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  #21  
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Evgeniy Grigoriev
 
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Default Re: Very simple question to relational theorists. - 09-08-2012 , 02:07 PM






Thank you Roy. I think almost the same.

But I'm sure that name is not only psychological issue. Just one example (no RM in the one).

I can write X+Y to sum two values. But in modern programm languages I can use very complex names to denote the values. These names are full of meaning, but they are not only psychology, because system translates them into setof operation to obtain the denoted values from memory.

I think that the same process is possible for relations too. Just imagine -a system can analyzes complex names (I write about) and calculate the virtual relation from real relations according to the given complex names and some predefined rules. (Of course the system uses relational operation only in all the calculations.) In this case it's not necessary to define the calculation expression by hand. Only complex names have to be given which correspond to predefined rules. If I use other combination of name I'll get other result. So the rules are equal to definitions of many virtual relation.In this case complex names are full of meaning and also are used by systemto obtain denoted relations.

My question again Is this system relational?

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  #22  
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Roy Hann
 
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Default Re: Very simple question to relational theorists. - 09-09-2012 , 01:51 PM






Evgeniy Grigoriev wrote:

Quote:
Thank you Roy. I think almost the same.

But I'm sure that name is not only psychological issue. Just one
example (no RM in the one).

I can write X+Y to sum two values. But in modern programm languages
I can use very complex names to denote the values. These names are
full of meaning, but they are not only psychology, because system
translates them into set of operation to obtain the denoted values
from memory.
Sounds physical to me.

Quote:
I think that the same process is possible for relations too. Just
imagine - a system can analyzes complex names (I write about) and
calculate the virtual relation from real relations according to the
given complex names and some predefined rules. (Of course the system
uses relational operation only in all the calculations.) In this
case it's not necessary to define the calculation expression by
hand. Only complex names have to be given which correspond to
predefined rules. If I use other combination of name I'll get other
result. So the rules are equal to definitions of many virtual
relation. In this case complex names are full of meaning and also
are used by system to obtain denoted relations.

My question again Is this system relational?
Relational theory explicitly eschews physical concerns. Provided your
physical machinations don't corrupt the logic you are free to do
whatever you want.

But if the physical implementation intrudes even slightly on the logic
(the allowable operations on the symbols in the formal system) then no,
it would not be relational. In fact one struggles to think it could be
anything else but junk.

--
Roy

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  #23  
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Evgeniy Grigoriev
 
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Default Re: Very simple question to relational theorists. - 09-10-2012 , 10:56 PM



Hi Philip,

1) My name is Evgeniy (russian Eugene).

2) You write a big text, but whem I see through it, I understand that you think that I use domain operations - because all these lists (like "Nut,Bolt,Screw,Cam,Cog") anв boolean values (YES, NO) in fact are results of operaions on domains (they are denoted by you as formatiing functions). I really don't understand what you write about, Philip. Defenitely It's not about my question.

I'm about a play with names of relations and their attributes and about combinations of the names. No domain operations! Ok, imagine two relation

Bill (#No, Date)


Bill.Lines ( #No, #Art, Qty)
// here "Bill.Lines" is just string, just name of relation!

In second relation the name sequense "Bill.Lines.Qty" is defined among other as combination of relation name "Bill.Lines" and attribute name "Qty" . Other name sequence is "Bill.Date": it is defined in first relation.

Imagine that system can analise the name sequences according to some mnemonic rules and automatically calculate new virtual relations just by "correct" name combinations given in them . E.g. system can calculate the next virtual relation just by name sequances (I shoudn't write JOIN and RENAME manually to obtain it)

'Bill (#No, Lines.#Art, Lines.Qty)
// here the "Lines.#Art" and "Lines.Qty" are just attribute names after RENAMing of #Art and Qty.

or

'Bill (#No, Date, Lines.#Art)

They are result of JOIN (by #No) and RENAME operations on source Bill and "Bill.Lines". The system "knows" how the relation should be calculated because of some rules and given "correct" name squences. (E.g. I cannot automatically obtain

'Bill (#No, Date, #Art)

because it's out of rules - sequence "Bill.#Art" is not allowed.)

The main rule is "keep the name sequense".

Roy's answer is clear for me

"Relational theory explicitly eschews physical concerns. Provided your
physical machinations don't corrupt the logic you are free to do
whatever you want."

but I'm still not sure if my mashinations is not physical.

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  #24  
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compdb@hotmail.com
 
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Default Re: Very simple question to relational theorists. - 09-11-2012 , 03:47 AM



On Monday, September 10, 2012 8:56:06 PM UTC-7, Evgeniy Grigoriev wrote:
Quote:
1) My name is Evgeniy (russian Eugene).
Sorry, I do know that Evgeniy. Frequently play hockey with an Evgeniy, Russian accent included.

Quote:
2) You write a big text,
Cutting and pasting was a fast way to explain more.

Quote:
but whem I see through it, I understand that you think that I use domain operations
I know what you are doing. I am just showing you two other proposals that encode information in relations inappropriately/non-relationally. In that sense they are similar to yours. (Observe that both also involve encoding into headings. Although that is coincidental.) Try to understand what they aredoing, what I say their problems are and what I say should be done instead.. I will send a message about yours in particular. If you want to learn stop focusing on why you are right and start focusing on how you might be wrong.

On Thursday, September 6, 2012 1:53:18 AM UTC-7, Evgeniy Grigoriev wrote:
Quote:
The OMG and O/R camps do not understand the relational model; SQL objectextensions are partly not relational, partly messily relational.
Your proposals are essentially these.
No.
See the Third-Generation Database System Manifesto for your O/R mapping anddotted name sequences in SQL style queries. Yes your using those sequencesas attributes is unique.

Quote:
I really don't understand what you write about, Philip. Definitely It's not about my question.
The point is that your use of relations is non-relational in a similar way.Seek the similarities.

Quote:
Roy's answer is clear for me
"Relational theory explicitly eschews physical concerns. Provided your physical machinations don't corrupt the logic you are free to do whatever you want."
but I'm still not sure if my mashinations is not physical.
(An inadvertent pun. You mean "machinations". Look up "mash". I like it.)

When he writes about the "physical" he means "implementation" or "representation".

On Monday, September 10, 2012 12:53:36 PM UTC-7, (unknown) wrote:
Quote:
But these are not relational. They are inadequately abstracted representations of formatted output types. You too are essentially using relations to represent a relation-plus-labels type that would likely best merely contain rather than be relations.
philip

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  #25  
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Evgeniy Grigoriev
 
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Default Re: Very simple question to relational theorists. - 09-11-2012 , 05:41 AM



Thank you Philip.

Once again to be clear.

"So if the wff R(A,B,...) denotes the predicate for R then the wff for R RENAME A TO N is R(N,B,...). If the wff S(X,Y,...,A,...) denotes the predicate for S then the wff for (R JOIN S) RENAME A TO N is R(N,B,...) AND S(X,Y,...,N,...). "

If a system calculates (X,Y,...,N,...) automatically (according to some predefined mnemonic rules) as (R JOIN S) RENAME A TO N is R(N,B,...), the system is not relational.

(Yes/No)?

But if a system allow the S relational to be calculated as the same as (R JOIN S) RENAME A TO N is R(N,B,...) by hand - the system is relational.

(Yes/No)?

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  #26  
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Evgeniy Grigoriev
 
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Default Re: Very simple question to relational theorists. - 09-11-2012 , 05:51 AM



CORRECTION(Dont reaв previous message)

I've formulated my question in wrong way.

"So if the wff R(A,B,...) denotes the predicate for R then the wff for R RENAME A TO N is R(N,B,...). If the wff S(X,Y,...,A,...) denotes the predicate for S then the wff for (R JOIN S) RENAME A TO N is R(N,B,...) AND S(X,Y,....,N,...). "

To be correst, let me first ask you about virtual view which is (I hope) possible for you. Can I define some virtual view Q as result of (R JOIN S)? Ok, "the wff for for (R JOIN S) RENAME A TO N is R(N,B,...) AND S(X,Y,...,N,....)" but was is name "Q" here?

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  #27  
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compdb@hotmail.com
 
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Default Re: Very simple question to relational theorists. - 09-13-2012 , 05:57 AM



On Tuesday, September 11, 2012 3:51:37 AM UTC-7, Evgeniy Grigoriev wrote:

Quote:
"So if the wff R(A,B,...) denotes the predicate for R then the wff for R RENAME A TO N is R(N,B,...). If the wff S(X,Y,...,A,...) denotes the predicate for S then the wff for (R JOIN S) RENAME A TO N is R(N,B,...) AND S(X,Y,...,N,...). "

To be correst, let me first ask you about virtual view which is (I hope) possible for you. Can I define some virtual view Q as result of (R JOIN S)?Ok, "the wff for for (R JOIN S) RENAME A TO N is R(N,B,...) AND S(X,Y,...,N,...)" but was is name "Q" here?
Do you mean, what is a wff for the predicate of Q?

Q has an associated relational expression R JOIN S. So a wff for Q's associated predicate is R(A,B,...) AND S(X,Y,...,A,...). This is because predicate-of-(R JOIN S) is predicate-of-R AND predicate-of-S. (We were given wffs for these two predicates.)

philip

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  #28  
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Evgeniy Grigoriev
 
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Default Re: Very simple question to relational theorists. - 09-14-2012 , 02:34 AM



No I mean how the name Q links to the wff? Where a info exists that wff forQ is R(A,B,...) AND S(X,Y,...,A,...)? I think the wff is given by the factthat Q is R JOIN S (once again - I cannot imagine how it is possibly to operate on relations without operations on their predicates). But instead of"Q" i can use any string in any language or any value of any sign or anything else. There is no link between NAME of a virtual relation and WFF of the relation.

My questin is about how posiible names of virtual relation can be used. So my question is - why do you tell me about predicates?

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