The Computable Reals (alpha version)

On Sep 2, 8:26 am, ju...@diegidio.name wrote:
> On 31 Aug, 19:49, ju...@diegidio.name wrote:
>
>
>
> > On 31 Aug, 18:33, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > ju...@diegidio.name writes:
> > > > On 30 Aug, 00:04, Barb Knox <s...@sig.below> wrote:
>
> > > > > So, given such an oracle, you can easily COMPUTE each digit of an
> > > > > anti-diagonal D.
>
> > > > > Got it now?
>
> > > > I am under the impression that it is you here off mark.
>
> > > > "So, given such an oracle, you can easily COMPUTE each digit of an
> > > > anti-diagonal D."
>
> > > > True, but the point is: the anti-diagonal is still in the list!
>
> > > > We can surely prove this by induction over a computable list.
>
> > > Well, if you think this is obvious, go on then and actually *do* it!
>
> > It's been under everybody's eyes for weeks now. If you cannot even
> > "see" it, then I must be putting it too "unmathematically". If you are
> > serious in this question (I have to ask) I can try to make it as
> > explicit as I can, after all is an argument by induction, but should
> > already be implied by the construction in beta version.
>
> > Otherwise I'd have these questions for you: Can you give an inductive
> > definition for "the" list? Can you then give an inductive definition
> > for the anti-diagonal of "the" list? Can you finally show that such
> > anti-diagonal is *not* a member of "the" list?
>
> An inductive definition for "the" list is in construction beta,
> namely, it is the list of all binary sequences over finite induction.
>
> Now, that list is obviously non-square, its size being in the order of
> n columns (the lenght of the sequences at each step) times 2^n rows
> (all combinations of the digits over the lenght at that step --
> although there is no need to calculate combinations here).
>
> Now, there are two ways of defining a "diagonal" over this list. One
> is to take the diagonal starting from the "upper-left corner" and stop
> the sequence at the "right border" (so, the diagonal has the same
> length as any sequence in the list); the other is take the diagonal
> and prolong it down to the "bottom" by wrapping around at the "right
> border" (that is, with a bit of modular arithmetic).
>
> The first option looks like -- say for n=3:
>
> list(3) = <
> [0] [0]0 0 ,
> [1] 0[0]1 ,
> [2] 0 1[0],
> [3] 0 1 1 ,
> [4] 1 0 0 ,
> [5] 1 0 1 ,
> [6] 1 1 0 ,
> [7] 1 1 1
> >
>
> Thus, the diagonal is the sequence: 0,0,0.
>
> Then, the anti-diagonal (by the mapping 0->1, 1->0) is the sequence:
> 1,1,1, which of course is a sequence of the list.
>
> The second option looks like -- again, for n=3:
>
> list(3) = <
> [0] [0]0 0 ,
> [1] 0[0]1 ,
> [2] 0 1[0],
> [3] [0]1 1 ,
> [4] 1[0]0 ,
> [5] 1 0[1],
> [6] [1]1 0 ,
> [7] 1[1]1
> >
>
> Thus, the diagonal is the sequence: 0,0,0,0,0,1,1,1.
>
> Then, the anti-diagonal (again, by the mapping 0->1, 1->0) is the
> sequence: 1,1,1,1,1,0,0,0, which of course is *not* a sequence of the
> list; however, it is a sequence of 'list(2^3)'.
>
> > > > With an uncomputable list but an oracle, I wouldn't know how it would
> > > > look like. Thanks for the puzzle.
>
> Still thinking about this.
>
> -LV

I'm not sure what you're trying to show with the construction but the
enumeration you're referring to should for instance be the
concatenation of list(k) for all k. I'm still a bit uncertain of
whether or not you still want the TM to output periods since you seem
to be referring to different ideas at different points. The one you
gave in the previous post, for example, contains only the dyadic
rationals, not every rational, but for any real in [0,1] and e > 0
there are numbers in your enumeration within e of that real.

Here are three lists with indicated diagonals, where I for simplicity
mark the relevant digit in the period.

L_a:
[0](0)
1([0])
00([0])
01([0]) (i.e. 010[0]000...)
10([0]) (i.e. 1000[0]000...)
11([0])
000([0])
....
Anti-diagonal: 1(1), not in the list.

L_b:
[0](0)
0([1])
1([0])
1([1])
00([0])
....
11([0])
11([1])
....
Anti-diagonal: 10(10), not in the list.

L_c:
[0](0)
0([1])
1([0])
1([1])
00([0])
00([1])
00([0]0)
00(0[1])
00([1]0)
00(1[1])
01([0])
....
000([0])
....
000(0[0]0)
....
Anti-diagonal: 10101010001010001010001010001011101010110111010101 ...
(i think). Is this anti-diagional in L_c? It is not clear by
inspection, but we can easily replicate Cantor's argument.

On 2 Sep, 08:50, Cenny Wenner <CWen...@gmail.com> wrote:
> On Sep 2, 8:26 am, ju...@diegidio.name wrote:
> > On 31 Aug, 19:49, ju...@diegidio.name wrote:
> > > On 31 Aug, 18:33, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > ju...@diegidio.name writes:
> > > > > On 30 Aug, 00:04, Barb Knox <s...@sig.below> wrote:
>
> > > > > > So, given such an oracle, you can easily COMPUTE each digit of an
> > > > > > anti-diagonal D.
>
> > > > > > Got it now?
>
> > > > > I am under the impression that it is you here off mark.
>
> > > > > "So, given such an oracle, you can easily COMPUTE each digit of an
> > > > > anti-diagonal D."
>
> > > > > True, but the point is: the anti-diagonal is still in the list!
>
> > > > > We can surely prove this by induction over a computable list.
>
> > > > Well, if you think this is obvious, go on then and actually *do* it!
>
> > > It's been under everybody's eyes for weeks now. If you cannot even
> > > "see" it, then I must be putting it too "unmathematically". If you are
> > > serious in this question (I have to ask) I can try to make it as
> > > explicit as I can, after all is an argument by induction, but should
> > > already be implied by the construction in beta version.
>
> > > Otherwise I'd have these questions for you: Can you give an inductive
> > > definition for "the" list? Can you then give an inductive definition
> > > for the anti-diagonal of "the" list? Can you finally show that such
> > > anti-diagonal is *not* a member of "the" list?
>
> > An inductive definition for "the" list is in construction beta,
> > namely, it is the list of all binary sequences over finite induction.
>
> > Now, that list is obviously non-square, its size being in the order of
> > n columns (the lenght of the sequences at each step) times 2^n rows
> > (all combinations of the digits over the lenght at that step --
> > although there is no need to calculate combinations here).
>
> > Now, there are two ways of defining a "diagonal" over this list. One
> > is to take the diagonal starting from the "upper-left corner" and stop
> > the sequence at the "right border" (so, the diagonal has the same
> > length as any sequence in the list); the other is take the diagonal
> > and prolong it down to the "bottom" by wrapping around at the "right
> > border" (that is, with a bit of modular arithmetic).
>
> > The first option looks like -- say for n=3:
>
> > list(3) = <
> > * * [0] *[0]0 0 ,
> > * * [1] * 0[0]1 ,
> > * * [2] * 0 1[0],
> > * * [3] * 0 1 1 ,
> > * * [4] * 1 0 0 ,
> > * * [5] * 1 0 1 ,
> > * * [6] * 1 1 0 ,
> > * * [7] * 1 1 1
>
> > Thus, the diagonal is the sequence: 0,0,0.
>
> > Then, the anti-diagonal (by the mapping 0->1, 1->0) is the sequence:
> > 1,1,1, which of course is a sequence of the list.
>
> > The second option looks like -- again, for n=3:
>
> > list(3) = <
> > * * [0] *[0]0 0 ,
> > * * [1] * 0[0]1 ,
> > * * [2] * 0 1[0],
> > * * [3] *[0]1 1 ,
> > * * [4] * 1[0]0 ,
> > * * [5] * 1 0[1],
> > * * [6] *[1]1 0 ,
> > * * [7] * 1[1]1
>
> > Thus, the diagonal is the sequence: 0,0,0,0,0,1,1,1.
>
> > Then, the anti-diagonal (again, by the mapping 0->1, 1->0) is the
> > sequence: 1,1,1,1,1,0,0,0, which of course is *not* a sequence of the
> > list; however, it is a sequence of 'list(2^3)'.
>
> > > > > With an uncomputable list but an oracle, I wouldn't know how it would
> > > > > look like. Thanks for the puzzle.
>
> > Still thinking about this.
>
> > -LV
>
> I'm not sure what you're trying to show with the construction

Obviously.

> but the
> enumeration you're referring to should for instance be the
> concatenation of list(k) for all k. I'm still a bit uncertain of
> whether or not you still want the TM to output periods since you seem
> to be referring to different ideas at different points. The one you
> gave in the previous post, for example, contains only the dyadic
> rationals, not every rational, but for any real in [0,1] and e > 0
> there are numbers in your enumeration within e of that real.
>
> Here are three lists with indicated diagonals, where I for simplicity
> mark the relevant digit in the period.
>
> L_a:
> [0](0)
> 1([0])
> 00([0])
> 01([0]) (i.e. 010[0]000...)
> 10([0]) (i.e. 1000[0]000...)
> 11([0])
> 000([0])
> ...
> Anti-diagonal: 1(1), not in the list.

Wrong, you just have -- as usual -- an undefined list. (And, of
course, you plainly ignore that, among other things, the list I have
given can be interpreted straight away as a list of periods, and then
we adopt a compound approach to express the computable numbers. But
let's assume I again am saying nothing at all here.)

So, the infinite list you are after *might* look like (as I have
already and of course and repeatedly shown):

+00)
+1:1(0)
+2:01(0)
+3:11(0)
+4:001(0)
....
-4:110(1)
-3:00(1)
-2:10(1)
-1:0(1)
-01)

And of course all anti-diagonals keep being into the list.

[I say "might" because nobody up to now has been able to give an
infinite recursion -- if that even makes sense. Anyway, this list is
just equivalent to the one over finite induction as per version beta,
modulo a remapping of the order (namely, here we order
lexicographically but right to left). We need transfinite induction
here, though.]

In any case, as of course it must be, any anti-diagonal is still there
and always will be.

> (i think). Is this anti-diagional in L_c? It is not clear by
> inspection, but we can easily replicate Cantor's argument.

Sure. I keep repeating the same arguments again and again. You keep
with your undefined notions again and again. And Cantor keeps proving
the whole thing in any case. Sure.

In the meantime, and again:

"Cantor's approach is like the oracle that spells out digits coming
straight from the hereafter. Then, however you define your anti-
diagonal sequence, you have no way to prove that it is not a sequence
already in the list -- that is a sequence that the oracle cannot
produce, as that would amount to solving the halting problem. Oops,
another disproof."

Have fun with this one too, which happens not to depend on any
construction or induction of any sort.

-LV

julio@diegidio.name writes:

> On 2 Sep, 01:42, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> >
> > Alan asked if you had a program that took (n,m) and output the n'th digit
> > of the m'th computable real. *When Alan asked this, the phrase "the m'th
> > computable real" meant "the m'th computable real from a COMPLETE LIST OF
> > ALL COMPUTABLE REALS".
>
> What Alan is asking does simply make no sense in this context.

Given *any* effective listing of computable reals, you should be
able to compute the nth digit of the mth computable real on the list;
otherwise you simply don't have an effective listing of computable reals.

Why you are unable to provide us with this, I simply don't know;
it's got to be easy, from everything you have posted so far;
if you can't do it, that's your problem.

>
> -LV

--
Alan Smaill

On Sep 2, 10:22 am, ju...@diegidio.name wrote:
> On 2 Sep, 08:50, Cenny Wenner <CWen...@gmail.com> wrote:
>
>
>
> > On Sep 2, 8:26 am, ju...@diegidio.name wrote:
> > > On 31 Aug, 19:49, ju...@diegidio.name wrote:
> > > > On 31 Aug, 18:33, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > > ju...@diegidio.name writes:
> > > > > > On 30 Aug, 00:04, Barb Knox <s...@sig.below> wrote:
>
> > > > > > > So, given such an oracle, you can easily COMPUTE each digit of an
> > > > > > > anti-diagonal D.
>
> > > > > > > Got it now?
>
> > > > > > I am under the impression that it is you here off mark.
>
> > > > > > "So, given such an oracle, you can easily COMPUTE each digit of an
> > > > > > anti-diagonal D."
>
> > > > > > True, but the point is: the anti-diagonal is still in the list!
>
> > > > > > We can surely prove this by induction over a computable list.
>
> > > > > Well, if you think this is obvious, go on then and actually *do* it!
>
> > > > It's been under everybody's eyes for weeks now. If you cannot even
> > > > "see" it, then I must be putting it too "unmathematically". If you are
> > > > serious in this question (I have to ask) I can try to make it as
> > > > explicit as I can, after all is an argument by induction, but should
> > > > already be implied by the construction in beta version.
>
> > > > Otherwise I'd have these questions for you: Can you give an inductive
> > > > definition for "the" list? Can you then give an inductive definition
> > > > for the anti-diagonal of "the" list? Can you finally show that such
> > > > anti-diagonal is *not* a member of "the" list?
>
> > > An inductive definition for "the" list is in construction beta,
> > > namely, it is the list of all binary sequences over finite induction.
>
> > > Now, that list is obviously non-square, its size being in the order of
> > > n columns (the lenght of the sequences at each step) times 2^n rows
> > > (all combinations of the digits over the lenght at that step --
> > > although there is no need to calculate combinations here).
>
> > > Now, there are two ways of defining a "diagonal" over this list. One
> > > is to take the diagonal starting from the "upper-left corner" and stop
> > > the sequence at the "right border" (so, the diagonal has the same
> > > length as any sequence in the list); the other is take the diagonal
> > > and prolong it down to the "bottom" by wrapping around at the "right
> > > border" (that is, with a bit of modular arithmetic).
>
> > > The first option looks like -- say for n=3:
>
> > > list(3) = <
> > > [0] [0]0 0 ,
> > > [1] 0[0]1 ,
> > > [2] 0 1[0],
> > > [3] 0 1 1 ,
> > > [4] 1 0 0 ,
> > > [5] 1 0 1 ,
> > > [6] 1 1 0 ,
> > > [7] 1 1 1
>
> > > Thus, the diagonal is the sequence: 0,0,0.
>
> > > Then, the anti-diagonal (by the mapping 0->1, 1->0) is the sequence:
> > > 1,1,1, which of course is a sequence of the list.
>
> > > The second option looks like -- again, for n=3:
>
> > > list(3) = <
> > > [0] [0]0 0 ,
> > > [1] 0[0]1 ,
> > > [2] 0 1[0],
> > > [3] [0]1 1 ,
> > > [4] 1[0]0 ,
> > > [5] 1 0[1],
> > > [6] [1]1 0 ,
> > > [7] 1[1]1
>
> > > Thus, the diagonal is the sequence: 0,0,0,0,0,1,1,1.
>
> > > Then, the anti-diagonal (again, by the mapping 0->1, 1->0) is the
> > > sequence: 1,1,1,1,1,0,0,0, which of course is *not* a sequence of the
> > > list; however, it is a sequence of 'list(2^3)'.
>
> > > > > > With an uncomputable list but an oracle, I wouldn't know how it would
> > > > > > look like. Thanks for the puzzle.
>
> > > Still thinking about this.
>
> > > -LV
>
> > I'm not sure what you're trying to show with the construction
>
> Obviously.

Perhaps you feel like clarifying?

>
>
>
> > but the
> > enumeration you're referring to should for instance be the
> > concatenation of list(k) for all k. I'm still a bit uncertain of
> > whether or not you still want the TM to output periods since you seem
> > to be referring to different ideas at different points. The one you
> > gave in the previous post, for example, contains only the dyadic
> > rationals, not every rational, but for any real in [0,1] and e > 0
> > there are numbers in your enumeration within e of that real.
>
> > Here are three lists with indicated diagonals, where I for simplicity
> > mark the relevant digit in the period.
>
> > L_a:
> > [0](0)
> > 1([0])
> > 00([0])
> > 01([0]) (i.e. 010[0]000...)
> > 10([0]) (i.e. 1000[0]000...)
> > 11([0])
> > 000([0])
> > ...
> > Anti-diagonal: 1(1), not in the list.
>
> Wrong, you just have -- as usual -- an undefined list. (And, of
> course, you plainly ignore that, among other things, the list I have
> given can be interpreted straight away as a list of periods, and then
> we adopt a compound approach to express the computable numbers. But
> let's assume I again am saying nothing at all here.)
>
> So, the infinite list you are after *might* look like (as I have
> already and of course and repeatedly shown):
>
> +00)
> +1:1(0)
> +2:01(0)
> +3:11(0)
> +4:001(0)
> ...
> -4:110(1)
> -3:00(1)
> -2:10(1)
> -1:0(1)
> -01)

This is one of the many reasons why you should be clear what about
what you're trying to say. The fact that the above is treated as a
list is because it is an enumerable set with a bijection from N (or a
subset thereof) to your set, e.g. iterleaving the two lists. As, I
think Virgil earlier said, any such choice of a bijection yields a
number not in the list through Cantor's. One such choice yields L_b
which, as stated before, has an anti-diagonal 10(10), which isn't in
the above set. In an earlier post, it was suggested to you how to
cover all of the rationals which would cover 10(10) but, e.g., leave
you with L_c which has another anti-diagonal.

(there is a detail left out due to the fact that we're dealing with
reals rather than binary strings though, but I imagine you feel like
making this stronger claim)

>
> And of course all anti-diagonals keep being into the list.
>
> [I say "might" because nobody up to now has been able to give an
> infinite recursion -- if that even makes sense. Anyway, this list is
> just equivalent to the one over finite induction as per version beta,
> modulo a remapping of the order (namely, here we order
> lexicographically but right to left). We need transfinite induction
> here, though.]
>
> In any case, as of course it must be, any anti-diagonal is still there
> and always will be.
>
> > (i think). Is this anti-diagional in L_c? It is not clear by
> > inspection, but we can easily replicate Cantor's argument.
>
> Sure. I keep repeating the same arguments again and again. You keep
> with your undefined notions again and again. And Cantor keeps proving
> the whole thing in any case. Sure.
>
> In the meantime, and again:.
>
> "Cantor's approach is like the oracle that spells out digits coming
> straight from the hereafter. Then, however you define your anti-
> diagonal sequence, you have no way to prove that it is not a sequence
> already in the list -- that is a sequence that the oracle cannot
> produce, as that would amount to solving the halting problem. Oops,
> another disproof."
>
> Have fun with this one too, which happens not to depend on any
> construction or induction of any sort.
>
> -LV

Can you prove that the list { 2k }_{k=0}^inf = 0,2,4,6,... does not
contain the number 3?

julio@diegidio.name writes:

> On 1 Sep, 15:29, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > ju...@diegidio.name writes:
> > > On 1 Sep, 10:31, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > ju...@diegidio.name writes:

> > > > > It's been under everybody's eyes for weeks now.
> >
> > > > You said "We can surely prove this ...", not "I have proved it".
> > > > There is a difference, isn't there? No proof so far ...
> >
> > > There is a difference if you take it literally.
> >
> > I was taking you at your word.
>
> So, despite my clarification, you couldn't care less to go beyond your
> plain literal and wrong interpretation.

If you mean that you want to withdraw your claim that "we can surely
prove this", then by all means do so.

Are you withdrawing your claim?

> > > Still that's simply
> > > not what I meant. What I meant is: I have given this as all other
> > > arguments under all possible shapes already, but most of what I get
> > > back is "where it even is??" so that it is quite hard to improve at
> > > all.
> >
> > When you say "We can surely prove ...", then I expect you have an idea
> > of what a proof would look like.
>
> Of course, so you confirm your objections here are just a waste of
> time. You don't even know what you are complaining about.

It's true that I don't know even the statement of what you might
have thought you could prove -- I did imagine that *you*
knew what you were talking about, though.

> > > > > If you cannot even
> > > > > "see" it, then I must be putting it too "unmathematically". If you
> > > > > are serious in this question (I have to ask) I can try to make it as
> > > > > explicit as I can, after all is an argument by induction, but should
> > > > > already be implied by the construction in beta version.
> >
> > > > No, I do not see it from what you have given so far.
> >
> > > Why not ask then? I am even inviting you.
> >
> > There is a difference between writing a program, and proving that the
> > program has a particular property. *It looked like you wanted to *prove*
> > that the anti-diagonal is still in the list. *I'm inviting you to
> > supply that proof.
>
> Cool. Let's see if I feel like it... no, really.

no, indeed ...

> > > > > Otherwise I'd have these questions for you: Can you give an inductive
> > > > > definition for "the" list?
> > > > > Can you then give an inductive definition
> > > > > for the anti-diagonal of "the" list? Can you finally show that such
> > > > > anti-diagonal is *not* a member of "the" list?
> >
> > > > You are the one makuing claims here, not me;
> > > > you are the one who said "We can surely prove this".
> >
> > > > How about just doing it??
> >
> > > I see, this is a too polite way to ask, isn't it? Just guessing, as my
> > > English is still veeery rookey.
> >
> > I thought your mother warned you about the communists?
>
> Mentioning my mum is worth more than a simple punch in the nose around
> here. Anyway, I thought you knew better. I was mistaken.

You were, indeed.

> > > > > > Can you give us a program that takes two natural numbers (n,m) as
> > > > > > inout, and outputs the nth digit in the mth computable real in your
> > > > > > listing of computable reals?
> > > > > > That might advance the discussion.
> >
> > > > > At the moment, I'd say I can surely give a program that takes *three*
> > > > > natural numbers (n,m,w) as input, and outputs the nth digit
> > > > > (character) of the mth expansion (string) of the wth list (the list
> > > > > of expansions whose length is w). That is because (in beta version)
> > > > > I end up having the list of the rationals.
> >
> > > > Not good enough, if you want a computational version of the list of
> > > > all computable reals. *Why on earth is it a problem to give what I
> > > > asked for? *That's got to be easy, if you *really* have all the
> > > > computable reals, as you think you do.
> >
> > > Not good enough for what?
> >
> > Not good enough to give us a computational version of the list of
> > computable reals, obviously.
>
> Not good enough is your inability to collaboration. In general.

Do you agree that having a whole bunch of lists is different from
having a single list?

Can you write a program to combine the lists into a single list?

> > > If you happen to formulate the definition or
> > > property you want me to ponder about, maybe I could even ponder it?
> > > For instance, which definition of computable reals you have in mind?
> >
> > I'd just like to see a program that takes two natural numbers (n,m) as
> > inout, and outputs the nth digit in the mth computable real in your listing
> > of computable reals. *If you have the list of computable reals, that's got
> > to be easy, whatever you mean by computable reals, isn't it?
>
> I do have a list of computable reals. You are just incapable of
> formulating a question in proper terms, including a little bit of
> basic politeness: then, just go to hell.

Can you please provide a program than given natural number n and m
returns the nth digit of the mth computable real in your list?
You may use whatever definition of computable real you choose.

>
> -LV

--
Alan Smaill

On 2 Sep, 14:59, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> ju...@diegidio.name writes:
> > On 2 Sep, 01:42, "Mike Terry"
> > <news.dead.person.sto...@darjeeling.plus.com> wrote:
>
> > > Alan asked if you had a program that took (n,m) and output the n'th digit
> > > of the m'th computable real. *When Alan asked this, the phrase "them'th
> > > computable real" meant "the m'th computable real from a COMPLETE LISTOF
> > > ALL COMPUTABLE REALS".
>
> > What Alan is asking does simply make no sense in this context.
>
> Given *any* effective listing of computable reals, you should be
> able to compute the nth digit of the mth computable real on the list;
> otherwise you simply don't have an effective listing of computable reals.

Right. So go on, show me how *you* do that, given *any* "effective
list" of the computables, under whichever definition you like.

> Why you are unable to provide us with this, I simply don't know;
> it's got to be easy, from everything you have posted so far;
> if you can't do it, that's your problem.

So *you* do it.

Ah, while you are there:

"Give me a list of tables that have two legs."

-LV

On 2 Sep, 15:10, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> ju...@diegidio.name writes:
> > On 1 Sep, 15:29, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > ju...@diegidio.name writes:
> > > > On 1 Sep, 10:31, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > > ju...@diegidio.name writes:
> > > > > > It's been under everybody's eyes for weeks now.
>
> > > > > You said "We can surely prove this ...", not "I have proved it".
> > > > > There is a difference, isn't there? No proof so far ...
>
> > > > There is a difference if you take it literally.
>
> > > I was taking you at your word.
>
> > So, despite my clarification, you couldn't care less to go beyond your
> > plain literal and wrong interpretation.
>
> If you mean that you want to withdraw your claim that "we can surely
> prove this", then by all means do so.
>
> Are you withdrawing your claim?

FUCK OFF!!!

Now we are to the PLAIN NON SEQUITUR.

NOW WE ARE TO THE PURE TROLLING.

> > > > Still that's simply
> > > > not what I meant. What I meant is: I have given this as all other
> > > > arguments under all possible shapes already, but most of what I get
> > > > back is "where it even is??" so that it is quite hard to improve at
> > > > all.
>
> > > When you say "We can surely prove ...", then I expect you have an idea
> > > of what a proof would look like.
>
> > Of course, so you confirm your objections here are just a waste of
> > time. You don't even know what you are complaining about.
>
> It's true that I don't know even the statement of what you might
> have thought you could prove -- I did imagine that *you*
> knew what you were talking about, though.

FUCK OFF!!!

Now we are to the PLAIN NON SEQUITUR.

NOW WE ARE TO THE PURE TROLLING.

> > > > > > If you cannot even
> > > > > > "see" it, then I must be putting it too "unmathematically". If you
> > > > > > are serious in this question (I have to ask) I can try to make it as
> > > > > > explicit as I can, after all is an argument by induction, but should
> > > > > > already be implied by the construction in beta version.
>
> > > > > No, I do not see it from what you have given so far.
>
> > > > Why not ask then? I am even inviting you.
>
> > > There is a difference between writing a program, and proving that the
> > > program has a particular property. *It looked like you wanted to *prove*
> > > that the anti-diagonal is still in the list. *I'm inviting you to
> > > supply that proof.
>
> > Cool. Let's see if I feel like it... no, really.
>
> no, indeed ...

FUCK OFF!!!

Bullshit on a thread and denial on the other....!!!

FUCK OFF!!!

NOW WE ARE TO THE PURE TROLLING.

> > > > > > Otherwise I'd have these questions for you: Can you give an inductive
> > > > > > definition for "the" list?
> > > > > > Can you then give an inductive definition
> > > > > > for the anti-diagonal of "the" list? Can you finally show that such
> > > > > > anti-diagonal is *not* a member of "the" list?
>
> > > > > You are the one makuing claims here, not me;
> > > > > you are the one who said "We can surely prove this".
>
> > > > > How about just doing it??
>
> > > > I see, this is a too polite way to ask, isn't it? Just guessing, asmy
> > > > English is still veeery rookey.
>
> > > I thought your mother warned you about the communists?
>
> > Mentioning my mum is worth more than a simple punch in the nose around
> > here. Anyway, I thought you knew better. I was mistaken.
>
> You were, indeed.

REALLY, JUST FUCK OFF YOU MORONIC BASTARD.

> > > > > > > Can you give us a program that takes two natural numbers (n,m) as
> > > > > > > inout, and outputs the nth digit in the mth computable real in your
> > > > > > > listing of computable reals?
> > > > > > > That might advance the discussion.
>
> > > > > > At the moment, I'd say I can surely give a program that takes *three*
> > > > > > natural numbers (n,m,w) as input, and outputs the nth digit
> > > > > > (character) of the mth expansion (string) of the wth list (the list
> > > > > > of expansions whose length is w). That is because (in beta version)
> > > > > > I end up having the list of the rationals.
>
> > > > > Not good enough, if you want a computational version of the list of
> > > > > all computable reals. *Why on earth is it a problem to give what I
> > > > > asked for? *That's got to be easy, if you *really* have all the
> > > > > computable reals, as you think you do.
>
> > > > Not good enough for what?
>
> > > Not good enough to give us a computational version of the list of
> > > computable reals, obviously.
>
> > Not good enough is your inability to collaboration. In general.
>
> Do you agree that having a whole bunch of lists is different from
> having a single list?

FUCK OFF YOU MORONIC BASTARD BASKET FULL OF SHIT.

> Can you write a program to combine the lists into a single list?

FUCK OFF YOU MORONIC BASTARD BASKET FULL OF SHIT.

> > > > If you happen to formulate the definition or
> > > > property you want me to ponder about, maybe I could even ponder it?
> > > > For instance, which definition of computable reals you have in mind?
>
> > > I'd just like to see a program that takes two natural numbers (n,m) as
> > > inout, and outputs the nth digit in the mth computable real in your listing
> > > of computable reals. *If you have the list of computable reals, that's got
> > > to be easy, whatever you mean by computable reals, isn't it?
>
> > I do have a list of computable reals. You are just incapable of
> > formulating a question in proper terms, including a little bit of
> > basic politeness: then, just go to hell.
>
> Can you please provide a program than given natural number n and m
> returns the nth digit of the mth computable real in your list?
> You may use whatever definition of computable real you choose.

JUST FUCK OFF YOU MORONIC BASTARD BASKET FULL OF SHIT.

-LV

On 2 Sep, 15:06, Cenny Wenner <CWen...@gmail.com> wrote:
> On Sep 2, 10:22 am, ju...@diegidio.name wrote:
> > On 2 Sep, 08:50, Cenny Wenner <CWen...@gmail.com> wrote:
> > > On Sep 2, 8:26 am, ju...@diegidio.name wrote:
> > > > On 31 Aug, 19:49, ju...@diegidio.name wrote:
> > > > > On 31 Aug, 18:33, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > > > ju...@diegidio.name writes:
> > > > > > > On 30 Aug, 00:04, Barb Knox <s...@sig.below> wrote:
>
> > > > > > > > So, given such an oracle, you can easily COMPUTE each digitof an
> > > > > > > > anti-diagonal D.
>
> > > > > > > > Got it now?
>
> > > > > > > I am under the impression that it is you here off mark.
>
> > > > > > > "So, given such an oracle, you can easily COMPUTE each digit of an
> > > > > > > anti-diagonal D."
>
> > > > > > > True, but the point is: the anti-diagonal is still in the list!
>
> > > > > > > We can surely prove this by induction over a computable list.
>
> > > > > > Well, if you think this is obvious, go on then and actually *do* it!
>
> > > > > It's been under everybody's eyes for weeks now. If you cannot even
> > > > > "see" it, then I must be putting it too "unmathematically". If you are
> > > > > serious in this question (I have to ask) I can try to make it as
> > > > > explicit as I can, after all is an argument by induction, but should
> > > > > already be implied by the construction in beta version.
>
> > > > > Otherwise I'd have these questions for you: Can you give an inductive
> > > > > definition for "the" list? Can you then give an inductive definition
> > > > > for the anti-diagonal of "the" list? Can you finally show that such
> > > > > anti-diagonal is *not* a member of "the" list?
>
> > > > An inductive definition for "the" list is in construction beta,
> > > > namely, it is the list of all binary sequences over finite induction.
>
> > > > Now, that list is obviously non-square, its size being in the orderof
> > > > n columns (the lenght of the sequences at each step) times 2^n rows
> > > > (all combinations of the digits over the lenght at that step --
> > > > although there is no need to calculate combinations here).
>
> > > > Now, there are two ways of defining a "diagonal" over this list. One
> > > > is to take the diagonal starting from the "upper-left corner" and stop
> > > > the sequence at the "right border" (so, the diagonal has the same
> > > > length as any sequence in the list); the other is take the diagonal
> > > > and prolong it down to the "bottom" by wrapping around at the "right
> > > > border" (that is, with a bit of modular arithmetic).
>
> > > > The first option looks like -- say for n=3:
>
> > > > list(3) = <
> > > > * * [0] *[0]0 0 ,
> > > > * * [1] * 0[0]1 ,
> > > > * * [2] * 0 1[0],
> > > > * * [3] * 0 1 1 ,
> > > > * * [4] * 1 0 0 ,
> > > > * * [5] * 1 0 1 ,
> > > > * * [6] * 1 1 0 ,
> > > > * * [7] * 1 1 1
>
> > > > Thus, the diagonal is the sequence: 0,0,0.
>
> > > > Then, the anti-diagonal (by the mapping 0->1, 1->0) is the sequence:
> > > > 1,1,1, which of course is a sequence of the list.
>
> > > > The second option looks like -- again, for n=3:
>
> > > > list(3) = <
> > > > * * [0] *[0]0 0 ,
> > > > * * [1] * 0[0]1 ,
> > > > * * [2] * 0 1[0],
> > > > * * [3] *[0]1 1 ,
> > > > * * [4] * 1[0]0 ,
> > > > * * [5] * 1 0[1],
> > > > * * [6] *[1]1 0 ,
> > > > * * [7] * 1[1]1
>
> > > > Thus, the diagonal is the sequence: 0,0,0,0,0,1,1,1.
>
> > > > Then, the anti-diagonal (again, by the mapping 0->1, 1->0) is the
> > > > sequence: 1,1,1,1,1,0,0,0, which of course is *not* a sequence of the
> > > > list; however, it is a sequence of 'list(2^3)'.
>
> > > > > > > With an uncomputable list but an oracle, I wouldn't know how it would
> > > > > > > look like. Thanks for the puzzle.
>
> > > > Still thinking about this.
>
> > > > -LV
>
> > > I'm not sure what you're trying to show with the construction
>
> > Obviously.
>
> Perhaps you feel like clarifying?

Parhaps you feel like asking what you'd like to be clarified? Of
course, never!

> > > but the
> > > enumeration you're referring to should for instance be the
> > > concatenation of list(k) for all k. I'm still a bit uncertain of
> > > whether or not you still want the TM to output periods since you seem
> > > to be referring to different ideas at different points. The one you
> > > gave in the previous post, for example, contains only the dyadic
> > > rationals, not every rational, but for any real in [0,1] and e > 0
> > > there are numbers in your enumeration within e of that real.
>
> > > Here are three lists with indicated diagonals, where I for simplicity
> > > mark the relevant digit in the period.
>
> > > L_a:
> > > [0](0)
> > > 1([0])
> > > 00([0])
> > > 01([0]) (i.e. 010[0]000...)
> > > 10([0]) (i.e. 1000[0]000...)
> > > 11([0])
> > > 000([0])
> > > ...
> > > Anti-diagonal: 1(1), not in the list.
>
> > Wrong, you just have -- as usual -- an undefined list. (And, of
> > course, you plainly ignore that, among other things, the list I have
> > given can be interpreted straight away as a list of periods, and then
> > we adopt a compound approach to express the computable numbers. But
> > let's assume I again am saying nothing at all here.)
>
> > So, the infinite list you are after *might* look like (as I have
> > already and of course and repeatedly shown):
>
> > +00)
> > +1:1(0)
> > +2:01(0)
> > +3:11(0)
> > +4:001(0)
> > ...
> > -4:110(1)
> > -3:00(1)
> > -2:10(1)
> > -1:0(1)
> > -01)
>
> This is one of the many reasons why you should be clear what about
> what you're trying to say.

Fuck off, Cenny.

> The fact that the above is treated as a
> list is because it is an enumerable set with a bijection from N (or a
> subset thereof) to your set, e.g. iterleaving the two lists. As, I
> think Virgil earlier said, any such choice of a bijection yields a
> number not in the list through Cantor's.

Sure!!! S-U-R-E!!!!!!!!!!!!!!!!!!!!

> One such choice yields L_b
> which, as stated before, has an anti-diagonal 10(10), which isn't in
> the above set. In an earlier post, it was suggested to you how to
> cover all of the rationals which would cover 10(10) but, e.g., leave
> you with L_c which has another anti-diagonal.

You mean in ***YOUR*** examples, based on a list that is undefined.

> (there is a detail left out due to the fact that we're dealing with
> reals rather than binary strings though, but I imagine you feel like
> making this stronger claim)

Then you don't even know what all this is about.

> > And of course all anti-diagonals keep being into the list.
>
> > [I say "might" because nobody up to now has been able to give an
> > infinite recursion -- if that even makes sense. Anyway, this list is
> > just equivalent to the one over finite induction as per version beta,
> > modulo a remapping of the order (namely, here we order
> > lexicographically but right to left). We need transfinite induction
> > here, though.]
>
> > In any case, as of course it must be, any anti-diagonal is still there
> > and always will be.
>
> > > (i think). Is this anti-diagional in L_c? It is not clear by
> > > inspection, but we can easily replicate Cantor's argument.
>
> > Sure. I keep repeating the same arguments again and again. You keep
> > with your undefined notions again and again. And Cantor keeps proving
> > the whole thing in any case. Sure.
>
> > In the meantime, and again:.
>
> > "Cantor's approach is like the oracle that spells out digits coming
> > straight from the hereafter. Then, however you define your anti-
> > diagonal sequence, you have no way to prove that it is not a sequence
> > already in the list -- that is a sequence that the oracle cannot
> > produce, as that would amount to solving the halting problem. Oops,
> > another disproof."
>
> > Have fun with this one too, which happens not to depend on any
> > construction or induction of any sort.
>
> > -LV
>
> Can you prove that the list { 2k }_{k=0}^inf = 0,2,4,6,... does not
> contain the number 3?

Trying to be smart? You aren't. BTW, can *you* prove it?

-LV

In article
<0c5b8377-0e3d-4550-852b-9c36c5214d9a@k13g2000hse.googlegroups.com>,
julio@diegidio.name wrote:

> On 2 Sep, 07:53, Virgil <Vir...@gmale.com> wrote:
> > In article
> > <bc67343b-73f2-4dc9-8434-826bd4eb3...@8g2000hse.googlegroups.com>,
> >
> > *ju...@diegidio.name wrote:
> > > > Otherwise I'd have these questions for you: Can you give an inductive
> > > > definition for "the" list? Can you then give an inductive definition
> > > > for the anti-diagonal of "the" list? Can you finally show that such
> > > > anti-diagonal is *not* a member of "the" list?
> >
> > > An inductive definition for "the" list is in construction beta,
> > > namely, it is the list of all binary sequences over finite induction.
> >
> > If it is not the "list of all binary sequences"
>
> Ah, how desperate! I'm touched. Unfortunately, no binary sequence
> escapes that definition -- over finite induction. Are you saying that
> finite induction is not enough?
>
> > it does not disprove Cantor.
>
> Cantor's approach is like the oracle that spells out digits coming
> straight from the hereafter. Then, however you define your anti-
> diagonal sequence, you have no way to prove that it is not a sequence
> already in the list --

Its presence in the list would be a proof that 0 = 1 for some n, after
which all arithmetic and anything relying on arithmetic, would
necessarily crash.

We prefer to insist that 0 =/= 1, and that an "anti-diagonal" binary
sequence differs in at least one place from any listed binary sequence.




> that is a sequence that the oracle cannot
> produce, as that would amount to solving the halting problem. Oops,
> another disproof.

A list of binary sequences is equivalent to f:NxN -> {0,1}, a function
from NxN = {(m,n) | min N and n in N} to {0,1} with f(m,n) giving the
nth value in the mth binary sequence.

Then g:N -> {0,1} : n |--> 1 - f(n,n) differs from every
f(m,_): N -> {0,1} : n |-> f(m,n)

julio@diegidio.name writes:

> On 2 Sep, 14:59, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > ju...@diegidio.name writes:
> > > On 2 Sep, 01:42, "Mike Terry"
> > > <news.dead.person.sto...@darjeeling.plus.com> wrote:
> >
> > > > Alan asked if you had a program that took (n,m) and output the n'th digit
> > > > of the m'th computable real. *When Alan asked this, the phrase "the m'th
> > > > computable real" meant "the m'th computable real from a COMPLETE LIST OF
> > > > ALL COMPUTABLE REALS".
> >
> > > What Alan is asking does simply make no sense in this context.
> >
> > Given *any* effective listing of computable reals, you should be
> > able to compute the nth digit of the mth computable real on the list;
> > otherwise you simply don't have an effective listing of computable reals.
>
> Right. So go on, show me how *you* do that, given *any* "effective
> list" of the computables, under whichever definition you like.

here's an example:
list of computable reals:

0.10000...
0.01000...
0.00100...
....

Now f(n,m) = 1 if n=m, and 0 otherwise.

All I'm asking here is that you come up with a similar computation for
the list of computable reals you have presented to us so far.

> > Why you are unable to provide us with this, I simply don't know;
> > it's got to be easy, from everything you have posted so far;
> > if you can't do it, that's your problem.
>
> So *you* do it.

Like I said, it's *your* problem.

I don't claim I can do it,
I don't claim it's easy.

You do.

> Ah, while you are there:
>
> "Give me a list of tables that have two legs."

Why do you ask?
Do you think I am asking for something impossible??



> -LV

--
Alan Smaill