The New Sudoku Players' Forum

[bat999]

... The way I wrote it...

I think we have gone about the problem differently.

You have thought...
X can only have values 4 and 2.
Found that X=4 is the result of A=5 and X=2 is the result of B=5.
So at least one of those two squares must be 5.

I have thought...
What will happen to X if BOTH squares A and B are NOT 5?
The result is incompatible.
So at least one of those two squares must be 5.

[SteveG48]

I think that sums it up well.

[eleven]

Yes, you can write it this or that way (or as AIC - then it is out of discussion).
X also could contain more than 2 candidates. Then to "reverse" it you have to do it with a strong link from X.

Code: Select all

 *--------------------------------------------------*
 | 5    4    3    | 7    2    9    | 1    6    8    |
 | 2    1   *69   | 4    8    36   | 59   7    35   |
 | 68   89   7    | 1    5    36   |B49   2    34   |
 *----------------+----------------+----------------|
 | 7    59   29   | 259  1    4    | 3    8    6    |
 | 68   589 X469  | 59   3    7    | 2    1   A45   |
 | 1    3    24   | 25   6    8    | 45   9    7    |
 *----------------+----------------+----------------|
 | 9    7    8    | 3    4    2    | 6    5    1    |
 | 4    2    5    | 6    7    1    | 8    3    9    |
 | 3    6    1    | 8    9    5    | 7    4    2    |
 *--------------------------------------------------*

A=5->X=4
B=9=>r2c3=9->X=6
=>r3c9,r6c7<>4

I don't see a BUG+3, but the UR is obvious.

Granted that BUG+3 isn't needed on this simple puzzle, it's just what struck me first. You don't agree that we start with 3 tri-valued cells and that if neither r5c2 nor r5c3 is 9, then we're reduced to BUG+1 and r4c4 must be 5?

I agree with Danny. If you have a BUG+X, and you remove the extra candidates, a BUG should remain, i.e. each candidate 2 times in the 3 units.
This is not the case with yours, especially look at 5r4c4.

[SteveG48]

OK, I see it now. If we remove the 9's in r5c23 then we have three 5's in row 5, not two.

[DonM]

Code: Select all

*--------------------------------------------------*
 | 5    4    3    | 7    2    9    | 1    6    8    |
 | 2    1    69   | 4    8    36   | 59   7    35   |
 | 68    89  7    | 1    5    36   | 49   2    34   |
 *----------------+----------------+----------------|
 | 7   *59   29   |*259  1    4    | 3    8    6    |
 | 68  *589  469  |*59   3    7    | 2    1    45   |
 | 1    3    24   | 25   6    8    | 45   9    7    |
 *----------------+----------------+----------------|
 | 9    7    8    | 3    4    2    | 6    5    1    |
 | 4    2    5    | 6    7    1    | 8    3    9    |
 | 3    6    1    | 8    9    5    | 7    4    2    |
 *--------------------------------------------------*

Back in the day, I used to talk about 'what is' patterns (vs. chains which are 'what may be') whereby all one had to do was declare the parameters of a previously proven pattern to justify the resulting eliminations. That included some obvious examples such as X-wings, but also can include simple/basic Broken Wings/Guardians and Empty Rectangles even though the latter two would ordinarily require net-like chains to express them.

Anyway, this puzzle can be solved with 2 simple 'what-is' patterns:
The UR: (59)r45c24 is an example of the well-known 'strong-corner' or 'strong-elbow' (Hi Luke!) pattern whereby there are two digits (59) at opposite corners and, in addition, 2 sides each containing a strong link on the digit 5 which meet at one of those same corners. Thus the digit at the corner that has the strong link on the 2 adjacent sides, ie. 5, can be immediately placed at r4c2.

This results in:

Code: Select all

*--------------------------------------------------*
 | 5    4    3    | 7    2    9    | 1    6    8    |
 | 2    1   *69   | 4    8    36   | A59  7   *35   |
 | 68  *89   7    | 1    5    36   | 49   2    34   |
 *----------------+----------------+----------------|
 | 7    5    29   | 259  1    4    | 3    8    6    |
 | 68   8-9  469  | B59  3    7    | 2    1   *45   |
 | 1    3    24   | 25   6    8    | 45   9    7    |
 *----------------+----------------+----------------|
 | 9    7    8    | 3    4    2    | 6    5    1    |
 | 4    2    5    | 6    7    1    | 8    3    9    |
 | 3    6    1    | 8    9    5    | 7    4    2    |
 *--------------------------------------------------*

Now, there is a W-wing (well-known to those here) that eliminates (9)r5c2 and places (8)r5c2: The (59) pairs at r2c7 and r5c4 'see' each other via the conjugate link at (5)r25c9. Also, the same (59) pairs both 'see' (9)r5c2 either directly or thru the conjugate link at (9)r2c3 and (9)r3c2 so (9)r5c2 can be eliminated.

stte.

The above explanation may make the solution look more complicated than it is, but really with a puzzle with so many givens like this, it just takes a quick look to see that the strong-corner UR places r4c2=5 and the resulting W-wing places r5c2=8.

[eleven]

If you want, the UR does it alone.
With the strong link for 9 in c4 also the 9 in r5c2 is eliminated.

[DonM]

If you want, the UR does it alone.
With the strong link for 9 in c4 also the 9 in r5c2 is eliminated.

You're right. Even eye-balling the (8)r5c2=(2)r4c4 would have made that pretty clear- which I didn't do.
Thanks.