The New Sudoku Players' Forum

by **ArkieTech** » Mon Jan 05, 2015 1:59 am

Code: Select all: *-----------* |.1.|...|..2| |4..|21.|.7.| |7.9|..3|...| |---+---+---| |.78|4.6|...| |..1|...|8..| |...|3.1|79.| |---+---+---| |...|8..|6.9| |.4.|.52|..7| |8..|...|.4.| *-----------*

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by **Leren** » Mon Jan 05, 2015 2:15 am

Code: Select all: *------------------------------------------------------------------------* | 36 1 356 | 79 479 8 | 49 356 2 | | 4 8 356 | 2 1 a59 | 359 7 356 | | 7 2 9 |b56 46 3 | 145 15 8 | |-----------------------+------------------------+-----------------------| | 359 7 8 | 4 29 6 | 1235 1235 135 | | 369 369 1 | 579 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | |-----------------------+------------------------+-----------------------| | 135 35 27 | 8 3-7 4 | 6 1235 9 | | 1369 4 36 |c169 5 2 | 13 8 7 | | 8 3569 2-7 |c1679 369-7 ca79 | 1235 4 135 | *-----------------------------------------------------------------------*

ALS XY Wing: (7=5) r29c6 - (5=6) r3c4 - (6=7) r8c4, r9c46 => - 7 r7c5, r9c35; stte

Leren

by **SteveG48** » Mon Jan 05, 2015 3:20 am

Code: Select all: *------------------------------------------------------------* | 36 1 356 |ah79 ah479 8 | 4-9 356 2 | | 4 8 356 | 2 1 59 | 359 7 356 | | 7 2 9 | a56 ah46 3 | 145 15 8 | *-------------------+--------------------+-------------------| |d359 7 8 | 4 29 6 | 1235 1235 135 | |c369 c369 1 | b579 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | *-------------------+--------------------+-------------------| | 135 35 27 | 8 g37 4 | 6 1235 9 | |e1369 4 36 | 169 5 2 | 13 8 7 | | 8 f3569 27 | 1679 g3679 g79 | 1235 4 135 | *------------------------------------------------------------*

(9=57)r13c45 - (57=9)r5c4 - r5c12 = r4c1 - r8c1 = r9c2 - (9=6)r7c5,r9c56 - (6=9)r1c45,r3c5 => -9 r1c7 ; stte

by **pjb** » Mon Jan 05, 2015 4:10 am

accidental duplicate

by **pjb** » Mon Jan 05, 2015 4:21 am

Code: Select all: 36 1 356 | 79 479 8 | 49 356 2 4 8 356 | 2 1 b59 | 359 7 356 7 2 9 |a56 4-6 3 | 145 15 8 ------------------------+----------------------+--------------------- g359 7 8 | 4 h29 6 | 1235 1235 135 369 369 1 | 579 279 579 | 8 236 4 2 f56 4 | 3 8 1 | 7 9 56 ------------------------+----------------------+--------------------- 135 e35 27 | 8 d37 4 | 6 1235 9 1369 4 36 | 19-6 5 2 | 13 8 7 8 3569 27 | 179-6 3679 c79 | 1235 4 135 (6=5)r3c4 - (5=9)r2c6 - (9=7)r9c6 - (7=3)r7c5 - (3=5)r7c2 - r6c2 = (5-9)r4c1 = r4c5-(9=6)r9c5 => -6 r3c5, r8c4, r9c4; stte \ _________\ ____________________________________________ /

Phil

by **gurth** » Mon Jan 05, 2015 1:53 pm

by **JC Van Hay** » Mon Jan 05, 2015 5:41 pm

Code: Select all: +-----------------+---------------------+-----------------+ | 36 1 356 | 79 479 8 | 49 356 2 | | 4 8 356 | 2 1 (59) | 359 7 356 | | 7 2 9 | (56) 4(6) 3 | 145 15 8 | +-----------------+---------------------+-----------------+ | 359 7 8 | 4 29 6 | 1235 1235 135 | | 369 369 1 | 579 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | +-----------------+---------------------+-----------------+ | 135 35 27 | 8 (37) 4 | 6 1235 9 | | 1369 4 36 | 169 5 2 | 13 8 7 | | 8 3569 27 | 1679 9-37(6) (79) | 1235 4 135 | +-----------------+---------------------+-----------------+ [6r9c5=6r3c5 - (6=59)r3c4.r2c6 - (9=73)r9c6.r7c5] - (73)r9c5; stte

by **DonM** » Tue Jan 06, 2015 7:02 pm

SteveG48 wrote:
Code: Select all
*------------------------------------------------------------* | 36 1 356 |ah79 ah479 8 | 4-9 356 2 | | 4 8 356 | 2 1 59 | 359 7 356 | | 7 2 9 | a56 ah46 3 | 145 15 8 | *-------------------+--------------------+-------------------| |d359 7 8 | 4 29 6 | 1235 1235 135 | |c369 c369 1 | b579 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | *-------------------+--------------------+-------------------| | 135 35 27 | 8 g37 4 | 6 1235 9 | |e1369 4 36 | 169 5 2 | 13 8 7 | | 8 f3569 27 | 1679 g3679 g79 | 1235 4 135 | *------------------------------------------------------------*

(9=57)r13c45 - (57=9)r5c4 - r5c12 = r4c1 - r8c1 = r9c2 - (9=6)r7c5,r9c56 - (6=9)r1c45,r3c5 => -9 r1c7 ; stte

Not to select Steve out for special punishment (given the 'other' thread), but I was made aware of this solution by professor Eleven as part of a (rejected) notation homework assignment: What really caught my interest was the opening move.

If I am interpreting it correctly the full ALS is candidates 4,5,6,7,9 in cells r13c45. At first glance the move breaks the rules of the use of an ALS whereby all target candidates of the ALS (in this case 57) must 'see' all likewise candidates in the ALS: Here there is a 7 in r1c5 which doesn't 'see' cell r5c4.

On the other hand, my guess is that Steve's logic is that if 'not (9)r1c45' then that leaves 7 in r1c4 which must leave 4 in r1c5 so the 7 in r1c5 is not operative and can be disregarded as far as the ALS target is concerned. If I'm guessing Steve's logic correctly then there's something potentially clever about the premise, but I'm not sure the logic is accurate.

I don't think I've ever come across the use of this particular ALS construct before. My concern is the fact that groups (ie. here (9)r1c45)) have their own set of logic rules and I'm not so sure that one can assume the premise that 'not(9)r1c45" can allow the assumption (that 7 is placed in r1c4 thus removing the 7 in r1c5) being made here. (Again: All of this being dependent on the fact that I'm not missing something entirely different here.)

If David is still here, I'd particularly appreciate his input.

by **SteveG48** » Tue Jan 06, 2015 7:26 pm

DonM wrote:
SteveG48 wrote:
Code: Select all
*------------------------------------------------------------* | 36 1 356 |ah79 ah479 8 | 4-9 356 2 | | 4 8 356 | 2 1 59 | 359 7 356 | | 7 2 9 | a56 ah46 3 | 145 15 8 | *-------------------+--------------------+-------------------| |d359 7 8 | 4 29 6 | 1235 1235 135 | |c369 c369 1 | b579 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | *-------------------+--------------------+-------------------| | 135 35 27 | 8 g37 4 | 6 1235 9 | |e1369 4 36 | 169 5 2 | 13 8 7 | | 8 f3569 27 | 1679 g3679 g79 | 1235 4 135 | *------------------------------------------------------------*

(9=57)r13c45 - (57=9)r5c4 - r5c12 = r4c1 - r8c1 = r9c2 - (9=6)r7c5,r9c56 - (6=9)r1c45,r3c5 => -9 r1c7 ; stte

Not to select Steve out for special punishment (given the 'other' thread), but I was made aware of this solution by professor Eleven as part of a (rejected) notation homework assignment: What really caught my interest was the opening move.

If I am interpreting it correctly the full ALS is candidates 4,5,6,7,9 in cells r13c45. At first glance the move breaks the rules of the use of an ALS whereby all target candidates of the ALS (in this case 57) must 'see' all likewise candidates in the ALS: Here there is a 7 in r1c5 which doesn't 'see' cell r5c4.

On the other hand, my guess is that Steve's logic is that if 'not (9)r1c45' then that leaves 7 in r1c4 which must leave 4 in r1c5 so the 7 in r1c5 is not operative and can be disregarded as far as the ALS target is concerned. If I'm guessing Steve's logic correctly then there's something potentially clever about the premise, but I'm not sure the logic is accurate.

That was, indeed, my logic. Given the discussion in the other thread, I think that in the future I will write the opening move as (9=4567)r13c45. However, I can't see any confusion about the rest of the logic. Surely if we eliminate the 9 from the ALS we are left with a locked set, and surely we then know that the positions of the 5 and 7 are in c4. In fact, we know where each of the candidates in the LS must be. Am I missing something here?

by **JC Van Hay** » Tue Jan 06, 2015 9:36 pm

Here is how I would write Steve's solution : either as a forbidding matrix or as a more or less condensed AIC-like network.

Code: Select all: +-------------------+---------------------+-----------------+ | 36 1 356 | (79) (479) 8 | 4-9 356 2 | | 4 8 356 | 2 1 5-9 | 359 7 356 | | 7 2 9 | (56) (46) 3 | 145 15 8 | +-------------------+---------------------+-----------------+ | 359 7 8 | 4 29 6 | 1235 1235 135 | | 369 36(9) 1 | (579) 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | +-------------------+---------------------+-----------------+ | 135 35 27 | 8 (37) 4 | 6 1235 9 | | 1369 4 36 | 169 5 2 | 13 8 7 | | 8 356(9) 27 | 1679 (3679) (79) | 1235 4 135 | +-------------------+---------------------+-----------------+ 9r1c4=7r1c4 9r1c5=7r1c5=4r1c5 4r3c5=6r3c5 6r3c4=5r3c4 7r5c4=============5r5c4=9r5c4 9r5c2=9r9c2 ... instead of 9r5c12=9r4c1-9r8c1=9r9c2 9r9c6=7r9c6 7r7c5=3r7c5 6r9c5=============9r9c5=7r9c5=3r9c5 Conclusion : [9r1c4==9r1c5]-9r1c7.r2c6

which can be read as such, like an nrczt-chain, or as

Code: Select all: [NP(97)r1c45=4r1c5-(4=6)r3c5-(6=5)r3c4-(5=*7)r5c4-(7=9)r1c4]=*9r5c4-9r5c2=9r9c2-9r9c56=NT(736)r9c6.r79c5-(6=4)r3c5-4r1c5=NP(79)r1c45 :=> [9r1c4==9r1c5]-9r1c7.r2c6 Death Blossom[NP(97)r1c45=4r1c5-(4=6)r3c5-6r3c4=*NT(579)r351c4]=*9r5c4-9r5c2=9r9c2-9r9c56=NT(736)r9c6.r79c5-(6=4)r3c5-4r1c5=NP(79)r1c45 :=> [9r1c4==9r1c5]-9r1c7.r2c6 Death Blossom[NP(97)r1c45=4r1c5-(4=6)r3c5-6r3c4=*NT(579)r351c4]=*9r5c4-9r5c2=9r9c2-9r9c56=NT(736)r9c6.r79c5-6r3c5=NT(479)r3c5.r1c45 :=> [9r1c4==9r1c5]-9r1c7.r2c6 9r1c45==[7r1c4 AND 5r3c4]-(75=9)r5c4-9r5c2=9r9c2-9r9c56=NT(736)r9c6.r79c5-6r3c5=NT(479)r3c5.r1c45 :=> [9r1c4==9r1c5]-9r1c7.r2c6 .... where == is used to mean a derived SIS.

by **eleven** » Tue Jan 06, 2015 9:37 pm

DonM wrote:Not to select Steve out for special punishment (given the 'other' thread), but I was made aware of this solution by professor Eleven as part of a (rejected) notation homework assignment

Don, i really did not want you to make trivial things, but to compare the notations in a real-life chain:
(9=57)r13c45-(57=9)r5c4...-- (6=9)r1c45,r3c5 => -9 r1c7
Classical e.g.
(469=57)r13c45-(57=9)r5c4...- (379=6)r7c5,r9c56-(6=479)r1c45,r3c5 => -9 r1c7 or
(9=4567)r13c45-(57=9)r5c4...- (39=76)r7c5,r9c56-(476=9)r1c45,r3c5 => -9 r1c7
In both cases you need the grid to verify, where the ALS candidates are to follow the deduction.
To check the link, you have to look at the cell candidates (and count them) in both cases too.
In the classical notation it is arbitrary, on which side you put the unused candidates. So in the first link you can have them with the 9, which makes it harder to see the conclusion, or with the 57, which makes it harder to understand the following weak link. So for more clararity it might be best to put them in braces, if you want.

btw I think, that it is out of question, that the logic is correct here.

Added: JC, are you joking ?

by **DonM** » Tue Jan 06, 2015 10:23 pm

eleven wrote:
DonM wrote:Not to select Steve out for special punishment (given the 'other' thread), but I was made aware of this solution by professor Eleven as part of a (rejected) notation homework assignment

Don, i really did not want you to make trivial things, but to compare the notations in a real-life chain:
(9=57)r13c45-(57=9)r5c4...-- (6=9)r1c45,r3c5 => -9 r1c7
Classical e.g.
(469=57)r13c45-(57=9)r5c4...- (379=6)r7c5,r9c56-(6=479)r1c45,r3c5 => -9 r1c7 or
(9=4567)r13c45-(57=9)r5c4...- (39=76)r7c5,r9c56-(476=9)r1c45,r3c5 => -9 r1c7
In both cases you need the grid to verify, where the ALS candidates are to follow the deduction.
To check the link, you have to look at the cell candidates (and count them) in both cases too.
In the classical notation it is arbitrary, on which side you put the unused candidates. So in the first link you can have them with the 9, which makes it harder to see the conclusion, or with the 57, which makes it harder to understand the following weak link. So for more clararity it might be best to put them in braces, if you want.

btw I think, that it is out of question, that the logic is correct here.

Added: JC, are you joking ?

To answer you regarding the notation above, it may seem arbitrary on which side you put the unused candidates, but IMO, it isn't if one is trying to keep as close as possible to the typical simple AIC chain structure whereby the starting digit of the strong link and the final digit (at the end of the final strong link) are both the target digit.

Thus, while both formats you give above are logically correct, the second fomat (9=4567)r13c45-(57=9)r5c4...- (39=76)r7c5,r9c56-(476=9)r1c45,r3c5 => -9 r1c7 is IMO more appropriate because the target 9 is what begins and ends the AIC. It makes it very clear to the reader what the objective is.

Otherwise, I don't understand why you find any of this confusing. Anyone who understands ALS patterns will be able to follow the logic present in the format above. Plus, I think we all refer to the grid to understand someone else's solution no matter what patterns are used. Still, perhaps ad nauseum, I emphasize that all components of an ALS need to be notated.

I actually tend to believe that Steve's logic is correct. The fact is that I likely would have missed the ALS move because of the group (7)r1c45 and I wanted to make sure that it is valid so I can keep an eye out for the same construct in the future. So, very clever Steve! A move like that often seems simple and can be taken for granted after it is made, but it actually is pretty difficult to pick up manually.

BTW: I was also going to ask JC if he was serious.

by **daj95376** » Wed Jan 07, 2015 12:35 am

SteveG48 wrote:
Code: Select all
*------------------------------------------------------------* | 36 1 356 |ah79 ah479 8 | 4-9 356 2 | | 4 8 356 | 2 1 59 | 359 7 356 | | 7 2 9 | a56 ah46 3 | 145 15 8 | *-------------------+--------------------+-------------------| |d359 7 8 | 4 29 6 | 1235 1235 135 | |c369 c369 1 | b579 279 579 | 8 236 4 | | 2 56 4 | 3 8 1 | 7 9 56 | *-------------------+--------------------+-------------------| | 135 35 27 | 8 g37 4 | 6 1235 9 | |e1369 4 36 | 169 5 2 | 13 8 7 | | 8 f3569 27 | 1679 g3679 g79 | 1235 4 135 | *------------------------------------------------------------*

(9=57)r13c45 - (57=9)r5c4 - r5c12 = r4c1 - r8c1 = r9c2 - (9=6)r7c5,r9c56 - (6=9)r1c45,r3c5 => -9 r1c7 ; stte

Congratulations for turning the ALS world on its ear with a compound elimination!!! Here's my interpretation of your results:

Code: Select all: (9)r1c45 =ALS= (7)r1c4 - \ (5)r3c4 - (57=9)r5c4 - r5c12 = r4c1 - r8c1 = r9c2 - (9=6)r7c5,r9c56 - (6=9)r1c45,r3c5 (4)r1c5 (6)r3c5 => -9 r1c7,r2c6

_

[Edit: removed (incorrect) lasso reference.]

by **SteveG48** » Wed Jan 07, 2015 12:41 am

Thanks, Danny. I should have seen that second elimination. I keep telling myself that I'll stop doing that, but.....

by **DonM** » Wed Jan 07, 2015 4:15 am

My apologies, but I realize that I misspoke above regarding the notation that I would use for Steve's chain. The answer I gave is not what I even use myself. When it comes to a chain containing ALS patterns, my preference has always been to have the 'body' of the ALS at the right since that seems to fit with right-to-left logic flow of the pattern: if not a, then bcde.

So, in Steve's chain, I would have used:
(9=4567)r13c45-(57=9)r5c4........-(6=479)r1c45,r3c5 => -9 r1c7

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