quote:

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Originally posted by GoldPenguin
what the fuck??????? :lol:

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what the fuck??????? :lol:

well no.2 i got as

R.( (B+Qa).(A+Qc) + B.Qa.Qc + Qb )

if thats any help

hope its right

i did this last year so it should be right as long as my memory aint playin tricks on me :o

in those questions above have you used + for an OR and . for an AND yeh mate?

I could have helped a year ago, but thankfully I've wiped all of DeMorgan's laws out of my mind :cool:

Commutative laws can seriously SMB.

Google for "Boolean algebra rules" or Boolean algebra laws

In the third one, i've used (+) as EXOR (exclusive OR)

Da = A.B.R.Qa.Qb + B.R.Qc + A.B.R.Qa + B.R.(Qa(+)Qc) + A.B.R.Qb

[Edited on 18-03-2004 by cdcool1]

the second one is:

Db = A.B.R.Qa.Qb.Qc + A.B.R.Qb.Qc + B.R.Qa.Qb.Qc + R.Qb

oops, i've simplified that to:

Dc = Qa.Qb.Qc + A.R.Qa + A.Qa + Qa.Qc

[Edited on 18-03-2004 by cdcool1]

[Edited on 18-03-2004 by cdcool1]

ok, this is what i've got, more or less straight from the karnaugh (sp) mapping

i know that i can group some, but i think there are some simplifications that can be done other than grouping:

ok, there is no way i can do a "NOT" function in the normal way, so i've used an underline, instead of a line on the top.....

Qa is one term, Qb is one term, Qc is one term

DC = A.R.Qa.Qb.Qc + A.R.Qa.Qb.Qc + A.R.Qa + A.Qa + Qa.Qc

thats the first one

post em up

just looked Boolean Algebra up on google, aint got a clue :( sorryyy

Is anyone any good at it?? I've got 3 different equations which i've simplified as much as i can and i cant get them any further but i'm sure they will simplify more.

Cant be arsed to post them up if no-one knows it, so if someone does, then i'll type them up

:thumbs: