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| DATE | 2014-12-10 |
| FROM | Ruben Safir
|
| SUBJECT | Subject: [LIU Comp Sci] HW 6
|
From owner-learn-outgoing-at-mrbrklyn.com Wed Dec 10 23:58:51 2014
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Date: Wed, 10 Dec 2014 23:58:55 -0500
From: Ruben Safir
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To: learn-at-nylxs.com, Ping-Tsai Chung ,
"ptchung-at-ieee.org" ,
"pingtsaichung-at-gmail.com >> Ping-Tsai Chung"
Subject: [LIU Comp Sci] HW 6
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With regard to Problem 1E in the homework
What is this? This can't be right. I don't know what is right, but
this aint it.
In order to do this as division the divisor has to be a SUBSET of of the
numerator.
(e) Retrieve the names of employees who
work on every project.
PROJ_EMPS(PNO,SSN) ? ? PNO,ESSN (WORKS_ON)
ALL_PROJS(PNO) ? ?PNUMBER (PROJECT)
EMPS_ALL_PROJS ? PROJ_EMPS รท ALLPROJS
RESULT ? ? FNAME,MINIT,LNAME (EMPLOYEE
* EMP_ALL_PROJS)
RESULT:
No Data Found.
To quote the text, and this is honestly a weird operation
Produces a relation R(X) that includes all tuples t[X]
in R1(Z) that appear in R1 in combination with every
tuple from R2(Y), where Z = X ? Y.
That means Y MUST BE a SUBSET of Z...
We are trying to find the one the single individal who is all in ALL the
project, and all the other individual lke him.
I have no idea how to solve this.
In general set theory terms,
we want
Inviduals who's Set of pno is equal to the set of pnumbers in the
works_on table
You can get the sets?
? pno (sigma essn = SSN (works_on))
where each SSn is in the set of
? ssn (employee)
where the set of pno is complete.
Don't ask me how to express this. I have no idea and don't see it in
the books or on line.
FRANKLY, this problem is not done with relational agrebra. This is
where you DITCH SQL and use a peal loop. I don't know of any way of
doing conditional looping in set theory.
You might be able to compare to sets, but that is not covered in the
book here.
In sql I would use count
select employee.ssn from employee, works_on
where employee.ssn = works_on.essn
having count( works_on.essn) == subquery(
select count (pnumber) from works_on )
That might not be legl either. I don't think you can put a count in a
having without a group by claus?
Ruben
Anyway - I am Ccing Dr Ping on this. Maybe he can shed light on this
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