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Answer 1

+5 A:

Try putting it into a Karnaugh Map.

Peter K. 2010-04-14 00:53:52

+1 K-Maps are really simple, and once you get used to them you can easily reduce any boolean function of four or less variables.

BlueRaja - Danny Pflughoeft 2010-04-14 01:33:11

Karnaugh maps are great for this. I just worked this problem out myself, and it does simplify down a bit.

David Johnstone 2010-04-14 02:07:42

yeah, we're just starting to learn K-maps, but this HW requires us to use boolean algebra and not K-maps

xbonez 2010-04-14 02:51:28

@xbonez: But K-maps ARE boolean algebra, or at least another way of visualizing it. Aha! HW = HomeWork, not HW = HardWare. I see. Well, I advise looking at what the K-map is telling you about how to do the algebra.

Peter K. 2010-04-14 17:47:57

Answer 2

+3 A:

Quine-McCluskey reduction is one of the strongest tools for this, although it can be labor-intensive.

Ignacio Vazquez-Abrams 2010-04-14 00:54:16

Answer 3

A:

Like this:

x'y + wxy + w'y'z

RBarryYoung 2010-04-14 02:19:13

We generally don't just give answers to homework questions, but I guess this is okay, because you can actually simplify it more than this (assuming I didn't make a mistake, which is not always a safe assumption to make) :-)

David Johnstone 2010-04-14 02:30:55

Yes, I am pretty sure that my answer is still one step away from done. :-)

RBarryYoung 2010-04-14 02:39:00

Wow I really f'-ed my answer up - I made a mistake in my K-map. This answer is correct, but it can be reduced by removing exactly one literal, leaving 3 terms and 7 literals as required.

BlueRaja - Danny Pflughoeft 2010-04-14 02:40:56

yep. _ _ (stupid char min!)

RBarryYoung 2010-04-14 02:48:38

sadly, we're not allowed to use K-maps in this particular HW. we have to use boolean algebra.and I understand you guys dont give out answers to HW questions, and that is reasonable, but I actually tried working this one out for hours, and we even asked our recitation TA and she could not figure it out too

xbonez 2010-04-14 02:53:02

You're in for quite a trip if you're expecting to solve this using only boolean algebra laws. I would try double-negating the whole expression, but I don't even know if that will work...

BlueRaja - Danny Pflughoeft 2010-04-14 03:04:02

Well, I got it this far using only boolean algebra. Four variables is only 16 possible branches, so it's really not that hard.

RBarryYoung 2010-04-14 11:28:29

Just to summarize: I have given a helpful answer (eliminated 40% of the original complexity), I have not violated the homework rule, because I have not completed the original question (has 8 instead of 7 constants) and because I have not shown my work (which will obviously be required of the OP), I have done it entirely using the method specified for the problem (boolean algebra), and the OP has confirmed its correctness. If any of this is incorrect please let me know. If however, this summary is correct, then please advise as to why nobody, including the OP, thinks it worthy of an Upvote?

RBarryYoung 2010-04-14 11:43:56

@RBarryYoung: With four variables, there are 16 possible quantifications which can evaluate to true or false under a given expression, meaning there are 65536 possible expressions. I think that's pretty complex.

BlueRaja - Danny Pflughoeft 2010-04-14 13:21:38

@BlueRaja: Well it would actually be 3^16 because variables can also be negated in the terms. But either way, that is *NOT* how you approach this for boolean analysis, because you can evaluate the terms effectiveness separated from the whole expression (because of how OR works vs. AND for boolean evaluation). You're definitely making this too hard.

RBarryYoung 2010-04-15 02:17:09

@RBarryYoung: There are only 2^16 boolean functions of four variables (as noted by the fact that there are only 2^16 possible K-maps). Negating one of the 2^16 functions will give you another one of the 2^16 functions. You also cannot, as you say, evaluate the terms in isolation - your very answer is an example of where a literal can be removed only by looking at all three terms.

BlueRaja - Danny Pflughoeft 2010-04-15 02:59:11

Raja: OK, you're absolutely right. I must have enumerated everyone of your supposed 65,536 variations in my head without realizing it and solved this problem by accident. Thank-you so much for straightening me out so that I didn't keep thinking that I knew how to solve this problem that I already solved.

RBarryYoung 2010-04-19 01:41:46

Answer 4

+1 A:

I'm a bit rusty with boolean algebra, but I think I've worked out how to do this. I'll let you do the working, but here are the basic steps:

1) Group the terms with y and eliminate what you can inside the brackets. Once expanded again, this will leave you with four terms and ten literals.

2) Eliminate the redundant term, leaving you with three terms and seven literals.

Hint: I first worked out the answer with a Karnaugh map, and then used regular boolean algebra to get to the solution :-)

David Johnstone 2010-04-14 04:44:14

Agreed! "Cheat" by using the K-map first. ;-) +1 for ethical "cheating".

Peter K. 2010-04-14 17:48:43

Answer 5

A:

Can we use groups?

w'z(x' + y') + y(x' + w)

500 - Internal Server Error 2010-04-15 00:08:18

(Let me know if you want to know how I got there)

500 - Internal Server Error 2010-04-15 01:10:14

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