Re: Principle of Orthogonal Design

On 29 jan, 13:33, "Brian Selzer" <br..._at_selzer-software.com> wrote:
> "Jan Hidders" <hidd..._at_gmail.com> wrote in message
>
> news:d4f7d641-ea3e-4c2b-8050-ae66c469416d_at_d4g2000prg.googlegroups.com...
> On 29 jan, 06:28, "Brian Selzer" <br..._at_selzer-software.com> wrote:
>
>
>
> > "Jan Hidders" <hidd..._at_gmail.com> wrote in message
>
> >news:12c08729-2bf7-4080-ba63-d3c90b8d95e0_at_1g2000hsl.googlegroups.com...
> > On 28 jan, 22:26, "Brian Selzer" <br..._at_selzer-software.com> wrote:
>
> > > > "Jan Hidders" <hidd..._at_gmail.com> wrote in message
>
> > > >news:43bf804b-4ed0-493c-8499-2b3e8a83ee52_at_z17g2000hsg.googlegroups.com...
> > > > On 28 jan, 15:29, "Brian Selzer" <br..._at_selzer-software.com> wrote:
> > > > > "Jan Hidders" <hidd..._at_gmail.com> wrote in message
>
> > > > >news:0a872b75-448e-4131-ba5b-6bcee88da815_at_e10g2000prf.googlegroups.com...
>
> > > > > > On 28 jan, 02:12, mAsterdam <mAster..._at_vrijdag.org> 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?
>
> (There's something up with my news reader: when I hit reply, it doesn't
> quote what you wrote.)
>
>
>
> 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
> <<<<<<<<<<<<<<<<<<<<<
>
> In the context of the existing inclusion dependencies, doesn't the value
> '123-45-6789' for SSN in PAYROLL represent the EMPLOYEE that exemplifies the
> values in the tuple {Badge# = 2011, SSN = '123-45-6789'}, and doesn't the
> value 2011 for Badge# in CLOCK represent that same EMPLOYEE?  

Yes.

> Wouldn't that
> mean that the tuple, {Badge# = 2011, Date = 1/28/08, Hours = 8} in the
> projection over CLOCK {Badge#, Date, Hours} means exactly the same thing as
> the tuple, {SSN = '123-45-6789', Date = 1/28/08, Hours = 8} in PAYROLL--that
> is, that the employee with Badge# 2011 /and/ SSN '123-45-6789' worked 8
> hours on 1/28/08?

Not necessarily. It could be that there are other ways of registering working hours outside the manufacturing plant or at home of the worker, and then it is allowed that the payed hours are more than those that are measured by the clock. If it does mean the same thing then you can model this by adding the corresponding inclusion dependencies.

• Jan Hidders