name for relation of this type
Ken Quirici
given any two distinct elements a,b, then aRb or bRa iff
there exists c such that aRc and bRc. c can be a or b but
need not be.
?
A simple example would be:
R is the simple substring relation where A is this subset of strings
of [English] alphabetic lower-case letters:
a,ab,abc,d,de,def
But ab,bc,abc does NOT have the property because
even tho ab R abc and bc R abc, it is not true that
ab R bc or bc R ab.
Regards,
ken
Arturo Magidin
> Is there a name for a relation R:A->A that has this property:
>
> given any two distinct elements a,b, then aRb or bRa iff
> there exists c such that aRc and bRc. c can be a or b but
> need not be.
>
> ?
It's the composition of R and R^{op}.
Given any relation R < A x B, R^{op} is the relation
R^{op} = { (b,a) in B x A | (a,b) in R }.
Given a relation R<A x B and a relation S<B x C, the composition of R
with S, SoR, is the relation
SoR = { (a,c) in A x C | there exists b in B with (a,b) in R and (b,c)
in T }.
--
Arturo Magidin
Ken Quirici
I assume you meant 'S' instead of 'T'?
The only problem is that, to paraphrase my favorite scene from
'Treasure of the Sierra Madre', 'there ain't no stinkin' B.
There's just an 'A'.
In the first example I gave, the set of strings
{a, ab, abc, d, de, def}, the relation R was
'substring', so R^ would be 'superstring'
and it would operate, just like R, on AxA?
so a R ab, ab R abc, d R de, de R def.
R^ would be
ab R^ a, abc R^ ac, de R^ d, and def R^ de?
I need to take baby steps here.
That's what you mean?
but the composition, let's call it R*, just says a R*a, ab R* ab, etc.
It seems the baby AND the bath-water have disappeared
namely the notions of sub and superstring, which I intended
to have content.
I'm missing something.
Regards,
Ken
Gerry Myerson
<48805a76-9985-49e6...@glegroupsg2000goo.googlegroups.com
>,
Ken Quirici <kqui...@yahoo.com> wrote:
And leaving out everything. All context, gone.
Post utterly incomprehensible on its own.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Ken Quirici
If you would Gerry. I appreciate your contribution so far.
Ken Quirici
The relation I was trying to define was hierarchical in some sense - if aRc and bRc then
aRb or bRa. That is, if there is a subordinate relation between a and c and b and c, then
there must be one between a and b.
Or maybe the relation I'm trying to define is a relation described as consisting of
non-intersecting chains.
Notice I said 'maybe'.
Ken Quirici
because the context that is getting left out here as my
post below indicates is the context of my examples.
This is my thread and I'll define the context thank
you very much :-)
Arturo Magidin
> Thanks Arturo.
>
> I assume you meant 'S' instead of 'T'?
Do learn to quote the messages you are replying to. Yes. The final one
was T.
>
> The only problem is that, to paraphrase my favorite scene from
> 'Treasure of the Sierra Madre', 'there ain't no stinkin' B.
>
> There's just an 'A'.
There is no problem. In mathematics, just because you say "let A and B
be sets" does not preclude the possibility that A=B.
The definition I gave was general and applies equally well to the case
where A=B.
Again:
Given any relation R < A x B, R^{op} is the relation
R^{op} = { (b,a) in B x A | (a,b) in R }.
Given a relation R<A x B and a relation S<B x C, the composition of R
with S, SoR, is the relation
SoR = { (a,c) in A x C | there exists b in B with (a,b) in R and (b,c)
in S}.
Now, where does it require A, B or C to be different? Newsflash!
Nowhere.
if A = B, the definition still applies, and it is precisely what lets
you compose R with R^{op}.
R^{op} o R = {(x,y) in A x A | there exists z in A such that (x,z) in
R and (z,y) in R^{op} } = {(x,y) in A x A | there exists z in A such
that (x,z) and (y,z) in R}.
This is *EXACTLY* the condition you describe. You want R to have this
condition, so you want R^{op} o R to be contained inside of R. That
is,
R^{op} o R <= R.
> In the first example I gave, the set of strings
> {a, ab, abc, d, de, def}, the relation R was
> 'substring', so R^ would be 'superstring'
> and it would operate, just like R, on AxA?
Your set A is {a, ab, abc, d, de, def}. This is also equal to the set
B.
The relation R is {(a,a), (a,ab), (a,abc), (ab,ab), (ab,abc),
(abc,abc), (d,d), (d,de), (d,def), (de,de), (de,def), (def,def)}.
The set R^op is therefore
R^{op} = { (a,a), (ab,a), (abc,a), (ab,ab), (abc,ab), (abc,abc),
(d,d), (de,d), (def,d), (de,de), (def,de), (def,def) }.
So R^{op} o R = {(a,a), (a,ab), (a,abc), (ab,a), (ab,ab), (ab,abc),
(abc,a), (abc,ab), (abc,abc), (d,d), (d,de), (d,def), (de,d), (de,de),
(de,def), (def,d), (def,de), (def,def)}
exactly "(x,y) in R^{op} o R if and only if x is a substring of y or y
is a substring of x".
--
Arturo Magidin
Arturo Magidin
> Actually I don't appreciate your contribution at all
> because the context that is getting left out here as my
> post below indicates is the context of my examples.
Actually, your utter inability to provide quotes renders your
monologues useless.
> This is my thread and I'll define the context thank
> you very much :-)
Right. Because it's better that it be yours and useless, than it be
useful, right?
Bye.
Michael Stemper
>Actually I don't appreciate your contribution at all
It's less than clear who you're addressing, since your post has
no quoted content, no "References:", no nuttin'
>because the context that is getting left out here as my
>post below indicates is the context of my examples.
Every that was "below" in your post is below this line.
>
>This is my thread and I'll define the context thank
>you very much :-)
That's not very useful context. Of course, if you're just tweeting
and not looking for anybody to answer whatever question you had,
that's your choice.
--
Michael F. Stemper
#include <Standard_Disclaimer>
Build a man a fire, and you warm him for a day. Set him on fire,
and you warm him for a lifetime.
Aatu Koskensilta
> Actually I don't appreciate your contribution at all because the
> context that is getting left out here as my post below indicates is
> the context of my examples.
What are you talking about?
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus