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bool arith
Posted: Fri Feb 10, 2006 8:51 pm
by stocktroll
symbols:
b' = not b
a.b = a and b
a + b = a or b
This is what i thought should be a simple problem got me stumped since I cant simplify to the given answer...
Simplify the boolean expression:
a.b’.c’.d + a.b’.c.d + a.b.c’.d + a.b.c.d’ + a.b.c.d
the most i can simplify is a.b'.d + a.c'.d + a.b.c
but the answer given is: a.d + a.b.c
anyone see how this is so?
Posted: Fri Feb 10, 2006 9:39 pm
by losCHUNK
you cant add letters you stupid cawk
Posted: Fri Feb 10, 2006 9:42 pm
by Guest
what class do you learn that in?
Posted: Fri Feb 10, 2006 9:51 pm
by MKJ
why, thinking of enrolling after the answer was given in this thread ?
Posted: Fri Feb 10, 2006 10:08 pm
by jester!
ToxicBug wrote:what class do you learn that in?
I learned the simple stuff with that and Venn diagrams in Logic.
Posted: Fri Feb 10, 2006 10:42 pm
by stocktroll
lol the hell i remember academic crap as i cruise controlled throughout college but this is what i suppose you should learn intro discrete math and probably in get some in intro course about computer architecture.... its the basis of taking 1's and 0's in circuits and doing all the cool crap computers can do
btw any of you noobs got a clue if they can simplify to the given answer?
Posted: Sat Feb 11, 2006 12:56 am
by Guest
Sorry, I haven't done that branch of mathematics.
Posted: Sat Feb 11, 2006 3:53 am
by Pext
rather easy. maybe you should learn what boolean arithmetic actually is:
http://mathworld.wolfram.com/BooleanArithmetic.html
i'll leave the . out. it's just like you can leave them out whith multiplication.
btw: 1+1 = 0
b' = 1+b = b-1 or 1-b... whatever.
~1
a(1+b)(1+c)d
= a(1+b)d + a(1+b)cd
= ad + abd + acd + abcd
~2
a(1+b)cd
= acd + abdc
~3
ab(1+c)d
= abd + abcd
~4
abc(1-d)
= abc + abcd
~adding it all together:
ad + abd + acd + abcd
+ acd + abdc
+ abd + abcd
+abc + abcd
= ad + abc
q.e.d.
Posted: Sat Feb 11, 2006 4:02 am
by Guest
owned