Sudoku Players' Forum (clone)

.
Hi Robert,
You have eliminated from your website and from the last version of your paper "Technique des pistes" (Décembre 2017) any reference to tracks and anti-tracks based on arbitrary sets of candidates.
What remains is only tracks and anti-tracks based on a single candidate and (track, anti-track) couples based on bivalue CSP-Variables. As a result, there remain only trivial elimination theorems (plus assertion ones in the latter case, corresponding to Forcing-T&E).

Did you find any irreparable error in your initial more general "theorems" for arbitrary pairs of sets of candidates?

[Edit 1]: I changed the title of this thread, because there is more to say about tracks and anti-tracks than about those based on arbitrary sets of candidates.

[Edit 2]: the reference for my forthcoming posts: the december 2017 original French version of the document "Théorie de la technique des pistes en sudoku. Par Robert Mauriès (*)". I'll cite the original and do the translations myself.

[Edit 3]: On the website's left part, there are indeed two buttons, leading to 2 different versions of the paper. So, the previous version has not disappeared. It remains that any reference to tracks and anti-tracks based on arbitrary sets of candidates have disappeared from the new paper.

[Edit 4]: for the main result, jump to http://forum.enjoysudoku.com/is-there-any-original-theory-or-any-theory-at-all-in-tdp-t39766-39.html

.
About resolution techniques.

Robert_Mauriès, p.2 wrote:- Une technique de résolution est une démarche (raisonnement) logique permettant de placer ou d'éliminer des candidats sur une grille en appliquant R.

Translator= me wrote:A resolution technique is a logical step/move/approach (reasoning) allowing to assert or eliminate candidates in a grid by applying R

Note: R is the 4 rules of Sudoku (see p.1).
Honestly, I'm re-assured. Robert will not speak of sacrificing virgins, drinking potions or eating brains of alive monkeys. Phew!!!!

But do I know what a resolution technique (RT) is? Of course not. Does this intend to restrict RTs to pattern-based solutions - which would exclude T&E...? We'll never know.
"Applying R" (the 4 rules of Sudoku) doesn't mean anything, as these 4 rules are not operational (they don't allow any assertion or elimination of a candidate).
That's all we'll get about an RT. Not even enough to make an educated guess. There are indeed several very different ways of interpreting this in unambiguous terms. But, before doing this, I'll give Robert a chance to state what he had in mind.

Robert_Mauriès, p.2 wrote:On désigne par TR l’ensemble des techniques de résolution permettant de résoudre une grille

Translator= me wrote:We notate by TR the set of resolution techniques allowing to solve a grid

As we still don't know what a resolution technique is, TR is not defined. Note that this "definition" of TR is the basis for all the rest of the paper. This seems to be very shaky grounds for a theory.
.

Bonjour Denis,
Ne comptez pas sur moi pour alimenter vos délires à l'encontre de la TDP, alors que depuis que je suis venu sur ce forum, vous ne cessez de la démonter. Et pour cela vous n'hésiter pas à dire n'importe quoi comme le fait maintenant que j'aurais retiré un document de mon site internet.
Si vous voulais traduire en anglais tout ce que j'ai écris à l'attention des lecteur d'Enjoy Sudoku, sachez que j'ai d'autres documents à vous proposer comme celui-ci (https://www.assistant-sudoku.com/Pdf/piste-antipiste.pdf) qui est le premier écrit sur les antipistes à la suite de la publication de mon livre (1ere édition) et à l'époque je ne connais pas François.
Robert

Hello Denis,
Don't count on me to feed your delusions against the TDP, while since I came on this forum, you don't stop dismantling it. And for that you don't hesitate to say anything like the fact now that I would have removed a document from my website.
If you wanted to translate into English everything I wrote for the readers of Enjoy Sudoku, you should know that I have other documents to propose to you like this one (https://www.assistant-sudoku.com/Pdf/piste-antipiste.pdf) which is the first one written about the antipists after the publication of my book (1st edition) and at the time I don't know François.
Robert

.
Robert,
I see you've found a translator.

I've opened this thread for technical discussions about TDP. My question about TR was and remains purely technical. You're now trying to put this on a conflictual ground in order to avoid answering. This is not what I expect and this is very unwelcome in this thread.

If you don't want to clarify what a TR is for you, don't hope that I'll stop here. I'll go on with my best interpretations of it and show that all of them quickly lead to deep inconsistencies. For this purpose, I'll need only the first 4 or 5 pages of the referenced document.

.
[Edit:added this paragraph for a better undersanding]
Tracks

Robert_Mauriès, p.3 wrote:Une piste P(Ak ) issue d’un candidat A k est l’ensemble des candidats A i ∈ G que l’on placerait avec les TR si A k était placé.

Translator= me wrote:A track P(Ak) based on a candidate Ak is the set of candidates that would be asserted as true by the TR if Ak was true.

Nothing to say here, except the previous remark that the TR is not defined and that it should be "the track..."



Resolution technique: first interpretation
In the absence of an answer, here is my first interpretation of what a TR is. This is the most obvious one, the one that any mathematician would consider as the most likely.
A TR is any way of proving logical consequences of the 4 axioms.

For a puzzle with a single solution, this has the unfortunate consequence that any track is:
- either the set of all the candidates (inconsistent puzzle [Edit: or false candidate as the starting point]);
- or the set of all the candidates that are true in the solution.
(The basis for this is the well-known theorem that you can find in the first pages of any Logic book: a formula is provable in a theory T if and only if it is true in all the models of T).
In both cases, a track depends only of whether its starting point is in the solution or not and the notion is devoid of any use.

.
Resolution technique: second interpretation
The second interpretation is suggested by Robert himself in his "simplified" paper.
Instead of "all the resolution techniques", which I've shown to have no meaning, use only a fixed set of resolutions rules - what he names the TB (basic techniques).
It is not an interpretation strictly speaking, as it is defined in the referenced document as a kind of special case, and the resulting tracks are defined as the traces of the full tracks.

In this case, which is the only one used in practice by Robert, there's nothing much to say about the definition at this point, except that:
- it puts to light that the general theory is a totally useless and irrelevant outgrowth of the special case; it suggests that this outgrowth is there only to make the "theory" look more like a theory; but my analysis shows the result is quite the opposite;
- I haven't delved into all the details, but if the TB don't have the confluence property, there may be difficulties with some definitions.

By itself, adopting this interpretation destroys the whole outgrowth.
In the next posts, I'll keep this interpretation, because this is the only meaningful one and obviously, Robert doesn't have anything else to propose.

Even so, this is not the end of the story. I'll expose later other deep problems.
.

.
Tracks and c°

Before going further, we'll need the definition of an "entity". I'll have more to say on that topic later, but for now let's take it as is.
Apart from the name, nothing weird here. Robert is not going to speak of ghosts or alien entities.

Robert_Mauriès, p.2 wrote:Une entité de G est l’ensemble formé de tous les candidats d’une même case, ou de tous les candidats de même occurrence d’une même zone

Translator = me wrote:An entity is the set of candidates for the same cell or the set of candidates with the same number in the same unit"

(I've translated to more usual Sudoku vocabulary.)
In my vocabulary, an "entity" is merely the content of a 2D-cell or CSP-Variable.

Robert_Mauriès, p.3 wrote:Définitions 2-1
1) Une piste P(Ak ) issue d’un candidat Ak est l’ensemble des candidats Ai ∈ G que l’on placerait avec les TR si Ak était placé.
2) Une antipiste P’(Ak ) issue d’un candidat Ak est l’ensemble des candidats Ai ∈ G que l’on placerait avec les TR si Ak était éliminé de la grille.
On dit que P’(Ak ) est l’antipiste de la piste P(Ak ).

3) Une piste P(E) issue d’un ensemble de candidats E est l’ensemble des candidats Ai communs à toutes les pistes issues de tous les candidats Ak ∈ E .
On a donc P(E) = ∩ E P(Ak ), soit P(E) ⊆ P(Ak ) ∀k.
4) Une antipiste P’(E) issue d’un ensemble de candidats E est l’ensemble des candidats Ai que l’on placerait avec les TR si on éliminait tous les candidats A k ∈ E .
On dit que P’(E) est l’antipiste de la piste P(E).

Translator = me wrote:Definitions 2.1
1) The track P(Ak) based on a candidate Ak is the set of candidates that would be asserted as true by the TR if Ak was true.
2) The antitrack P'(Ak) based on a candidate Ak is the set of candidates that would be asserted as true by the TR if Ak was false.

3) The track P(E) based on a set of candidates E is the set of candidates common to all the tracks P(Ak), Ak ∈ E.
We therefore have P(E) = ∩E P(Ak ), i.e. P(E) ⊆ P(Ak ) ∀k.

4) The antitrack P'(E) based on a set of candidates E={Ak...} is the set of candidates that would be asserted as true by the TR if all the Ak ∈ E were false.
We say that P'(E) is the anti track of the track P(E).

Edit: I've improved the translation of 4) so that there is no ambiguity about the meaning of "true"

Nothing to say here (except always the same remarks about "the TR". Just a bunch of definitions. Whether they are useful will be sen later.

Ah, yes, something more; just a remark on the fly: the end of definition 3 - "P(E) = ∩E P(Ak ), i.e. P(E) ⊆ P(Ak ) ∀k" - must be a new fundamental theorem in Set Theory, making Robert a potential candidate for a Fields Medal.
Starting from today, P(E) = ∩E P(Ak ) is equivalent to P(E) ⊆ P(Ak ) ∀k; you don't need to prove the reverse inclusion.
That'll make a lot of students happy; they can replace half of their home work by playing with their Nintendo.

Sometimes, you think something is a fly, but it turns out to be a hornet.
.

.
Theorem 2.1

This is where it becomes hot. No, no, no, not hot in that sense!

A pair of sets of candidates, E1 and E2, is defined as two disjoint complementary subsets of an "entity". Said otherwise E1 and E2 form a partition of the "entity".

Robert_Mauriès, p.5 wrote:Théorème 2-1 :
Si E1 et E2 forment une paire d’ensembles, la piste P(E1) issue de E1 est identique à l’antipiste P’(E2) issue de E2, et réciproquement.
En particulier, si deux candidats forment une paire, la piste issue de l’un et identique à l’antipiste issue de l’autre.

Translator = me wrote:Theorem 2-1 :
If E1 and E2 form a pair of sets, the track P(E1) based on E1 is identical to the antitrack P’(E2) based on E2, and conversely.
In particular, if two candidates make a bivalue/bilocal pair, the track based on one of them is identical to the antitrack based on the other.

Note that the particular case is totally trivial and requires only the notion of a track based on a single candidate.


Here is now Robert's "proof":

Robert_Mauriès, p.5 wrote: l’antipiste P’(E2) est construite en supposant que les candidats de E2 sont éliminés, donc que seuls les candidats de E1 subsistent puisque E1 et E2 forment une paire d’ensembles.
Comme un des candidats Ak de E1 doit être placé pour appliquer R, on peut dire que un des Ak de E1 est un candidat de P’(E2), donc que pour ce Ak on a P(Ak ) ⊆ P’(E2), donc que P(E1) ⊆ P(Ak ) ⊆ P’(E2)....

Translator = me wrote: the antitrack P’(E2) is built by supposing that the candidates in E2 are eliminated; therefore only the candidates in E1 remain, because E1 and E2 form a pair of sets.
As one of the candidates Ak in E1 must be true for applying R, one can say that one of the Ak's in E1 is a candidate of P’(E2); for this Ak one has P(Ak ) ⊆ P’(E2), and therefore P(E1) ⊆ P(Ak ) ⊆ P’(E2)....

The last sentence is false.
Indeed, in any resolution state RS where all the candidates in E1 are sill present, there's no reason at all to suppose that no Sudoku rule can apply unless a candidate in E1 has first been proven to be true. On the contrary, most of the resolution rules (such as Naked Pairs) apply without any condition on decided values.

This proves that the "proof" of Theorem 2.1 has a fundamental flaw at its very start.

Now, I'll let some time before going further. Can anyone find a direct counter-example to the "theorem"?

Indeed, the proof of this theorem is not valid. Robert knows this since I told him in 2019 !

DEFISE wrote:Indeed, the proof of this theorem is not valid. Robert knows this since I told him in 2019 !

Sorry that you also put!
If it is true that you told me, in your opinion, that this demonstration was false in 2019, I maintain that it is true and that my demonstration is correct. This is because of a point that we had discussed at length and that you had admitted, which is that an invalid lead can, because of its multiform nature, reach any candidate in the puzzle.
But as I wrote above, I will not comment further or participate in Denis' delirium.
If you want us to continue this discussion as we both did in its time, let's do it by live mail.
Sincerely
Robert

DEFISE wrote:Indeed, the proof of this theorem is not valid. Robert knows this since I told him in 2019 !

Why am I not surprised?
This explains why he found a pretext for not answering my request for precisions at the start of this thread.

Mauriès Robert wrote:I maintain that it is true and that my demonstration is correct.

When a fundamental flaw is found in a proof, "maintaining" that the demonstration is correct or insulting people has no value. The only acceptable answer is a correct proof or a retractation.


Back to serious matters:
François, how did you reach the conclusion in 2019 that the theorem is false? Did you find something similar to me or another flaw in the proof or did you have a counter-example?

Hi Denis,
I think you are wasting your time. It's like telling Picasso that he doesn't respect the proportions.
Nevertheless, I sent you what I said to Robert in 2019, including a counter-example. His answer had been "Joker for now".

DEFISE wrote:I think you are wasting your time. It's like telling Picasso that he doesn't respect the proportions.
Nevertheless, I sent you what I said to Robert in 2019, including a counter-example. His answer had been "Joker for now".

OK. Thanks for the counter-example. I let you publish it if you see fit to do so.

I know I'm wasting my time with Robert and I don't plan to waste more of it, but I think having debunked the useless or false parts of his "theory" will spare time for those trying to understand what's not worth the effort.

There remains the restricted theory, with TB instead of TR, and with tracks based on a single candidate. I don't think there is anything false in that part, nor is there anything very original:
- as defined (as sets of candidates), tracks are just T&E, as is obvious from the definition;
- as used (as sequences), tracks are dynamic contradiction chains without any notion of complexity.
There's a major gap between the "theory" and what Robert does in practice (such as minimising the lengths and number of chains).

DEFISE wrote:Hi Denis,
I think you are wasting your time. It's like telling Picasso that he doesn't respect the proportions.
Nevertheless, I sent you what I said to Robert in 2019, including a counter-example. His answer had been "Joker for now".

La correction François, serait que vous m'adressiez exactement ce que vous avez adressé à Denis.
Merci

The correction, François, would be for you to address to me exactly what you addressed to Denis.
Thank you

Robert

.
Robert,
Je ne pense pas que vous soyez en bonne position pour donner des leçon de correction.
La correction pour vous serait de reconnaître votre erreur évidente dans le théorème 2.1. Ça arrive à tout le monde de se planter.
Quant au contre-example, le mieux serait de le publier. Je l'ai lu ; il est imparable.

I don't think you're in a position to give lessons in correctness.
Correctness for you would be to acknowledge your obvious error in the proof of theorem 2.1. Everyone happens to make errors.
As for the counter-example, the best thing would be to publish it. I've read it; it is foolproof.