The New Sudoku Players' Forum

[vanhezz]

Hi, I've just created this account because I need your help in finding my next step, only one. I think I've tried everything...
I'd be grateful if you could give me a hint and explain your way of thinking.

https://files.fm/u/cmw4bnrqyu

[P.O.]

not really an easy solution but it solves the puzzle in one step

Code: Select all

5  2  3  6  9  4  8  7  1
.  .  7  2  8  5  9  6  3
9  8  6  .  .  .  2  5  4
.  5  8  .  6  7  3  4  .
6  .  .  .  4  .  5  8  7
3  7  4  .  5  .  6  1  .
8  6  .  4  1  .  7  .  5
7  3  1  5  2  6  4  9  8
.  .  5  .  .  .  1  .  6

523694871..7285963986...254.58.6734.6...4.587374.5.61.86.41.7.5731526498..5...1.6

intersection:
((((9 0) (5 2 4) (1 9)) ((9 0) (5 3 4) (2 9))))

5     2     3     6     9     4     8     7     1             
14    14    7     2     8     5     9     6     3             
9     8     6     137   37    13    2     5     4             
12    5     8     19    6     7     3     4     29             
6     19    29    13    4     123   5     8     7             
3     7     4     89    5     289   6     1     29             
8     6     29    4     1     39    7     23    5             
7     3     1     5     2     6     4     9     8             
24    49    5     3789  37    389   1     23    6             

c3n9{r5 r7} - r7c6{n9 n3} - r3c6{n3 n1} - r5c6{n13 n2} => r5c3 <> 2
ste.


[eleven]

You can solve it with unique rectangles:

Code: Select all

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

In the starred cells r39c5 are 3/7, so r39c4 can't be 3/7. One of 3 and 7 must be outside in the columns (same for rows and boxes), here r456c4, and the only possibility is r5c4=3.

Code: Select all

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

Now because of a possible 13 UR in the starred cells, r3c6 cannot be 3, because it would force r3c3 and r5c6 to 1 (no other 1 is possible in the row/column). so 3 must be in r3c5.
Note, that this is only a valid deduction, because the 3 in r5c4 was not a given.

Without the uniqueness moves you would need e.g. xy-chains to solve it.

[Maxito_Bahiense]

If you like colouring, you might want to try 3D Medusa: candidate 2 is a good seed or starting point: you colour alternate conjugate 2 candidates with each colour. In bivalue cells, you may start extending the colour pattern to, say, first candidate 9:

Code: Select all

.------------------.-------------------.---------------.
| 5    2     3     |  6      9    4    |  8   7   1    |
| 14   14    7     |  2      8    5    |  9   6   3    |
| 9    8     6     |  137    37   13   |  2   5   4    |
:------------------+-------------------+---------------:
| 12B   5     8    |  19A     6    7   |  3   4   2A9B |
| 6    19    2A9B  |  13     4    12B3 |  5   8   7    |
| 3    7     4     |  89     5    2A89 |  6   1   2B9A |
:------------------+-------------------+---------------:
| 8    6     2B9A  |  4      1    39   |  7   2A3  5   |
| 7    3       1   |  5      2    6    |  4   9    8   |
| 2A4   49B    5   |  3789!  37   389  |  1   2B3  6   |
'------------------'-------------------'---------------'

Candidate 9 on cell r9c4 sees both colours, thus can be eliminated.

In this way, you may continue extending the colour cluster, eliminating 9 r6c6. Extending colouring to candidate 8, you will find a collision in cell r9c6 (both 8 and 9 tagged with the same colour), which eliminates lots of candidates and solves to singles.

[vanhezz]

Thank you all. I used the unique rectangle technique.
But now, another, I think harder one. I don't see any unique rectangles to be used or any other techniques...

https://files.fm/u/9q9ke6cedn

Maybe you could tell me the name of a technique I should implement, without indicating numbers. I'd be grateful.

[Hajime]

But now, another, I think harder one. I don't see any unique rectangles to be used or any other techniques...

The original puzzle is:

Code: Select all

                     
  5 6 8 . . . . 9 .   
  . . . . . . 2 . .   
  . . . . 6 . 7 . .   
  . . . 8 . . 1 . .   
  . . 7 4 . . . 5 6   
  . . 3 9 . . . . .   
  . . 9 . . 1 . . .   
  4 . . . . . . 3 .   
  1 . . 6 9 . . . .   
                     
568....9.......2......6.7.....8..1....74...56..39.......9..1...4......3.1..69....


You got until:

Code: Select all

+-----------+----------------+-------------+   
| 5   6   8 | 1    2      7  | 4    9    3 |   
| 7   39  1 | 35  3458  34589| 2    6   58 |   
| 39  2   4 | 35   6    3589 | 7    1   58 |   
+-----------+----------------+-------------+   
| 26  4   5 | 8    37    36  | 1   27    9 |   
| 89  89  7 | 4    1      2  | 3    5    6 |   
| 26  1   3 | 9    57    56  | 8   247  247|   
+-----------+----------------+-------------+   
| 38  5   9 | 7   348     1  | 6   248  248|   
| 4   7   6 | 2    58    58  | 9    3    1 |   
| 1   38  2 | 6    9     348 | 5   478  478|   
+-----------+----------------+-------------+   


First some Pointing, Claiming   (8)r8,b8 => (-8)r7c5 (-8)r9c6 | (3)c4,b2 => (-3)r2c5 (-3)r2c6 (-3)r3c6 | (5)c4,b2 => (-5)r2c5 (-5)r2c6 (-5)r3c6 | (8)c8,b9 => (-8)r7c9 (-8)r9c9
But than somewhat difficult AIC Type 2 called M3-Wing [5 links] (9=3)r2c2-(3)r9c2=(3-4)r9c6=(4)r2c6 => (-9)r2c6
STTE (Singles to the end)


PS. In stead of a picture within IMG tags, could you post your subject within code-tags? So you type over the image into digits, in stead of me

[P.O.]

from your pm there is a xy-chain starting in r3c1 that solves the puzzle.

Code: Select all

5    6    8    1    2    7    4    9    3             
7    39   1    35   48   489  2    6    58           
39   2    4    35   6    89   7    1    58           
26   4    5    8    37   36   1    27   9             
89   89   7    4    1    2    3    5    6             
26   1    3    9    57   56   8    247  47           
38   5    9    7    34   1    6    48   2             
4    7    6    2    58   58   9    3    1             
1    38   2    6    9    34   5    478  47   


[eleven]

Code: Select all

 +-------+-------+--------+
 | 5 6 8 | 1 2 7 | 4 9 3  |
 | 7 . 1 | . . . | 2 6 58 |
 | . 2 4 | . 6 . | 7 1 58 |
 +-------+-------+--------+
 | . 4 5 | 8 . . | 1 . 9  |
 | . . 7 | 4 1 2 | 3 5 6  |
 | . 1 3 | 9 . . | 8 * 47 |
 +-------+-------+--------+
 | . 5 9 | 7 . 1 | 6 . 2  |
 | 4 7 6 | 2 . . | 9 3 1  |
 | 1 . 2 | 6 9 . | 5 * 47 |
 +-------+-------+--------+

There is a (hidden) unique rectangle 47 in r69c89. r69c9 are 4/7 and 7 is restricted in r9 to c89: 4r6c8->7r6c9->4r9c9->7r9c8 (impossible, if unique) => -4r6c8. (Same for 7r9c8, because 4 is restricted to r6c89).
Another way to see it: in columns 89 only 7r4c8 and 4r6c8 can avoid the 47 UR pattern. Since 4r7c8->7r9c9, one of r4c8 and r9c9 must be 7.

Then you arrive at a grid, which only has a singe cell with 3 candidates (the rest has 2 or are solved). You can solve it with BUG.

A remark: Hidden unique rectangles are weak techniques in the sense, that they seldom make real progress in solving, Maybe accidentally you found 2 nice examples.

[Leren]

Code: Select all

*-------------------------------------*
| 5  2   3   | 6    9   4   | 8 7  1  |
| 14 14  7   | 2    8   5   | 9 6  3  |
| 9  8   6   | 137  37 b13  | 2 5  4  |
|------------+--------------+---------|
| 12 5   8   | 19   6   7   | 3 4  29 |
| 6  19 a29  | 13   4  b123 | 5 8  7  |
| 3  7   4   | 89   5   289 | 6 1  29 |
|------------+--------------+---------|
| 8  6   2-9 | 4    1  b39  | 7 23 5  |
| 7  3   1   | 5    2   6   | 4 9  8  |
| 24 49  5   | 3789 37  389 | 1 23 6  |
*-------------------------------------*

ALS XZ Rule: X = 2, Z = 9: (9=2) r5c3 - (2=9) r357c6 => - 9 r7c3; stte

If vanhezz is still listening this move is described here.

Leren

[vanhezz]

I found out in a program that it must be an xy chain now, so how is it possible that one of you found a unique rectangle?

As for xy chain, I must learn the theory about it first before I go on and finish my puzzle.

Btw where can I read about the code you're using on this forum?

[Leren]

Code: Select all

I think you mean the Code button which is a the top of your reply window.

If you want to post something in easy to read format select the text click on the Code button and the selected text will appear green in a fixed with font, probably courier.

I used this on the first two sentences in this post. Leren

[eleven]

I found out in a program that it must be an xy chain now, so how is it possible that one of you found a unique rectangle?

Many solvers have not implemented hidden unique rectangles (see e.g. here).
I wonder, how you found r7c9=2. Have you used the hidden unique rectangle 24 in r67c89 (otherwise you would need a chain) ?

[eleven]

I give a clearer explanation, because this is, what smart manual solvers would do, before searching for xy-chains, when there are so many bivalue cells, and therefore so much possibilities to follow chains through 10 or 20 cells without success.

Any pattern of 2 digits only in non given cells of a rectangle in 2 boxes, 2 rows, 2 columns is impossible in a unique puzzle, because in the solution you could switch the 2 digits to get a second solution for the original puzzle.

Example:
If you have

Code: Select all

a . . | b . .
. . . | . . .
. . . | . . .
--------------
b . . | a . .

in the solution, then also

Code: Select all

b . . | a . .
. . . | . . .
. . . | . . .
--------------
a . . | b . .

would be a solution, because it would not have any impact to the rest of the solution.
So you can eliminate any digit, which forces such a pattern.

In a hidden unique rectangle a digit can be eliminated, if it would force such a pattern by bivalue or bilocal conditions (single candidate left or strong link).
In the original puzzle after the 2 pairs you have

Code: Select all

+----------------+----------------+----------------+
| 5    6    8    | 1    2    7    | 4    9    3    |
| 7    39   1    | 35   48   489  | 2    6    58   |
| 39   2    4    | 35   6    89   | 7    1    58   |
+----------------+----------------+----------------+
| 26   4    5    | 8    37   36   | 1    27   9    |
| 89   89   7    | 4    1    2    | 3    5    6    |
| 26   1    3    | 9    57   56   | 8   *247 *247  |
+----------------+----------------+----------------+
| 38   5    9    | 7    34   1    | 6   *248 *2-4  |
| 4    7    6    | 2    58   58   | 9    3    1    |
| 1    38   2    | 6    9    34   | 5    478  47   |
+----------------+----------------+----------------+

The 4r7c9 forces 2 in r6c9 and r7c7, because there is no other 2 left in the column/row (strong link). Then the 2r6c9 forces a 4 in r6c8 (no other in the row), and you have the deadly pattern => -4r7c9.

Now there is the 47 hidden UR, as described above, and you get here:

Code: Select all

+----------------+----------------+----------------+
| 5    6    8    | 1    2    7    | 4    9    3    |
| 7    39   1    | 35   48  *49+8 | 2    6    58   |
| 39   2    4    | 35   6    89   | 7    1    58   |
+----------------+----------------+----------------+
| 26   4    5    | 8    37   36   | 1    27   9    |
| 89   89   7    | 4    1    2    | 3    5    6    |
| 26   1    3    | 9    57   56   | 8    27   4    |
+----------------+----------------+----------------+
| 38   5    9    | 7    34   1    | 6    48   2    |
| 4    7    6    | 2    58   58   | 9    3    1    |
| 1    38   2    | 6    9    34   | 5    48   7    |
+----------------+----------------+----------------+

This is a BUG+1 (1 unsolved cell with 3 candidates, all others with 2).
Without the 8r2c6 you have each candidate exactly twice in the row, column and box. This is an invalid pattern in a unique puzzle (the proof is complex).
If you have not used advanced techniques before (like a kite or chain) there is an easy rule, how to check, which candidates can be eliminated. Just look, which one of the 3 appears 3 times in box/row/column (here 8r2c6). This one must be true.