The New Sudoku Players' Forum

[daj95376]

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25......14.923........75...8....91...4..5..3...18....5...49........185.23......84

 *-----------------------------------------------------------*
 | 2     5     367   | 69    8     4     | 367   679   1     |
 | 4     678   9     | 2     3     1     | 678   5     78    |
 | 1     68    368   | 69    7     5     | 24    24    389   |
 |-------------------+-------------------+-------------------|
 | 8     3     5     | 7     24    9     | 1     24    6     |
 | 679   4     267   | 1     5     26    | 278   3     789   |
 | 679   279   1     | 8     246   3     | 247   79    5     |
 |-------------------+-------------------+-------------------|
 | 5     2678  2678  | 4     9     26    | 367   1     37    |
 | 679   79    4     | 3     1     8     | 5     67    2     |
 | 3     1     26    | 5     26    7     | 9     8     4     |
 *-----------------------------------------------------------*


Setting [r6c2]=7 results in a contradiction using 7s only. Thus, [r6c2]<>7. Is there another way, using 7s only, to derive this elimination?

[ronk]

Setting [r6c2]=7 results in a contradiction using 7s only. Thus, [r6c2]<>7. Is there another way, using 7s only, to derive this elimination?

Mike Barker's solver would probably find a "frankenfish" or an "almost fish", but I explain the deduction like this ...

Code: Select all

  .  .  7  |  .  .  .  |  7  7  .
  . *7  .  |  .  .  .  | *7  . *7
  .  .  .  |  .  .  .  |  .  .  .
-----------+-----------+----------
  .  .  .  |  .  .  .  |  .  .  .
 #7  * #7  |  .  .  .  | *7  . *7
  7 -7  .  |  .  .  .  |  7  7  .
-----------+-----------+----------
  . *7 #7  |  .  .  .  | *7  . *7
  7  7  .  |  .  .  .  |  .  7  .
  .  .  .  |  .  .  .  |  .  .  .



Either r5c1=7, r5c3=7 and r7c3=7 (#) are all false OR at least one of them is true.
If all false, then a row swordfish (*) in r257 and c279 implies r6c2<>7.
If r5c1=7 or r5c3=7, then r6c2<>7.
If r7c3=7, then the strong link in box 1 causes r6c2<>7.
In all four cases, r6c2<>7.

I suspect Nishio -- which I've yet to learn -- would do the same.

[daj95376]

Thanks ronk !!! An impressive set of logic. I just knew that a fish of some kind was hiding in there. Unfortunately, I'm terrible at spotting them. It's ironic that I had most of the answer from setting [r6c2]=7. The contradiction resulted when just your (*) marked cells for <7> in the right stack were all that remained after other eliminations.

BTW: Templates is how I ran across this ornery elimination. Normally, I can explain them with Double Implication Chains; but, if the Swordfish existed during my DIC searches, then I must have missed it.

[Carcul]

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 *------------------------------------------------------*
 | 2     5     367  | 69    8     4   | 367   679   1   |
 | 4     678   9    | 2     3     1   | 678   5     78  |
 | 1     68    368  | 69    7     5   | 24    24    389 |
 |------------------+-----------------+-----------------|
 | 8     3     5    | 7     24    9   | 1     24    6   |
 | 679   4     267  | 1     5     26  | 278   3     789 |
 | 679   2679  1    | 8     246   3   | 247   79    5   |
 |------------------+-----------------+-----------------|
 | 5     2678  2678 | 4     9     267 | 367   1     37  |
 | 679   679   4    | 3     1     8   | 5     67    2   |
 | 3     1     267  | 5     26    267 | 9     8     4   |
 *------------------------------------------------------*


r5c3=7 or r9c3=7, and so r1c3<>7, which solves the puzzle.

Carcul

[ravel]

r5c3=7 or r9c3=7 ...

Do you mean this ?
r9c3<>7 => r9c6=7 => r7c6<>7 => remote pair 26 in r9c35,r57c6 => r5c3<>26 => r5c3=7.

[Carcul]

Do you mean this ?
r9c3<>7 => r9c6=7 => r7c6<>7 => remote pair 26 in r9c35,r57c6 => r5c3<>26 => r5c3=7.

No. If r5c3 and r9c3 are not "7" then {TILA(6): r5c3|r5c6|r6c5|r9c5|r9c3}.
So at least one of them must be "7".

Carcul

[ravel]

Ah, thanks, this is simpler:
If r59c3<>7:
r5c3=6 => r5c6=2 => r6c5=6 => r9c5=2 => r9c3=6

[ronk]

Nice deduction -- but ignores the thread title Elimination Using 7s Only