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by **marek stefanik** » Wed May 05, 2021 8:32 am

Code: Select all: +-------+-------+-------+ | . 4 . | . . . | . . . | | 5 . . | 9 . 8 | . 7 3 | | . . 7 | . 5 . | . . . | +-------+-------+-------+ | . 6 . | . . . | . 3 . | | . . 4 | . 6 . | 8 9 . | | . 9 . | . . 7 | . . 1 | +-------+-------+-------+ | . . . | . 9 . | . . . | | . 7 . | 2 8 . | . 4 . | | . 3 . | . . 1 | . 6 2 | +-------+-------+-------+ .4.......5..9.8.73..7.5.....6.....3...4.6.89..9...7..1....9.....7.28..4..3...1.62

I created this puzzle with a 1-step solution in mind. It's a neat trick that I haven't seen before. Can you find it?

Marek

Website · by **denis_berthier** » Mon May 10, 2021 7:24 am

Hi Marek
Welcome on this forum

Because you're new here, your post took some time before being visible.

Code: Select all: Resolution state after Singles and whips[1]: +-------------------+-------------------+-------------------+ ! 1239 4 1239 ! 1367 1237 236 ! 1259 1258 5689 ! ! 5 12 6 ! 9 124 8 ! 124 7 3 ! ! 1239 8 7 ! 1346 5 2346 ! 1249 12 469 ! +-------------------+-------------------+-------------------+ ! 1278 6 1258 ! 158 12 9 ! 2457 3 457 ! ! 1237 125 4 ! 135 6 235 ! 8 9 57 ! ! 238 9 2358 ! 3458 234 7 ! 6 25 1 ! +-------------------+-------------------+-------------------+ ! 1246 125 125 ! 34567 9 3456 ! 1357 158 578 ! ! 169 7 159 ! 2 8 356 ! 1359 4 59 ! ! 489 3 589 ! 457 47 1 ! 579 6 2 ! +-------------------+-------------------+-------------------+

There is a solution in W6, with nothing noticeable: Show

I haven't found any 1-step solution.
After initial Singles and whips[1], there's only one W1-anti-backdoor (n8r4c4), which would ensure a single-step solution (counting as 0-step any Single or whip[1]).
But it can't be eliminated by any whip, g-whip, braid or g-braid of any length.
This doesn't imply there couldn't be a 1-step solution with a pattern that would eliminate several candidates at once.

I'm very curious to see your 1-step solution.

[Edit:] Indeed, I have a 1-step solution, but a very dirty one:
FORCING[3]-T&E(S) applied to trivalue candidates n4r9c1, n8r9c1 and n9r9c1 :
===> 16 values decided in the three cases: n7r9c5 n8r7c9 n1r7c8 n3r8c7 n7r7c7 n2r3c8 n5r6c8 n8r1c8 n4r3c6 n3r3c1 n8r6c4 n2r6c1 n4r6c5 n3r6c3 n1r8c3 n3r1c5
===> 79 candidates eliminated in the three cases: n2r1c1 n1r1c3 n3r1c3 n6r1c4 n1r1c5 n2r1c5 n7r1c5 n2r1c7 n9r1c7 n1r1c8 n2r1c8 n5r1c8 n8r1c9 n9r1c9 n1r2c5 n4r2c5 n2r2c7 n1r3c1 n2r3c1 n9r3c1 n3r3c4 n4r3c4 n2r3c6 n3r3c6 n6r3c6 n2r3c7 n4r3c7 n1r3c8 n4r3c9 n1r4c1 n2r4c1 n1r4c3 n2r4c3 n1r4c4 n8r4c4 n5r4c7 n7r4c7 n5r4c9 n2r5c1 n3r5c1 n2r5c2 n1r5c4 n5r5c6 n5r5c9 n8r6c1 n2r6c3 n5r6c3 n8r6c3 n4r6c4 n5r6c4 n2r6c5 n3r6c5 n2r6c8 n1r7c1 n2r7c1 n1r7c2 n1r7c3 n5r7c4 n7r7c4 n4r7c6 n5r7c6 n1r7c7 n3r7c7 n5r7c7 n5r7c8 n8r7c8 n5r7c9 n7r7c9 n1r8c1 n5r8c3 n9r8c3 n3r8c6 n1r8c7 n5r8c7 n9r8c7 n5r9c3 n7r9c4 n4r9c5 n7r9c7
stte
As you can see, n8r4c4 is among the candidates eliminated, so that it should be possible to find some kind of Forcing-S-braid (or skyscraper) based on r9c1{n4 n8 n9) that is simpler than the above Forcing-T&E and that eliminates it.

by **marek stefanik** » Mon May 10, 2021 2:12 pm

Hi Denis,
Thank you for your solutions.

I'm not surprised that finding a 1-stepper via antibackdoors wasn't easy (I tried my best not to give any, the one remaining requires many chains to reach a contradiction when assumed true, according to Andrew's solver).
It's interesting that the forcing T&E comes from a non-backdoor cell. I'm assuming it uses the X-Wing on 3s.
As for the other candidates, I'm curious whether your solver just ignores them once it finds a contradiction (which I think would just produce a solution, rather than 16 placements) or finds a narrow path which actually proves all of the placements before reaching a contradiction.

I'll give the pattern a few days, for now I'll just say that all of the required truths are cells.

Marek

Website · by **denis_berthier** » Mon May 10, 2021 5:58 pm

marek stefanik wrote:It's interesting that the forcing T&E comes from a non-backdoor cell. I'm assuming it uses the X-Wing on 3s.
As for the other candidates, I'm curious whether your solver just ignores them once it finds a contradiction (which I think would just produce a solution, rather than 16 placements) or finds a narrow path which actually proves all of the placements before reaching a contradiction.

Forcing-T&E is a procedure that starts from a bivalue or (in the present case) trivalue cell (r9c1) and develops 2 (or 3) branches in parallel. The result is what's common to the 2 (or 3) branches. As I said, it's a dirty solution, cumulating the dirtiness of T&E with that of OR branching. I've defined this procedure to show what's really under the hood of Robert's "conjugated tracks". I haven't considered adding it to the public release of CSP-Rules.
One doesn't have to keep all of these candidates. Here one is enough for stte (n8r4c4) and I could have written:
FORCING[3]-T&E(S) applied to trivalue candidates n4r9c1, n8r9c1 and n9r9c1 : ===> r4c4 ≠ 8
stte

There are several possibilities of having a 1-step solution that way. Here is another one (where n8r4c4 is not eliminated and more than one candidate has to be kept for stte - I should check the anti-backdoor-pairs to see if 2 candidates are enough, but I'm lazy).
FORCING[3]-T&E(W1) applied to trivalue candidates n1r2c5, n2r2c5 and n4r2c5 :
===> 5 values decided in the three cases: n7r4c1 n7r5c9 n7r9c5 n8r7c9 n8r1c8
===> 65 candidates eliminated in the three cases: n2r1c1 n3r1c1 n1r1c3 n9r1c3 n1r1c4 n3r1c4 n2r1c5 n7r1c5 n3r1c6 n2r1c7 n9r1c7 n1r1c8 n2r1c8 n5r1c8 n8r1c9 n1r2c7 n1r3c1 n9r3c1 n3r3c4 n4r3c4 n2r3c6 n6r3c6 n4r3c7 n9r3c9 n1r4c1 n2r4c1 n8r4c1 n2r4c3 n5r4c3 n1r4c4 n7r4c7 n5r4c9 n7r4c9 n7r5c1 n1r5c2 n5r5c4 n3r5c6 n5r5c9 n3r6c1 n2r6c3 n8r6c3 n3r6c4 n5r6c4 n2r6c5 n1r7c1 n2r7c1 n1r7c3 n5r7c4 n7r7c4 n5r7c6 n1r7c7 n5r7c7 n5r7c8 n8r7c8 n5r7c9 n7r7c9 n9r8c1 n5r8c3 n5r8c7 n9r8c7 n9r9c1 n5r9c3 n7r9c4 n4r9c5 n7r9c7
stte

by **eleven** » Mon May 10, 2021 8:22 pm

There is an interesting constellation in the #-marked cells.

Code: Select all: *-----------------------------------------------------------------------* | 1239 4 1239 | 1367 1237 236 | 1259 1258 5689 | | 5 #12 6 | 9 #12+4 8 | 124 7 3 | | 1239 8 7 | 1346 5 2346 | 1249 12 469 | |----------------------+------------------------+-----------------------| | 1278 6 1258 | 158 #12 9 | 2457 3 457 | | 1237 #12+5 4 | #135 6 #235 | 8 9 7-5 | | 238 9 2358 | 3458 #234 7 | 6 25 1 | |----------------------+------------------------+-----------------------| | 1246 125 125 | 34567 9 3456 | 1357 158 578 | | 169 7 159 | 2 8 356 | 1359 4 59 | | 489 3 589 | 457 7-4 1 | 579 6 2 | *-----------------------------------------------------------------------*

1. remote pair 12 r52c2,r24c5 => (-1|2=35)r5c46 & 124r246c5 => -5r5c9, -4r9c5
2. 4r2c5 => 1235 r46c5, r5c46 => -4r9c5, -5r5c9
3. 5r5c2 => 1234 r46c5, r5c46 => -5r5c9, -4r9c5

=> -4r9c4, -5r5c9, stte

by **marek stefanik** » Mon May 10, 2021 10:35 pm

Hi eleven,
Thank you for your solution. Well done. That's the pattern I had in mind.

There is another approach to proving it, which is imo much easier to generalize.

Notice how the marked cells form an interesting constellation of their own:
X: r25c2 r246c5 => -Xr5c46
Thanks to this Finned X-Wing, each digit can only appear twice in the marked cells.

Now these seven cells contain at most: two 1s, two 2s, one 3, one 4 and one 5 => the rest of the pattern is of Rank 0.
Eliminations on 345 can therefore be made.

Marek

by **eleven** » Tue May 11, 2021 6:50 pm

marek stefanik wrote:X: r25c2 r246c5 => -Xr5c46
Thanks to this Finned X-Wing, each digit can only appear twice in the marked cells.

I don't understand that notation (1 and 2 also can be in both sets), but obviously both can't be in the marked cells more than 2 times.

So in my words i would say:
5 digits in 7 cells, at least two must be twice (since none can be triple), this must be 1 and 2, therefore 3,4,5 must be in the marked cells.
(Also 3 can be eliminated from r6c4).

I always stopped that counting with 6 cells, but it would have been useful for this pattern.

by **marek stefanik** » Wed May 12, 2021 9:26 am

Hi eleven,
That notation is indeed confusing, I should have made it clearer that by 'X' I meant 'any digit'.
I don't think there is a convenient way of notating the inner Finned X-Wing, as it doesn't have any truths by itself and only exists as a contradiction between three potential links of the main pattern.

The logic is as follows:
If you put digit X on the pattern in c2 (Xr25c2) and c5 (Xr246c5), you cannot place it in the seemingly uncovered cells (-Xr5c46), as you have created a Finned X-Wing which prevents that.
Therefore digit X can only appear in the pattern two times.

From each of the links Xc2, Xc5 and Xr5 my approach only considers the part that covers the pattern (Xr25c2, Xr246c5, Xr5c46; note that r5c2 has already been covered, so we can ignore it, unless there are overlapping truths, which is not the case) and then finds a contradiction between them (a pattern that eliminates one of them while using the others as truths).
This can be extended to more complex patterns, but I don't think that anything other than Almost MSLS + finned fish combos is feasible for manual solvers.

If you (or anyone else) have a better notation for these cases, I will appreciate any suggestions.

Marek

PS: I don't really understand what you mean by 'both sets', there are many combinations of r25c25b5 that cover the entire pattern, you're not limited to rows and columns like with classical MSLS or lines and boxes with SK Loops. That's why I think that such patterns are persistent against multi-floor analysis (you can't just ignore for example 8 and 9 and have the same pattern with 9r2c6 instead, as you could with two sets).

Edit: I realised multi-floor analysis would take into account the absence of 89 in r2c2, therefore it could find the pattern, but I think it would have to be done on 5 digits (6 if it had to work with 6b2 instead of 6r5c5), which still makes the pattern quite persistent.

by **eleven** » Wed May 12, 2021 7:01 pm

marek stefanik wrote:The logic is as follows:
If you put digit X on the pattern in c2 (Xr25c2) and c5 (Xr246c5), you cannot place it in the seemingly uncovered cells (-Xr5c46), as you have created a Finned X-Wing which prevents that.

Ok, so by X: r25c2 r246c5 you mean, that the digit is in both columns.

This can be extended to more complex patterns, but I don't think that anything other than Almost MSLS + finned fish combos is feasible for manual solvers.

Agreed, at least for most manual solvers.

If you (or anyone else) have a better notation for these cases, I will appreciate any suggestions.

I don't know one (but i am no expert in notations).

by **Cenoman** » Wed May 12, 2021 10:59 pm

What about this presentation of eleven's solution ?
Almost-almost RP:
RP(12)r2c25,r5c2,r4c5 - (1|2=345)b5p468*^
||
(4)r2c5* - (4=1235)b5p2468^
||
(5)r5c2^ - (5=1234)b5p2468*
---------
=> -4 r9c5*, -5 r5c9^; ste

by **marek stefanik** » Thu May 13, 2021 2:28 pm

eleven wrote:
marek stefanik wrote:The logic is as follows:
If you put digit X on the pattern in c2 (Xr25c2) and c5 (Xr246c5), you cannot place it in the seemingly uncovered cells (-Xr5c46), as you have created a Finned X-Wing which prevents that.

Ok, so by X: r25c2 r246c5 you mean, that the digit is in both columns.

Yes, since if the digit in one of the columns wasn't on the pattern, it couldn't appear three times elsewhere. So in order to prove that it cannot do that at all, I only need to prove that it cannot be on the pattern in those two columns and r5c46 at the same time.

Cenoman wrote:What about this presentation of eleven's solution ?
Almost-almost RP:
RP(12)r2c25,r5c2,r4c5 - (1|2=345)b5p468*^
||
(4)r2c5* - (4=1235)b5p2468^
||
(5)r5c2^ - (5=1234)b5p2468*
---------
=> -4 r9c5*, -5 r5c9^; ste

As this pattern has multiple eliminations, notating which steps produce which eliminations makes it a bit more readable (especially if the chains were longer, in this case I would say that both presentations are fine).
I'm still trying to devise a notation that would together with the set notation prove part of this pattern rank 0. The tricky part are the digits 1 and 2.
How would you notate that no digit can appear on the pattern three times?

Marek

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