# How to write a boolean function using only XOR and NOT gates?

Discussion in 'Homework Help' started by thermistor, Mar 6, 2013.

1. ### thermistor Thread Starter New Member

Mar 6, 2013
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Can we represent the boolean function of a truth table using only XOR and NOT? And if so, how can this be done? Is there any technique or algorithm?

Thanks

Last edited: Mar 6, 2013
2. ### eng.mustafasalah Member

Nov 10, 2011
41
2
what is that function ?

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3. ### takao21203 AAC Fanatic!

Apr 28, 2012
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I'd take a piece of paper and try and at first make clear: What is XOR and how does XOR work.

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4. ### WBahn Moderator

Mar 31, 2012
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Is the question asking if you can represent ANY boolean function using only XOR and NOT? Or is it asking about a SPECIFIC boolean function?

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5. ### thermistor Thread Starter New Member

Mar 6, 2013
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It asks for a specific function:
Code ( (Unknown Language)):
1. A'B'C'+A'BC+AB'C'+AB'C
Where X' means negation of X and + means OR

6. ### WBahn Moderator

Mar 31, 2012
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That's good, because, in general, you can't code an arbitrary boolean function using only XOR (the NOT is extraneous because you CAN make a NOT out of an XOR).

What you need to do is try to rewrite your boolean function into combinations of the form (A'B+AB') or (AB+A'B'). The former is an XOR and the latter is an XNOR, which is an XOR with any odd number of inputs/outputs inverted.

If you are familiar with Karnaugh maps, those can help you identify the XOR/XNOR relationships.

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7. ### thermistor Thread Starter New Member

Mar 6, 2013
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I have tried to find a solution. I have tried to use http://www.eetimes.com/design/programmable-logic/4015081/Logic-101--Part-3--Reed-Muller-Logic/
in order to find a checkboard pattern in order to solve it but I have not found something. Can you help me find if this problem has solution in order not to try without purpose? I give the full truth table of the function that should be written using only NOT and EXOR gates.

INPUTS OUTPUT
A B C | E
--------------------
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0

What I have found is that the upper half table is B XNOR C (1) and the other
half A XOR B (2). But I can't find a way to tell if A=0 use function (1) else function (2).

8. ### WBahn Moderator

Mar 31, 2012
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I could be wrong, but I don't think that function can be done using XOR gates. I don't think there is the necessary antisymmetry across the mid line.

If it turns out that it can be done, please post the solution once you know it.

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9. ### thermistor Thread Starter New Member

Mar 6, 2013
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0
This subject was given in the entry examination for the National School of Public Administration. So I do not have the answer. I asked some other contenstants and they told me that they could not find a solution. So I suspect it is wrong too. I was just trying to solve it in order to confirm this fact. After spending a day to solve it and considering your answer, I give up. Thanks for your help.

10. ### MrChips Moderator

Oct 2, 2009
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Maybe the expression was incorrectly stated and should have been:

A'B'C' + A'BC + ABC' + AB'C

11. ### WBahn Moderator

Mar 31, 2012
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I was wondering if it shouldn't have been:

A'B'C'+A'BC+AB'C'+AB'C'

MrChips caught a typo above. I meant:

A'B'C'+A'BC+AB'C'+ABC

Which would be

A'(B'C'+BC)+A(B'C'+BC)

A XOR (B XNOR C)

Last edited: Mar 10, 2013
12. ### WBahn Moderator

Mar 31, 2012
23,214
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National School of Public Administration?

What kind of school is this? What is meant by "Public Administration"? Here, that term generally refers to government bureaucrats of one ilk or another. A foundation in logic, of any sort, is generally not high on their list of core competencies. But I'm guessing that it has a broader meaning where you are.

13. ### MrChips Moderator

Oct 2, 2009
17,372
5,366
You made a typo. Your last two terms are the same.

Here's my guess:

A'B'C'+A'BC+ABC'+AB'C

14. ### WBahn Moderator

Mar 31, 2012
23,214
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Yep, sure did. Fixed it in the post above. Thanks!