Robert's puzzles 2019-11-16

Post puzzles for others to solve here.

Robert's puzzles 2019-11-16

Postby Mauriès Robert » Sat Nov 16, 2019 9:43 am

Hi,

In this unpretentious section, I will propose puzzles that are not created by me but that seem interesting to me.

Here is a fairly easy first one.

..59...2..7.4..5....9...4.34.2...7.......1.5..3...68......2.6..2..1.8..5..7.3....

Hidden Text: Show
Image

What advanced technique do you propose to solve it?
This is my resolution by the TDP (see ici).
Once the puzzle is reduced by the basic techniques, I consider the tracks P(8r5c2) and P(69r5c2) which are conjugated.
P(8r5c2)∩P(69r5c2)={3r4c4, ...} solution, end.

Hidden Text: Show
P(8r5c2) is marked in yellow, P(69r5c2) is marked in blue.
The yellow candidates surrounded by blue are those of P(8r5c2)∩P(69r5c2).

Image

Robert
Last edited by Mauriès Robert on Sat Nov 16, 2019 3:17 pm, edited 6 times in total.
Mauriès Robert
 
Posts: 594
Joined: 07 November 2019
Location: France

Re: November 16, 2019

Postby Ajò Dimonios » Sat Nov 16, 2019 11:30 am

Hi Robert.

The theory of the tracks allows to find a solution in a step after basic working on the conjugate tracks P (5R4C6) and P (9R4C6) or P (3R8C3) and P (3R8C8) or P (6R8C2) and P (9R8C2).

Ciao a Tutti
Paolo
Ajò Dimonios
 
Posts: 213
Joined: 07 November 2019

Re: November 16, 2019

Postby eleven » Sat Nov 16, 2019 12:54 pm

Hi Robert,

a problem with your presentation is, that we cannot follow it step by step, and don't know, when to stop.

After placing 8r5c2 we have to see the 59 pair in row 4 to get 38 in r4c45, but it is not mentioned. Having that, the puzzle could be solved to the end with singles.

Similar with the other path. You did not mention the 69 pair r58c2 and 589 triple then in r4c256 to get the 3r4c4 - singles from here too.

So the question always is, where to stop the development of the paths.
The shortest elimination -6r4c2 from the starting candidates (8r5c2->6r5c3 or 69r58c2) is not even mentioned, though it directly leads to the triple 589 and 3r4c4 (for both paths).

In AIC: (68=9)r5c23 - (9=6)r8c2 => -6r4c2
To get a stte solution you can add the triple on both ends
(589=6)r4c256 - (68=9)r5c23 - (9=6)r8c2 - (6=958)r4c256 => -58r4c4, stte
eleven
 
Posts: 3152
Joined: 10 February 2008

Re: November 16, 2019

Postby SpAce » Sat Nov 16, 2019 1:21 pm

If one insists on a singles solution:

Code: Select all
.--------------------.-------------------.------------------------------.
| 68     4       5   | 9     678    3    | 1        2            678    |
| 1368   7       368 | 4     168    2    | 5        689          689    |
| 168    2       9   | 5678  15678  57   | 4        678          3      |
:--------------------+-------------------+------------------------------:
| 4     b5689    2   | 358   589    59   | 7    ae([16])-39  ae([16])-9 |
| 6789  c986    c86  | 2378  4789   1    | 239      5           d6249   |
| 579    3       1   | 257   4579   6    | 8        49           249    |
:--------------------+-------------------+------------------------------:
| 3589   1589    348 | 57    2      4579 | 6        134789       14789  |
| 2     b69      346 | 1     679    8    | 39       3479         5      |
| 5689   15689   7   | 56    3      459  | 29       1489         12489  |
'--------------------'-------------------'------------------------------'

(16)r4c89 = (69)r48c2 - (9=86)r5c23 - r5c9 = (61)r4c89 => -39 r4c89; stte

...which seems to be just a hidden-pair variant of eleven's stte-solution.

Otherwise the obvious (and my preferred) answer is the simple XYZ-Wing also mentioned by eleven:

(68)r5c23 = (96)r58c2 => -6 r4c2; btte

The longer version is just an extension of that to get an stte finish, so it's really the core of the solution. The same cells obviously include your starting point as well, though you seem to use them a bit differently.

--
FYI:
  • stte = singles to the end
  • btte = basics to the end (in this case a single hidden pair (16)r4c89 or a naked triple (589)r4c256)

Btw, reference links to: AICs and the Eureka notation, in case you're not familiar with them. The latter is not quite up to date, though, but it explains the basics. The weirdest thing probably is that '=' is not equality or assignment but a strong link (OR). Similarly '-' is a weak link (NAND). The basic idea of an AIC is that at least one or the other end point of the chain must be true, i.e. they have a derived strong link, so anything that sees both of them can be eliminated. In my coloring those are seen as trap eliminations, so there's a direct relationship with AICs.
Last edited by SpAce on Sat Nov 16, 2019 2:18 pm, edited 2 times in total.
-SpAce-: Show
Code: Select all
   *             |    |               |    |    *
        *        |=()=|    /  _  \    |=()=|               *
            *    |    |   |-=( )=-|   |    |      *
     *                     \  ¯  /                   *   

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017

Re: November 16, 2019

Postby eleven » Sat Nov 16, 2019 2:09 pm

Robert,

please don't use the date as title in this format.
For us it is reserved for ArchieTech, who provides us with (almost) daily puzzles since years.
You can change it, if you would be so kind.
eleven
 
Posts: 3152
Joined: 10 February 2008

Re: November 16, 2019

Postby Mauriès Robert » Sat Nov 16, 2019 3:28 pm

Hi Eleven,

eleven wrote:In AIC: (68=9)r5c23 - (9=6)r8c2 => -6r4c2
To get a stte solution you can add the triple on both ends
(589=6)r4c256 - (68=9)r5c23 - (9=6)r8c2 - (6=958)r4c256 => -58r4c4, stte


Thank you for these remarks.
I will try to be more explicit.

For this puzzle, the path I followed for both colors is as follows:
Yellow : 8r5c2=>6r5c3=>589r4c256=>3r4c4
Blue : 69r5c2=>589r4c256=>3r4c4

How you write this with your writing convention ?

For the name of the section I propose, without pretension, Robert's puzzles.

Robert
Mauriès Robert
 
Posts: 594
Joined: 07 November 2019
Location: France

Re: November 16, 2019

Postby Mauriès Robert » Sat Nov 16, 2019 4:04 pm

Hi SpAce,

SpAce wrote:FYI:
  • stte = singles to the end
  • btte = basics to the end (in this case a single hidden pair (16)r4c89 or a naked triple (589)r4c256)

Btw, reference links to: AICs and the Eureka notation, in case you're not familiar with them. The latter is not quite up to date, though, but it explains the basics. The weirdest thing probably is that '=' is not equality or assignment but a strong link (OR). Similarly '-' is a weak link (NAND). The basic idea of an AIC is that at least one or the other end point of the chain must be true, i.e. they have a derived strong link, so anything that sees both of them can be eliminated. In my coloring those are seen as trap eliminations, so there's a direct relationship with AICs.


Thank you for this information.
Look also at my answer to Eleven.

Robert
Mauriès Robert
 
Posts: 594
Joined: 07 November 2019
Location: France

Re: November 16, 2019

Postby eleven » Sat Nov 16, 2019 4:42 pm

Mauriès Robert wrote:Yellow : 8r5c2=>6r5c3=>589r4c256=>3r4c4
Blue : 69r5c2=>589r4c256=>3r4c4

How you write this with your writing convention ?

It's a perfect notation for me, apart from the missing pair 69r58c2.
In AIC you would write it something like
Yellow : 8r5c2 - (8=6)r5c3 - (6=5893)r4c2345
Blue : 69r58c2 - (6=5893)r4c2345
=> 3r4c4, stte
eleven
 
Posts: 3152
Joined: 10 February 2008

Re: November 16, 2019

Postby SpAce » Sat Nov 16, 2019 5:07 pm

Hi Robert,

Mauriès Robert wrote:For this puzzle, the path I followed for both colors is as follows:
Yellow : 8r5c2=>6r5c3=>589r4c256=>3r4c4
Blue : 69r5c2=>589r4c256=>3r4c4

That's valid implication chain (or forcing chain) notation, and I'm sure we all can understand that. It shows the logic clearly, so it's definitely an improvement. It's also easy to convert into an AIC, and I'll show how shortly. I would suggest just a couple of minor tweaks first. Here's how I'd write it as a forcing chain:

8r5c2 -> 6r5c3 -> 589r4c256 -> 3r4c4
69r58c2 -> 589r4c256 -> 3r4c4

=> +3 r4c4; stte

One minor change is that '->' is used for the minor implications and '=>' for the final conclusion. It makes the latter easier to read, and it's important to show so we can more easily see what you wanted to prove. I also added spaces between the nodes for readability. Last but not least, I switched your 69r5c2 into a locked set in two cells 69r58c2 because that's how you probably intended it to work but it's not shown.

Furthermore, if you write 69 we interpret that to be a shorthand for 6&9, i.e. 6 AND 9. Obviously 6 and 9 can't both be true in one cell, so the correct interpretation is 6 OR 9, which we would write (6|9)r5c2. There's no real ambiguity with a single cell, but it's still preferable that way. I know it caused me a bit of a headache when I first read your solutions because I naturally saw it as 6 AND 9 (which works for the two-cell locked set but not for a single cell).

How you write this with your writing convention ?

Like I said, we can all understand that previous notation, so don't take any pressure about using AICs until you feel comfortable with them. We're happy to help, however, if you want to learn. It's actually quite easy, especially if we add the negative implications to the forcing chain:

8r5c2 -> -8r5c3 -> 6r5c3 -> -6r4c2 -> 589r4c256 -> -58r4c4 -> 3r4c4
69r58c2 -> -6r4c2 -> 589r4c256 -> -58r4c4 -> 3r4c4

Now we can just turn the implications into weak and strong inferences which must be alternating in an AIC (alternating inference chain). The first thing to notice is that your starting nodes are strongly linked (together they form an almost locked set, if we use my little change). Horizontally that's notated with the '=', but since we keep the two chains in two lines at first, we'll use '||' to denote a vertical strong link:

(8)r5c2 ...
||
(96)r58c2 ...

Note that I also added parentheses around the digits because it's standard Eureka (and I find it more readable). Some prefer not to use them and that's fine too. They're mandatory if we use internal linking within a term, as in: (8=96)r58c2. I also reordered the digits 69 -> 96 for readability because our next linking digit is the 6. Just a nuance, but I prefer it like that. So, let's then add the rest. Since the links must be alternating, the next links are weak, and after that strong again, and so on:

(8)r5c2 - (8=6)r5c3 - (6=958)r4c256 - (5|8=3)r4c4
||
(96)r58c2 - (6=958)r4c256 - (5|8=3)r4c4

=> +3 r4c4; stte

That's actually an AIC, just written on two lines. If you "pull" it straight into a single line, it's a normal AIC. It proves that one or the other end must be 3r4c4, so it must be true and can be placed. However, it has a bit of redundancy, and we prefer things as simple as possible. We don't in fact need to prove a placement as long as we prove an elimination that leads to it. So, if we cut out the last term from both ends (colored red) and straighten it out, we have this:

(589=6)r4c256 - (6=8)r5c3 - (8)r5c2 = (96)r58c2 - (6=958)r4c256 => -58 r4c4, stte

One last tweak is to combine the colored (blue) nodes into a single ALS term:

(589=6)r4c562 - (6=8)r5c3 - (8=96)r58c2 - (6=958)r4c256 => -58 r4c4, stte

That's it. As you can see, it's almost the same as eleven's chain, and the relevant logic is exactly the same. Mine is almost the same as well, except that I used the hidden pair (16)r4c89 to prove -3r4c8. Both lead to the same placement +3r4c4 which solves the puzzle.

Did that help? Btw, if you really want to conclude with a placement, you can add that cell to the ALS at both ends as eleven suggested:

(3=5896)r4c4562 - (6=8)r5c3 - (8=96)r58c2 - (6958=3)r4c2564 => +3 r4c4, stte

For the name of the section I propose, without pretension, Robert's puzzles.

Seems good to me!
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017

Re: Robert's puzzles 2019-11-16

Postby Ajò Dimonios » Sat Nov 16, 2019 5:40 pm

Hi Space

This is my second post in this forum. I am very interested in the discussions you report on sudoku schemes.

SpAce wrote:


(16)r4c89 = (69)r48c2 - (9=86)r5c23 - r5c9 = (61)r4c89 => -39 r4c89; stte

...which seems to be just a hidden-pair variant of eleven's stte-solution.

Otherwise the obvious (and my preferred) answer is the simple XYZ-Wing also mentioned by eleven:

(68)r5c23 = (96)r58c2 => -6 r4c2; btte






I would like clarification on the use of the Eureka nomenclature.In the AIC that you reported the first strong inference is not clear to me. In fact, when the pair 16 in R4C89 is false, different combinations of truths can occur but only some are strong inferences.
The possibilities are as follows:
1) both numbers 1 and 6 are not present in R4C89, in this case the presence of 3 and 9 always in R4C89 is certain and consequently the inference is strong.
2) only one number between 1 and 6 is present in R4C89 and the other is either 3 in R4C8 or 9 in R4C9, in this case we cannot speak of strong inference because when the premise (16) R4C8 is false it does not imply that (39) R4C89 is true and therefore it is not certain that at least one of the two hypotheses is true.

Ciao a tutti
Paolo
Ajò Dimonios
 
Posts: 213
Joined: 07 November 2019

Re: November 16, 2019

Postby Mauriès Robert » Sat Nov 16, 2019 6:01 pm

Hi SpAce,

All this seems complicated to me, even if I understand that it is a universal language (Eureka) in the world of sudoku. It is also used in France, but I am a little bit of a special sudokist who is as uncomfortable as possible with obligations. :D
Actually, I don't think like you. I do not think in terms of chain but by considering a track as a puzzle in which the candidates are placed with the basic techniques as soon as the generator (candidate or group of candidates) has been placed.
Each track sometimes has a lot of candidates, so it is tedious to write all the construction sequences.
I also believe that to understand what I do, you have to understand the definitions I have given (Here): track, anti-track, generator, etc... It's another vision than yours..
Without using exactly the Eureka language, I will try to be understandable.
Thank you for your efforts to integrate me into your methods.

Robert
Mauriès Robert
 
Posts: 594
Joined: 07 November 2019
Location: France

Re: November 16, 2019

Postby SpAce » Sat Nov 16, 2019 7:40 pm

Hi Robert,

I sort of understand your point of view, but...

Mauriès Robert wrote:Actually, I don't think like you. I do not think in terms of chain but by considering a track as a puzzle in which the candidates are placed with the basic techniques as soon as the generator (candidate or group of candidates) has been placed.

I don't see any fundamental difference. That sounds like chaining to me, and your chosen term "track" implies a chain as well. It's a bit misleading if you're actually thinking something else. Nevertheless, anything you do can be expressed with chains, and they depict very clearly how those candidates get placed in the different tracks. Your original way of showing your logic hid those details, which made it very difficult to follow without doing it yourself. In that sense it's annoyingly similar to POM, which just basically states the results but doesn't show how it happened. The point of writing chains is to show what happens and why so the reader doesn't have to do it.

It's true that the philosophy behind AICs is a bit different from implication chains, because they're in fact boolean logic constructs, but they can just as easily be seen as implication chains (and I bet most do without ever even realizing what they actually are) and vice versa. I just see them as two sides of the same coin, and I'm comfortable with both. That's why it's pretty easy for me to see how you probably think and see the same logic.

Each track sometimes has a lot of candidates, so it is tedious to write all the construction sequences.

Most of them aren't usually needed. Do you think my chains depict the full GEM coloring I might have made to arrive at a conclusion? Thank goodness, no. The idea is to extract the relevant parts of the logic and write that as a chain, or a kraken, or a net, or a fish, or whatever captures the logic in a concise way that can be followed. In other words, those chains don't (often) depict how the logic was found, but that's not their purpose either. If that is requested, I can post a screenshot of my coloring or explain it in other ways. Writing the logic in a concise way is also an additional challenge that I happen to enjoy.

I also believe that to understand what I do, you have to understand the definitions I have given (Here): track, anti-track, generator, etc...

Well, as I've said before, you've made the presentation so complicated that they're not exactly easy to understand. I would suggest that you'd add a simple practical example of each and every term right next to it. Also, I don't really understand why you start by defining "anti-track" before you've defined "track". That doesn't seem like a logical order. It would also help if you made a concluding list of the different but similar-looking terms and explained how they're related and what makes them different (such as anti-track, opposite track, conjugated track...). Now they're all over the place, which makes it kind of hard to grasp the big picture.

It would also help tremendously if you tried to relate those concepts to the terms that are used more widely in normal chaining. Like I've said, it's much easier to learn something if people can easily see how it's related to something they already know. Now you've made it seem like your concepts are something totally new and have powers that no other technique does, which is simply not true, as far as I see. Make no mistake, I'm definitely not saying that you're purposefully making such claims, just that it's easy to interpret that way when there's no accurate comparison to other techniques.

It's another vision than yours..

Based on what I've seen and what I think I understand of your terms, I simply can't agree with that. There are minor philosophical differences, perhaps, but mostly I think you're just not seeing the fundamental similarities. Not to take anything away from your original ideas, but I think your special terms are just other names for the same concepts that have been used for ages in chaining. AICs are just a concise way of expressing those concepts, or most of them at least.

Without using exactly the Eureka language, I will try to be understandable.

Like I said, anything that we can understand without jumping through hoops is fine. I suggest using the forcing chain notation with the minor tweaks I proposed, because I think it depicts your way of solving quite closely. Writing fully correct AICs is harder, and like I said, it's philosophically a bit different.

Thank you for your efforts to integrate me into your methods.

No problem. Thanks for sharing your methods with us! Like I said, I still think they're not as far from ours than you think. There aren't that many fundamentally different ways to solve sudokus or express that logic. In fact, I only know two: chains (including nets) and fishes (including complicated ones like Allan Barker's). POM might be a third fundamentally different concept, but mostly it has no manual applicability. Yours is clearly in the chain department, although the way you present and apparently think about it, it does have some similarity to very much simplified POM like StrmCkr suggested.
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017

Re: Robert's puzzles 2019-11-16

Postby Mauriès Robert » Sat Nov 16, 2019 9:29 pm

Hi SpAce,

As for the technical remarks, I did not give a pedagogical presentation of TDP. I'll try to correct it.

SpAce wrote:Now you've made it seem like your concepts are something totally new and have powers that no other technique does, which is simply not true, as far as I see. Make no mistake, I'm definitely not saying that you're purposefully making such claims, just that it's easy to interpret that way when there's no accurate comparison to other techniques.


I have never written anything that says that, it is not my state of mind. I only presented the technique that I use in my own way and that I wrote in this form.
When I conceived this personal way of doing things, I had no knowledge of GEM and Hodoku I use does not mention it. Since 2013 that my website is online, no one has contacted me to tell me that TDP exists in equivalent forms.

So everyone thinks what they want and uses TDP or not as I do. I remain available for those who find it interesting.

Sincerely
Robert
Mauriès Robert
 
Posts: 594
Joined: 07 November 2019
Location: France

Re: Robert's puzzles 2019-11-16

Postby SteveG48 » Sun Nov 17, 2019 1:55 am

Code: Select all
 *-----------------------------------------------------------------------------*
 | 68      4       5       | 9       678     3       | 1       2       678     |
 | 1368    7       368     | 4       168     2       | 5       689     689     |
 | 168     2       9       | 5678    15678   57      | 4       678     3       |
 *-------------------------+-------------------------+-------------------------|
 | 4     cf5689    2       | 358     589     59      | 7     bg1369  bg169     |
 | 6789   e689    e68      | 2378    4789    1       |a3-29    5      h2469    |
 | 579     3       1       | 257     4579    6       | 8      h49     h249     |
 *-------------------------+-------------------------+-------------------------|
 | 3589    1589    348     | 57      2       4579    | 6       134789  14789   |
 | 2      d69      346     | 1       679     8       | 39      3479    5       |
 | 5689    15689   7       | 56      3       459     | 29      1489    12489   |
 *-----------------------------------------------------------------------------*


3r5c7 = (31-6)r4c89 = 6r4c2 - (6=9)r8c2 - (9=86)r5c23 - 6r4c2 = r4c89 - (6=249)b6p689 => -29 r5c7 ; stte

Or simpler, if we allow proof by contradiction:

3r5c7 - (31-6)r4c89 = (689)b4p256 - (6|9)r8c2 contradiction => r5c7 = 3
Steve
User avatar
SteveG48
2019 Supporter
 
Posts: 4481
Joined: 08 November 2013
Location: Orlando, Florida

Re: Robert's puzzles 2019-11-16

Postby SpAce » Sun Nov 17, 2019 2:53 am

Hi Steve,

SteveG48 wrote:3r5c7 - (31-6)r4c89 = (689)b4p256 - (6|9)r8c2 contradiction => r5c7 = 3

I like the idea, but I think it needs a strong link at the start and a comma. How about a bit shorter:

(31-6)r4c89 = (6,89)b4p265 - (6|9=!)r8c2 => -3 r4c8; stte

Even shorter and nicer if reversed and using Dan's compact kraken style:

(6|9)r8c2 - (98,6)b4p562 = (61)r4c89 => -39 r4c89; stte

I think that would actually be my preferred stte solution for this puzzle, avoiding the redundant start and end nodes which I hate. It's actually pretty close to Robert's paradigm as well. Related to that, I guess the ultimate shortening (that I would really hate to see) would be:

(6|9)r8c2 -> (61)r4c89 => -39 r4c89; stte
User avatar
SpAce
 
Posts: 2671
Joined: 22 May 2017

Next

Return to Puzzles