The New Sudoku Players' Forum

[denis_berthier]

unless youre counting the locked 1s

I don't count whips[1] / intersections...

i usually consider ...hidden/naked tuples as not-a-step

I count them as a step because they generally eliminate many candidates and not counting them can be justified on no serious ground.
Moreover, generally, people who don't count them, do nevertheless count fish - which are the same thing.

at this point why not break down the last step into something more palatable (im not a fan of how the chain implies a 3 in r2c9 and r2c6 simultaneously, that much would be better split into two)
[...]

Code: Select all

.-------------.---------------.------------------.
| 1   8    4  |#39   29   6   | 5     2379 #237  |
| 7   6    5  | 4    289 #13  | 1239  1239  128  |
| 2   9    3  | 18   7    5   | 18    6     4    |
:-------------+---------------+------------------:
| 5   7    8  | 6    13   4   | 123   123   9    |
| 9   4    2  | 7    13   8   | 136   5    #136  |
| 36  13   16 | 2    5    9   | 48    47    78   |
:-------------+---------------+------------------:
|#36  5    7  |#39   46   12  | 49    8     12   |
| 8   123 #16 | 5    49   12-3| 7     49   #1236 |
| 4   123  9  | 138  68   7   | 1236  123   5    |
'-------------'---------------'------------------'

 3r1c9 - 3r1c4 ========================== 3r2c6
   ||                                         \
(3-6)r5c9 = 6r8c9 - 6r8c3 = (6-3)r7c1 = 3r7c4  \
   ||                                        \  \
 3r8c9 -------------------------------------  3r8c6

kraken column (3)r158c9
-3r8c6 stte

It's certainly more readable that way. Starting from the same PM:

Code: Select all

   +----------------+----------------+----------------+
   ! 1    8    4    ! 39   29   6    ! 5    2379 237  !
   ! 7    6    5    ! 4    289  13   ! 1239 1239 128  !
   ! 2    9    3    ! 18   7    5    ! 18   6    4    !
   +----------------+----------------+----------------+
   ! 5    7    8    ! 6    13   4    ! 123  123  9    !
   ! 9    4    2    ! 7    13   8    ! 136  5    136  !
   ! 36   13   16   ! 2    5    9    ! 48   47   78   !
   +----------------+----------------+----------------+
   ! 36   5    7    ! 39   46   12   ! 49   8    12   !
   ! 8    123  16   ! 5    49   123  ! 7    49   1236 !
   ! 4    123  9    ! 138  68   7    ! 1236 123  5    !
   +----------------+----------------+----------------+

n3r8c6 can be eliminated by a z-chain that uses the same 5 CSP-Variables:
z-chain[5]: r7n3{c4 c1} - b7n6{r7c1 r8c3} - c9n6{r8 r5} - c9n3{r5 r1} - c4n3{r1 .} ==> r8c6≠3
stte

[shye]

I count them as a step because they generally eliminate many candidates and not counting them can be justified on no serious ground.
Moreover, generally, people who don't count them, do nevertheless count fish - which are the same thing.

thats a fair stance, the best reason i have for it would be personal preference. while solving a puzzle tuples are usually very common and easy to find, they worst they come as would be quadruples or hidden sets in a row/column. but fish take more involved scanning for (to me), are often far less common, and also are generally less intuitive to think about despite logically being no different to a tuple. we could extend the argument past fish and say CNLs or any rank0 pattern is logically the same thing, but of course that starts to get ridiculous since these patterns take a lot more thinking and scanning for. but yea, i agree with the sentiment of your argument, though i still prefer to leave tuples out of the step count. ill always mention them if its important, at least

[denis_berthier]

I count them as a step because they generally eliminate many candidates and not counting them can be justified on no serious ground.
Moreover, generally, people who don't count them, do nevertheless count fish - which are the same thing.

thats a fair stance, the best reason i have for it would be personal preference. while solving a puzzle tuples are usually very common and easy to find, they worst they come as would be quadruples or hidden sets in a row/column. but fish take more involved scanning for (to me), are often far less common, and also are generally less intuitive to think about despite logically being no different to a tuple.

Yes, I know this is very controversial. There are even some people and some solvers that count Fish as simpler than Naked or Hidden Subsets of same size.

[we could extend the argument past fish and say CNLs or any rank0 pattern is logically the same thing, but of course that starts to get ridiculous since these patterns take a lot more thinking and scanning for.

The problem with "rank0' patterns (of a fixed size) is, their number grows up exponentially with size and no solver has ever been written that could find them all.
A very simple illustration in Sudoku is the complexity of the extended Fish patterns, when you mix up the types of base elements.

but yea, i agree with the sentiment of your argument, though i still prefer to leave tuples out of the step count. ill always mention them if its important, at least

I don't know how your solver works to minimise the number of steps, but if you minimise it without taking Subsets into account what you get may not be the minimum when they are afterwards taken into account.

[marek stefanik]

not counting them [subsets] can be justified on no serious ground.

If there is no universally established definition of a step, how can you say that with such certainty?
I think that in order to get a consistent theory, you shouldn't allow '1-steps' to contain other '1-steps'.
In this sense your theory is very consistent, since your patterns only contain the equivalent of singles and intersections (either you have a single rlc or they're part of the same mini-line).
But others like to use different techniques and you cannot quite call an ALS chain a 1-step, if you count subsets themselves as steps.
Do you want us to count them as 1+n-steps, where n is the number of ALSs (>1) included?
I don't think that makes much sense, since many paradoxes would arise.
SDC would be counted as four steps (a loop with three (n)ALSs), but a different POV – (n)ALS linked to (m)ALS by n + m links – would only count as 3 (loop with two (n)ALS).
That completely kills off the consistency we were after.

I don't know how your [shye's] solver works...

Sorry to interrupt you two's conversation, but I think you should talk to a neuroscientist if you're interested in this topic.

Marek

[denis_berthier]

not counting them [subsets] can be justified on no serious ground.

If there is no universally established definition of a step, how can you say that with such certainty?

For the exact reason you're mentioning: if there was a universally accepted definition, it would have to rely on some serious ground.
To be more specific, the only reason why some people on this forum consider Naked and Hidden Pairs but not X-Wings or larger Subsets as "Basic" (and count them as 0-step) relies on the existence of a totally outdated solver (simple sudoku) that had only these techniques.

I think that in order to get a consistent theory, you shouldn't allow '1-steps' to contain other '1-steps'.
In this sense your theory is very consistent, since your patterns only contain the equivalent of singles and intersections (either you have a single rlc or they're part of the same mini-line).

Not exactly; S-whips can have inner Subsets as right-linking patterns; W-whips can even have inner whips as right-linking patterns .
But you're right if you consider only my most common (and useful) patterns.

But others like to use different techniques and you cannot quite call an ALS chain a 1-step, if you count subsets themselves as steps.
Do you want us to count them as 1+n-steps, where n is the number of ALSs (>1) included?
I don't think that makes much sense, since many paradoxes would arise.

For more than 10 years, I've had a very clear and consistent way to count a Subset appearing inside a chain (always as a right-linking pattern in my approach): instead of increasing the global length of the chain by 1 (as it would for a right-linking candidate or g-candidate), increase it by the size of the Subset (2 for Pairs, 3 for Triplets...).
In case the Subset can itself be seen as a chain and made to appear as such directly the global chain, the two definitions of length coincide.
The simplest example is a (Naked, Hidden or Super-Hidden) Pair, which becomes a mere bivalue-chain[2].
The number of steps is not increased by the presence of a complicated inner pattern in a chain. What's increased is the length of that chain.

I don't know how your [shye's] solver works...

Sorry to interrupt you two's conversation, but I think you should talk to a neuroscientist if you're interested in this topic.

If you mean shye doesn't use a solver, he said himself he does sometimes.

[shye]

I don't know how your [shye's] solver works...

Sorry to interrupt you two's conversation, but I think you should talk to a neuroscientist if you're interested in this topic.

If you mean shye doesn't use a solver, he said himself he does sometimes.

"she" please. i use YZF_Sudoku to assist in creating and analysing puzzles

[denis_berthier]

"she" please.

My apologies. I should have guessed from the pic.

[marek stefanik]

For the exact reason you're mentioning: if there was a universally accepted definition, it would have to rely on some serious ground.

I thought you were implying that counting them could be justified. My bad.

For more than 10 years, I've had a very clear and consistent way to count a Subset appearing inside a chain...

Code: Select all

   +---------+----–
   | 12 .  -1| 1234
   | .  .  . | 34
   | .  .  . | 34

If I understand it correctly, you would find an S-whip[4] eliminating 1r1c3 and count it as one step, even though it contains the entirety of the two subsets (ok, it contains one, the other can be found with singles if we assume 1r1c3).
It doesn't make sense to me to say that this path is in some sense shorter, simpler, than the two subsets.
Also it seems that solving a puzzle using DFS would count as one step, since you don't count anything that happens during the 'step' as a step.
I don't think this definition gives a meaningful way to measure the length of a solution path.

Marek

[denis_berthier]

For the exact reason you're mentioning: if there was a universally accepted definition, it would have to rely on some serious ground.

I thought you were implying that counting them could be justified. My bad.

If not counting them is not justified, is counting them justified? Does not not A = A in this case?

If I understand it correctly, you would find an S-whip[4] eliminating 1r1c3 and count it as one step, even though it contains the entirety of the two subsets (ok, it contains one, the other can be found with singles if we assume 1r1c3).
It doesn't make sense to me to say that this path is in some sense shorter, simpler, than the two subsets.
Also it seems that solving a puzzle using DFS would count as one step, since you don't count anything that happens during the 'step' as a step.
I don't think this definition gives a meaningful way to measure the length of a solution path.

Wow, much confusion here. What is DFS doing here? When has DFS become a resolution rule? How can you confuse a step in DFS with the number of steps in a resolution path? Are you still thinking of a chain as a chain of inferences?

A resolution path is a sequence of steps, each step is made of one resolution rule (be it a simple chain, a Subset, a chain with inner Subsets, or whatever), based on a pattern with its own length or size (defined as I explained in my previous post).
If you allow each step to use more complex or larger patterns, it's very likely you can find a solution with fewer steps. That's why I always say that 1- or 2- step solutions with very complicated step(s), as some of those presented on this forum, are pure nonsense for puzzles that can be solved with simple patterns but more steps. I think I've given sufficiently many examples to make my position totally unambiguous.

There's no need to consider very complex patterns to see how I count steps:
Take 2 independent Pairs. In my view, it's two Pairs and therefore 2 steps (each of size 2), because I don't allow degenerated patterns. But for some, it can be one Quad and 1 step (of size 4). [So, that's the exact opposite of what you supposed I was doing.]
You can easily generalise this to solutions based on very large base/cover sets.


The minimum length of chains (of a specific type) necessary for solving a puzzle is an intrinsic, pure logic property of this puzzle (whence my various ratings).
But the minimum number of steps to solve a puzzle is not even defined if you don't specify beforehand what kind of steps you allow.
There is a trade-off between the maximum length you allow for chains (of a certain type) and the minimum possible number of steps of a solution.

[marek stefanik]

What is DFS doing here?

My bad again, if you choose a candidate, assume it and eliminate it based on DFS (this way we get nested T&E instead of repeated nested T&E; and you've used T&E yourself), it would count as a 1-step. This seems weird to me.

Are you still thinking of a chain as a chain of inferences?

That's the only way I can think about your chains, especially since you only show one llc, even if the csp-variable has more candidates than the llc and possible rlcs, as if you were looking at a grid with those candidates already deleted and only remembered the last one you rubbed out.

I don't allow degenerated patterns.

OK, then it's maybe fine, but if the cost is to ignore patterns that can justify their eliminations, is it really worth it?

Marek

[denis_berthier]

What is DFS doing here?

My bad again, if you choose a candidate, assume it and eliminate it based on DFS (this way we get nested T&E instead of repeated nested T&E; and you've used T&E yourself), it would count as a 1-step. This seems weird to me.

There's no way DFS or T&E can be written as a resolution rule, so this is totally irrelevant.

Are you still thinking of a chain as a chain of inferences?

That's the only way I can think about your chains, especially since you only show one llc, even if the csp-variable has more candidates than the llc and possible rlcs, as if you were looking at a grid with those candidates already deleted and only remembered the last one you rubbed out.

Obviously, you haven't read the definitions of my chains with enough care, so any discussion about them is useless at this point. Understanding why z- and t- candidates are not part of the chains requires some level of abstraction, but it's not rocket science. Only point I repeat is, they are not chains of inferences, but patterns, exactly in the same sense as Subsets.

I don't allow degenerated patterns.

OK, then it's maybe fine, but if the cost is to ignore patterns that can justify their eliminations, is it really worth it?

I have no idea of what you're talking. I never ignore any pattern.

[marek stefanik]

What does DFS to T&E even mean? How do you add another layer of T&E after DFS?

I don't think arbitrarily choosing which candidate of an intersection is part of the chain (llc) and which is ignored as a t-candidate, can be classified as abstraction.

Now, for your third point:

I never ignore any pattern.

I don't allow degenerated patterns.

I don't quite understand your definitions of 'allow' and 'ignore' (or at least one of them), since the way I understand those words, this is a contradiction.

Also, this doesn't seem to make sense:

Code: Select all

   +---------+---------+---------+
   | –1 .  . | .  12 . | .  . 123|
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  34|
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  578 .|
   | .  .  . | .  .  . |578 578 .|
   | 16 .  . | .  56 . | .  . 345|
   +---------+---------+---------+

In this resolution state you can eliminate 1r1c1 with one step (hopefully I got the syntax right):
whip[6]: r1c5{n1 n2} – r1c9{n2 n3} – r5c9{n3 n4} – r9c9{n4 n5} – r9c5{n5 n6} – r9c1{n6 .} ==> r1c1≠1
After applying the eliminations of the 578 triple in b9, you can no longer do that – the pattern would be degenerated and you don't allow those.
So you have no other way than to apply the pairs one-by-one, which is two steps.
How can the number of steps required to eliminate a candidate increase in the process of solving the puzzle?

Marek

[denis_berthier]

What does DFS to T&E even mean? How do you add another layer of T&E after DFS?

Tell me. I never do this.

I don't think arbitrarily choosing which candidate of an intersection is part of the chain (llc) and which is ignored as a t-candidate, can be classified as abstraction.

If that's all you understand of whips, you have a problem with abstraction.
If that can make you feel better with it, you don't have to choose arbitrarily, you can develop several whips in parallel instead. (Of course, that'd be very dumb in practice.)

INow, for your third point:

I never ignore any pattern.

I don't allow degenerated patterns.

I don't quite understand your definitions of 'allow' and 'ignore' (or at least one of them), since the way I understand those words, this is a contradiction.

Only if you absolutely want to find a contradiction somewhere. A degenerated pattern is not a pattern: 2 NPs are not a NQ.

Also, this doesn't seem to make sense:

Code: Select all

   +---------+---------+---------+
   | –1 .  . | .  12 . | .  . 123|
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  34|
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  578 .|
   | .  .  . | .  .  . |578 578 .|
   | 16 .  . | .  56 . | .  . 345|
   +---------+---------+---------+

In this resolution state you can eliminate 1r1c1 with one step (hopefully I got the syntax right):
whip[6]: r1c5{n1 n2} – r1c9{n2 n3} – r5c9{n3 n4} – r9c9{n4 n5} – r9c5{n5 n6} – r9c1{n6 .} ==> r1c1≠1
After applying the eliminations of the 578 triple in b9, you can no longer do that – the pattern would be degenerated and you don't allow those.
So you have no other way than to apply the pairs one-by-one, which is two steps.

If you choose to apply the NT in b9 before the whip[6] (which the simplest-first strategy would do), the RS after the RT is:

Code: Select all

   +---------+---------+---------+
   | –1 .  . | .  12 . | .  . 123|
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  34|
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  578 .|
   | .  .  . | .  .  . |578 578 .|
   | 16 .  . | .  56 . | .  . 34 |
   +---------+---------+---------+

The original whip[6] disappears, but what's left of it is a whip[4] that still eliminates n1r1c1:
whip[4]: r1c5{n1 n2} – r1c9{n2 n3} – r5c9{n3 n4} – r9c9{n4 .} ==> r1c1≠1
(But of course, there are simpler eliminations by NPs available before that.)

How can the number of steps required to eliminate a candidate increase in the process of solving the puzzle?

As "the number of steps required to eliminate a candidate" doesn't mean anything, it can neither increase nor decrease. The only thing that makes sense (within a fixed resolution theory) is the number of steps to solve a puzzle.

[marek stefanik]

A degenerated pattern is not a pattern: 2 NPs are not a NQ.
[...]
whip[4]: r1c5{n1 n2} – r1c9{n2 n3} – r5c9{n3 n4} – r9c9{n4 .} ==> r1c1≠1

So, is whip[4] what two NPs are?

Marek

[denis_berthier]

A degenerated pattern is not a pattern: 2 NPs are not a NQ.
[...]
whip[4]: r1c5{n1 n2} – r1c9{n2 n3} – r5c9{n3 n4} – r9c9{n4 .} ==> r1c1≠1

So, is whip[4] what two NPs are?

In the resolution state we're talking about:

Code: Select all

   +---------+---------+---------+
   | –1 .  . | .  12 . | .  . 123|
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  34|
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  578 .|
   | .  .  . | .  .  . |578 578 .|
   | 16 .  . | .  56 . | .  . 34 |
   +---------+---------+---------+

you have two ways of eliminating n1r1c1:
- either you apply two NPs
- or you apply the whip[4]

The two ways correspond to opposite goals: using the simplest patterns (normal way of working in SudoRules) vs trying to find fewer steps (within a fixed set of rules, here S+W4)

I don't see how this makes the whip[4] identical to 2 NPs.

[Edit]: However, this is an interesting example, because it clearly shows the tradeoffs when one asks for 1-step or 2-step solutions.