The New Sudoku Players' Forum

[qiuyanzhe]

I met a puzzle where I realized a new(?) technique.
Original puzzle:(Puzzle 80 of the book 数独游戏技巧 by 蓝天)

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.38..1......7..9.2.4...8..59.6....4.....1.....5....8.13..8...2.5.4..7......3..51.

As of XYZ-Wing:

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+-------------------+-------------------+-------------------+
| 2     3     8     | 59    59    1     | 46    7     46    |
| 1     6     5     | 7     3     4     | 9     8     2     |
| 7     4     9     | 26    26    8     | 1     3     5     |
+-------------------+-------------------+-------------------+
| 9     1     6     | 25    8     235   | 237   4     37    |
| 8     27    237   | 4     1     2369  | 236   5     369   |
| 4     5     23    | 269   7     2369  | 8     69    1     |
+-------------------+-------------------+-------------------+
| 3     79    1     | 8     456   56    | 467   2     4679  |
| 5     289   4     | 1     269   7     | 36    69    3689  |
| 6     278   27    | 3     49    29    | 5     1     4789  |
+-------------------+-------------------+-------------------+


posted here.

It can be easily deduced that r13c5, r46c4, r79c6 are the same pair of numbers, two of 2569.
They cannot be 26 or 59(r13c5), 29 or 56(r79c6), or 69(r4c4), so they can only be 25.

The similar shape also appear in some almost-buglite puzzles, and I have seen another example using this somewhere, but couldn't find it now.
*EDIT: Puzzle Source

[StrmCkr]

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+--------------+------------------------+----------------+
| 2  3     8   | (+9-5)   5-9   1       | 46    7   46   |
| 1  6     5   | 7        3     4       | 9     8   2    |
| 7  4     9   | (+6-2)   2-6   8       | 1     3   5    |
+--------------+------------------------+----------------+
| 9  1     6   | (+5-2)   8     (+3-25) | 27-3  4   7-3  |
| 8  27    237 | 4        1     (69-23) | 236   5   369  |
| 4  5     3-2 | (+2-69)  7     (69-23) | 8     69  1    |
+--------------+------------------------+----------------+
| 3  79    1   | 8        46-5  (+5-6)  | 467   2   4679 |
| 5  289   4   | 1        69-2  7       | 36    69  3689 |
| 6  78-2  7-2 | 3        49    (+2-9)  | 5     1   4789 |
+--------------+------------------------+----------------+


Almost Locked Set XZ-Rule: A=r36c4 {269}, B=r4567c6 {23569}, X=9, Z=2 => r4c4<>2
Almost Locked Set XZ-Rule: A=r146c4 {2569}, B=r569c6 {2369}, X=6, Z=2 => r4c6<>2
Almost Locked Set XZ-Rule: A=r34c4 {256}, B=r4569c6 {23569}, X=5, Z=6 => r6c4<>6
Almost Locked Set XZ-Rule: A=r14c4 {259}, B=r4567c6 {23569}, X=2, Z=9 => r6c4<>9

all combined then the extended eliminations also included for the after affects of these 4.

or u can look at it in chain form and realize it causes eliminations that degenerate the chain to self solving

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Grouped AIC: 2 2- r4c4 -5- r1c4 -9- r1c5 =9= r89c5 -9- r9c6 -2 => r456c6<>2

[SpAce]

It can be easily deduced that r13c5, r46c4, r79c6 are the same pair of numbers, two of 2569.

An example of braiding. (Of course it can be seen other ways too.)

They cannot be 26 or 59(r13c5), 29 or 56(r79c6), or 69(r4c4), so they can only be 25.

Very elegant! I can't think of a simpler way to express that piece of logic off-hand.