The New Sudoku Players' Forum

[SpAce]

whip[13]: r7c4{n7 n9} - r7c1{n9 n2} - c2n2{r9 r5} - r7c2{n2 n8} - r9n8{c2 c5} - r7c5{n8 n1} - r7c9{n1 n3} - r5c9{n3 n9} - r6n9{c7 c3} - c3n4{r6 r4} - r4c9{n4 n1} - r2c9{n1 n6} - r2c5{n6 .} ==> r9c6 ≠ 7

So you did find a whip to do it. Good. Here's the matrix for it:

14x14 TM: Show

Code: Select all

 7r7c4 9r7c4
 . . . 9r7c1 2r7c1
 . . . 9r7c1 . . . 2r7c1
 . . . . . . 2r7c2 2r9c2 2r5c2
 . . . . . . . . . . . . 2r7c2 8r7c2
 8r9c6 . . . . . . . . . . . . 8r9c2 8r9c5
 . . . 9r7c5 . . . . . . . . . . . . 8r7c5 1r7c5
 . . . . . . . . . 2r7c9 . . . . . . . . . 1r7c9 3r7c9
 . . . . . . . . . . . . 2r5c9 . . . . . . . . . 3r5c9 9r5c9
 . . . . . . . . . . . . . . . . . . . . . . . . . . . 9r6c7 9r6c3
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4r6c3 4r4c3
 . . . . . . . . . . . . . . . . . . . . . . . . 3r4c9 . . . . . . 4r4c9 1r4c9
 . . . . . . . . . . . . . . . . . . . . . . . . 3r2c9 9r2c9 . . . . . . 1r2c9 6r2c9
 . . . . . . . . . . . . . . . . . . 8r2c5 . . . . . . . . . . . . . . . . . . 6r2c5
====================================================================================
-7r9c6

The gap (normally indicating a braid) and the 13+1 size are due to breaking the group node, so they don't count against the whip[13] classification.

For practical purposes, I think that's actually way more complex than Robert's braid[13]. Most importantly it has 1 x 4-SIS and 6 x 3-SISs, meaning that over half (7/13) of its SISs (or CSP-variables) have more than two candidates. Robert's braid has only four 3-SISs (4/13). Consequently the latter has fewer candidates (30 vs 33) too. For a manual solver those are more meaningful differences than the continuity of the core chain, which makes little difference except as an aesthetic value in some forms of notation.

That said, I do (kind of) like the theoretical distinction between whips and braids because they look different in the matrix form (whips being prettier). I just don't think braids are necessarily more complex or difficult to find for a manual solver. In the practical solving context it's more or less an artificial distinction, because both are found with the same exact process. The biggest practical difference is that whips are easier to notate as linear memory chains, but that isn't a great way to notate nets anyway. In most complete and readable forms of net notations there are hardly any meaningful differences.

[denis_berthier]

The gap (normally indicating a braid) and the 13+1 size are due to breaking the group node, so they don't count against the whip[13] classification

If there's a group node, it's not a braid but a g-braid and it's meaningless to "break" it. If you do have to break it because it doesn't link to anything further in the chain, it's not even a g-braid.

For practical purposes, I think that's actually way more complex than Robert's braid[13]. Most importantly it has 1 x 4-SIS and 6 x 3-SISs, meaning that over half (7/13) of its SISs (or CSP-variables) have more than two candidates. Robert's braid has only four 3-SISs (4/13). Consequently the latter has fewer candidates (30 vs 33) too. For a manual solver those are more meaningful differences than the continuity of the core chain, which makes little difference except as an aesthetic value in some forms of notation.

You declared several times that you don't understand whips and braids and once more, your random claims show you were right.
Obviously, you have no notion of the difference in complexity between a whip, a braid and a g-braid. The continuity condition is an essential difference when looking for them, be it by hand or by machine, because it avoids a lot of possibilities for extending the partial chains; conversely, the number of z- or t- candidates is almost totally irrelevant in terms of complexity. But, as usual, instead of trying to understand, you have to feed your narrative, otherwise you wouldn't have your daily quota of posts. Sure, it's easier to move your fingers on a keyboard than your neurones.

No new comment is needed on your repeated rantings about my notation or your inability to understand the difference between a chain and a net.
As for matrices, I've never seen any matrix drawn on a Sudoku grid; but a whip or braid can easily be represented on the grid in exactly the same way as a mere bivalue-chain and in a way that strictly corresponds to what is written in my notation.

[SpAce]

If there's a group node, it's not a braid but a g-braid and it's meaningless to "break" it. If you do have to break it because it doesn't link to anything further in the chain, it's not even a g-braid.

<sigh> Let me quote yourself back to you:

Sure, it's easier to move your fingers on a keyboard than your neurones.

Apparently. You see, what you wrote above had nothing to do with what I wrote. I thought I clearly agreed that what you found was indeed a whip[13] even though the matrix didn't make it as obvious as usual. Not a whip[14]. Not a g-whip. Not a braid. Not a g-braid. What part did you not understand? Are you compelled to argue even about things we agree on?

FYI 1: I've known for a good while that left-linking group nodes don't exist (visibly) in your whips/braids, because they're handled as t-candidates. Only right-linking ones are visible and cause an upgrade into a g-whip/g-braid. It was admittedly unintuitive at first, but it's obviously the only consistent way to do it, so I don't have any problem with it. Surprised?

FYI 2: The "breaking" was necessary to write the matrix without the left-linking group node. I did it to make it match the logic of your whip more completely. That way the matrix shows precisely that 2r7c2 is considered a t-candidate in your system. Surprised again?

Next time before you respond with completely irrelevant and scornful comments, please read and try to understand what was actually said. You've made the same mistake several times.

You declared several times that you don't understand whips and braids

Lol. Yes, I said that when I actually didn't have a clue about them and asked for your help (which you denied). It was in May. That's like an eternity ago in my cognitive cycle. If you don't want to appear extra-stupid, you probably shouldn't keep assuming that my understanding is still at the same level as half a year ago. But since you obviously do, there's no point in discussing anything. Believe whatever the hell you want if it makes you happy

[denis_berthier]

FYI 1: I've known for a good while that left-linking group nodes don't exist (visibly) in your whips/braids, because they're handled as t-candidates. Only right-linking ones are visible and cause an upgrade into a g-whip/g-braid. It was admittedly unintuitive at first, but it's obviously the only consistent way to do it, so I don't have any problem with it. Surprised?

Yes, surprised that you've at last understood a little bit of whips logic, in spite of your strong prejudices. It seems you needed some whipping for that.

FYI 2: The "breaking" was necessary to write the matrix without the left-linking group node. I did it to make it match the logic of your whip more completely. That way the matrix shows precisely that 2r7c2 is considered a t-candidate in your system. Surprised again?

Not surprised at all: matrices fail to capture the essential characteristics of chains. AND they can't be drawn on the grid.

You declared several times that you don't understand whips and braids

Lol. Yes, I said that when I actually didn't have a clue about them and asked for your help (which you denied). It was in May.

Which didn't prevent you, at that time, from making absurd critics. Criticising is your usual way of trying to hide your lack of understanding, while still filling your daily quota of posts. You reap what you sow.

That's like an eternity ago in my cognitive cycle. If you don't want to appear extra-stupid, you probably shouldn't keep assuming that my understanding is still at the same level as half a year ago.

Your current comments about complexity prove that you still have a long way to go before you understand whips or braids. See you in a few more eternities, at the pace of your cognitive cycle.


All that being said, let's keep in mind the only important thing that should be remembered wrt to this puzzle, a puzzle that can be solved with the most elementary patterns (as in my first solution): all the overly complicated solutions (including my two ones with long braids) are totally absurd.
One-step solutions suppose a player knows the backdoors in advance (which is a complex problem for a manual solver). This is hidden in the proposed solutions, whose apparent simplicity is then fake.
Note that I'm not totally against adding a requirement for a 1-step solution for pre-digested puzzles that allow one. Anyone is free to have fun as he likes. But it should be clear that this is not a standard Sudoku requirement.

[Mauriès Robert]

Hi,
To get out of these personal jousts that pollute the fundamental debate, I give my opinion on the following remark by Denis :

All that being said, let's keep in mind the only important thing that should be remembered wrt to this puzzle, a puzzle that can be solved with the most elementary patterns (as in my first solution): all the overly complicated solutions (including my two ones with long braids) are totally absurd.
One-step solutions suppose a player knows the backdoors in advance (which is a complex problem for a manual solver). This is hidden in the proposed solutions, whose apparent simplicity is then fake.

I do not agree with this opinion as regards, as I do, chains (tracks or anti-tracks) developed recursively for which no particular target is set in advance, it is useful to extend the development beyond the first elimination encountered to find others. In this way it is possible to obtain the eliminations obtained by several short successive steps in one go.
Furthermore, I find it interesting to find out which resolution is possible with a minimum of chain (track or anti-track), which requires not limiting oneself in length, as this reveals the level of difficulty of a puzzle. This puzzle solved in one step is for me level 1. A level 2 puzzle cannot be solved with only one step.
Cordialy
Robert

[SpAce]

Hi Robert,

I have a question for you. As we all know, you are a very good and accomplished manual solver. You've also demonstrated interest in Denis' system, so you probably know things like whips and braids better than most people here. (In fact, it was you who made me understand how to interpret whips in the first place, thanks for that.) For the same reason I presume that you have no prejudices against them or contradiction patterns in general, so you should be pretty objective. Wouldn't you agree?

My question: do you agree with any or all of this:

Obviously, you have no notion of the difference in complexity between a whip, a braid and a g-braid. The continuity condition is an essential difference when looking for them, be it by hand or by machine, because it avoids a lot of possibilities for extending the partial chains; conversely, the number of z- or t- candidates is almost totally irrelevant in terms of complexity.

When you build a track to find a direct contradiction for an assumption, the resulting pattern is an AND-net that could be written as either a whip or a braid (or their g- or S-variant). The same is true if you build an anti-track, though it's not so obvious because it's not a contradiction pattern natively (but can be seen as one, like anything). With conjugate tracks you build two such AND-nets that are ORed with each other, which corresponds with Denis' forcing braids.

Thus, all of your basic tracks are in fact whips or braids of some kind, unless they're simple chains. Realizing that, have you ever considered the "continuity condition" essential when building them? Do you actively (or intuitively) use it as a limiter to how you continue the track? In other words, do you agree with Denis' claim that whips are easier to find than braids, and the continuity condition is an essential measure of complexity? Have you ever analyzed what percentages of your tracks are whip-like and braid-like? (As a side, do you agree with Denis' claim that the number of z- and t-candidates is irrelevant?)

[denis_berthier]

my opinion on the following remark by Denis :

All that being said, let's keep in mind the only important thing that should be remembered wrt to this puzzle, a puzzle that can be solved with the most elementary patterns (as in my first solution): all the overly complicated solutions (including my two ones with long braids) are totally absurd.
One-step solutions suppose a player knows the backdoors in advance (which is a complex problem for a manual solver). This is hidden in the proposed solutions, whose apparent simplicity is then fake.

I do not agree with this opinion as regards, as I do, chains (tracks or anti-tracks) developed recursively for which no particular target is set in advance, it is useful to extend the development beyond the first elimination encountered to find others. In this way it is possible to obtain the eliminations obtained by several short successive steps in one go.

You missed my point: this puzzle can essentially be solved by Subsets, which is the first thing every puzzle solver learns to use, after singles and intersections. What's needed in addition to these elementary rules is a mere bivalue-chain - the next thing beginners learn.

[Mauriès Robert]

Hi Space,

I have a question for you. As we all know, you are a very good and accomplished manual solver. You've also demonstrated interest in Denis' system, so you probably know things like whips and braids better than most people here. (In fact, it was you who made me understand how to interpret whips in the first place, thanks for that.) For the same reason I presume that you have no prejudices against them or contradiction patterns in general, so you should be pretty objective. Wouldn't you agree?

Thank you for giving me as much quality as I don't have. I agree on only one point, that of showing interest in de Denis' colossal work, therefore in the whips and braids, and of having no prejudices on this subject.
But make no mistake, not being sufficiently qualified in abstract mathematics, I find it very difficult to understand the subtleties of the theory, all the more so as my "very very poor" English complicates my task.
Having said that, what I have understood about whips and braids is that they are "logical sequences" revealing a contradiction and therefore one or more eliminations. I found that they could be translated in an equivalent way either by an antitrack from the first candidate of the sequence or by a track from the target.
I also understood (this is clearly stated in Denis's book) that the essential difference between whip and braid is the condition of continuity which is no longer necessary in braids.

My question: do you agree with any or all of this:
When you build a track to find a direct contradiction for an assumption, the resulting pattern is an AND-net that could be written as either a whip or a braid (or their g- or S-variant). The same is true if you build an anti-track, though it's not so obvious because it's not a contradiction pattern natively (but can be seen as one, like anything). With conjugate tracks you build two such AND-nets that are ORed with each other, which corresponds with Denis' forcing braids.

Thus, all of your basic tracks are in fact whips or braids of some kind, unless they're simple chains. Realizing that, have you ever considered the "continuity condition" essential when building them? Do you actively (or intuitively) use it as a limiter to how you continue the track? In other words, do you agree with Denis' claim that whips are easier to find than braids, and the continuity condition is an essential measure of complexity? Have you ever analyzed what percentages of your tracks are whip-like and braid-like? (As a side, do you agree with Denis' claim that the number of z- and t-candidates is irrelevant?)

Tracks and anti-tracks are not, in general, chains but sequences (to use a term from Denis), so continuity is not a necessary condition and I don't use it. From this point of view they are therefore rather equivalent to braids, I won't go any further on the comparison because, as I said before, I haven't understood all the subtleties of Denis' theory.
That said, while a braid (or a whip) is limited in length, a track is not necessarily limited in length, which allows it to be developed in a greater amount generating more simultaneous elimination, which made me tell Denis that I did not agree with him about the absurdity of developing longer whips or braids.
In my opinion, yes, whips are easier to find than braids, precisely because of the condition of continuity.

Have you ever analyzed what percentages of your tracks are whip-like and braid-like? (As a side, do you agree with Denis' claim that the number of z- and t-candidates is irrelevant?)

No, I haven't done any research on this subject, but I quite often look at the solutions proposed by Denis to study their equivalent in terms of anti-tracks, very rarely in terms of tracks. Besides, I do it in the same way also with the AICs.
Cordialy
Robert

[Mauriès Robert]

Hi Denis,

You missed my point: this puzzle can essentially be solved by Subsets, which is the first thing every puzzle solver learns to use, after singles and intersections. What's needed in addition to these elementary rules is a mere bivalue-chain - the next thing beginners learn.

I understood you very well Denis, let's say it's the word "absurd" that makes me react.
As far as I'm concerned, after learning the basic techniques (TB), the simple ones, the intersections, but also alignments and closed sets, the only technique to know is the use of tracks. Therefore, whether a track is short or long, the work is the same (up to a reasonable length), but in the second case you get more simultaneous results. It is therefore not absurd to work on long tracks. It seems to me then that it would not be absurd to work with longer braids and fewer braids. That's just my opinion!
Cordialy
Robert

[denis_berthier]

You missed my point: this puzzle can essentially be solved by Subsets, which is the first thing every puzzle solver learns to use, after singles and intersections. What's needed in addition to these elementary rules is a mere bivalue-chain - the next thing beginners learn.

As far as I'm concerned, after learning the basic techniques (TB), the simple ones, the intersections, but also alignments and closed sets, the only technique to know is the use of tracks.

I think you're still missing my point, as you're not applying this approach to the present puzzle. To be clear, what tracks would you find if you first applied all the elementary rules to this puzzle?

[SpAce]

You missed my point: this puzzle can essentially be solved by Subsets, which is the first thing every puzzle solver learns to use, after singles and intersections. What's needed in addition to these elementary rules is a mere bivalue-chain - the next thing beginners learn.

As far as I'm concerned, after learning the basic techniques (TB), the simple ones, the intersections, but also alignments and closed sets, the only technique to know is the use of tracks.

I think you're still missing my point, as you're not applying this approach to the present puzzle. To be clear, what tracks would you find if you first applied all the elementary rules to this puzzle?

As usual, you're expecting other people to understand and abide by your non-standard definitions. Fishes are not subsets in normal sudoku parlance, and you know it very well. Fishes are also not considered elementary techniques by the conventions of this section, which you should know very well too. The same conventions also declare a preference to the shortest possible solutions, which you think is absurd.

Whether you agree with any of that is irrelevant. Those conventions are accepted by everyone else, which makes different solutions interesting and comparable, and accurate communication possible. You just don't think any rules or behavioral norms apply to you. If you respected other people at all, you'd adjust your own language to that of the majority instead of expecting others to adjust to a minority of one. For the same reasons you think you're entitled to throw any insults you like.

[denis_berthier]

As usual, you're expecting other people to understand and abide by your non-standard definitions. Fishes are not subsets in normal sudoku parlance, and you know it very well. Fishes are also not considered elementary techniques by the conventions of this section, which you should know very well too.

There are no "conventions of this section" except in your imagination.
Fish and Subsets are the same pattern, viewed in different spaces. They have always been considered as belonging to the basic patterns.

The same conventions also declare a preference to the shortest possible solutions, which you think is absurd.

A handful of players have recently (1 or 2 years ago) chosen to look for 1-step solutions. It's their right and it may be interesting for some puzzles. But it has never been a constraint of Sudoku and it has never been a condition for participating in this section. Sudoku solving is not limited to such techniques, which generally lead to overly complicated patterns and which are only possible for pre-digested puzzles.

Finally, "shortest possible" depends on one's interpretation of "short". No consistent definition of size based on the full resolution path has ever been given.

[Mauriès Robert]

Hi Denis,

I think you're still missing my point, as you're not applying this approach to the present puzzle. To be clear, what tracks would you find if you first applied all the elementary rules to this puzzle?

I gave a first resolution with 4 steps at the very beginning of this thread. Does this answer your question?
My remarks on the interest of making longer sequences are of a general nature. I'll give you an example to make me understand better.
On Robert's puzzles 2020-10-26 for which you give your solution, you have these two steps:
z-chain [3]: c1n3{r2 r9} - c1n4{r9 r3} - r1c3{n4 .} ==> r2c1 ≠ 7
z-chain[4]: b1n3{r2c1 r1c2} - b1n9{r1c2 r3c1} - c1n4{r3 r9} - c1n3{r9 .} ==> r2c1 ≠ 5
The first can be written in terms of TDP :
(-3r2c1)=>3r9c1->4r78c3->7r1c3 => -7r2c1
But if we extend this sequence like this
(-3r2c1)=>3r9c1->(4r78c3->7r1c3)->5r2c3 => -7r2c1 and -5r2c1
This avoids your second step.
Couldn't you write a longer sequence that would give these two eliminations at once?
This example is very simple here, but very often a sequence that allows one elimination can be developed is giving others, sometimes many others. This usefully reduces the number of steps, and manually this is not negligible. See my intervention in your diagonals+centres thread.
Cordialy
Robert

[SpAce]

There are no "conventions of this section" except in your imagination.

Really?

Fish and Subsets are the same pattern, viewed in different spaces.

A fact known to advanced players and probably coders of solvers. Not to average sudoku players. Most of them probably know subsets, but significantly fewer know (or can find) fishes, and almost none see them as manifestations of the same thing. Even fewer have ever heard of "different spaces".

They have always been considered as belonging to the basic patterns.

Where, except in your system? That's not true in this section, nor in common public software solvers, nor in Robert's TDP.

I also happen to agree with that decision. As a manual solver, I find fishes much harder to spot than normal subsets -- especially if solving on paper or without digit highlighting. They definitely don't belong to the same level, no matter what your theory claims. If I've understood correctly, I don't think Robert uses fishes at all, and I rarely use them either. They're easy for software solvers, not so easy for most humans (mith is an exception).

A handful of players have recently (1 or 2 years ago) chosen to look for 1-step solutions.

I've been around for over three years, and all of that time this convention has been in effect. More importantly, it doesn't take much effort to check the old posts and find out that the same policy has applied at least since 2012 (no continuous older records available). Didn't you bother to check, or are you gaslighting on purpose?

Added. Here's documentation of how long it took me to figure it out.

But it has never been a constraint of Sudoku

I don't think anyone is confused about that. I don't see even the slightest risk.

and it has never been a condition for participating in this section.

You're right, it's not a condition. Anyone can freely post multi-steppers and even include singles and basics in them, as long as they realize that no one probably reads them.

Nothing prevents one from posting multiple solutions either. For example, you could post both your "normal" solution for beginners (preferably hidden if very long) and also a short solution for the rest of us (preferably without the clutter of basics).

I'd have nothing against that, because it would provide a wider variety of examples of your notation for those who might want to learn it. Right now there's probably little to nothing for me to learn from your solutions, even notationally, so I'd rather see examples of more complicated patterns (starting with the S-variants). If you only use those when they're absolutely necessary, no one has a chance to learn how you notate them.

Sudoku solving is not limited to such techniques, which generally lead to overly complicated patterns and which are only possible for pre-digested puzzles.

There are very few patterns that have such limited applicability, mostly restricted to very hard puzzles. It's really nice that some people pre-digest and post such puzzles occasionally so that the rest of us get to practice finding those rare patterns. You just did that too (by accident or not, who knows), which I appreciated. (I don't think anyone minds that it didn't solve the puzzle. It's pretty irrelevant, at least for me.)

Added. In fact, the only reason why I'm still here despite my recent promise to leave is the resurgent interest in those "exotic patterns". My prior experience with them was so minimal that I couldn't pass up the learning opportunity. Don't worry, though. Probably sooner than later you shall have peace.

Other than those rarities, all of the techniques I use are applicable in "real world" puzzles. This is just a great way to practice them and to learn from others' ingenious moves. Added. Apparently I thought so even when I didn't have the skills to participate:

My solving style and skills don't support one-stepper hunting at this point, but seeing others' imaginative and highly varying solutions has been extremely instructive.

Btw, I was still solving purely on paper at the time. I started participating when I switched to Hodoku. It was so much faster, allowing to skip basics and to see immediately whether a found elimination was stte or not. That said, for a long time I felt crippled without the unique visual aids I'd developed for paper-solving, because they were (and are) more sophisticated than anything Hodoku (or anything else) could offer. Just slower to use, obviously, until the day comes that I get inspired to code my own helper.

Finally, "shortest possible" depends on one's interpretation of "short". No consistent definition of size based on the full resolution path has ever been given.

You're right, but here the convention is to simply count the non-basic steps (including fishes). Again, aiming for the smallest count is not in any way a hard rule. Just a general preference. If the smallest step-count requires some totally incomprehensible mess of a net, a simpler and more elegant multi-step solution is obviously preferred (though the former might be an interesting curiosity).

[denis_berthier]

Hi Robert,

My remarks on the interest of making longer sequences are of a general nature. I'll give you an example to make me understand better.
On Robert's puzzles 2020-10-26 for which you give your solution, you have these two steps:
z-chain [3]: c1n3{r2 r9} - c1n4{r9 r3} - r1c3{n4 .} ==> r2c1 ≠ 7
z-chain[4]: b1n3{r2c1 r1c2} - b1n9{r1c2 r3c1} - c1n4{r3 r9} - c1n3{r9 .} ==> r2c1 ≠ 5
The first can be written in terms of TDP :
(-3r2c1)=>3r9c1->4r78c3->7r1c3 => -7r2c1
But if we extend this sequence like this
(-3r2c1)=>3r9c1->(4r78c3->7r1c3)->5r2c3 => -7r2c1 and -5r2c1
This avoids your second step.

It doesn't avoid the second step. It fuses it with the first, making the first elimination more complicated than necessary. And I can't see there your own solution of that puzzle.

Sticking to the "general nature" of your remarks, my general resolution paradigm is "simplest first". What is yours? In particular:
- in your example, why do you fuse 2 chains and not 3?
- how do you decide where to stop the combination of tracks (which level of DFS)?
- how do you know there isn't a "shorter" solution?
- what's your definition of "shorter" for the full resolution path?