Boolean question

[macca]

Guys I need help with this problem if you'd be so kind?

Q. Make a circuit comprising four inputs, output is TRUE if an odd number of inputs are TRUE.

My solution => A+B+C+D+ABC+ABD+ACD+BCD

So, how do I simplify that?

Thank you.

[Janus]

Guys I need help with this problem if you'd be so kind?

Q. Make a circuit comprising four inputs, output is TRUE if an odd number of inputs are TRUE.

My solution => A+B+C+D+ABC+ABD+ACD+BCD

So, how do I simplify that?

Thank you.

First off, your solution is incorrect, since it would simplify to A+B+C+D.

Secondly, the real solution doesn't simplify using AND or OR gates. (but you can simplify the circuit.)

[Hurkyl]

I think he's using + for or, not xor. Silly having 4 ways to write the same thing. :frown:

[e(ho0n3]

I think he's using + for or, not xor. Silly having 4 ways to write the same thing. :frown:
I thought the symbol for xor is a circle-enclosed plus sign. What "thing" are you talking about?

macca: Draw yourself a Karnaugh map of the circuit behaviour. It shouldn't be too difficult.

[Hurkyl]

In logic, xor is \veebar. In circuits, xor is \oplus. In GF(2n), xor is +. In C, xor is ^.

[Alkatran]

Assuming + is OR and multiplication is AND

A+B+C+D+ABC+ABD+ACD+BCD is wrong because:

ABD + ABC = AB(D + C)
A+B+C+D+ABC+ABD+ACD+BCD = A(True + BC + BD + CD) + B(True + CD) + C +D
True+BC+BD+CD = true and True + CD = true
A + B + C + D
If any of them is true...

(A xor BCD) + (B xor ACD) + (C xor ABD) + (D xor ABC)
would make more sense, either one is true or three are true. I'm not sure how to simplify this because.. well I haven't ever "done" boolean math except minimally in programming.

[master_coda]

The simplest representation for this circuit that I know of is:

A\oplus(B\oplus(C\oplus D))

[Alkatran]

The simplest representation for this circuit that I know of is:

A\oplus(B\oplus(C\oplus D))

Right, because every true NOTs the final answer (starting with false)... right?

[master_coda]

Right, because every true NOTs the final answer (starting with false)... right?

If you mean what I think you mean, than yes.

Actually, I made it unnecessarily complicated. A\oplus B\oplus C\oplus D does the same thing, now that I think about it. I forgot that xor is associative.

[Janus]

The simplest representation for this circuit that I know of is:

A\oplus(B\oplus(C\oplus D))


This what I meant when I said it could be simplfied. I was trying to give him a push in the right direction without giving him the answer outright, just in case this was a homework assignment.

[paul11273]

This question has me curious. By putting it into a truth table, then transferring to a K-Map doesn't help. You are unable to make any groups.
So reading directly from the truth table, I end up with:
A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AB'C'D' + AB'CD + ABC'D + ABCD'

By beginning to group and simplify I get:


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


A'B'(C \oplus D) + AB(C \oplus D) + C'D'(A \oplus B) + CD(A \oplus B)


(A \oplus B) + (C \oplus D)

As you can see by this, I am not getting down to the simplified answer given by you guys: A \oplus B \oplus C \oplus D

But, I do agree with you answer, because when I figure a truth table using your answer, I get the desired output results of true when there are an odd number of true inputs.

How would one go about simplifying this step by step?
Where did I make an error in my algebra?

[master_coda]

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


A'B'(C \oplus D) + AB(C \oplus D) + C'D'(A \oplus B) + CD(A \oplus B)


(A \oplus B) + (C \oplus D)


The third line does not follow from the second. You should instead be getting:


(AB+A'B')(C \oplus D) + (A \oplus B)(CD+C'D')


(A \oplus B)'(C \oplus D) + (A \oplus B)(C \oplus D)'


(A \oplus B)\oplus(C \oplus D)

[TenaliRaman]

XOR gives 1 when odd number of inputs are high .... (works for n number of inputs and its bcos as noticed by master coda .... XOR is associative)

-- AI

[paul11273]

Cool, I now see where my error was. Thanks master_coda.