The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |..1|.7.|4..|
 |.4.|9.8|..7|
 |...|..5|..2|
 |---+---+---|
 |..2|..4|36.|
 |...|...|...|
 |.93|2..|7..|
 |---+---+---|
 |9..|8..|...|
 |4..|1.6|.7.|
 |..7|.9.|1..|
 *-----------*

Play/Print this puzzle online

[JC Van Hay]

Code: Select all

+------------------+---------------------+-----------------------+
| 23568  23568  1  | (36)   7       (23) | 4    9        58(36)  |
| 23     4      56 | 9      3-2(1)  8    | 56   1(3)     7       |
| 3678   3678   9  | 346    36(14)  5    | 68   18(3)    2       |
+------------------+---------------------+-----------------------+
| 1      578    2  | 57     58      4    | 3    6        9       |
| 5678   5678   4  | 3567   3568    9    | 258  258      1       |
| 568    9      3  | 2      568     1    | 7    458      458     |
+------------------+---------------------+-----------------------+
| 9      1      56 | 8      25(34)  7    | 256  245(3)   45(36)  |
| 4      235    8  | 1      235     6    | 9    7        35      |
| 2356   2356   7  | 35(4)  9       23   | 1    235(48)  35(468) |
+------------------+---------------------+-----------------------+


Triangular Matrix

Code: Select all

1r2c5 1r3c5
      4r3c5 4r7c5
            4r9c4 4r9c8 4r9c9
                  8r9c8 8r9c9
2r1c6                         3r1c6
                              3r1c4 6r1c4
                        6r9c9       6r1c9 6r7c9
            3r7c5                         3r7c9 3r7c8
                              3r1c9             3r23c8 => [1r2c5==2r1c6]-2r2c5; stte


or

Double Kraken 6C9 + 3R7 => [1r2c5==2r1c6]-2r2c5; stte
6r1c9-(6=3)r1c4-(3=2)r1c6
||
6r7c9-3r7c9=*[1r2c5=(1-4)r3c5=(4-3)r7c5=*3r7c8-3r23c8=3r1c9-(3=2)r1c6]
||
6r9c9-HP(48)r9c89=4r9c4-4r7c5=(4-1)r3c5=1r2c5

or

Bidirectional chain :
3r7c9=*[1r2c5=(1-4)r3c5=(4-3)r7c5=*3r7c8-3r23c8=3r1c9-(3=2)r1c6]
|
6r7c9=*[1r2c5=(1-4)r3c5=4r7c5-4r9c4=HP(48)r9c89-6r9c9=*6r1c9-(6=3)r1c4-(3=2)r1c6] => [1r2c5==2r1c6]-2r2c5; stte

Number of constraints : 9; Number of steps : 1

[Cenoman]

Code: Select all

 +----------------------+--------------------+-----------------------+
 | 23568   23568   1    | 36     7      23   | 4     9       3568    |
 | 23      4       56   | 9      123    8    | 56    13      7       |
 | 3678    3678    9    | 346    1346   5    | 68    138     2       |
 +----------------------+--------------------+-----------------------+
 | 1       578     2    | 57    b58     4    | 3     6       9       |
 | 5678    5678    4    | 3567  b3568   9    | 258   258     1       |
 | 568     9       3    | 2     b568    1    | 7     458     458     |
 +----------------------+--------------------+-----------------------+
 | 9       1       56   | 8      245-3  7    | 256   2345    3456    |
 | 4       235     8    | 1     b235    6    | 9     7       35      |
 | 2356    2356    7    | 345    9     a23   | 1     23458   34568   |
 +----------------------+--------------------+-----------------------+

Not found any stte finish else than JC' double kraken, so:
ALS XZ rule (3=2)r9c6 - (2=3)r4568c5 => -3 r7c5; lcls to 81

Edit: corrected typo (3=2)r9c6 instead of (3=2)r8c6

Cenoman

[eleven]

Code: Select all

 *---------------------------------------------------------------*
 | 23568  23568  1   | ga36    7    ga23  | 4    9       g3568   |
 | 23     4      56  |   9     13-2   8   | 56   13       7      |
 | 3678   3678   9   |  a346   1346   5   | 68   138      2      |
 |-------------------+--------------------+----------------------|
 | 1      578    2   |   57    58     4   | 3    6        9      |
 | 5678   5678   4   |   3567  3568   9   | 258  258      1      |
 | 568    9      3   |   2     568    1   | 7    458     f458    |
 |-------------------+--------------------+----------------------|
 | 9      1      56  |   8    d2345   7   | 256 e2345    e3456   |
 | 4      235    8   |   1     235    6   | 9    7       f35     |
 | 2356   2356   7   |  b345   9      23  | 1   c23458  fc34568  |
 *---------------------------------------------------------------*

Note, that if r9c4 is not 4, then there is a hidden pair 48 in r9c89, and with 4r7c5 the 3 must be in r7c89, i.e r8c9=5
This gives a triple 458 with r6c9.
(2=4)b2p137-r9c4=(hp48r9c89 & [(4-3)r7c5=r7c89-(3=5)r8c9])-(3|5|6=458)r689c9-(5|8=2)r1c469 => 2r1c6, stte

[Cenoman]

Could someone give me a reference how to use square brackets in Eureka ?
Thanks in advance,

Cenoman

[eleven]

Me not, to my knowledge there are no definitions. That's why i explained in words, what is meant.

Alternative:
(*) 4r9c4=(4-3)r7c5=r7c89-(3=5)r8c9
... -4r9c4=(hp48r9c89 & (*)5r8c9) ...

I saw a different ("lasso") use of square brackets here. Might be confusing, i see.

[StrmCkr]

cant get the puzzle to crack to singles , but this move brings it down to locked candidates and 1 naked pair

{size 5 barn}
Almost Locked Set XZ-Rule: A=r9c6 {23}, B=r4568c5 {23568}, X=2, Z=3 => r7c5<>3

[Cenoman]

StrmCkr wrote:
Almost Locked Set XZ-Rule: A=r9c6 {23}, B=r4568c5 {23568}, X=2, Z=3 => r7c5<>3

Hi StrmCkr, that was exactly my path in my first post above (with a typo, though)

Best regards, Cenoman.

[JC Van Hay]

Could someone give me a reference how to use square brackets in Eureka ?
Thanks in advance,

Cenoman

Does these references answer your question, as I assumed you should know them
1. First example
2. Formalization

[JC Van Hay]

An equivalent shorter stte single-step [Number of constraints : 8] :

Code: Select all

+------------------+--------------------+-----------------------+
| 23568  23568  1  | (36)   7       2-3 | 4    9        58(36)  |
| 23     4      56 | 9      123     8   | 56   1(3)     7       |
| 3678   3678   9  | (346)  136(4)  5   | 68   18(3)    2       |
+------------------+--------------------+-----------------------+
| 1      578    2  | 57     58      4   | 3    6        9       |
| 5678   5678   4  | 3567   3568    9   | 258  258      1       |
| 568    9      3  | 2      568     1   | 7    458      458     |
+------------------+--------------------+-----------------------+
| 9      1      56 | 8      25(34)  7   | 256  245(3)   45(36)  |
| 4      235    8  | 1      235     6   | 9    7        35      |
| 2356   2356   7  | 35(4)  9       23  | 1    235(48)  35(468) |
+------------------+--------------------+-----------------------+


Bidirectional chain or AAIC-XsZ Rule(?) :
3r7c9=*[3r1c9=3r23c8-3r7c8=*(3-4)r7c5=4r3c5-(4=63)r13c4]
|
6r7c9=*[(3=6)r1c4-6r1c9=*6r9c9-HP(48)r9c89=4r9c4-(4=63)r13c4] => [3r1c9==NP(36)r13c4]-(3=2)r1c6; stte

[Cenoman]

Thank you eleven and JC Van Hay for your responses.

I have been fighting to understand eleven's chain in its first version. It is much better with the addition of the missing 5r8c9 !

The logic is understandable with the words. I have drawn it with the following net:

Code: Select all

(2=4)b2p137-r9c4|=                    (hp48r9c89)          |
                |=(4-3)r7c5=r7c89-3r8c9=5r8c9|-5r6c9=48r6c9|-8r1c9=|2r1c6(ALS r1c469)
                                             |-          5r1c9    =|

Note this net was for my own understanding. I'm saying it is the right way to write it down.
Eleven's chain has the format ...4r9c4=(A AND B)-(C OR D OR E)=(4D8)r689c9...
Chaining the two parentheses is valid if
((A-C) OR (B-C)) AND ((A-D) OR (B-D)) AND ((A-E) OR (B-E)) is demonstrated. The symbol "-" is used here with its sudoku meaning of a weak link (a NAND operator, more or less)
I have two concerns:

1) in this chain, the term B is a sub-chain between brackets [...]
My understanding how to use square brackets is:
- the brackets contain a chain written following Eureka format,
- the strong link on one side links an additional term to a strong node of the chain inside (in other terms, the chain inside is an almost "something")
- as a consequence, the weak link on the other side links the candidate(s) that would be eliminated by the inside chain, if assumed stand-alone; i.e. it links each end of the inside chain to a conflicting term.
As far as I read up to now, this JC's practise.

Here, none of the terms C, D, E is in a weak link with 4r7c5, head of the "bracketted" chain.

2) I find a bit confusing to find the 5 on either side of the strong link symbol = in the term (3|5|6=458)r689c9. Of course (458)r689c9 has only one possible position for the 5, i.e. 5r8c9 AND (48)r69c9. By inference, the 5 on the other side of "=" is 5r6c9... Such writing requires an effort of interpretation to th readers!

Eventually, I would propose something close to eleven's alternative:
(2=4)b2p137-r9c4=(hp48r9c89 AND (4-3)r7c5=r7c89-(3=5*)r8c9-(5=48)r6c9)-(5*|8=2)r1c469

This chain shows everything: the partial chain conflicting with 5r1c9 (tagged with "*") and the derived hidden pair (48)r69c9 conflicting with 8r1c9.
No need for brackets: the AND operator plays also the role of a separator between two parallel terms. r9c4 is in a strong link with the head of each term and the terms behind the weak link are conflicting with at least one end of the parallel terms, or with both.

Best regards, Cenoman.
PS message written before I have read JC's messages. Further analysis is needed.