Technical of tracks (TDP)

Advanced methods and approaches for solving Sudoku puzzles

Technical of tracks (TDP)

Postby Mauriès Robert » Thu Nov 07, 2019 2:31 pm

Hello to all sudoku enthusiasts who frequent this forum on which I have just registered.
Being French and not fluent in English, I use an electronic translator to do my translations, so I apologize for my poor English, but I hope I will be understandable.
On this forum, I will explain the technique I have been using since 2011 and that I have perfected down to the last detail. I call it "technique des pistes" in French (technical of tracks in English) and in summary TDP.
To be followed in other articles....
Robert
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Re: Technical of tracks (TDP)

Postby SpAce » Fri Nov 08, 2019 1:39 pm

Hi Robert,

Mauriès Robert wrote:On this forum, I will explain the technique I have been using since 2011 and that I have perfected down to the last detail. I call it "technique des pistes" in French (technical of tracks in English) and in summary TDP.
To be followed in other articles....

I guess this is it? My first impression is that it seems to be quite similar to what I already use via coloring. The idea is explained here, though it's somewhat difficult to follow because David explained it in terms of text-based grid markings that no one but himself would ever use. Nevertheless, in practice it's very easy and powerful. I would be very surprised (and impressed) if your technique provided any additional powers! Seems unlikely (I'd rather bet the opposite), but I guess we'll find out if you do follow up with those other articles.

Btw, some time ago someone posted about a similar technique that also follows two tracks simultaneously. Funny enough, his post quickly disappeared after I mentioned the similarity with David's GEM. So... I don't want to sound discouraging, but that idea is far from new. I don't claim to know who invented it first, though. (It's also possible that I've misunderstood something and your idea is about something else.)

Either way, you're welcome to the forum! Like I told the other guy (who disappeared), you're very welcome to participate in the daily Puzzles as well. We have another skilled Frenchman (Cenoman) there too.
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Re: Technical of tracks (TDP)

Postby Mauriès Robert » Fri Nov 08, 2019 4:28 pm

Hello SpAce,
Thank you for your feedback. Indeed the TDP (https://www.top-sudoku.com/sudoku/fr/techniques-des-pistes.php) is similar to the coloring technique you use on Hodoku and probably to that of David Bird (GEM) which I have never studied in depth. It certainly differs only in the way it is formulated. So I don't claim to have invented this method of resolution, but I have been using it since 2011 as a self-taught sudoku without having tried to find out if it already existed elsewhere. In France, there is another technique called "Virtual Coloring" very similar developed by Bernard Borrelly.
Nevertheless, I would like to give my vision of this technique and hope to bring some new features. So I will do the articles announced, because the article you mention that is mine is only a popularization.
As for the article you mention (here), it is only from 2019 and totally ignores my work published since 2013.
Sincerely
Robert
Last edited by Mauriès Robert on Fri Nov 08, 2019 5:35 pm, edited 1 time in total.
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Re: Technical of tracks (TDP)

Postby Mauriès Robert » Fri Nov 08, 2019 5:12 pm

TDP Part 1

Introduction

As announced in my previous article on the forum, I present here the method of solving sudoku grids (puzzle) that I call "Technical of tracks", TDP in short (TDP = "Technique Des Pistes" in French).
This method is inspired by "Forcing chains" but differs from it by the fact that it treats the puzzle in a global way, and allows to develop a theory with its definitions, its properties and its theorems. It is a global approach to resolution in the same way as that of Allan Barker, or that of Denis Berthier.
From a practical point of view, it allows the sudokist to memorize only one way of doing things, which is simpler than knowing different advanced techniques that are not always easy to implement. In France, where I published a book and developed a website (http://www.assistant-sudoku.com), many sudokists use this method.
The purpose of this article, and the following ones, is to present the TDP and its arguments in as simple a way as possible. A complete document on this method can be found on my website, but alas in French.

What I present in the 4-part forum is gathered in this PDF : http://www.assistant-sudoku.com/Pdf/TDP-anglais.pdf

Basic techniques

The TDP uses basic techniques (TB) for its implementation, so I specify here what I consider to be basic techniques. These are the ones that every sudokist knows in principle: unique candidates (singles), closed sets (pairs, triples, etc...) and alignments (pointing, box line).
We can eventually add X-wing for its simplicity, but nothing prevents those who want to add other techniques that they master well like fishs.

Anti-track

Let's start with the notion of anti-track and its definition.

Definition :
E = {Ai, i=1,2,...,n} being a set of n candidates Ai of the puzzle G, an anti-track P'(E) = {Bj, j=1,2,...,m} is the set of m candidates Bj that would be placed with the basic techniques (TB) in the cells of G as (true) solution IF the candidates of E were eliminated (false) from G.
It is said that E generates P'(E).


Of course, by doing this, three situations can occur:
- or, the placement of candidates Bi encounters a contradiction (no candidate Bj possible in a box, two candidates Bj in the same area, etc...). It is then said that P'(E) is invalid.
- or, the candidates of P'(E) are all solution candidates. It is said that P'(E) is valid. We'll see later when this situation occurs.
- or, nothing can be concluded at this stage on the status of P'(E), because it cannot be proven that P'(E) is valid or invalid.

An obvious first property to be used is as follows:

Property (P1):
If P'(E) is invalid, then at least one candidate Ai of E is a solution in his box.

Let us give other definitions that will make it possible to establish a first interesting theorem to illustrate the use that can be made of anti-tracks.

K1, K2,... the boxes that contain the candidates Ai of any set E={{Ai, i=1,p} and by Ei the set of candidates of E contained in Ki. We have E=UEi. E'i refers to the complementary set of Ei in Ki.
The Ei sets are called the components of E. The E'i sets are called the counter-components of E.

Definition:
An anti-track P'(E) is said to contain a set of candidates E1={B1j, j=1,2,...} when a candidate of each component E1i of E1 is necessarily a candidate belonging to P'(E).

Let us then demonstrate the following theorem.

Theorem 1 :
In the same zone Z (block, row or column), E1 and E2 are two distinct sets with two components each consisting only of candidates of occurrence a and b.
If any of the conditions H1, H2 or H3 below are met, then all candidates of occurrence a or b that are not contained in E1UE2 can be eliminated from zone Z.
- (H1) : P'(E1) and P'(E2) are invalid.
- (H2) : P'(E2) is invalid and P'(E1) contains E2.
- (H3) : P'(E1) contains E2 and P'(E2) contains E1.


Demonstration :

1)- According to H1 and the property (P1), E1 and E2 each contain a solution candidate, so the two candidates of Z of occurrence a and b are in E1UE2 => elimination of candidates of Z of occurrence a or b who are not in E1UE2 .

2)- According to H2 and the property (P1), E2 contains at least one candidate solution and two cases are possible:
- P'(E1) valid => E2 contains two solution candidates, one of which is occurrence to the other of occurrence b => elimination of candidates from Z of occurrence a or b who are not in E1UE2.
- P'(E1) invalid and this returns to case 1) => elimination of candidates from Z of occurrence a or b who are not in E1UE2.

3- According to H3, four cases are possible:
- P'(E1) valid => E2 contains two solution candidates, one of which is occurrence to the other of occurrence b => elimination of candidates from Z of occurrence a or b who are not in E1UE2.
- P'(E1) invalidates which leads to case 2) => elimination of candidates from Z of occurrence a or b who are not in E1UE2.
- Same reasoning and conclusion with P'(E2) depending on whether it is valid or invalid.
In all possible cases of H1, H2 and H3 candidates from Z of occurrence a or b that are not in E1UE2 are eliminated.

Here is an example on the famous Easter Monster puzzle of an application of all this.
G = 1.......2.9.4...5...6...7...5.9.3.......7.......85..4.7.....6...3...9.8...2.....1
Let's take E1={38r2c13} and E2={38r2c79}.
P'(E1) ={2r2c1, 7r2c1, 1r7c2, 6r9c2, 2r8c7, 7r8c9, 1r3c8, 6r1c8, ...} => P'(E1) contains E2.
P'(E2) ={1r2c7, 6r2c9, 2r7c8, 7r9c8, 1r8c3, 6r8c1, 7r1c2, 2r3c2 ...} => P'(E2) contains E1.
According to the theorem, 38r2c5 and 8r2c6 can be eliminated.

It is understood that the theorem also applies following the sets of type E1 and E2 which constitute the well-known Sk-loop of this puzzle and all eliminations are deduced from it.

But this theorem also applies to puzzles that do not count SK-loop, for partial eliminations.

On the practical side, the representation of an anti-track is easier to do by marking the candidates on the puzzle in colour than in the form of an abstract set. This can be done as below, for example with the Hodoku application or the application of my website.
These are in blue the candidates of P'(E1) studied previously. We see that if we continue the marking we obtain a contradiction in the r5c8 box, so P'(E1) is invalid.

Image

I will stop there for this first article, but the TDP has many other aspects that I will present gradually.

Thank you for your possible remarks and excuse me for my approximate English.

Robert
Last edited by Mauriès Robert on Sun Nov 17, 2019 10:17 am, edited 2 times in total.
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Re: Technical of tracks (TDP)

Postby eleven » Sun Nov 10, 2019 12:21 am

Did i understand it right, that your approach is a kind of "grouped Medusa coloring" with it's own elimination rules ?
There you start with an either/or condition of single digits (e.g. bivalue/bilocation) and follow both ways. Then you can eliminate all candidates, which would be eliminated from both paths (or if a path leads to a contradiction, you can eliminate the starting digit).
You seem to start with an either this group of digits or that.
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Re: Technical of tracks (TDP)

Postby StrmCkr » Sun Nov 10, 2019 6:30 am

It reads as if its doing multi digit template conditional testing via cycling multiple tiles {46656 templates} together that either are true as a n set, or contain a contradiction thus false. Manually instead a program that can compare all p.o.m tiles for each cycling the (1..9) digit combination set together to form eliminations.

1 digit - template delete/ most fish.
2 sets: ie multi coloring,
2-3sets for 3d medusa,
4 sets for sk-loop.

partial template sets for combinations of digits via comparison for completed N-sets {ie subsets}.

if i am correct, this has been around since pre: 2005/2006 long before the name i listed above came out usually these ideas fall under POM/template delete.
{ which some hints to bifurcation{nisho} / traveling pairs and braids theorem which is based on studying pom. which is in the 2006-2007 time frame}

not to knock anything you are presenting,
but if it it something new to us i am sure
champagne whom is from the french Sudoku forms can preamble on, or mayhaps help with translating for you.
Some do, some teach, the rest look it up.
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Re: Technical of tracks (TDP)

Postby Mauriès Robert » Sun Nov 10, 2019 9:04 am

Hello Eleven, I'm glad you read my article and thank you for your question.
The TDP I present uses in a certain way, such as Medusa 3D, Forcing Chains, etc... the "true/false" principle. So from this point of view there is no novelty and it is normal that we achieve the same results.
What I originally wanted to do in 2011 (without knowing Medusa at the time) was to build a global theory where, in everything I read, things were presented case by case without identifying a general principle. This gave rise to multiple techniques. I happen to have published my work only in France and in French, and as a result they are totally ignored by the Anglo-Saxon world. Medusa 3D has only been on display since 2015, I believe (http://www.sudokuwiki.org/3D_Medusa). Without seeking parternity, I still want to expose the TDP as a global theory of sudoku with elements that I have never read elsewhere.
The foundation of TDP is based on a network of two suites of candidates, one of which is necessarily true and the other false, and this generates the well-known properties found in Medusa3D on the elimination or validation of candidates. The notion of anti-track seems to me not to have been used elsewhere, and it is on the basis of this notion that I build the whole TDP. I hope that with my next articles you will better understand my approach, knowing that these articles will not cover all the aspects that can be developed, the forum being not a pedagogical book.
You can also consult my publication in French through this link.
With kind regards
Robert
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Re: Technical of tracks (TDP)

Postby Mauriès Robert » Sun Nov 10, 2019 9:21 am

Hello StrmCkr, thank you for your intervention on the subject of TDP.
I know the POM method (http://www.sudokuwiki.org/Pattern_Overlay) invented by Myth Jellies. It seems to me that it is different from TDP, POM not being from my point of view a global theory.
I suggest you read my answer to Eleven above, as well as my next articles.
With kind regards.
Robert
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Re: Technical of tracks (TDP)

Postby SpAce » Sun Nov 10, 2019 9:42 am

Hi Robert,

Thanks for your reply and the initial presentation. It's a bit heavy on theory for my taste, but hopefully we'll see more practical examples later.

Mauriès Robert wrote:Medusa 3D has only been on display since 2015, I believe (http://www.sudokuwiki.org/3D_Medusa).

That's completely false. As far as I know, 3D Medusa is a very old technique, preceding the likes of GEM (which fully contains 3D Medusa, but is much more powerful). According to David, he's used GEM since 2006. Added: I just found the first reference to Medusa on this forum. It's from 2005, and I believe by the original inventor (Bob Hanson). Here's the link to his site.

StrmCkr knows the history of the various techniques better. In any case, it makes no sense to compare your technique with 3D Medusa, as the latter is obviously much much weaker (at least the very basic version presented on SudokuWiki -- but I don't think any of the more powerful extensions are called 3D Medusa these days). The only relevant comparison is between your technique and GEM, and I would be very interested in seeing an actual comparison of those two.

Here's 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?
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Re: Technical of tracks (TDP)

Postby eleven » Sun Nov 10, 2019 10:38 am

Hi Robert,

since it was a bit hard to understand for me, what you are doing, i wanted to clarify, what the similarities and differences to known techniques in the english world are.

Medusa coloring goes back to at least 2008, with variations, see e.g. this link.

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).
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Re: Technical of tracks (TDP)

Postby SpAce » Sun Nov 10, 2019 10:50 am

eleven wrote:Medusa coloring goes back to at least 2008, with variations, see e.g. this link.

I just proved that it goes back to at least 2005. However, your link proves that GEM has also been known in 2008:

It is akin to a watered down GEM and has surely been done by others.

It ("watered down GEM") also implies that GEM was probably considered the most powerful version of the advanced coloring techniques at the time. I haven't seen anything more powerful since then either, except maybe multi-digit multi-coloring (like champagne's "full tagging") but that's not humanly applicable. Until proven otherwise, I consider GEM the gold standard for manual coloring.
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Re: Technical of tracks (TDP)

Postby champagne » Sun Nov 10, 2019 11:07 am

SpAce wrote: I haven't seen anything more powerful since then either, except maybe multi-digit multi-coloring (like champagne's "full tagging") but that's not humanly applicable.


Hi SpAce,
your remark push me back years ago.
Undoubtly, the "full tagging" belongs to the coloring family. It worked however in a different way. The first step was to find all kinds of subchains made of bi-values (and equivalences when specific properties as "symmetry of given" were applied), then to apply a classic chaining step to the sub chains.
As you noticed, this is not the way skill players like to work, so, I turned to other ways to attack puzzles. SKFR has been written using the full tagging code as basis. Next versions are much closer to Sudoku Explainer view. My feeling here is that we are closer to the attack done in Sudoku explainer, but with a start using a set of candidates instead of a unique candidate. In Sudoku Explainer, the expansion is hidden and only chains of interest are produced.
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Re: Technical of tracks (TDP)

Postby Mauriès Robert » Sun Nov 10, 2019 12:13 pm

TDP Part 2

Introduction

In this second part, I discuss the notion of conjugated tracks and their properties. Some of the terms used in this Part 2 have been defined in Part 1 to which I refer the reader.

Track and anti-track

Part 1 has been used to define an anti-track and the counter-components of a set E.
Then let's define a track P(E).

Let's start with a single candidate Ak, and designate E'k as Ak's complementary in Ak's box.

Definition:
The track P(Ak) is the set of candidates of the anti-track P'(E'k), so P(Ak)=P'(E'k).
It is the same to say that:
P(Ak)={Bj, j=1.2, ...} is the set of candidates Bj who would be placed with the basic techniques (TB) in the cells of G as (true) solution SI Ak was placed (true) in its cell.


Then let's define a track P(E).

Definition :
E ={Ak, k=1,2,...n} being any set of candidates Ai of the puzzle G, a track P(E)={Bj, j=1,2, ...} is the set of candidates Bj common to the tracks P(Ak), i. e. obtained by intersection of the antitracks P(Ak), so P(E)=∩P(Ak).
It is said that E generates P(E).


As with anti-track, several situations can occur when building P(E):
- or, the placement of candidates Bj encounters a contradiction (no candidate Bj possible in a box, two candidates Bj in the same area, etc...). It is then said that P(E) is invalid.
- or, P(E) candidates are all solution candidates. It is said that P(E) is valid. We will look further ahead when this situation occurs.
- or, nothing can be concluded at this stage on the status of P(E), because we do not know how to prove that P(E) is valid or invalid.

Note that we can write P(E)=∩P(Ei), where Ei are the components of E.
As it can be shown that P(Ei)=P'(E'i), where E'i are the counter-components of E, it can be deduced that P(E)=∩P'(Ei').

We can already state a property that is easy to demonstrate.

Property (P2):
If P(E) is invalid, then none of the candidates Ai of E is a solution in his box, so all candidates of E can be eliminated.

Indeed:
P(E) invalid means that the placement of candidates Bj common to the antitracks P'(E'i) encounters a contradiction, therefore it is the same with the placement of candidates of each P'(E'i) which are therefore invalid => For all i, E'i contains a solution candidate (see property P1 in part 1) => For all i, the components Ei of Ei do not contain any solution candidate => E contains no solution candidate.

Finally, for a set E of candidates from G, there are two types of candidate sets generated by E, a track P(E) and an anti-track P'(E) that satisfy the following property.

Property (P3):
P(E) and P'(E) cannot be disabled simultaneously.
If P(E) is invalid => P'(E) is valid, and if P'(E) is invalid => P(E) is valid.


Indeed:
if P(E) and P'(E) were simultaneously invalid, then according to the property (P2) no candidate of E would be solution and according to the property (P1) at least one candidate of E would be solution, which is absurd.

Conjugated tracks

Language convention:
It will then be appropriate to write "Track P" to designate a track or an anti-track when it is not necessary to specify whether it is a track or an anti-track.

The track P(E) and the anti-track P'(E) are part of a set of pairs of "Tracks P" which are referred to as "onjugated tracks" and whose definition is as follows, in which the concepts of invalidity and validity are those previously given for the tracks and anti-tracks:

Definition :
A P1 Track and a P2 Track are conjugated when they cannot be disabled simultaneously.
If Track P1 is invalid, then Track P2 is valid, and vice versa.


Thus, P(E) and P'(E) are conjugated according to the property (P3), but there are other pairs of conjugated P Tracks as shown in the following theorem :

Theorem 1 :
Let E1 and E2 be two distinct sets ( E1∩E2 ≠ Ø) of candidates from G.
If P'(E1UE2) is invalid, then P(E1) and P(E2) are conjugated.


Indeed:
if P(E1) and P(E2) were disabled simultaneously, then no candidate of E1 and E2, therefore of E1UE2, would be solution, which is absurd since P'(E1UE2) disabled implies that a candidate of E1UE2 at least is solution.

For these pairs of conjugated tracks that generalize the couple P(E)/P'(E) we can state the following two theorems.

Theorem 2 :
P1/P2 being a pair of conjugated tracks, any candidate from G who sees* both a candidate from P1 Track and a candidate from P2 Track can be eliminated.

* It is said that candidate A sees candidate B, if A and B belong to the same area (block, row or column).

Indeed:
Let M be a candidate who sees both A ∈ Track P1 and B ∈ Track P2. If M is solution in its box, then A and B are not solutions => P1 Track and P2 Track are invalid, which is impossible. So M cannot be a solution and can therefore be eliminated.

This theorem has an obvious corollary that is very useful in practice and is as follows:

Theorem 3:
P1/P2 being a couple of conjugated tracks, any candidate of G common to P1 Track and P2 Track is a solution in his box.

Indeed:
In the box of this candidate A common to P1 Track and P2 Track , all the other candidates in the box see A and can therefore be eliminated.

Here is a very simple example to illustrate these results.

G=....9..5.9.5..346.......9.871.6...4...4.5.6...6...7.218.1.......574..1...2..7....

On the puzzle simplified by the basic techniques (TB), we choose:

E1={4r1c2} and E2={2r12c4}, so E1UE2={4r1c2, 2r12c4}
Obviously, we have P'(E1UE2) invalid (rectangle prohibited 78r12c24) => P(E1) and P(E2) is conjugated tracks.
P(E1) ={4r1c2, 4r3c4, ....}.
P(E2)=P'(78r1c4) ∩P'(78r2c4)=P(2r1c4) ∩P(2r2c4) = {6r3c6, ...}.
6r3c6 can be eliminated because it sees 6r3c5 and 4r3c6 (theorem 2).
Image
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Re: Technical of tracks (TDP)

Postby eleven » Sun Nov 10, 2019 1:11 pm

Your image did not work. Here is an example, how to post one:
Code: Select all
[img]http://i56.tinypic.com/xlwn7b.jpg[/img]

Image

This is the grid.
Code: Select all
+-------------------------+-------------------------+-------------------------+
| 1      A4'78*   268     |B2'78*   9      A268'4   | 27      5       3       |
| 9       78*     5       |B2'78*   1       3       | 4       6       27      |
| 2346    2347    236     | 257    B6'2    A4'25-6  | 9       1       8       |
+-------------------------+-------------------------+-------------------------+
| 7       1       2389    | 6       238     289     | 358     4       59      |
| 23      389     4       | 12389   5       1289    | 6       3789    79      |
| 5       6       389     | 389     4       7       | 38      2       1       |
+-------------------------+-------------------------+-------------------------+
| 8       349     1       | 2359    236     2569    | 27      379     245679  |
| 36      5       7       | 4       2368    2689    | 1       389     269     |
| 346     2       369     | 13589   7       15689   | 358     389     4569    |
+-------------------------+-------------------------+-------------------------+
UR 78r12c24: 4r1c2 = 2r12c4
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Re: Technical of tracks (TDP)

Postby StrmCkr » Sun Nov 10, 2019 1:30 pm

POM not being from my point of view a global theory.
- interesting that you don't think/consider its a global theory when its a n-set satisfaction problem and pom is able to cycle all 46656 templates for each digit in full combinations and partitions of uniform sections to solve n-set satisfaction. template delete/pom is the bases for and confirmation for pretty much every technique developed to-date. the biggest problem for pom is how long it take it to find/confirm eliminations in very large sets of information.
tabling

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).
agreed many of us have multiple overlapping discoveries. I will contest misinformation.

do not use this source as a date reference: to http://www.sudokuwiki.org/3D_Medusa

Andrew stewards source material is often not actuate and only updated to the dates he publicized or adapted it to his solver.

the real sudokuwiki was lost and Andrew conveniently acquired the rights to the name to draw more people to his page.
his original website can still be referenced and entered via googling scanraid

we have a recovered copy of some of the original content on this sudopedia

I happen to have published my work only in France and in French, and as a result they are totally ignored by the Anglo-Saxon world.
that's also a misconceptions, many of the contributes on this site spend hours combing multiple multi-langue sites for information to adapt/translate directly and we reference all source material back to the original webpages.
many of those that are here are not Anglo-Saxon.

so where exactly is your predated work/research not a recently publish paper.

which by the way has a publication date of: sunday, March 18, 2018, 4:18:24 PM

the biggest headache we will acknowledge is the loss of 3 of the major original source material. aka sudoku.org, sudoku programmers forum, our original hosting site
a lot of the stuff on this page has been restored by very dedicated people from data base backups from our own web-crawlers. because of this some topics dates of publication are only verifiable via the waybackmachine even then its an approximation based on multiple tracing sources.

closing remarks, as i said before do not get me construed i will digest what you post and critique and evaluate it for merit.

interesting als a & als b with strong links on 7, transport on digit 4 eliminates digit 2&6.
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+---------------------+----------------------+-------------------+
| 1       (478)  268  | (278)  9     2468    | 27   5     3      |
| 9       (78)   5    | (278)  1     3       | 4    6     27     |
| 236(4)  3(47)  236  | 25(7)  (26)  5-26(4) | 9    1     8      |
+---------------------+----------------------+-------------------+
| 7       1      2389 | 6      238   289     | 358  4     59     |
| 23      389    4    | 12389  5     1289    | 6    3789  79     |
| 5       6      389  | 389    4     7       | 38   2     1      |
+---------------------+----------------------+-------------------+
| 8       349    1    | 2359   236   2569    | 27   379   245679 |
| 36      5      7    | 4      2368  2689    | 1    389   269    |
| 346     2      369  | 13589  7     15689   | 358  389   4569   |
+---------------------+----------------------+-------------------+

or you know we could do a simple....
Finned X-Wing: 2 c14 r35 fr1c4 fr2c4 => r3c56<>2 ->> R3C5 = 6 ....

and both really avoid the unique rectangle arguments.

for giggles ALS - W - RING
Code: Select all
+-------------------+----------------------+-------------------+
| 1     (478)  26-8 | (278)   9     (2468) | 27   5     3      |
| 9     (78)   5    | (278)   1     3      | 4    6     27     |
| 2346  34(7)  236  | 5-2(7)  (26)  45-26  | 9    1     8      |
+-------------------+----------------------+-------------------+
| 7     1      2389 | 6       238   289    | 358  4     59     |
| 23    39-8   4    | 12389   5     1289   | 6    3789  79     |
| 5     6      389  | 389     4     7      | 38   2     1      |
+-------------------+----------------------+-------------------+
| 8     349    1    | 2359    236   2569   | 27   379   245679 |
| 36    5      7    | 4       2368  2689   | 1    389   269    |
| 346   2      369  | 13589   7     15689  | 358  389   4569   |
+-------------------+----------------------+-------------------+


Last edited by StrmCkr on Mon Nov 11, 2019 10:14 am, edited 6 times in total.
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