The New Sudoku Players' Forum

[daj95376]

Code: Select all

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


Play this puzzle online at the Daily Sudoku site

[JC Van Hay]

Code: Select all

+-----------------+--------------------+---------------+
| 3  6      2     | 4     19    5      | 8  7     19   |
| 7  19     19    | 23    6     8      | 5  23    4    |
| 8  5      4     | 79    1379  129    | 6  23    19   |
+-----------------+--------------------+---------------+
| 1  39(2)  39(6) | 8     359   9-2(6) | 4  5(6)  7    |
| 5  9(2)   8     | 79    479   2469   | 3  1     (26) |
| 4  7      (36)  | (23)  135   126    | 9  8     256  |
+-----------------+--------------------+---------------+
| 6  4      7     | 5     2     3      | 1  9     8    |
| 9  13     135   | 6     8     7      | 2  4     35   |
| 2  8      35    | 1     49    49     | 7  56    356  |
+-----------------+--------------------+---------------+

Chain[5] : (2=3)r6c4-(3=6)r6c3-6r4c3=[6r4c6=]=6r4c8-(6=2)r5c9-2r5c2=2r4c2 :=> 2r6c4=6r4c6=2r4c2 :=> -2r4c6; ste

[Leren]

Two moves:

Code: Select all

*--------------------------------------------------------------*
| 3     6     2      | 4     19    5      | 8     7     19     |
| 7     19    19     | 23    6     8      | 5     23    4      |
| 8     5     4      | 79    1379  129    | 6     23    19     |
|--------------------+--------------------+--------------------|
| 1     239   369    | 8     35-9  269    | 4     56    7      |
| 5     29    8      |a79   b479   2469   | 3     1     26     |
| 4     7     36     | 23    135   126    | 9     8     256    |
|--------------------+--------------------+--------------------|
| 6     4     7      | 5     2     3      | 1     9     8      |
| 9     13    135    | 6     8     7      | 2     4     35     |
| 2     8     35     | 1    c49    49     | 7     56    356    |
*--------------------------------------------------------------*

xyz-wing (47-9) Cells abc => -9 r4c5;

Code: Select all

*--------------------------------------------------------------*
| 3     6     2      | 4     19    5      | 8     7     19     |
| 7     19    19     | 23    6     8      | 5     23    4      |
| 8     5     4      | 79    1379  129    | 6     23    19     |
|--------------------+--------------------+--------------------|
| 1     239   369    | 8    a35    269    | 4    b56    7      |
| 5     29    8      | 79    479   469-2  | 3     1    b26     |
| 4     7     36     |a23    135   126    | 9     8     56-2   |
|--------------------+--------------------+--------------------|
| 6     4     7      | 5     2     3      | 1     9     8      |
| 9     13    135    | 6     8     7      | 2     4     35     |
| 2     8     35     | 1     49    49     | 7     56    356    |
*--------------------------------------------------------------*

als-xz-rule: (2=5) r6c4, r4c5 - (5=2) r4c8, r5c9 => -2 r5c6, r6c9; stte

Leren

[daj95376]

Chain[5] : (2=3)r6c4-(3=6)r6c3-6r4c3=[6r4c6=]=6r4c8-(6=2)r5c9-2r5c2=2r4c2 :=> 2r6c4=6r4c6=2r4c2 :=> -2r4c6; ste

JC,

Did you ever consider treating it as a discontinuous loop where the initial assumption is remembered and used to create a quasi-SI on <6> in [r4]?

Code: Select all

(2)r4c6 - (2=3)r6c4 - (3=6)r6c3 - r4c3 *= r4c8 - (6=2)r5c9 - r5c2 = r4c2 - (2)r4c6


This might be an easy way to deal with a Kraken Line with three candidates ... sometimes.

[Marty R.]

Code: Select all

+-----------+--------------+----------+
| 3 6   2   | 4  19   5    | 8 7  19  |
| 7 19  19  | 23 6    8    | 5 23 4   |
| 8 5   4   | 79 1379 129  | 6 23 19  |
+-----------+--------------+----------+
| 1 239 369 | 8  359  269  | 4 56 7   |
| 5 29  8   | 79 479  2469 | 3 1  26  |
| 4 7   36  | 23 135  126  | 9 8  256 |
+-----------+--------------+----------+
| 6 4   7   | 5  2    3    | 1 9  8   |
| 9 13  135 | 6  8    7    | 2 4  35  |
| 2 8   35  | 1  49   49   | 7 56 356 |
+-----------+--------------+----------+


Play this puzzle online at the Daily Sudoku site

Also two moves.

UR 35 r89c39 (1r8c3=6r9c9)-(6=2)r5c9-(2=9)r5c2-(9=1)r2c2=>r2c3,r8c2<>1
XY-Wing (39-2), pivot r4c5, with transport from r6c4 to r5c9=>r5c2<>2

[Leren]

daj 95376 wrote: Did you ever consider treating it as a discontinuous loop where the initial assumption is remembered and used to create a quasi-SI on <6> in [r4]?

I think I can combine the 2 moves in my previous post into 1 move.

(2=3) r6c4 [-3 r46c5 = (159) r146c5 - 19 r359c5 = 4 r9c5 - (49=7) r5c5 - (7=9) r5c4 - r4c5] -(39=5) r4c5 - (5=2) r4c8, r5c9 => -2 r5c6, r6c9; stte

The initial assumption - 2 r6c4 => -9 r4c5 (essentially the xyz-wing move in the square brackets) and I've "remembered" this result to carry out the als=xz-rule move under the same initial assumption.

The only difference from the 2 move approach is that -9 r4c5 is a consequence of the initial assumption, so doesn't count as a direct elimination.

That's 1 move isn't it ?

Leren

[daj95376]

Leren, you can condense the []'d term.

(2=3) r6c4 [-3 r46c5 = (3-7) r3c5 = 7 r5c5 - (7=9) r5c4] - (*39=5) r4c5 - (5=6) r4c8 - (6=2) r5c9 => -2 r5c6, r6c9; stte

[daj95376]

Also two moves.

UR 35 r89c39 (1r8c3=6r9c9)-(6=2)r5c9-(2=9)r5c2-(9=1)r2c2=>r2c3,r8c2<>1
XY-Wing (39-2), pivot r4c5, with transport from r6c4 to r5c9=>r5c2<>2

Marty, those are the same two moves that I selected from my solver's results, but w/o the transport. Congratulations !!!

[Marty R.]

Also two moves.

UR 35 r89c39 (1r8c3=6r9c9)-(6=2)r5c9-(2=9)r5c2-(9=1)r2c2=>r2c3,r8c2<>1
XY-Wing (39-2), pivot r4c5, with transport from r6c4 to r5c9=>r5c2<>2

Marty, those are the same two moves that I selected from my solver's results, but w/o the transport. Congratulations !!!

It's so easy to see things after they've been pointed out. I now can easily see that the transport wasn't necessary.

[Leren]

Leren, you can condense the []'d term.

(2=3) r6c4 [-3 r46c5 = (3-7) r3c5 = 7 r5c5 - (7=9) r5c4] - (*39=5) r4c5 - (5=6) r4c8 - (6=2) r5c9 => -2 r5c6, r6c9; stte

That's nicely tidied up. I'm declaring the final scoreline to read: Leren/Danny: 1 Puzzle: 0



Leren

[SudoQ]

A network solution:

Code: Select all

|-------------|----------------|------------|
| 3  6    2   | 4   19    5    | 8  7   19  |
| 7  19   19  | 23  6     8    | 5  23  4   |
| 8  5    4   | 79  1379  129  | 6  23  19  |
|-------------|----------------|------------|
| 1  239  369 | 8   359   69-2 | 4  56  7   |
| 5  29   8   | 79  479  249(6)| 3  1  (26) |
| 4  7    36  |3(2) 3(15)(16/2)| 9  8  6(25)|
|-------------|----------------|------------|
| 6  4    7   | 5   2     3    | 1  9   8   |
| 9  13   135 | 6   8     7    | 2  4   35  |
| 2  8    35  | 1   49    49   | 7  56  356 |
|-------------|----------------|------------|

r6c9=2  -> r5c9=6 -> r5c6<>6 ->
        -> r6c5=5 -> r6c6<>6 -> r4c6=6
r6c46=2 ->                             => r4c6<>2

/SudoQ

[pjb]

A net approach, not sure of notation: the r6c9 uses the previous r5c9 = 6, and the final r4c5 uses the previous r4c8 = 5:

(9=2) r5c2 - (2=6) r5c9 - (6=5) r4c8 - (56=2) r6c9 - (2=3) r6c4 - (35=9) r4c5 => r4c23, r5c456 <> 9; stte

Phil

[daj95376]

A net approach, not sure of notation: the r6c9 uses the previous r5c9 = 6, and the final r4c5 uses the previous r4c8 = 5:

(9=2) r5c2 - (2=6) r5c9 - (6=5) r4c8 - (56=2) r6c9 - (2=3) r6c4 - (35=9) r4c5 => r4c23, r5c456 <> 9; stte

FWIW: Using multiple lines is still my preference for networks because it identifies any splits precisely. However, my quasi-discontinuous loop example (above) does seem acceptable as a special case for using a single line.

Code: Select all

(9=2)r5c2 - (2=6)r5c9 - (6=5)r4c8 \
              =2 r6c9 - (2=3)r6c4  - (35=9)r4c5 => r4c23,r5c456<>9; stte


=== === === ===

If we're going to continue using a single lines for networks, then we might want to start a thread on standardizing the notation.

Using Phil's network, here's a suggestion for notating a split/degeneration in a box:

(9=2)r5c2 - 2r5c9 = [6r5c9,5*r4c8,2r6c9] - (2=3)r6c4 - (*53=9)r4c5 => r4c23,r5c456<>9; stte

Notice that I've also dropped the parens where a single digit is used. This seems to be the "trend" now, anyway.

[JC Van Hay]

Chain[5] : (2=3)r6c4-(3=6)r6c3-6r4c3=[6r4c6=]=6r4c8-(6=2)r5c9-2r5c2=2r4c2 :=> 2r6c4=6r4c6=2r4c2 :=> -2r4c6; ste

JC,

Did you ever consider treating it as a discontinuous loop where the initial assumption is remembered and used to create a quasi-SI on <6> in [r4]?

Code: Select all

(2)r4c6 - (2=3)r6c4 - (3=6)r6c3 - r4c3 *= r4c8 - (6=2)r5c9 - r5c2 = r4c2 - (2)r4c6


This might be an easy way to deal with a Kraken Line with three candidates ... sometimes.

Yes I did but as a Triangular Matrix :

Code: Select all

2r4c6
2r6c4=3r6c4
      3r6c3=6r6c3
6r4c6=======6r4c3=6r4c8
                  6r5c9=2r5c9
2r4c2===================2r5c2

However, on a single line, I would have written

Code: Select all

(2)r4c6 - (2=3)r6c4 - (3=6)r6c3 - r4c3 *= r4c8 - (6=2)r5c9 - r45c2 = .

where the initial assumption is also remembered in the last SIS, thus emptying 2C2 (-> .).
In "Eurekanizing" the Triangular Matrix, I dropped the first row and reversed the order in the SIS of the last row, getting what has been called a Transfer Matrix by some players :

Code: Select all

2r6c4=3r6c4
      3r6c3=6r6c3
6r4c6=======6r4c3=6r4c8
                  6r5c9=2r5c9
                        2r5c2=2r4c2

:=> Chain[5]

Further examples.
1. XYZ Wing

Code: Select all

+------------+----------+
| xyz  .z  . | yz  .  . |
| .    xz  . | .   .  . |
| .    .   . | .   .  . |
+------------+----------+
zr1c2
zr2c2=xr2c2
zr1c1=xr1c1=yr1c1
zr1c4=======yr1c4

(z=x)r2c2-(x=[z=]y)r1c1-(y=z)r1c4 -> zr2c2=zr1c14 :=> -zr1c2
or

Code: Select all

zr1c2
zr2c2=xr2c2
zr1c4=======yr1c4
zr1c1=xr1c1=yr1c1

[(z=x)r2c2 and (z=y)r1c4]-(xy=z)r1c1 -> zr2c2=zr1c14 :=> -zr1c2 (Kraken Cell xyzr1c1 in "reverse" order)

Of course, if the two last rows in the TM are grouped than one gets the usual presentation :
(z=x)r2c2-(x=yz)r1c14 -> zr2c2=zr1c14 :=> -zr1c2
2. SudoQ's network solution.

Code: Select all

r6c9=2  -> r5c9=6 -> r5c6<>6 ->
        -> r6c5=5 -> r6c6<>6 -> r4c6=6
r6c46=2 ->                             => r4c6<>2
+-------------+---------------------+--------------+
| 3  6    2   | 4     19     5      | 8  7   19    |
| 7  19   19  | 23    6      8      | 5  23  4     |
| 8  5    4   | 79    1379   129    | 6  23  19    |
+-------------+---------------------+--------------+
| 1  239  369 | 8     359    9-2(6) | 4  56  7     |
| 5  29   8   | 79    479    249(6) | 3  1   (26)  |
| 4  7    36  | 3(2)  3(15)  (126)  | 9  8   6(25) |
+-------------+---------------------+--------------+
| 6  4    7   | 5     2      3      | 1  9   8     |
| 9  13   135 | 6     8      7      | 2  4   35    |
| 2  8    35  | 1     49     49     | 7  56  356   |
+-------------+---------------------+--------------+
2r4c6
2r6c46=2r6c9
       2r5c9=6r5c9
       5r6c9=======5r6c5
                   1r6c5=1r6c6
6r4c6========6r5c6=======6r6c6


2r6c46=2r6c9-[2r5c9=6r5c9-6r5c6 and 5r6c9=(5-1)r6c5=1r6c6-6r6c6]=6r4c6 -> 2r6c46=6r4c6 :=> -2r4c6
3. pjb'net approach

Code: Select all

+---------------+--------------------+----------------+
| 3  6     2    | 4     19     5     | 8  7     19    |
| 7  19    19   | 23    6      8     | 5  23    4     |
| 8  5     4    | 79    1379   129   | 6  23    19    |
+---------------+--------------------+----------------+
| 1  23-9  36-9 | 8     (359)  269   | 4  (56)  7     |
| 5  (29)  8    | 7-9   47-9   246-9 | 3  1     (26)  |
| 4  7     36   | (23)  135    126   | 9  8     (256) |
+---------------+--------------------+----------------+
| 6  4     7    | 5     2      3     | 1  9     8     |
| 9  13    135  | 6     8      7     | 2  4     35    |
| 2  8     35   | 1     49     49    | 7  56    356   |
+---------------+--------------------+----------------+
9r4c23.r5c456
9r5c2=2r5c2
      2r5c9=6r5c9
            6r4c8=5r4c8
            6r6c9=5r6c9=2r6c9
                        2r3c4=3r6c4
9r4c5=============5r4c5=======3r4c5

(9=2)r5c2-(2=6)r5c9-[(6=5#)r4c8-5r4c5 and (65#=2)r6c9-(2=3)r6c4-3r4c5]=9r4c5 -> 9r5c2=9r4c5 :=> -9r4c23.r5c456

Using multiple lines is still my preference for networks because it identifies any splits precisely.

I agree and this is more generally achieved by Forbidding Matrices of which the Triangular Matrix is only a particular case.