Hi bildy-ga,
Here are the answers to the questions you requested:
1. In a small town, trains arrive at the train station every hour on
the hour every day of the week. They arrive on the half-hour, as well,
only during rush hours. They also arrive on the half-hour all day
Saturday but never on Sunday or holidays.
Assume the following variable assignments:
A = It is rush hour
B = It is Saturday
C =It is a holiday
D = It is Sunday
Write, in terms of A, B, C, and D, the Boolean Expression for
F = Trains arrive on the half-hour.
ANSWER:
F = C'(A+B)
This is basically stating that trains arrive on the half-hour as long
as it is not a holiday and either it is rush hour or a Saturday.
2.
Simplify the following Boolean Expressions algebraically. Show your
work.
a. A+ABC+A'BC+A'B
A + BC + A'B // combining ABC+A'BC as they are collectively
exhaustive
A + B(C + A') // factoring out a common term of B
b. (AB+C+D)(C'+D)(C'+D+E)
(AB+C+D)(C'+D+E) // removing redundant term
(AB+C)(C'+E)+D // factoring out common term of D
3. Put the following Boolean Expressions into Disjunctive Normal Form
(DNF)
a. A + B
ANSWER:
Assuming that this expression is only a function of A and B the DNF
form is:
f(A,B) = A(B+B')+(A+A')B
b. AB' + C(A' + B')
ANSWER:
f(A,B,C) = AB'(C+C') + A'(B+B')C + (A+A')B'C
4.Determine whether the following Boolean Expression can be rewritten
using only the XOR operator. If it can, rewrite it using only the XOR
operator.
A'B'C'D+A'B'CD'+A'BC'D'+AB'C'D'+ABCD'+ABC'D+AB'CD+A'BCD
ANSWER: Yes, the expression can be written only using the XOR operator
as follows:
A'B'C'D (XOR) A'B'CD' (XOR) A'BC'D' (XOR) AB'C'D' (XOR) ABCD' (XOR)
ABC'D (XOR) AB'CD (XOR) A'BCD
The reason this conversion is possible is that for any given values of
A,B,C,D at most one of the "ANDed" terms can be true. It is important
to notice, however, that although these terms are mutually exclusive,
they are NOT collectively exhaustive (there are 8 other combinations
that have not been represented in the original equation).
I hope you have found this helpful - since I know you needed the
actual answers quickly I aimed to provide a concise set of solutions
rather than a comprehensive one. However, please do post a
clarification if any of the answers require further explanation :)
Cheers!
answerguru-ga
Google Answers Researcher |
Request for Answer Clarification by
bildy-ga
on
30 Mar 2003 09:54 PST
I need a few clarifications as soon as possible:
1. Why isnt the term D = It is Sunday used in your boolean
expression, when i did this proble it looked as follows: F =
(C'+D')(A+B). Please clarify.
2. Same as mine, looks good.
3. Same as mine, looks good.
4. Shouldnt using the exclusive or operator simplify the equation
down.
Please get back to me asap
|
Clarification of Answer by
answerguru-ga
on
30 Mar 2003 10:30 PST
Hi again,
For the first question, I did not include a "D'" term like you did
because it seemed redundant. The reason for this is that from my
understanding, rush hour only occurs at certain times on weekdays. By
including rush hour as a term, Sunday is automatically excluded unless
it is stated explicitly (like Saturday in this case). Regardless, I
believe either statement can be deemed accurate based on the
interpretation of the term's real meaning.
For the fourth question, I'm not quite sure where you are having a
problem - did you understand the written explanation I provided under
the answer? Essentially, the reason XOR can be used here is because
XOR=OR as far as functionality in this equation. The reason for this
is that at most one of these terms is possible at any one time.
Hope this helps...
answerguru-ga
|
Request for Answer Clarification by
bildy-ga
on
30 Mar 2003 10:52 PST
Ok, I believe I understand where these answers are coming from. But
when i do the truth tables for #1, using your answer as well as mine,
I come up with two different answers. For your answer I get:
A B C D f
0000 0
0001 0
0010 0
0011 0
0100 1
0101 1
0110 0
0111 0
1000 1
1001 1
1010 0
1011 0
1100 1
1101 1
1110 0
1111 0
When you create a truth table from my boolean expresion, you come up
with the following:
A B C D f
0000 0
0001 0
0010 0
0011 0
0100 1
0101 1
0110 1
0111 0
1000 1
1001 1
1010 1
1011 0
1100 1
1101 1
1110 1
1111 0
Please verify these results, and let me know which one you feel to be
more correct. Thanks again for the clarification.
|
Request for Answer Clarification by
bildy-ga
on
30 Mar 2003 11:00 PST
Also, for #3, for the answer to be considered to be in Disjunctive
normal form, doesnt it need to be simplified further? Or is that
enough? Look forward to hearing back from you one these last 2
clarifications. Thanks
|
Clarification of Answer by
answerguru-ga
on
30 Mar 2003 11:18 PST
In response to your two previous requests:
The assumption I made for #1 does actually result in different truth
tables, and that is not a problem because we both interpreted the
question differently. As I said before, either answer is correct given
that the assumption made is stated clearly. The assumption I made to
arrive at my answer for this question resulted in a more simplified
answer, so don't think there really is a "most correct solution". The
safest bet would be to provide both solutions and explain why they
could both be right.
Disjunctive normal form is defined as meeting the following
conditions:
"A Boolean expression is in disjunctive normal form (DNF) if:
1. the variables within each term are ANDed together,
2. the terms are ORed together, and
3. every variable or its complement is represented in every term
(i.e. either A or ~A is in each term, B or ~B is in each term,
etc.).
4. No parentheses or other Boolean operations appear in the
expression."
http://mathforum.org/library/drmath/view/51857.html
Hope that clears everything up :)
answerguru-ga
|
Request for Answer Clarification by
bildy-ga
on
30 Mar 2003 11:30 PST
You have parentheses in your answers for this problem, which you state
are incorrect.
Please clarify
|
Clarification of Answer by
answerguru-ga
on
30 Mar 2003 11:51 PST
Yes, you're right - the expression just needs to be expanded one step further.
3. Put the following Boolean Expressions into Disjunctive Normal Form
(DNF)
a. A + B
ANSWER:
Assuming that this expression is only a function of A and B the DNF
form is:
f(A,B) = A(B+B')+(A+A')B
= AB + AB' + A'B
b. AB' + C(A' + B')
ANSWER:
f(A,B,C) = AB'(C+C') + A'(B+B')C + (A+A')B'C
= AB'C + AB'C' + A'BC + A'B'C
answerguru-ga
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