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While i think, that POM is a global theory (and not applicable without a program), i don't think, that Robert's approach is really related to it.

Hello everyone and thank you for your many comments.
That's a lot to say, so I'll say it in the order of comments.

SpAce wrote:a recent example of an SE 9.0 puzzle that I solved quite effortlessly with GEM using Hodoku coloring tools. My rough solving steps are found later in the thread. How would you approach that puzzle with TDP?

Indeed, it will be nice to show examples of TDP resolution, but for puzzles at this level, I still need to give other TDP tools, which I will do in Part 3 and Part 4. After that I will give you my resolution of this puzzle.

Sincerely
Robert

eleven wrote:Hi Robert,
Medusa coloring goes back to at least 2008, with variations.
But i had a quick look at your site and saw a small 2-digit path sample, which cannot be done with that (however i would have done it another way). So you must have developed a bigger theory around it.
I don't care much about originator claims, and i don't doubt, that you found that on your own (many of us have found the basic and some advanced techniques on our own, before reading about them).

Thank you Eleven for that answer and for your interest in my work.
Sincerely
Robert

Mauriès Robert wrote:Indeed, it will be nice to show examples of TDP resolution, but for puzzles at this level, I still need to give other TDP tools, which I will do in Part 3 and Part 4. After that I will give you my resolution of this puzzle.

Excellent, looking forward to that!

About one thing you mentioned in your first reply:

It is a global approach to resolution in the same way as that of Allan Barker, or that of Denis Berthier.

Does that mean it can basically solve any puzzle? That's not something GEM promises, nor is it any kind of complete resolution theory. While it can be used at any level as long as there's something to feed it some kind of strong links, recognizing such possibly complex strong links is up to the player.

A mistake on my part. See the following message.

Hello StrmCkr, thank you for all the information you are giving me.
- As far as POM is concerned, I do not consider it as a global theory on a manual level. If I understood correctly, it is designed for IT developments. But I admit it, I know very little about POM.
- Regarding the publication date of my work that you indicate, it is an update date. This document was first published in 2005, but I have been developing TDP since 2011. But it doesn't matter who did what first. What I hope is that you will find some originality in the TDP.
- The example I have chosen in Part 2 is intended only to illustrate theorem 2 simply. I am well aware that this result can be found with other techniques. What I can tell you is that the TDP, with only one way of proceeding, finds all the results of advanced techniques (except perhaps the MSLS type techniques). This is why it is a global technique and my work has been to give it a rigorous framework.
- What distinguishes TDP from other colouring techniques is that we do not care about the nature of the links (strong or weak), but we construct a track as a set of candidates determined by the basic techniques (TB) and from a starting hypothesis.
I hope that Parts 3 and 4 will give you a complete overview of the TDP.
Sincerely
Robert

Hii Robert,
Just to let you know that I posted a message in your private box

TDP Part 3
See TDP Theory (part 3) which replaces the original text of this post.
Robert

If i see it right, you would not need the 4r9c2 at all. You can try 4r8c1, and when you are stuck, eliminate 2r8c4, to get to the solution.
Am i missing something ?

You mentioned the uniqueness argument. If a puzzle is unique, and you try a backdoor candidate, which solves it with singles, it is a logically correct solution method. However it is not considered as a regular solution technique (in the english sudoku community). Different to other uniqueness techniques like UR you have not proved, that the starting assumption is a must to get a unique solution.

Hi Robert,

Mauriès Robert wrote:To illustrate this part here is an example of a resolution with the SE 9.0 puzzle proposed by SpAce.

Thanks! That was quick work.

We draw two combined tracks P(4r8c1) and P(4r9c2) directly on the puzzle, the first with the blue candidates and the second with the yellow candidates.

Good! I used the same seeding cluster to begin with, so this allows for a direct comparison.

This allows for some elimination and validation (8r9c3).

Darn, I missed that +8r9c3! That's completely my own fault and not a weakness of GEM, though. I guess I just didn't bother to color the 4r8c1 parity all the way through or simply missed some continuation. That's a risk in manual coloring, especially when it gets so tedious as with such an almost-backdoor like this. Anyway, despite the different result due to my manual error, there seems to be no difference between TDP and GEM so far. However...

Then the track P(2r8c4) is built. As this track is opposed to P(4r9c2), all candidates of P(4r8c1) are candidates of P(2r8c4), so it is easy to trace the few candidates (in green) of P(2r8c4) to see that this track meets a contradiction in block 4, so it is invalid and 2r8c4 can be eliminated.

This part is interesting! As far as I know, that trick is not explained as part of standard GEM, and I've never thought of using it like that. However, GEM does have built-in support for this feature! In GEM terms, the 2r8c4 would be considered a "sub-grade" candidate in P(4r8c1), meaning that it must be false if the opposite parity P(4r9c2) is true but otherwise it's undefined. Thus it follows that if it's assumed true, then its own parity P(4r8c1) must be true too. As you just demonstrated, that could be quite handy for some nested tracking, since it automatically inherits all the already colored candidates of its nesting parity. (Unfortunately, there's no practical way I could currently add a nested track without ruining the main coloring. Too bad Hodoku doesn't support saving or copying colorings, as far as I know.)

Anyway, thanks for that! Now you've already taught me at least one new trick! (There are possibly others hidden in the earlier theory, but like I said, I learn better with concrete examples when I can easily see how it relates to something I already know.)

Guessing is a new trick for you, SpAce ?

eleven wrote:Guessing is a new trick for you, SpAce ?

Indeed it is! As I already mentioned here:

SpAce wrote:I should try that some time! ... I need education on this topic!

That said, I don't necessarily see how this is related to pure guessing. On the whole it does have a taste of it, because it seems to me that TDP is happy once it finds a solution in one or the other track (I'm not). The trial of the 2r8c4 is not guessing, though. It's T&E because it produces a contradiction. Well, I don't normally like that approach either (testing an individual candidate), but here it's a bit more interesting form of it because of the already established coloring that it inherits and what makes it easy to find the contradiction quickly. There's a reason why the 2r8c4 is tested specifically, so it's not a random candidate.

I can see how that could be useful with my favorite GEM coloring, because I already have potential candidates for such trials marked as sub-candidates for both parities. It means that in cases when the main coloring no longer produces results, something can be still squeezed out of it before scrapping it and starting with a new seed.

Theoretically the same idea could work with the super-candidates, which are the opposite of subs. They're fully defined only if their own parity is true (then they're true too), but in the opposite case they're undefined. If one of those is assumed false it means that the opposite parity must be true, and its established coloring could be reused similarly for the nested assumption. I don't know if that case would be as useful, but at least such a possibility exists.

In any case, I need to test this in real puzzles and see if it works in practice.

Bonjour Eleven,

eleven wrote:If i see it right, you would not need the 4r9c2 at all. You can try 4r8c1, and when you are stuck, eliminate 2r8c4, to get to the solution.
Am i missing something ?

Yes, you didn't understand the interest of theorem 4, but I probably took a bad example to see its interest.
Contrary to what you suggest about SpAce in another comment, there is no coincidence in my choices. I chose the pair of 4s from block 7 because the two tracks are conjuguated (theorem 1 part 2) and because the simple observation of the puzzle shows that one of the two tracks (the blue one) will develop well. From then on, it will be used (theorem 4 part 3) as a support for any track opposite to the other track (yellow). There is no longer any left, with the observation to choose the start of the opposite track carefully. This is obviously the 1 or 2 of r8c4.
For me, there is no more chance than to discover in a corner of the puzzle that you have some kind of fish or other disposition of candidates to apply an advanced technique.
I will therefore give you another more convincing example.

eleven wrote:You mentioned the uniqueness argument. If a puzzle is unique, and you try a backdoor candidate, which solves it with singles, it is a logically correct solution method. However it is not considered as a regular solution technique (in the english sudoku community). Different to other uniqueness techniques like UR you have not proved, that the starting assumption is a must to get a unique solution.

Theorem 4 part 3 is even proven for multi-solution puzzles (see my publication in French page 15 here), so it is still valid. I shouldn't have made that remark about uniqueness, so I deleted it. But here, in the context of the grids you are processing, they are always single-solution and that's the only reason you can use URs.

Sincerely
Robert

Hello SpAce, thank you for your encouraging comments.
I made an answer to Eleven, I suggest you read it.
Sincerely
Robert

Hi Robert,

Mauriès Robert wrote:Hello SpAce, thank you for your encouraging comments.
I made an answer to Eleven, I suggest you read it.

I did, and I agree with you.

Contrary to what you suggest about SpAce in another comment, there is no coincidence in my choices. I chose the pair of 4s from block 7 because the two tracks are conjuguated (theorem 1 part 2) and because the simple observation of the puzzle shows that one of the two tracks (the blue one) will develop well.

The same reasons why I chose them as my initial coloring seed. Edit: I would, however, prefer that both tracks develop well. That would give more chances for trap eliminations.

From then on, it will be used (theorem 4 part 3) as a support for any track opposite to the other track (yellow). There is no longer any left, with the observation to choose the start of the opposite track carefully. This is obviously the 1 or 2 of r8c4.

Yes, this is very obvious in my GEM-colored version as well. The 1 and 2 in r8c4 are the only sub-candidates left for the 4r8c1 parity, so they're the only ones that can be tested in that context (i.e. within the existing 4r8c1 coloring). If you don't mind, I could post some images of what my process looks like. I don't mean to hijack your thread with GEM stuff, but I think it might visualize this concept quite well.