The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |...|.8.|...|
 |..2|4.3|5..|
 |.9.|...|.2.|
 |---+---+---|
 |3..|...|..7|
 |.2.|7.6|.1.|
 |..8|...|4..|
 |---+---+---|
 |.6.|.1.|.9.|
 |..4|...|8..|
 |...|.6.|...|
 *-----------*


Play/Print this puzzle online

[SteveG48]

Code: Select all

 *-----------------------------------------------------------------------*
 |  14567    13457  13567  | 12569  8      12579  | 13679  367    1369   |
 |  167     b17     2      | 4      79     3      | 5      8      169    |
 |  8        9      13567  | 156    57     157    | 1367   2      4      |
 *-------------------------+----------------------+----------------------|
 |  3      ac145    156-9  | 1259   2459   8      | 269    56     7      |
 |ac459      2    ac59     | 7      345-9  6      | 3-9    1      8      |
 |  1567-9 ac157    8      | 12359  2359   1259   | 4      356    23569  |
 *-------------------------+----------------------+----------------------|
 |  257      6      357    | 8      1      4      | 237    9      235    |
 |  12579    1357   4      | 2359   23579  2579   | 8      3567   12356  |
 |  12579    8      13579  | 2359   6      2579   | 1237   4      1235   |
 *-----------------------------------------------------------------------*


(9=1457)b4p2468 - (1=7)r2c2 - (7=1459)b4p2468 => -9 r4c3,r5c57,r6c1 ; stte

[999_Springs]

(9=1457)b4p2468 - (1=7)r2c2 - (7=1459)b4p2468 => -9 r4c3,r5c57,r6c1 ; stte

that's in fact a sue-de-coq pattern, so in addition to this, you can also get rid of 1,7 from the rest of column 2, and 5 from the rest of block 4. good job spotting that, i haven't seen one of those in years

[SCLT]

I had almost the same - viewing r246c2 and r5c13 as two doubly-linked ALSes. I'm on mobile so I won't post a grid, but the eliminations are the same as the Sue-de-Coq.

Are all SDC patterns just doubly-linked ALSes?

[999_Springs]

Are all SDC patterns just doubly-linked ALSes?

i remember having this revelation a while ago that most sue-de-coq patterns were doubly linked als's that could be viewed in 2 different ways depending on which als you put the intersection cells into

e.g. in this example you could have r246c2 and r5c13 with 4,5 in common, or you could have r2c2 and r4c2r5c13r6c2 with 1,7 in common. in this case you both get the same eliminations as the sue-de-coq whichever way you do it

in answer to your question, i haven't seen a sue-de-coq in a real puzzle that wasn't just a pair of doubly linked als's (not that i know of), but i'm pretty sure this hypothetical pattern counts as a sue-de-coq but i don't see how to get it as doubly linked als's

Code: Select all

123456 456 # | . . . | . . .
123456 456 # | . . . | . . .
@#     #   # | . . . | . . .
-------------+-------+-------
123    .   . | . . . | . . .
123    .   . | . . . | . . .
@      .   . | . . . | . . .
-------------+-------+-------
@      .   . | . . . | . . .
@      .   . | . . . | . . .
@      .   . | . . . | . . .

@ =/= 123
# =/= 456

can someone who's been active in the sudoku universe in the last few years confirm or deny my suspicions please, i'm pretty rusty on this

[SteveG48]

(9=1457)b4p2468 - (1=7)r2c2 - (7=1459)b4p2468 => -9 r4c3,r5c57,r6c1 ; stte

that's in fact a sue-de-coq pattern, so in addition to this, you can also get rid of 1,7 from the rest of column 2, and 5 from the rest of block 4. good job spotting that, i haven't seen one of those in years

Thanks. I saw an interesting pattern there, but I didn't recognize it. I'm somewhat familiar with the sue-de-coq, but not enough so to remember it when I see it.

[SCLT]

in answer to your question, i haven't seen a sue-de-coq in a real puzzle that wasn't just a pair of doubly linked als's (not that i know of), but i'm pretty sure this hypothetical pattern counts as a sue-de-coq but i don't see how to get it as doubly linked als's

Code: Select all

123456 456 # | . . . | . . .
123456 456 # | . . . | . . .
@#     #   # | . . . | . . .
-------------+-------+-------
123    .   . | . . . | . . .
123    .   . | . . . | . . .
@      .   . | . . . | . . .
-------------+-------+-------
@      .   . | . . . | . . .
@      .   . | . . . | . . .
@      .   . | . . . | . . .

@ =/= 123
# =/= 456

True, but it is an AALS triply-linked to an ALS, which can be seen as a variation on doubly-linked ALSes leading to the stated eliminations

[Cenoman]

Are all SDC patterns just doubly-linked ALSes?

According to the definitions of each pattern, the answer is No.

ALS: subset of n unsolved cells in a sector (row, column, box) with n+1 digits, AALS: subset of n unsolved cells in a sector with n+2 digits
Sue de Coq: one AALS doubly linked to two ALS's, therefore possibly three different sectors for the pattern.
Doubly linked ALS's: two different sectors at most.

If a Sue de Coq spreading over three sectors exists, the conjecture would be demonstrated False. I have not in mind such a counter example.

The reciprocal conjecture "Can all doubly-linked ALS's be split into a SDC pattern ?" could also be raised.
I fear that not any ALS can be considered as the result of a chain of an AALS doubly linked to an ALS in the same sector... No demo of that, and no time to spend on it.

[999_Springs]

Sue de Coq: one AALS doubly linked to two ALS's, therefore possibly three different sectors for the pattern.
Doubly linked ALS's: two different sectors at most.

If a Sue de Coq spreading over three sectors exists, the conjecture would be demonstrated False.

thanks for providing the definitions, so you mean something like this hypothetical pattern?

Code: Select all

12345 123 @ | 12 @ @ | @ @ @
345   .   . | .  . . | . . .
#     .   . | .  . . | . . .
------------+--------+-------
45    .   . | .  . . | . . .
#     .   . | .  . . | . . .
#     .   . | .  . . | . . .
------------+--------+-------
#     .   . | .  . . | . . .
#     .   . | .  . . | . . .
#     .   . | .  . . | . . .

@ =/= 12
# =/= 45

The reciprocal conjecture "Can all doubly-linked ALS's be split into a SDC pattern ?" could also be raised.

pretty sure that isn't true, here's my favourite puzzle of all time top1465#2 with an extra clue 8r8c2 i don't see how this could look like a sue-de-coq

Code: Select all

     7     126        8 |   459   4569    4569 |     3   1456   1259
   469      36     3469 |     2  34569       1 |  4679  45678   5789
     5    1236   123469 |    78   3469      78 | 12469    146    129
------------------------+----------------------+---------------------
   189       4     1579 |  3579     59    3579 |   178      2      6
     3    1267    12679 |   479      8   24679 |   147   1457    157
   268    2567     2567 |     1   2456   24567 |   478      9      3
------------------------+----------------------+---------------------
    12       9    12357 |     6    125    2358 |   127   1378      4
  1246       8    12346 |   349      7    2349 |     5    136    129
  1246  123567  1234567 | 34589  12459  234589 | 12679  13678  12789

als a = r2c12357
als b = r7c157
restricted common 5,7
eliminations -5r1469c5 -7r4569c7 -46r2c8 -9r2c9 -1r7c38 -2r7c36

[SpAce]

Are all SDC patterns just doubly-linked ALSes?

According to the definitions of each pattern, the answer is No.

ALS: subset of n unsolved cells in a sector (row, column, box) with n+1 digits, AALS: subset of n unsolved cells in a sector with n+2 digits
Sue de Coq: one AALS doubly linked to two ALS's, therefore possibly three different sectors for the pattern.
Doubly linked ALS's: two different sectors at most.

If a Sue de Coq spreading over three sectors exists, the conjecture would be demonstrated False. I have not in mind such a counter example.

First a disclaimer: I want to make clear that I know almost nothing about SDCs. I've never spotted one on my own, and have trouble following all but the simplest examples. Thus, anything I say of this topic is of limited value.

However, I must ask if a three-sector version would be called a Sue de Coq (whose original name was Two-Sector Disjoint Subset)? I've thought that the SDC is still limited to two sectors, and any extensions to more sectors aren't covered by that name. If that's true, would it change the answer to the original question? (Also, as a different side-note: I think SDC cores can be AAALS as well as AALS).

This link seems to have examples of Distributed Disjoint Subsets with more than two sectors:
http://forum.enjoysudoku.com/distributed-disjoint-subsets-t5423.html

[Added: I tested the theory (of SDCs being doubly-linked ALS-XZs) and it seemed to hold at least for all six Hodoku SDC examples, including the listed extended types. It probably means that for all practical purposes the rule holds. That's really helpful as it means I don't have to learn SDC as a distinct technique - or doubly-linked ALSs either because I just see them as ALS-loops. Thanks SCLT and 999_Springs for pointing out this possibility! ]

[SteveG48]

Sue de Coq: one AALS doubly linked to two ALS's, therefore possibly three different sectors for the pattern.

I find myself getting more interested in the SDC, so I'd like to make sure that I understand this definition.

When we say one AALS doubly linked to two ALS's, we mean that the AALS has double links to each of the ALS's. That is, two candidated in the AALS "see" two candidates in one of the ALS's and two candidates in the AALS "see" two candidates in the other ALS. Is that correct? Given that, my understanding of the SDC has an additional requirement that the 2 candidates in one of those double links both need to be different from both candidates in the other double link. Is that also correct?

[Cenoman]

When we say one AALS doubly linked to two ALS's, we mean that the AALS has double links to each of the ALS's. That is, two candidated in the AALS "see" two candidates in one of the ALS's and two candidates in the AALS "see" two candidates in the other ALS. Is that correct? Given that, my understanding of the SDC has an additional requirement that the 2 candidates in one of those double links both need to be different from both candidates in the other double link. Is that also correct?

To me, yes that is also correct.
Steve's requirement is the application of the general rule for chaining ALS's, AALS's, AAALS's...
General rule: each of these "not locked sets" must have at least as many restricted commons as freedom degrees, and all its RCs must be distinct.

I think SDC cores can be AAALS as well as AALS).

What makes SDC eliminate any candidate in sight of its digit groups is its freedom degree equal to 0. If you select an AAALS as a core, you must have one more restricted common to keep this character (In total an AAALS triply linked with an ALS and doubly linked with another ALS)

you mean something like this hypothetical pattern?

Code: Select all

12345 123 @ | 12 @ @ | @ @ @
345   .   . | .  . . | . . .
#     .   . | .  . . | . . .
------------+--------+-------
45    .   . | .  . . | . . .
#     .   . | .  . . | . . .
#     .   . | .  . . | . . .
------------+--------+-------
#     .   . | .  . . | . . .
#     .   . | .  . . | . . .
#     .   . | .  . . | . . .

@ =/= 12
# =/= 45

The idea is somehow like this. But this example is also a triply-linked AALS with an ALS
(12345)r1c124 -345- (345)r24c1 or (12345)r124c1 -123-(123)r1c24

Rather something like this:

Code: Select all

123456 125 @ | 12 @ @ | @ @ @      r1c1 could be 156, 256, 356 or 456 only
345    56  & | .  . . | . . .
#      &   & | .  . . | . . .
-------------+--------+-------
34     .   . | .  . . | . . .
#      .   . | .  . . | . . .
#      .   . | .  . . | . . .
-------------+--------+-------
#      .   . | .  . . | . . .
#      .   . | .  . . | . . .
#      .   . | .  . . | . . .

@ =/= 12
# =/= 34
& =/= 56


[SpAce]

When we say one AALS doubly linked to two ALS's, we mean that the AALS has double links to each of the ALS's. That is, two candidated in the AALS "see" two candidates in one of the ALS's and two candidates in the AALS "see" two candidates in the other ALS. Is that correct? Given that, my understanding of the SDC has an additional requirement that the 2 candidates in one of those double links both need to be different from both candidates in the other double link. Is that also correct?

To me, yes that is also correct.

I think it's clearly stated in the original definition of the SDC pattern as well. The digits drawn from the core (intersection) must not overlap in the two (box and line) ALSs. However, any extra digits used in the two ALSs can overlap.

Steve's requirement is the application of the general rule for chaining ALS's, AALS's, AAALS's...
General rule: each of these "not locked sets" must have at least as many restricted commons as freedom degrees, and all its RCs must be distinct.

Makes sense.

I think SDC cores can be AAALS as well as AALS).

What makes SDC eliminate any candidate in sight of its digit groups is its freedom degree equal to 0. If you select an AAALS as a core, you must have one more restricted common to keep this character (In total an AAALS triply linked with an ALS and doubly linked with another ALS)

Also makes sense, and seems to hold with all the AAALS examples I've seen. A good rule too, as it makes it easier to understand the SDC logic.

This has been a good discussion. I doubt I'll be spotting many SDCs on my own, but at least I now understand them better if seen as solution steps. For example, the standard Hodoku solution for this puzzle http://forum.enjoysudoku.com/all-to-all-t34912.html uses five SDCs, and I just ignored them the first time I saw them because they weren't part of my understandable techniques list. Now I have no problem seeing how they work, though it's still highly unlikely I would spot them on my own. So, thanks for that, everyone!

[SteveG48]

Thank you both.