Computer exam question(s)

Dell19

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From my exam today, just want to check the answer

¬ = Not
x = And
+ = Or

F = ¬YxZ + (X + ¬Y)xZ

(X + ¬Y)xY has one long line above it negating this part of the expression.

I won't say the answer I got in case its wrong... Feel free to post your solutions, I have more questions to come if I get a response :)

Either simplify the expression or post a Karnaugh map.
 
You done anything on Boolean algebra?
 
F= "not" YxZ or XxY or "not" YxY
which leads me to the strange conclusion, that F does not equal any of those to the right of the equation sign. And that confuses me quite a lot

I am way off I think, but I am studying economics, so I am supposed to handle mathematics poorly :p
 
¬YxZ means NOT Y AND Z. so Y would be 0 and Z 1 for the statement to be true. Yeah we have done Boolean Algeba its just that I got a strange answer.
 
Ok - I'll have a go. I last did Boolean stuff over 30 years ago!
Gimme five mins. ;)
 
F= NOT Y and Z or NOT (X OR Not Y) and Y

Expand last bracket

F = Y'.Z + NOT(X.Y + Y'Y)
F= Y'Z + NOT (X.Y) since Y'Y = 0
F= Y'Z + (X'+Y') using de morgans theorem
F= Y'(Z+1) + X'
F= Y'+X'
or F = NOT (X.Y) using de morgans theorem again
 
I can see how you got your answer but it doesn't support my answer from splitting the formula into parts and calculating whether it would be a 1 or 0 depending on each of the 8 different combinations of X,Y and Z. Thanks anyway though :)

Just to be clear that we are talking about the same formula:
 

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Dell19 said:
¬YxZ means NOT Y AND Z. so Y would be 0 and Z 1 for the statement to be true. Yeah we have done Boolean Algeba its just that I got a strange answer.

So the x represented a multiplication sign?
Well that would explain why I made such a silly conclusion...
I thought the little x was a variable, different from the big X
->since putting XY together implies X multiplied by Y

edit:
from the looks of col's post, I assume he is using derivation, but I cant really tell what you guys are doing at all, since the < textarea > tags do not support ordinary mathematical symbols
 
oops I copied it down wrong - I've got a Y at the end instead of Z

Hmm - so have you in your first post.

Whixh version is correct Dell?
 
It should have been a Z... Sorry.

x represents AND, which in boolean algebra tends to function like multiplication but AxB means:

(Truth table with inputs A and B and the result of the AND operation AxB)
A B AxB
0 0 0
0 1 0
1 0 0
1 1 1
 
I think the second expression reduces to

F= X'+Y'+Z'
or
F= not (X.Y.Z)

where ' means inverse. + means OR . means AND.
 
Can you see where this truth table goes wrong?

1 = ¬Y.Z
2 = X + ¬Y
3 = 2.Z
4 = ¬4

X Y Z 1 2 3 4 F
0 0 0 0 0 0 1 1
0 0 1 1 0 0 1 1
0 1 0 0 0 0 1 1
0 1 1 0 0 0 1 1
1 0 0 0 1 0 1 1
1 0 1 1 1 1 0 1
1 1 0 0 0 0 1 1
1 1 1 0 0 0 1 1
 
Column 2 is wrong. You ANDED instead of ORed

should be 1 1 0 0 1 1 1 1
 
Oh. Good. For a little while there I wasnt sure if Boolean algebra was the English name for algebra...

That explains why I didnt get what you were talking about
ah yes - the Y' does not necessarily mean the derivative of Y, but could mean 1/Y...

This was a good learning experience for me. Unfortunately, now my head hurts :crazyeye:
 
yup it works now.

:smug:
 
Is column 3 also wrong then since it should be OR instead of AND and then there was no need for column 4?

As you might be able to tell I am now thorougl confused... Fortunately this was only about half of one of 5 questions on a paper that I had to get about 30% on to pass... And the first question was lots of number conversions which I am hopeful that they are correct.
 
I loved Boolean algebra - did it for A level Computing. Developed by George Boole in the 18th century. It deals with binary quantities and OR AND and NOT operations. Its much faster and neater than using Venn diagrams.

There are rules similar but not the same as 'ordinary' algebra.
 
col 1 01000100
col 2 11001111
col 3 01000101
col 4 10111010
F 11111110
 
Ah now I see where I went wrong on question two, I wrote down the right thing but did the wrong operation :mad:

It would be interesting to see whether I gain any marks for this question part for getting the rest right as the question was out of 8, although the Karnaugh map I did would be wrong and the simple equation would be wrong. Also there was another question worth 4 marks that was based on this one.
 
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