The New Sudoku Players' Forum

Website · by **denis_berthier** » Fri Jan 29, 2021 4:58 am

This is a puzzle from the same "fish on wave" pattern as mith's puzzle here: http://forum.enjoysudoku.com/a-rather-silly-fish-thermo-t38336.html

Code: Select all: . . X X . . . . X . X . . X . . X . X . . . . X X . . X . . . . X X . . . X . . X . . X . . . X X . . . . X X . . . . X X . . . X . . X . . X . . . X X . . . . X

This is one of the two hardest puzzles (among 1000) I found for this pattern: SER = 9.3. This one may involve a lot of g-candidates.

Code: Select all: +-------+-------+-------+ ! . . 4 ! 2 . . ! . . 6 ! ! . 6 . ! . 7 . ! . 3 . ! ! 2 . . ! . . 6 ! 1 . . ! +-------+-------+-------+ ! 1 . . ! . . 4 ! 6 . . ! ! . 5 . ! . 9 . ! . 1 . ! ! . . 7 ! 1 . . ! . . 9 ! +-------+-------+-------+ ! 9 . . ! . . 3 ! 7 . . ! ! . 8 . ! . 4 . ! . 5 . ! ! . . 3 ! 7 . . ! . . 8 ! +-------+-------+-------+ ..42....6.6..7..3.2....61..1....46...5..9..1...71....99....37...8..4..5...37....8 # 95617 FNBTHWXYKG C27.m/M1.44.3 27 givens, SER = 9.3

This is a challenge for AIC solvers.

by **DEFISE** » Fri Jan 29, 2021 2:03 pm

With my basic technics = (singles, alignments, naked & nude pairs, naked triplets) I found:
W = 15
B = 13
g-W = 11
S2-W = 9

So, according to your statistics, Denis, that would be a pretty rare case, since simple whips are much less effective than other patterns. Isn't it ?

Website · by **denis_berthier** » Fri Jan 29, 2021 2:41 pm

DEFISE wrote:With my basic technics = (singles, alignments, naked & nude pairs, naked triplets) I found:
W = 15
B = 13
g-W = 11
S2-W = 9
So, according to your statistics, Denis, that would be a pretty rare case, since simple whips are much less effective than other patterns. Isn't it ?

Yes, rather rare case.
I'm curious to see what people get with AICs.

by **mith** » Fri Jan 29, 2021 5:03 pm

SE gets the singles and then immediately jumps up to a 9.1 dynamic chain. Quite hard, yes.

by **coloin** » Fri Jan 29, 2021 7:01 pm

Code: Select all: +---+---+---+ |..4|..2|..6| |.6.|.7.|.3.| |2..|6..|1..| +---+---+---+ |..7|..1|..9| |.5.|.9.|.1.| |1..|4..|6..| +---+---+---+ |..3|..7|..8| |.8.|.4.|.5.| |9..|3..|7..| +---+---+---+

This is a morphed version ....
27 clues all in diagonal pattern

Its the same as the first pattern [#1] in this thread.

There are only 15 ED of these 27 clue diagonal patterns ... and interestingly only one hasn't got a symmetric form ....

Trying to see if any of those 14 patterns have been played in the patterns game, should really have done that ! ....

EDIT This #1 pattern was played in Patterns game , way back, at Game no. 016
In Patterns Game no. 150 the diagonal clue C27 pattern #2 was played. [Highest rated by SE was 10.7/10.7/8.9 eleven]

Website · by **denis_berthier** » Sat Jan 30, 2021 6:32 am

mith wrote:SE gets the singles and then immediately jumps up to a 9.1 dynamic chain. Quite hard, yes.

To be more precise, here's the first elimination after the two singles, in SE hyper-verbose style (a single elimination takes as much space as a full resolution path in SudoRules):

Hidden Text: Show

Dynamic Contradiction Forcing Chains
With this solving technique, we will prove the two following assertions:
If R2C3 contains the value 5, then R9C8 must contain the value 9
If R2C3 contains the value 5, then R9C8 cannot contain the value 9
Because the same assumption yields to contradictory results, we can conclude that the assumption is false, that is, R2C3 cannot contain the value 5.
Each assertion is proved by a different chain of simple rules. The chains can be dynamic, which means that the conclusions of multiple sub-chains must be combined in some cases.
The details of each chain are given below. Use the view selector below the grid to switch between the graphical illustrations of the two different chains.
Chain 1: If R2C3 contains the value 5, then R9C8 cannot contain the value 9 (View 1): (1) If R2C3 contains the value 5, then R2C1 cannot contain the value 5 (the value can occur only once in the block) (2) If R2C1 does not contain the value 5, then R2C1 must contain the value 8 (only remaining possible value in the cell) (3) If R2C1 contains the value 8, then R3C3 cannot contain the value 8 (the value can occur only once in the block) (4) If R2C3 contains the value 5 (initial assumption), then R3C3 cannot contain the value 5 (the value can occur only once in the block) (5) If R3C3 does not contain the value 5 and R3C3 does not contain the value 8 (3), then R3C3 must contain the value 9 (only remaining possible value in the cell) (6) If R3C3 contains the value 9, then R3C8 cannot contain the value 9 (the value can occur only once in the row) (7) If R2C3 contains the value 5 (initial assumption), then R2C3 cannot contain the value 1 (the cell can contain only one value) (8) If R2C3 does not contain the value 1, then R1C2 must contain the value 1 (only remaining possible position in the block) (9) If R1C2 contains the value 1, then R1C2 cannot contain the value 7 (the cell can contain only one value) (10) If R1C2 does not contain the value 7, then R1C8 must contain the value 7 (only remaining possible position in the row) (11) If R1C8 contains the value 7, then R1C8 cannot contain the value 9 (the cell can contain only one value) (12) If R1C8 does not contain the value 9 and R3C8 does not contain the value 9 (6), then R9C8 must contain the value 9 (only remaining possible position in the column)
Chain 2: If R9C8 must contain the value 9, then R9C8 cannot contain the value 9 (View 2): (1) If R2C3 contains the value 5, then R2C3 cannot contain the value 1 (the cell can contain only one value) (2) If R2C3 does not contain the value 1, then R2C6 must contain the value 1 (only remaining possible position in the row) (3) If R2C6 contains the value 1, then R9C6 cannot contain the value 1 (the value can occur only once in the column) (4) If R2C3 does not contain the value 1 (1), then R1C2 must contain the value 1 (only remaining possible position in the block) (5) If R1C2 contains the value 1, then R9C2 cannot contain the value 1 (the value can occur only once in the column) (6) If R9C2 does not contain the value 1 and R9C6 does not contain the value 1 (3), then R9C5 must contain the value 1 (only remaining possible position in the row) (7) If R9C5 contains the value 1, then R9C5 cannot contain the value 6 (the cell can contain only one value) (8) If R2C3 contains the value 5 (initial assumption), then R2C1 cannot contain the value 5 (the value can occur only once in the block) (9) If R2C3 contains the value 5 (initial assumption), then R1C1 cannot contain the value 5 (the value can occur only once in the block) (10) If R1C1 does not contain the value 5 and R2C1 does not contain the value 5 (8), then R9C1 must contain the value 5 (only remaining possible position in the column) (11) If R9C1 contains the value 5, then R9C1 cannot contain the value 6 (the cell can contain only one value) (12) If R9C1 does not contain the value 6 and R9C5 does not contain the value 6 (7), then R9C8 must contain the value 6 (only remaining possible position in the row) (13) If R9C8 contains the value 6, then R9C8 cannot contain the value 9 (the cell can contain only one value)

Depending on which rules are activated, SudoRules will have an elimination of the same n5r2c3, which seems to be a necessary point, but with possibly different rules. Moreover, in each case, simpler rules will make other eliminations before this one.

1) Solution with whips, in W15. n5r2c3 is eliminated by a whip[11], after 3 simpler eliminations:

Hidden Text: Show

2) Solution with braids, in B13. n5r2c3 is eliminated by a braid[8], after 2 simpler eliminations:

Hidden Text: Show

3) And finally, my preferred solution, with g-whips, in gW11. As I have mentioned many times, g-whips are often more effective than braids for reducing the maximal length of chains, when long whips are necessary to solve a puzzle - a situation very rare in itself.
n5r2c3 is eliminated by the same whip[11] as before, but after 4 simpler eliminations:

Hidden Text: Show

So, yes, quite hard. Thats why I'd like to see if someone can find an AIC solution.

by **Mauriès Robert** » Sat Jan 30, 2021 7:15 pm

Hi Denis,
The 5r2c3 can be eliminated from the beginning with an anti-track built on a sequence of 8 entities, like this :
E={5r7c3,1r2c3}
P'(E) : {-E}=>[ [ [5r9c1 and (1r1c2->1r2c6)->7r1c8)]->1r9c5 ]->6r9c8]->9r3c8->9r2c3 => -5r2c3
where the [] nested indicate the order of construction of the sequence.
So I think we can do the same thing with a braid, right from the beginning.
Robert

Website · by **denis_berthier** » Sun Jan 31, 2021 3:36 am

Mauriès Robert wrote:Hi Denis,
The 5r2c3 can be eliminated from the beginning with an anti-track built on a sequence of 8 entities, like this :
E={5r7c3,1r2c3}
P'(E) : {-E}=>[ [ [5r9c1 and (1r1c2->1r2c6)->7r1c8)]->1r9c5 ]->6r9c8]->9r3c8->9r2c3 => -5r2c3
where the [] nested indicate the order of construction of the sequence.
So I think we can do the same thing with a braid, right from the beginning.

As ordered anti-tracks are a degraded version of braids, this shouldn't be a surprise.

How to obtain this in SudoRules?
- First, load braids.
- Second, init the puzzle: (init-sudoku-string "..42....6.6..7..3.2....61..1....46...5..9..1...71....99....37...8..4..5...37....8")
- Third, ask for the wanted elimination: (try-to-eliminate-candidates 523); SudoRules gives:
braid[8]: r2n1{c3 c6} - r1n1{c6 c2} - r9n1{c2 c5} - c1n5{r1 r9} - r1n7{c2 c8} - r9n6{c1 c8} - c8n9{r1 r3} - b1n9{r3c3 .} ==> r2c3 ≠ 5

There appears to be some common points between your anti-track and the rlcs of my braid, but the difference is, almost all the useful information is missing in the anti-track.

Anyway, in such hard puzzles, I think the first elimination has to be as simple as possible. Here, a braid[6] and a whip[8] are available before this braid[8].
Moreover, there is no justification for trying n5r2c3 first.

by **Mauriès Robert** » Sun Jan 31, 2021 8:21 am

denis_berthier wrote:As ordered anti-tracks are a degraded version of braids, this shouldn't be a surprise.

I indicated this elimination, with an anti-track or a braid [8], to emphasize the oversized aspect of the SE elimination that you were recalling, whereas this one is quite direct.
This said, an anti-track is not comparable to a braid but rather to an S-braid where S is either an alignment (g) or a closed set (doublet, triplet, etc ...).
Moreover, as it is possible to make nested anti-tracks, one could also make comparisons with the B-braids.

denis_berthier wrote:How to obtain this in SudoRules?

I don't have Sudorules, and as you know I make resolutions by hand.

denis_berthier wrote:There appears to be some common points between your anti-track and the rlcs of my braid, but the difference is, almost all the useful information is missing in the anti-track.

I know that I'm being criticized on this site for not indicating the elements on the left in the sequence. I have explained this to myself. My writing is only a textual representation of a marking (coloring) on the puzzle where the left elements are never marked. If I make a representation of the type (-E)=>(a,A)->(b,B)-> etc... instead of (-E)=>A->B-> etc..., to avoid writing (E)=>-a->A->-b->B->etc...which is not representative of the marking, some people will find that I'm plagiarizing you when I don't have any. I conceived the notion of anti-track several years ago when I did not know of the existence of your work on whips and braids, nor AICs for that matter.

denis_berthier wrote:Anyway, in such hard puzzles, I think the first elimination has to be as simple as possible. Here, a braid[6] and a whip[8] are available before this braid[8].
Moreover, there is no justification for trying n5r2c3 first.

Of course, but by hand you don't necessarily see the simplest first. We have already discussed this.
On this puzzle, what attracts my attention is the pair 1b1. It is therefore natural that I am interested in the anti-track P'(1r2c3). This means that the target to be eliminated is one of the candidates who see 1r2c3 without being part of the anti-track. At this stage we have :
P'(1r2c3): (-1r2c3)=>(1r1c2->1r2c6)->1r9c5->...
Here is the reasoning I made.
There remain in r9 two 5's and two 6's that would allow further development of the anti-track. To rely on the 5s, we must eliminate the 5r7c3, which suggests taking E={1r2c3, 5r7c3}, but then the target is perfectly identified, it is the 5r2c3. The rest is done as I indicated in my comment.
It is generally this type of reasoning that I use: I start with a pair of candidates and progressively identify a target by constituting E.
Robert

Website · by **denis_berthier** » Sun Jan 31, 2021 9:36 am

Mauriès Robert wrote:
denis_berthier wrote:As ordered anti-tracks are a degraded version of braids, this shouldn't be a surprise.

I indicated this elimination, with an anti-track or a braid [8], to emphasize the oversized aspect of the SE elimination that you were recalling, whereas this one is quite direct.
This said, an anti-track is not comparable to a braid but rather to an S-braid where S is either an alignment (g) or a closed set (doublet, triplet, etc ...).
Moreover, as it is possible to make nested anti-tracks, one could also make comparisons with the B-braids.

OK, like T&E, anti-tracks are a hotpot where you can mix any ingredients.
But, in the present case, I don't see any Subset or g-candidate in yours; so, the proper comparison is with braids.

Mauriès Robert wrote:
denis_berthier wrote:There appears to be some common points between your anti-track and the rlcs of my braid, but the difference is, almost all the useful information is missing in the anti-track.

I know that I'm being criticized on this site for not indicating the elements on the left in the sequence. I have explained this to myself. My writing is only a textual representation of a marking (coloring) on the puzzle where the left elements are never marked.

What's missing most for the reader is not the llcs, but the csp-variables.

by **Mauriès Robert** » Sun Jan 31, 2021 10:50 am

denis_berthier wrote:OK, like T&E, anti-tracks are a hotpot where you can mix any ingredients.

No you can't call the anti-track "hodpot", nor can you say it's like the T&E. It is to deny the principles of TDP.
TDP is based on the construction of sequences using basic techniques (TB =singles, alignments, closed sets). The symbol A->B should be understood like this : If A is true, then in the resolution state obtained by placing A the TB implies that B is true too.
I know this is oversized for you, but it is not a catch-all. I could talk about S-TDP by limiting the TB to singles, and the antitracks would be comparable to your whips and braids. But that's not what I'm looking for, because my goal is not the same as yours.
We will have the opportunity to talk about it again, this thread is not the place for this debate.

denis_berthier wrote:What's missing most for the reader is not the llcs, but the csp-variables.

Ok, so I should write (-E)=>CSP1(a1,A2)->CSP2(a2,A2)->... instead of (-E)=>A1->A2-.... so that it is understandable to everyone?
For example for the puzzle of this thread :
(-E)=>1b1(c2r3,r1c2)->1b2(r1c34,r2c6)->... instead of (-E)->1r1c2->1r2c6->...
If so, why not, if it allows the readers to understand better without looking at the marking on the puzzle.
Robert

Website · by **denis_berthier** » Sun Jan 31, 2021 12:11 pm

Mauriès Robert wrote:
denis_berthier wrote:OK, like T&E, anti-tracks are a hotpot where you can mix any ingredients.

No you can't call the anti-track "hodpot", nor can you say it's like the T&E. It is to deny the principles of TDP.
TDP is based on the construction of sequences using basic techniques (TB =singles, alignments, closed sets). The symbol A->B should be understood like this : If A is true, then in the resolution state obtained by placing A the TB implies that B is true too.

Which is exactly what T&E(S) does, except that you keep only the assertion part of it.

by **Mauriès Robert** » Sun Jan 31, 2021 2:39 pm

denis_berthier wrote:
Mauriès Robert wrote:
denis_berthier wrote:OK, like T&E, anti-tracks are a hotpot where you can mix any ingredients.

No you can't call the anti-track "hodpot", nor can you say it's like the T&E. It is to deny the principles of TDP.
TDP is based on the construction of sequences using basic techniques (TB =singles, alignments, closed sets). The symbol A->B should be understood like this : If A is true, then in the resolution state obtained by placing A the TB implies that B is true too.

Which is exactly what T&E(S) does, except that you keep only the assertion part of it.

You're not going to make me believe that for a whip [15] you "see" the model otherwise than by constructing a sequence that leads to invalidity as one does with T&E(S). So you're doing what I do, except that it's the program that does it.
I understand, as we've talked about in other threads, that it's not the target that is the starting point for a whip or a braid. It's the same thing with the anti-tracks, it's a whip-partial [1] for you, a pair for me.
Robert

by **Mauriès Robert** » Sun Jan 31, 2021 3:23 pm

denis_berthier wrote:How to obtain this in SudoRules?
1- First, load braids.
2- Second, init the puzzle: (init-sudoku-string "..42....6.6..7..3.2....61..1....46...5..9..1...71....99....37...8..4..5...37....8")
3- Third, ask for the wanted elimination: (try-to-eliminate-candidates 523); SudoRules gives:
braid[8]: r2n1{c3 c6} - r1n1{c6 c2} - r9n1{c2 c5} - c1n5{r1 r9} - r1n7{c2 c8} - r9n6{c1 c8} - c8n9{r1 r3} - b1n9{r3c3 .} ==> r2c3 ≠ 5

What would have happened if at stage 3 Sudorules had found nothing?
Robert

Website · by **denis_berthier** » Sun Jan 31, 2021 3:58 pm

Mauriès Robert wrote:
denis_berthier wrote:How to obtain this in SudoRules?
1- First, load braids.
2- Second, init the puzzle: (init-sudoku-string "..42....6.6..7..3.2....61..1....46...5..9..1...71....99....37...8..4..5...37....8")
3- Third, ask for the wanted elimination: (try-to-eliminate-candidates 523); SudoRules gives:
braid[8]: r2n1{c3 c6} - r1n1{c6 c2} - r9n1{c2 c5} - c1n5{r1 r9} - r1n7{c2 c8} - r9n6{c1 c8} - c8n9{r1 r3} - b1n9{r3c3 .} ==> r2c3 ≠ 5

What would have happened if at stage 3 Sudorules had found nothing?

Nothing.
Obviously, if the elimination was not available, SudoRules wouldn't do it. Function "try-to-eliminate-candidates" doesn't do more than its name suggests; in particular, it doesn't try to eliminate any other candidate than those listed as arguments.

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