The New Sudoku Players' Forum

[SpAce]

i do not take part in your notation games.

Yet you just did, like many times before. If you really don't want to play that game, then why even comment my solution? On my part I definitely welcome such quality control, except for the fact that the resulting discussions tend to fall into déjà vu spirals which never end well. I'd rather break that cycle if only I knew how. I'm tired of useless and especially repeating conflicts.

So, what should I do if I can't honestly agree with you? Seems like you don't really appreciate me talking back to you and defending my points of view. Do you expect me to automatically concede because you're one of the top senior members of this forum and have superior sudoku skills in general? If we were talking about MUGs, for example, I'd be happy to do that (and have). Not likely with this particular topic, though.

When i see a solution in whatever notation, i try to find out the logic contained in it, and as far as i remember i always succeeded, if the solution was correct. Not in this case.

That's not quite accurate. You've failed to see the logic in several of my solutions, mostly due to similarly tricky links. Yet you've almost always accepted them as correct after my explanations, even though you've hated them without exception. (Of course that's not counting any careless mistakes which I've gratefully corrected immediately, or my oldest chains when I couldn't write even basic ALS terms correctly.)

The difference this time is that neither one of us has conceded even after several rounds of explanations. I can't remember but a couple of other cases when more than one round was needed. That makes this a pretty interesting case, so I'm definitely glad I wrote it, even if you aren't.

You mangled 3 cases into one AIC style link

No, just 2 into one AIC style link. I never even considered 3 cases, or thought about the actual combinations in the cells. What I saw was simply this:

Code: Select all

(3,8)r3c6,r7c9 = (-3r3c6 | -8r7c9)

Now, I must admit that this discussion made me doubt that logic for a while. Was it too simplistic? Did I miss valid cases? I don't actually think so, even though it was partly luck. In fact, after pondering this for about ten minutes I'm conjecturing that it's a general rule for similar situations, and it works no matter how many candidates those cells have. Technically it should have a third case: (-3r3c6 & -8r7c9) but since it's a union of the other two we can dismiss it for any practical purposes.

Like I said, I just came up with that rule. Am I overlooking something obvious?

[Added. Actually I think I was, but apparently it changes nothing. To really cover all cases systematically, I think they should be these:

Code: Select all

(3,8)r3c6,r7c9 = ((-3r3c6 & 8r7c9) | (3r3c6 & -8r7c9) | (-3r3c6 & -8r7c9))

Lucky for me, that reduces exactly to what I conjectured anyway. I used this boolean calculator to check the result:

Code: Select all

(~a * b) + (a * ~b) + (~a * ~b) <=> (~a + ~b)

So, despite my slightly incomplete approach originally, it seems that my logical instincts were correct nevertheless. Does anyone disagree?

Added more. Actually I think result should be pretty self-evident. To prove a verity we need show that A and its logical complement -A both prove the same thing, right? Well, if A is (3,8)r3c6,r7c9, i.e. (3r3c6 & 8r7c9), we have:

(3r3c6 & 8r7c9) = -(3r3c6 & 8r7c9) <-> (-3r3c6 | -8r7c9)

Basic De Morgan's theorem, right?
]

Btw, if you insist on using the digit combos, how about this:

(38=33|88)r3c6,r7c9 - (8)r7c9&r8c6 = (8-6)r7c5 = RP(38)r38c6,r7c59 => -38 r3c9

and when i tried to verify the first one, it turned out to be wrong. Now i see, that it would be right, if i had used the strong links for 8 instead of the candidates links in the cells.

To be honest, I can't see why it wasn't obvious to you that the bilocation strong links were the key. Nothing in what I wrote indicated that there was any bivalue linking happening in the r3c6 and r7c9 cells. The only bivalue strong link used in the whole logic was in r8c6 (and that's buried in the internal logic of the 38b8p26 node). Are you sure that your own preconceptions didn't make you blind to what was actually written in the notation?

The conclusion is, that the link is both right and wrong, which i cannot accept.

No. It's one or the other but not both. This is not a Schrödinger's cat situation.

--
Edits. Added the boolean calculation and some ponderings.

[denis_berthier]

No need of any artificially complicated rule and no need of any debate on any artificially complicated notations:

Hidden Text: Show

***********************************************************************************************
*** SudoRules 20.1.s based on CSP-Rules 2.1.s, config = W+SFin
*** using CLIPS 6.32-r764
***********************************************************************************************
naked-pairs-in-a-row: r9{c5 c7}{n3 n6} ==> r9c9 ≠ 3, r9c6 ≠ 3, r9c3 ≠ 6
naked-single ==> r9c6 = 4
hidden-single-in-a-row ==> r8c3 = 4
hidden-single-in-a-block ==> r9c3 = 7
naked-single ==> r9c9 = 5
hidden-single-in-a-block ==> r8c9 = 7
finned-x-wing-in-columns: n8{c7 c6}{r8 r1} ==> r1c5 ≠ 8
whip[1]: r1n8{c9 .} ==> r3c9 ≠ 8
biv-chain[3]: r1c7{n3 n8} - r8n8{c7 c6} - c6n3{r8 r3} ==> r3c9 ≠ 3, r1c5 ≠ 3
stte

[SpAce]

No need of any artificially complicated rule and no need of any debate on any artificially complicated notations

Maybe not. Yet you needed two non-basic steps to solve it (fewer than usual, I give you that). I know it's because of your "simplest first" strategy, but it still takes luck or skill to miss a single-step solution here In this case both of your steps are actually needed too, because your W-Wing is one of the three that doesn't solve it alone. The other four would (like rjamil's). Like I said, well aimed!

I can only speak for myself, but I bet that the main reason for any artificial complexity in the guaranteed manual solutions (mine and both Steves', I presume) was because this was too easy otherwise. In fact, it's one of the most trivial puzzles ever seen here. Normally these aren't supposed to be solvable in a single step with a basic W-Wing (one of the easiest patterns to spot), much less four different ones (two of which are also XY-Chains[4]). Many other "one-letter wings" (chains of length 3) also solve it in a single step, as shown by Leren's and Phil's L3-Wings. For those reasons some of us made it a bit more interesting for ourselves with some artificial complexity.

(That said, SteveC's solution is actually very reasonable despite the apparent complexity of a kraken. That deadly pattern is relatively easy to spot for a human, and the kraken branches are very simple in this case. So it might be one of the easiest yet most interesting manual solutions, which is typical of uniqueness solutions in general. Mine is also very easy if executed as two steps (Skyscraper and W-Wing). The only complication and interesting bit is in combining them into a single step.)

[SpAce]

No need of any artificially complicated rule

You're right. If you look at the third edition of that rule, and agree with the logic there, there really is no need for any artificially complicated rule to accept my original solution (or at least its base logic). Basic De Morgan's theorem is enough.

[denis_berthier]

Real players don't care about single step solutions. This is an artificial constraint that a handful of players in this forum have added. If it's fun for them, I have no problem with this. But what I don't agree with is letting new players believe that this is really Sudoku solving. Systematically presenting artificially complicated patterns as if they were normal is misleading for new players.

[SteveG48]

Real players don't care about single step solutions. This is an artificial constraint that a handful of players in this forum have added. If it's fun for them, I have no problem with this. But what I don't agree with is letting new players believe that this is really Sudoku solving. Systematically presenting artificially complicated patterns as if they were normal is misleading for new players.

I agree that our little "one-step" game is not the same as regular Sudoku solving, and I hope that we're not misleading anyone. However, the one-step game is very educational when it comes to learning about logic chains. There's plenty of room for both activities.

[SpAce]

Real players don't care about single step solutions.

Are we fake players then?

This is an artificial constraint that a handful of players in this forum have added. If it's fun for them, I have no problem with this.

I think it's a perfect constraint for daily puzzles which should take little time yet be challenging enough for a wide audience. It makes them fun yet quick for advanced players, while the puzzles themselves are still relatively easy to solve "normally" for others.

Without that constraint the puzzles should be much harder to be interesting for advanced players, and as such they would be totally beyond the skills of most players. They would also take a lot of time, so few of even the best would bother to solve them regularly or to read others' mostly boring solutions either. (Of course none of that applies to totuan and his beautiful solutions.)

On the other hand, everyone can focus on a single step both in their own and others' solutions, so those have a much higher chance of being creative and interesting.

But what I don't agree with is letting new players believe that this is really Sudoku solving.

It isn't?

Systematically presenting artificially complicated patterns as if they were normal is misleading for new players.

I don't buy that. Why does everyone seem to assume that new players are automatically idiots who can't think for themselves? One has to be pretty stupid to not realize that players posting single-step solutions for these are probably pretty good (or using software solvers, or both). For the most dedicated newbies that should provide both faith and motivation to get better. Exactly that happened to me at least.

You can read my comments here about the very first puzzle I ever looked at and solved here. (Too bad I didn't post my solution, though it wouldn't have been very interesting; besides I only knew the horrible Nice Loop notation back then.)

I didn't at first know the single-step nature of the challenge, so I solved it normally (pure p&p at that time) with around six chains, it seems. I was pretty happy with that, considering that less than six months earlier I'd been a mediocre basic solver who didn't know any techniques beyond X-Wing. However, when I saw the posted one-steppers I realized two things: 1) they were way beyond my skill level, and 2) I wanted to get there. Less than a year later I was posting solutions regularly.

Without that reality check, I would have probably thought I was already a hotshot solver and my skills would have stagnated. Finding this avenue made me see how much more I had yet to learn, and it also provided excellent examples from which to learn. I'm still regularly marveling and learning from others' ingenious solutions even though my own skills have improved quite a bit since then. I can't really think of a better and more fun way to learn either. Everything one learns here is also transferable to much harder puzzles that actually require such advanced techniques (and more).

[eleven]

No need of any artificially complicated rule

You're right. If you look at the third edition of that rule, and agree with the logic there, there really is no need for any artificially complicated rule to accept my original solution (or at least its base logic). Basic De Morgan's theorem is enough.

So you really do not admit, that your link is ambigous and therefore a BAD notation ?

I immediatlely saw, that (3r3c6->8r8c6) & 3r7c9 kill 38 in r7c5.
You also can directly see, that the two 3's see both r8c6 and r7c5, so no 3 is possible in the 2 cells. No strong link for 8 needed.
Having seen that, i know, that the link is wrong and stop.

Why should i try to find another explanation, that could save it ??
Because i know, that you are the one, who writes such silly things, just to get a chain shorter ?

A good and useful notation is never ambigous. Give me one sample of other notations, which are.

Your tries to reformulate it maybe make it more probable, that the implications with the 8-links are used to understand it, but as long as this not part of the notation, they are as bad as the original, because a reader still can prove it wrong using the other links.

[SpAce]

I agree that our little "one-step" game is not the same as regular Sudoku solving

Personally I don't fully agree with even that. My "regular" solving isn't really different at all from how I attack these. I approach these just like any other puzzle, and most of the time I find some single-step eliminations as an inevitable side-effect pretty quickly. If anything, solving these has taught me to think much more strategically when attacking any puzzle.

Very rarely the critical eliminations are so well hidden that my normal approach doesn't work, and only then do I cheat and look them up to know what I need to kill. That's the only difference, as in normal solving I obviously don't do that. There I just try to find the shortest path by making educated guesses about the puzzle's potential weaknesses. Usually I find multiple options before making a strategic decision about the next move. (In very difficult puzzles I just take what eliminations I find and hope they lead to more openings.)

(That said, for a long time now this has been pretty much the only sudoku solving I even do, except for some basic paper puzzles. In fact, the couple of puzzles I've solved here in the past few weeks have been all but the first non-basics all year for me. Thus, this is my "regular" solving anyway. That doesn't invalidate my point.)

[SpAce]

So you really do not admit, that your link is ambigous and therefore a BAD notation ?

No, I don't. I agree with the implication and the conclusion but not with the premise. I've admitted from the get-go that my chain is hard to read, which makes it bad notation for communication purposes (it's a pretty chain, though!). No argument about that. However, ambiguity is a much more serious accusation because it would mean my logic is incorrect. That's where I draw the line.

Once you prove to my satisfaction that the link is truly ambiguous instead of just hard to understand, I'll fix it. Problem is, I still don't see it that way, which means that one of us is not understanding something very fundamental. That's why I hope you bear with me until this is truly settled. If I have such a fundamental misunderstanding, I really want to fix that.

To get to the bottom of this, I need better explanations for these claims:

I immediatlely saw, that (3r3c6->8r8c6) & 3r7c9 kill 38 in r7c5.
You also can directly see, that the two 3's see both r8c6 and r7c5, so no 3 is possible in the 2 cells. No strong link for 8 needed.
Having seen that, i know, that the link is wrong and stop.
...
...because a reader still can prove it wrong using the other links
...
... and it quickly eliminates 38 from r7c5, proving the link wrong.

What exactly do you mean that those contradictions prove the link wrong? How do you draw that conclusion? If you get a contradiction it only proves the assumption wrong, nothing else. In this case you can see easily that (33) is an impossible combo. It's even easier to see that (88) is too. That leaves (38) as the only possibility which solves the puzzle, so we don't even need to worry about any links. What has any of that to do with my chain and its links???

In particular, how does any of that prove the strong link (3,8)r3c6,r7c9 = (38)b8p26 wrong? Or anything about it at all? This is the fundamental question. At least one of us has the wrong answer to it, and I think it's pretty important to get it right (especially if it's me).

To me those contradictions don't prove anything about the link, and it's totally irrelevant if you happen to see one that appears to do so. It certainly doesn't mean the (33) combo can't imply (38)b8p26. All three combos can do that just fine, and it doesn't matter one bit if you spot a contradiction that seems to make it impossible. Just find a different route in that case. Like I said from the get-go, a false premise can imply anything.

The only way you could prove the link wrong is if one of those combos simply couldn't imply (38)b8p26 or commit suicide by contradiction. The only way that can happen is if that combo is true and 38b8p26 is false. Not the case here:

Code: Select all

(3,8)r3c6,r7c9
||
(8,3)r3c6,r7c9 - 8r7c9 = (8,3)b8p26
||
(3,3)r3c6,r7c9 - 8r7c9 = (8,3)b8p26
||
(8,8)r3c6,r7c9 - 8r8c6 = (8,3)b8p26

=> (3,8)r3c6,r7c9 == (38)b8p26

In my mind that proves conclusively that the link is valid, even using your complicated point of view. Do you not agree? Where's the ambiguity? It's totally irrelevant if any (or all but one) of the branches produce contradictions with some other routes (it just proves them false). As long as all live branches can be made to agree with the conclusion we want, we're good, since at least one of them is guaranteed to be true and hence the conclusion as well. False branches can always be made to agree with anything or killed with a contradiction, so they're nothing to worry about.

That said, in case you missed the memo, my chain doesn't use those combos anyway. They're all in your head, not in my chain, so why do you keep bringing them up? They're certainly one complement of (3,8)r3c6,r6c9, and as can be seen it proves my link just as well, but it's not the one I used. Thus it's pretty irrelevant if you want to argue something about my actual chain. Why don't you take a look at the simpler logic I really used, and what the chain is supposed to depict? I thought I just proved with De Morgan's theorem that it's perfectly valid, and the link is much easier to understand that way too. Here it is:

Code: Select all

(3,8)r3c6,r7c9
||
(-3)r3c6 -> (3-8)r8c6 = (8)r7c5 // <-> (3,8)b8p62
||
(-8)r7c8 -> (8)r7c5 - (8=3)r8c6 // <-> (8,3)b8p26

=> (3,8)r3c6,r7c9 == (38)b8p26

Thus, in my point of view I've now proved in two different ways that the link is in fact perfectly correct. If it's correct it can't be ambiguous either. Do you have convincing counterarguments? If not, the only thing I might agree to change in the link is to mark it '==' since it's more or less derived. Incorrect or ambiguous it's not, unless better proof of that is provided.

Why should i try to find another explanation, that could save it ??

Perhaps because your original explanation doesn't make sense in the first place?

Because i know, that you are the one, who writes such silly things, just to get a chain shorter ?

Or perhaps because you should learn to appreciate beauty in all things?

A good and useful notation is never ambigous.

I fully agree. Like I said, I consider true ambiguity to be incorrect. I don't knowingly write incorrect chains. Ever. So, please read what I've said above, and tell me exactly where I get it wrong. Or not.

PS. Please don't mix in any subjective comments about my style. I know what you think about that. All I care about is whether you finally agree that the link is correct or not. That's the only interesting part.

[denis_berthier]

I don't think I need to write more about it. You're proving my points by yourselves.

[SpAce]

I don't think I need to write more about it. You're proving my points by yourselves.

Denis, instead of your constant and pointless criticism, why don't you make yourself useful and adjudicate our case? You're supposed to be a hotshot logician so it should be a piece of cake for you to see whose proof makes more (or less) sense. Right? I'm not being sarcastic. I'm actually interested in what you think.

The question is really simple. Is this a valid implication:

if NOT (3r3c6 AND 8r7c9) then (3 AND 8)r7c5,r8c6

...or not? The puzzle state with the relevant cells marked:

Code: Select all

.------------.------------------.---------------.
| 7  59   25 | 4    238     1   | 38   6   389  |
| 3  1    8  | 9    7       6   | 4    5   2    |
| 4  69   26 | 5    238   a[3]8 | 1    7   389  |
:------------+------------------+---------------:
| 8  4    1  | 3    9       5   | 7    2   6    |
| 5  2    3  | 6    1       7   | 9    8   4    |
| 6  7    9  | 8    4       2   | 5    3   1    |
:------------+------------------+---------------:
| 1  356  56 | 7  b(38)6    9   | 2    4  a3[8] |
| 2  36   4  | 1    5     b(38) | 368  9   7    |
| 9  8    7  | 2    36      4   | 36   1   5    |
'------------'------------------'---------------'


(3,8)r3c6,r7c9 == (38)b6p26 ???

[denis_berthier]

Is this a valid implication:

if NOT (3r3c6 AND 8r7c9) then (3 AND 8)r7c5,r8c6

...or not? The puzzle state with the relevant cells marked:

[code].------------.------------------.---------------.
| 7 59 25 | 4 238 1 | 38 6 389 |
| 3 1 8 | 9 7 6 | 4 5 2 |
| 4 69 26 | 5 238 a[3]8 | 1 7 389 |
:------------+------------------+---------------:
| 8 4 1 | 3 9 5 | 7 2 6 |
| 5 2 3 | 6 1 7 | 9 8 4 |
| 6 7 9 | 8 4 2 | 5 3 1 |
:------------+------------------+---------------:
| 1 356 56 | 7 b(38)6 9 | 2 4 a3[8] |
| 2 36 4 | 1 5 b(38) | 368 9 7 |
| 9 8 7 | 2 36 4 | 36 1 5 |
'------------'------------------'---------------'

NOT (3r3c6 AND 8r7c9) is equivalent to (NOT n3r3c6) OR (NOT n8r7c9); so, you already have branching in your conditions. Was that your intention?
Given your PM, this is also equivalent to n8r3c6 OR n3r7c9
Obviously:
- the first case (n8r3c6) implies NOT n8r8c6 , which implies n3r8c6
- the second case implies NOT n3r7c5, which implies n8r7c5 OR n6r7c5 and nothing more


" (3 AND 8)r7c5,r8c6" Sorry, but I don't understand what that means.

[eleven]

What exactly do you mean that those contradictions prove the link wrong? How do you draw that conclusion? If you get a contradiction it only proves the assumption wrong, nothing else. In this case you can see easily that (33) is an impossible combo. It's even easier to see that (88) is too. That leaves (38) as the only possibility which solves the puzzle, so we don't even need to worry about any links. What has any of that to do with my chain and its links???[

When i follow your argumentation, why do you not write
38r3c6,r7c9 = 9r3c9 => -36r3c9, stte

And if someone says, with both 3's in the 2 cells also 8 is possible in r3c9, you give the above answer.

[SpAce]

Hi Denis,

thanks for accepting the mission!

Is this a valid implication:

if NOT (3r3c6 AND 8r7c9) then (3 AND 8)r7c5,r8c6

NOT (3r3c6 AND 8r7c9) is equivalent to (NOT n3r3c6) OR (NOT n8r7c9);

Exactly. That was my point about the De Morgan's law earlier, and the rest of my logic follows from that.

so, you already have branching in your conditions. Was that your intention?

Yes. My logic proves (I hope) that both of those two branches imply that there's a pair (n3 & n8) in the cells r7c5 and r8c6 (without specifying the order, though it could be done in this case). That's the contested part.

If the previous logic is accepted, that pair then guarantees that n6r7c5 can't be true (hidden pair logic) and thus we have a Remote Pair (n3 & n8) in the cells r3c6 and r7c9, linked by r7c5 and r8c6, again without specifying any order of the digits. This part of the logic is not contested (I hope).

Given your PM, this is also equivalent to n8r3c6 OR n3r7c9

Yes, though my logic uses a different equivalence: (n3r8c6 OR n8r7c5). Do you agree with that too? It's based on the column and row CSPs instead of the cells: c6n3{r3 r8} and r7n8{c9 c5}. Both should work, so let's use yours here:

Obviously:
- the first case (n8r3c6) implies NOT n8r8c6 , which implies n3r8c6

Yes, but it doesn't stop there:

n8r3c6 -> -n8r8c6 -> n3r8c6 -> -n8r8c6 -> n8r7c5

or more directly:

n8r3c6 -> -n8r8c6 -> (n3r8c6 & n8r7c5)

Either way: n8r3c6 -> (n3r8c6 & n8r7c5).

- the second case implies NOT n3r7c5, which implies n8r7c5 OR n6r7c5 and nothing more

That's true but it doesn't help, as you noted. What helps is noting this other route:

n3r7c9 -> -n8r7c9 -> n8r7c5 -> -8r8c6 -> n3r8c6

Thus: n3r7c9 -> (n8r7c5 & n3r8c6).

Thus, as far I'm concerned, both branches imply the same thing: (n3r8c6 & n8r7c5), which is what I wanted. Do you agree?

(3 AND 8)r7c5,r8c6" Sorry, but I don't understand what that means.

It just means that there's a pair with the digits n3 and n8 in those two cells, without specifying the exact cells of each digit. Either way there can't be any other digits (n6r7c5 specifically) as per hidden pair logic. In this case I could have also used exact cells because my logic implies (n3r8c6 & n8r7c5) specifically, both ways. It might have been easier to read that way, but I avoid overspecifying things. The rest of the logic doesn't depend on that order, as it simply states that if n6r7c5 is not true then we have a Remote Pair.