The New Sudoku Players' Forum

[udosuk]

From this thread, RW has posted a challenging 8-clue odd-even puzzle, based on a grid discovered by JPF. The odd part is releatively easy to solve, but the even part is very tricky. Hope somebody can find an elegant way to solve from the following position:

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 7     2468  3     | 268   1     5     | 2468  2468  9     
 1     2468  5     | 9     248   2468  | 2468  3     7     
 248   9     468   | 7     3     2468  | 1     5     2468 
-------------------+-------------------+-------------------
 248   1     468   | 268   7     3     | 2468  9     5     
 5     2468  7     | 268   248   9     | 3     2468  1     
 3     2468  9     | 1     5     2468  | 2468  7     2468 
-------------------+-------------------+-------------------
 9     3     2     | 4     68    7     | 5     1     68   
 48    7     48    | 5     26    1     | 9     26    3     
 6     5     1     | 3     9     28    | 7     248   248   

Thanks!

PS: The original puzzle is like this:

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 *-----------*
 |,.,|.,5|..,|
 |,.,|,..|.,,|
 |.,.|,,.|1,.|
 |---+---+---|
 |.,.|.,,|.,,|
 |,.,|..,|,.,|
 |,.,|,,.|.,.|
 |---+---+---|
 |,,2|4.7|,,.|
 |.,.|,.,|,.3|
 |65,|,,.|,..|
 *-----------*
, = odd
. = even

Which is equivalent to this starting position:

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 1379   2468   1379   | 268    1379   5      | 2468   2468   79     
 13579  2468   13579  | 1379   2468   2468   | 2468   3579   579   
 248    379    468    | 379    379    2468   | 1      3579   2468   
----------------------+----------------------+----------------------
 248    1379   468    | 268    13579  139    | 2468   13579  1579   
 13579  2468   13579  | 268    2468   139    | 3579   2468   1579   
 13579  2468   13579  | 13579  13579  2468   | 2468   13579  2468   
----------------------+----------------------+----------------------
 139    139    2      | 4      68     7      | 59     159    68     
 48     179    48     | 159    268    19     | 579    2468   3     
 6      5      1379   | 139    139    28     | 79     248    248   

[daj95376]

This isn't elegant, and it doesn't solve the puzzle, but I think it qualifies as interesting. A kind of Broken Wing approach where guardian cells exist.

If all guardian (#/@) cells are false, then X-Wing r34\c13 leads to [r8c13]<>4, which is invalid. So, let's look at the effect of the guardian cells.

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(4) [r3c6]-[r6c6]=[r5c5]-[r5c8]   (exo)
(4) [r3c9]-[r9c9]=[r9c8]-[r5c8]   (exo)
(4) [r4c7]-              [r5c8]   (endo)
*-----------------------------------*
|  .  4  .  |  .  .  .  |  4  4  .  |
|  .  4  .  |  .  4  4  |  4  .  .  |
| *4  . *4  |  .  . @4  |  .  . @4  |
|-----------+-----------+-----------|
| *4  . *4  |  .  .  .  | #4  .  .  |
|  .  4  .  |  .  4  .  |  . -4  .  |
|  .  4  .  |  .  .  4  |  4  .  4  |
|-----------+-----------+-----------|
|  .  .  .  |  4  .  .  |  .  .  .  |
| ~4  . ~4  |  .  .  .  |  .  .  .  |
|  .  .  .  |  .  .  .  |  .  4  4  |
*-----------------------------------*


[udosuk]

daj95376, in this post RW has listed out a few template eliminations plus a short loop to solve it. To summarise:

r6c6<>2 (template on 2s)
r5c8<>4 (template on 4s)
r6c6<>6 (template on 6s)
r2c5<>8 (otherwise r7c5=6, r8c5=2, r9c6=8, r6c6=4 => no candidate for r5c5)
r5c5<>8 (template on 8s)

I'm sure you'll enjoy working out those 4 template eliminations.

[daj95376]

Yes, I saw RW's post. He does great work. However, Templates alone will crack the puzzle. As you probably already know, it's sufficient to get [r1c4]=2 and SSTS will complete the puzzle.

While looking for a solution, I ran across the anomaly in stack 1 for <4> and discovered that it lead easily to [r5c8]<>4. The same anomaly exists for <8> in stack 1, but the exo-guardian cells lead to complicated elimination patterns for [r1c4]<>8.

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r6c6    <> 2     Templates (A: 1)

r5c8    <> 4     Templates (A: 1)

r1c4    <> 6     Templates (A: 1)
r6c6    <> 6     Templates (A: 1)

r1c4    <> 8     Templates (A: 1) -or-
r5c5    <> 8     Templates (A: 1)
r6c6    <> 8     Templates (A: 1)


[ronk]

Hope somebody can find an elegant way to solve from the following position:

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 7     2468  3     | 268   1     5     | 2468  2468  9     
 1     2468  5     | 9     248   2468  | 2468  3     7     
 248   9     468   | 7     3     2468  | 1     5     2468 
-------------------+-------------------+-------------------
 248   1     468   | 268   7     3     | 2468  9     5     
 5     2468  7     | 268   248   9     | 3     2468  1     
 3     2468  9     | 1     5     2468  | 2468  7     2468 
-------------------+-------------------+-------------------
 9     3     2     | 4     68    7     | 5     1     68   
 48    7     48    | 5     26    1     | 9     26    3     
 6     5     1     | 3     9     28    | 7     248   248   

A loop with an odd number of conjugate links is illegal. Therefore, at least one "guardian" that can prevent such a loop must be true. Illustrating the illegal turbot fish (*) and its guardians (# and @) for digit 2 ...

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 . @2  . |  2  .  . | @2  2  .
 . *2  . |  . @2 @2 | *2  .  .
 2  .  . |  .  .  2 |  .  .  2
---------+----------+----------
 2  .  . |  2  .  . | *2  .  .
 . *2  . | #2 #2  . |  . *2  .
 . #2  . |  .  . -2 | #2  . #2
---------+----------+----------
 .  .  . |  .  .  . |  .  .  .
 .  .  . |  .  2  . |  .  2  .
 .  .  . |  .  .  2 |  .  2  2

r6c6 directly sees all the # guardians ... and indirectly sees all the @ guardians via the grouped strong link in c4. Therefore, r6c6<>2.

Except for an absence of a digit 6 guardian in r2c5, the identical illegal turbot fish and guardians exist for digit 6 and digit 8. Therefore, r6c6<>6 and r6c6<>8 ... leaving r6c6=4.

The puzzle is then solved with simple coloring (turbot fish), locked candidates, and singles.

[udosuk]

Thanks guys... Impressive fishing work!

[daj95376]

One more try. This time with chains.

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(6) [r3c69]=[r3c3]-[r4c3]=[r4c47] =>
    one of four cells in [r3c69],[r4c47] must be <6>

(6) [r3c6]-[r1c4]
(6) [r4c4]-[r1c4]

(6) [r3c9] =>  [r4c3 ]* => [r6c7] => [r2c6] => ![r1c4]
(6) [r4c7] => ![r3c69]*           => [r1c8] => ![r1c4]

(*) means 'from first inference chain for <6>'


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(8) [r3c69]=[r3c13]-[r12c2]=[r56c2]-[r4c13]=[r4c47] =>
    one of four cells in [r3c69],[r4c47] must be <8>

(8) [r3c6]-[r1c4]
(8) [r4c4]-[r1c4]

(8) [r3c9] =>  [r4c13]*,[r9c8] => [r6c7] => [r2c6] => ![r1c4]
(8) [r4c7] => ![r3c69]*                  => [r1c8] => ![r1c4]

(*) means 'from first inference chain for <8>'


Thus, [r1c4]=2 ... and SSTS completes with one Naked Pair, one Colors, and Singles.