'D+A'B'CD'+A'BC'D'+AB'C'D' is SOP

This: A'B'(C+D')+A'BC'D'+AB'C'D' is not SOP

12. ### Freak80MC Thread Starter New Member

Feb 26, 2015
6
0
Well I only used the SOP to set up my equation, but for simplifying it, I wanted the equation to be as simplified as possible and of course you are most likely not going to get that with SOP.

13. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
What does "simplified as possible" mean? Given two expressions, what metric do you use to decide if one expression is "simpler" than the other? Is your metric the number of literals? If so, then you started with six literals and ended up with six literals. But you've added an additional layer of computation in the process.