Re: Principle of Orthogonal Design

On 29 jan, 06:28, "Brian Selzer" <br...@xxxxxxxxxxxxxxxxxxx> wrote:
"Jan Hidders" <hidd...@xxxxxxxxx> wrote in message

news:12c08729-2bf7-4080-ba63-d3c90b8d95e0@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 28 jan, 22:26, "Brian Selzer" <br...@xxxxxxxxxxxxxxxxxxx> wrote:



"Jan Hidders" <hidd...@xxxxxxxxx> wrote in message

news:43bf804b-4ed0-493c-8499-2b3e8a83ee52@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 28 jan, 15:29, "Brian Selzer" <br...@xxxxxxxxxxxxxxxxxxx> wrote:
"Jan Hidders" <hidd...@xxxxxxxxx> wrote in message

news:0a872b75-448e-4131-ba5b-6bcee88da815@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

On 28 jan, 02:12, mAsterdam <mAster...@xxxxxxxxxxx> wrote:
Jan Hidders wrote:
mAsterdam wrote something very much like:
Pragmatical redefinitions must be temporary and tracked.
Sure, we agree on that.

<unsnip>

Wether the relation between heading and tuples goes
via names or ordering is relevant or not.

If it is not I want it out of scope.

</unsnip>

I don't think that it is possible to get it out of scope. If you
think
it is, then by all means provide an equivalent and complete
definition
where it is. I'm also not sure what your problem exactly is. We have
a
definition that works for the named perspective, which is arguably
the
most appropriate for the relational model anyway, so can we now
please, please, please, pretty please, move on with the discussion?

I don't think the definition is sufficient even for the named
perspective:

Certainly. I think I already said earlier that there is a stronger
version that removes even more redundancy. But I wanted to wait a
little until mAsterdam had gotten his head around the current one.

consider

R1 {J, K}
KEY {J}
KEY {K}

R2 {J, A}
KEY {J}
FOREIGN KEY {J} REFERENCES R1

R3 {K, A}
KEY {K}
FOREIGN KEY {K} REFERENCES R1

Supposing that J, K and A have different types and discounting any
meaning
attributed by relation names, there is overlap between R2 and R3.

J and K are both keys for R1, so J --> K and K --> J.

And due to the foreign keys between R2 and R1 and R3 and R1:

from J --> K and K --> A, J --> A can be inferred;
from K --> J and J --> A, K --> A can be inferred.

You can reason like that about FDs in the context of a single
relation, but you seem to do it here at schema level. What exactly
does it mean that J --> K holds at schema level? The only way I can
make sense of your statements is if you are working under the
universal relation assumption. Are you? Or are you perhaps assuming a
few extra dependencies you haven't told us about? Dependencies like
(R1 NJN R2)[K,A] = R3?

I didn't have the universal relation assumption in mind. The foreign key
constraints require that whenever there is a tuple in either R2 or R3,
there
must also be a tuple in R1, so since there must be a tuple in both R2 and
R1
or both R3 and R1, it stands to reason that

whenever R2 then R2 JOIN R1,
or
whenever R3 then R3 JOIN R1.

From that it follows that

the FDs from R2 and the FDs from R1 must also hold in R2 JOIN R1
and
the FDs from R3 and the FDs from R1 must also hold in R3 JOIN R1

because the foreign keys ensure that

(R2 JOIN R1) {J, A} = R2
and
(R3 JOIN R1) {K, A} = R3

That actually follows even if there are no inclusion dependencies /
foreign keys.
<<<<<<<<<<<<<

Actually, without the foreign key, (R2 JOIN R1) {J, A} is not necessarily
R2, since there could be one ore more tuples in R2 with a value for J that
does not appear in R1.

True, but that "that" in my sentence referred to the FDs holding in
the result of the natural joins.

But so, yes, it follows that K --> A holds on R2 JOIN R1 and J --> A holds
on R3 JOIN R1. Why then do you think it follows that if there are
tuples {j1, k1} in R1, {j1, a1} in R2, and {k1, a2} in R3, the
database is inconsistent? I really don't
see the contradiction here. Sure, we now have two A values associated
with a J value and a K value, but so what? That certainly doesn't
contradict the fact that K --> A holds on R2 JOIN R1 and J --> A holds
on R3 JOIN R1.

-- Jan Hidders
<<<<<<<<<<<<<<

Since K --> A holds in R2 JOIN R1 and also in R3 JOIN R1, shouldn't it hold
in (R2 JOIN R1) JOIN (R3 JOIN R1), too?  

It does.

But the information in the tuples,
{j1, k1}, {j1, a1}, {k1, a2} disappears when (R2 JOIN R1) is joined to (R3
JOIN R1), even though there is a tuple with the same value for J in each of
R2 and R1 and a tuple with the same value for K in each of R3 and R1.  The
information disappears because the database is inconsistent.

I have no idea what you mean by that. I don't see any inconsistencies.
And the information disappears because your are computing the
intersection of two different access paths from J/K to A and this
intersection happens to be empty. This is entirely valid. If there is
more than one way to get from A to B then these have apparently more
than one relationship, and there is no reason why these relationships
cannot be completely unrelated.

Perhaps a concrete example would be in order.  Suppose that you have an
EMPLOYEE relation with two keys, Badge# and SSN.  Suppose also that you have
a CLOCK relation that contains the Badge#, Date, TimeIn, TimeOut and Hours
for each employee that worked for each day.  Now suppose that you have a
PAYROLL relation that contains the SSN and Hours for each employee for each
day.  Note that in the PAYROLL relation, there is a tuple for each employee
for each workday even if there isn't one in the CLOCK relation.  An employee
may be on paid vacation, for example, and therefore should still get paid
for the hours he otherwise would have worked.

What you have is three relations, (other attributes not relevant to the
discussion omitted)

EMPLOYEE {Badge#, SSN}
    KEY {Badge#}, KEY {SSN}

CLOCK {Badge#, Date, TimeIn, TimeOut, Hours}
    KEY {Badge#, Date}
    FOREIGN KEY {Badge#} REFERENCES EMPLOYEE

PAYROLL {SSN, Date, Hours}
    KEY {SSN, Date}
    FOREIGN KEY {SSN} REFERENCES EMPLOYEE

Now, if the employee with badge number 2011 worked 9 hours on 1/28/08 as
represented in a tuple in CLOCK, then wouldn't it be a bad thing for there
to be a tuple in PAYROLL with that employee's SSN, '123-45-6789,'
indicating that he only worked 8 hours on 1/28/08?

If you don't want that you need to specify an extra dependency /
database constraint that states this. Probably an inclusion dependency
from the join of EMPLOYEE and CLOCK to PAYROLL. The ones that you
specified so far don't imply this. So I don't see how this supports
your point.

-- Jan Hidders
.

Relevant Pages

  • Re: Principle of Orthogonal Design
    ... FOREIGN KEY REFERENCES R1 ... EMPLOYEE relation with two keys, Badge# and SSN. ... PAYROLL relation that contains the SSN and Hours for each employee for each ...
    (comp.databases.theory)
  • Re: Principle of Orthogonal Design
    ... EMPLOYEE relation with two keys, Badge# and SSN. ... for each workday even if there isn't one in the CLOCK relation. ...
    (comp.databases.theory)
  • Re: Principle of Orthogonal Design
    ... EMPLOYEE relation with two keys, Badge# and SSN. ... for each workday even if there isn't one in the CLOCK relation. ...
    (comp.databases.theory)
  • Re: Principle of Orthogonal Design
    ... EMPLOYEE relation with two keys, Badge# and SSN. ... Note that in the PAYROLL relation, there is a tuple for each ... for each workday even if there isn't one in the CLOCK relation. ...
    (comp.databases.theory)
  • Re: Principle of Orthogonal Design
    ... EMPLOYEE relation with two keys, Badge# and SSN. ... Note that in the PAYROLL relation, there is a tuple for each ... for each workday even if there isn't one in the CLOCK relation. ...
    (comp.databases.theory)