The New Sudoku Players' Forum

[marek stefanik]

I think I know what the problem is. We have different ideas of when a pattern is degenerated.
I would call this whip[4] degenerated, since it contains the same variables as the pairs and one of them creates an elimination inside the pattern by itself, but that's probably not what you mean.
Can you give me your definition, please?

Marek

[denis_berthier]

I think I know what the problem is. We have different ideas of when a pattern is degenerated.
I would call this whip[4] degenerated, since it contains the same variables as the pairs and one of them creates an elimination inside the pattern by itself, but that's probably not what you mean.
Can you give me your definition, please?

There is a precise definition of whips that doesn't leave any place for degenerated whips.
A NP can in no way be an inner pattern of a whip (it could be in a S-whip but not in a whip).

[marek stefanik]

So, if you can just cut part of it off (here the first two variables) and still get a valid pattern (with a different elimination inside the original pattern), that doesn't count as degenerated?
The pair isn't its inner pattern as it would be in an S-whip, but it contains both variables that form the pair.

Marek

[DEFISE]

Hi Denis and Marek Stefanik,

In my opinion I would consider as a step:
either a chain (Ex: whip or AIC) or other complex pattern
either a sequence of basics (Single, block/line, subset).
Example of my resolution of “Gata de mar” :

Step 1 :
14 singles
Block/line : 3b4r6 => -3r6c4 -3r6c7 -3r6c8 -3r6c9
Block/line : 3b5c5 => -3r1c5 -3r2c5 -3r7c5 -3r8c5 -3r9c5
Hidden pairs: 49r8c58 => -1r8c5 -2r8c5 -6r8c5 -1r8c8 -2r8c8 -3r8c8
Hidden pairs: 49b9p15 => -1r7c7 -2r7c7 -3r7c7 -6r7c7
Hidden triplets: 478r6c789 => -1r6c7 -2r6c7 -6r6c7 -1r6c8 -2r6c8 -1r6c9 -2r6c9 -6r6c9
Block/line : 6r6b4 => -6r5c3

Step 2 :
whip[7]: r9n8{c4 c5}- c5n6{r9 r7}- r7n4{c5 c7}- r6n4{c7 c8}- c8n7{r6 r1}- r1n3{c8 c9}- b9n3{r7c9 .} => -3r9c4

Step 3 :
Hidden pairs: 39c4r17 => -2r1c4 -1r7c4 -2r7c4

Step 4 :
whip[8]: c5n8{r2 r9}- r9n6{c5 c7}- r5n6{c7 c9}- r8n6{c9 c3}- r7c1{n6 n3}- r9n3{c2 c8}- c9n3{r7 r1}- c4n3{r1 .} => -8r2c9

Step 5 :
Singles to the end.

But if there are only basics in a resolution, then we can consider each one as a step (or perhaps only subsets). This is the case with mith's puzzles.

[denis_berthier]

So, if you can just cut part of it off (here the first two variables) and still get a valid pattern (with a different elimination inside the original pattern), that doesn't count as degenerated?
The pair isn't its inner pattern as it would be in an S-whip, but it contains both variables that form the pair.

whip[4]: r1c5{n1 n2} – r1c9{n2 n3} – r5c9{n3 n4} – r9c9{n4 .} ==> r1c1≠1
If I cut off the first two CSP-Vars, what I get is r5c9{n3 n4} – r9c9{n4 .} ==> r1c1≠1 which doesn't mean anything
I can't see how by any stretch of meaning the intended NP (with a different target) could be assimilated with this.

The situation for a NQ vs 2 Pairs is quite different. Take 4 cells with 12 12 34 34 in the same row. Whether you consider them as two NPs or 1 degenerated NQ, the targets are the same. The only things that change are the number of steps and the complexities of the patterns involved.

[denis_berthier]

Hi François.
It's hard to understand why a Pair would sometimes be 1 step and sometimes only a small part of a step.
I think my approach is safer. Define what you choose to count as 0-step (whatever you want*) and keep the same definition for the whole resolution path.

As I said several times, the only patterns that I consider as really 0-step is Singles and whips[1]. As for Subsets, one could ask why the patterns that eliminate the more candidates should not be counted.

[marek stefanik]

Hi François,
It's an interesting compromise, but as Denis says, it seems weird to count a technique as a step only sometimes.

Denis, when I said to delete the first two variables, I also said to change the elimination.
r5c9{n3 n4} – r9c9{n4 .} ==> r1c9≠3
This way you get an elimination inside the original pattern, which is why I would consider it degenerated.

It seems weird to consider a NQ which contains NPs degenerated, and not think the same way about that whip[4], but I think I get your point now.
If the NQ were 12 123 34 34, as in the pattern I am talking about, you wouldn't consider it degenerated, because you cannot get the eliminations on 1s and 2s with just a part of it, whereas in your example every elimination has a smaller pattern (while 1s and 2s have a different one than 3s and 4s).

If that's the case, I still don't understand what seemed degenerated to you about my first example.
Can you explain the difference between the two (degenerated S-whip[4] example versus valid whip[4] example)?

I am now realising I misread one of your previous posts:

There's no way DFS or T&E can be written as a resolution rule...

Why are you then occassionally using T&E (even forcing T&E) in your solutions?

Marek

[denis_berthier]

when I said to delete the first two variables, I also said to change the elimination.
r5c9{n3 n4} – r9c9{n4 .} ==> r1c9≠3
This way you get an elimination inside the original pattern, which is why I would consider it degenerated.

This elimination is unrelated to the whip[4] elimination - "an elimination inside a whip" doesn't mean anything.
Anyway, with the simplest-first strategy, this can never happen.

It seems weird to consider a NQ which contains NPs degenerated, and not think the same way about that whip[4], but I think I get your point now.
If the NQ were 12 123 34 34, as in the pattern I am talking about, you wouldn't consider it degenerated, because you cannot get the eliminations on 1s and 2s with just a part of it, whereas in your example every elimination has a smaller pattern (while 1s and 2s have a different one than 3s and 4s).

Actually, it is degenerated because the 34 34 part is an independent NP. This NQ degenerates into two successive NPs that have the same eliminations as the NQ.
What's different in the whip example is, the NP is not part of the whip structure (they are different as patterns) and it has an elimination that the whip[4] doesn't have.

If that's the case, I still don't understand what seemed degenerated to you about my first example.
Can you explain the difference between the two (degenerated S-whip[4] example versus valid whip[4] example)?

I'm lost: which two S-whip[4] ???

I am now realising I misread one of your previous posts:

There's no way DFS or T&E can be written as a resolution rule...

Why are you then occassionally using T&E (even forcing T&E) in your solutions?

I don't think I ever use T&E in my solutions. For very hard puzzles, I may write "it is easy to check that this puzzle is in T&E and therefore (thanks to the T&E vs braids theorem), it has a braids solution ..." when I'm lazy to compute the effective braids resolution path; but I never write the T&E solution (which is only the output of a procedure, not a resolution path in the strict sense I give to this concept). This is generally followed by "but here is a more interesting g-whip solution".
As for Forcing-T&E , I occasionally use it, always with some ironical remark, to show that a 1-step requirement without any condition on the step is highly ridiculous.

[marek stefanik]

What's different in the whip example is, the NP is not part of the whip structure (they are different as patterns) and it has an elimination that the whip[4] doesn't have.

Can you specify what part of the NP is not part of the whip structure (both r5c9 and r9c9 are)?
I thought you considered NPs and NQs also different, the same way you consider different two whips of different length.
In a NQ 12 123 34 34, the 34 NP also has an elimination the NQ doesn't have – 3 in the 123 cell.

When I asked you about a possible S-whip[4] in this sub-puzzle, you said you didn't allow degenerated patterns.

Code: Select all

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

I don't know how to notate S-whips, but hopefully you'll get the intention.
r1c1{n1 n2} – r12c4{n2 n34} – r3c4{n4 .} ==> r1c3≠1
It confuses me, because it's almost the same structure as in the whip[4] that you say isn't degenerated.

Marek

[denis_berthier]

What's different in the whip example is, the NP is not part of the whip structure (they are different as patterns) and it has an elimination that the whip[4] doesn't have.

Can you specify what part of the NP is not part of the whip structure (both r5c9 and r9c9 are)?

By definition, a whip is a continuous sequence of candidates, with additional conditions, of course, but that's enough here. There's no place in the structure of a whip for any embedded NP.

In a NQ 12 123 34 34, the 34 NP also has an elimination the NQ doesn't have – 3 in the 123 cell.

ah, sorry for this example; I had overlooked the extra elimination.
So, what would happen in SudoRules in this case? With the standard simplest-first strategy, first the 34 NP, then the resulting 12 NP will be applied (2 steps).

When I asked you about a possible S-whip[4] in this sub-puzzle, you said you didn't allow degenerated patterns.

Code: Select all

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

I don't know how to notate S-whips, but hopefully you'll get the intention.
r1c1{n1 n2} – r12c4{n2 n34} – r3c4{n4 .} ==> r1c3≠1
It confuses me, because it's almost the same structure as in the whip[4] that you say isn't degenerated.

S2-whip[4]: r1c1{n1 n2} – c4{r1n2 r12n34} – r3c4{n4 .} ==> r1c3≠1
It can't be the same structure as a whip, because it isn't a sequence of candidates. In formal definition, there's no place for any "almost".
About degenerated, I can't see how this S2-whip[4] woud be degenerated.

[Edit]:
Why the notation can only be: S2-whip[4]: r1c1{n1 n2} – c4{r1n2 r12n34} – r3c4{n4 .} ==> r1c3≠1
- a left-linking elements are always a mere candidate => llc2 has to be n1r2c4
- the second rlc is a NP in c4; the only thing common to llc2 and rlc2 is c4 => c4{r1n2 xxx}, where xxx must be the NP
- the support of the NP are c4r1 and c4r2, so the r12 comes before the n34 in the notation
Note: this is using the short notation for NPs. The full notation for this NP would be c4{r1 r2}{n3 n4}