cdcool1
Member
Registered: 9th Jun 02
Location: Scunny
User status: Offline
|
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
|
Craig6682
Member
Registered: 8th Apr 03
User status: Offline
|
I have to do this at uni but i haven't got a clue. Gonna have to blag the exam some how 
Sorry mate!
|
RichR
Premium Member
Registered: 17th Oct 01
Location: Waterhouses, Staffordshire
User status: Offline
|
as in boolean equations - I mightbe able to help - did them last year as part of my engineering maths - so if u can be arsed stick them up I might be arsed give them a go
|
Dan B
Member
Registered: 25th Feb 01
User status: Offline
|
Boolean logic, I got asked the following question for my interview for my current job:
Source IP: 1.2.3.4
Subnet: 255.255.255.240
Gateway:1.2.3.10
Destination: 1.2.3.20
Using Boolean logic, set up a logic-statement of the following format to prove if the destination is on the same subnet as the source:
if ( <something> = <something else> )
then: destination is local
otherwise: destination is remote
Solved it, eventually, despite not having done Boolean for about 5 years! 
|
cdcool1
Member
Registered: 9th Jun 02
Location: Scunny
User status: Offline
|
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 = Qa.Qb.Qc + A.R.Qa + A.Qa + Qa.Qc
Db = A.B.R.Qa.Qb.Qc + A.B.R.Qb.Qc + B.R.Qa.Qb.Qc + R.Qb
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
|
cdcool1
Member
Registered: 9th Jun 02
Location: Scunny
User status: Offline
|
and then there was silence
|
cdcool1
Member
Registered: 9th Jun 02
Location: Scunny
User status: Offline
|
ttt
|
groom
Member
Registered: 19th Apr 03
Location: In front of my pc
User status: Offline
|
quote:
Originally posted 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 = Qa.Qb.Qc + A.R.Qa + A.Qa + Qa.Qc
Db = A.B.R.Qa.Qb.Qc + A.B.R.Qb.Qc + B.R.Qa.Qb.Qc + R.Qb
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
6?
|
cdcool1
Member
Registered: 9th Jun 02
Location: Scunny
User status: Offline
|
almost
i've found a program on the net that does it all for you and its freeware too
cant install it on uni computers tho, so will do it when i get home
|
AdiSRI
Member
Registered: 1st May 02
Location: Berkshire
User status: Offline
|
i did that in my first year at uni, cant remember anything about it now tho.
|
NormSXi
Member
Registered: 26th Sep 03
User status: Offline
|
You done it yet then?? I might have a go 2moro at/after work mate, i think i can still remember stuff, there are obvious groupings, the K - map should do most of the minimisation tbh..
|
vibrio
Banned
Registered: 28th Feb 01
Location: POAH
User status: Offline
|
I'll stick to my DNA and proteins
|
