The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |..4|9..|...|
 |76.|.2.|5.9|
 |...|...|8..|
 |---+---+---|
 |..6|24.|9..|
 |..1|...|..2|
 |.95|.6.|.4.|
 |---+---+---|
 |...|..4|72.|
 |.8.|...|4..|
 |...|..3|.68|
 *-----------*


Play/Print this puzzle online

[pjb]

Code: Select all

 135     135     4      | 9      8      57     | 2      17     6     
 7       6       8      | 4      2      1      | 5      3      9     
 159     125     29     |*357   *357    6      | 8      17     4     
------------------------+----------------------+---------------------
 38      37      6      | 2      4      57     | 9      58     1     
 48      47      1      |*357   *357    9      | 6      58     2     
 2       9       5      | 1      6      8      | 3      4      7     
------------------------+----------------------+---------------------
 6       15      39     | 8      19     4      | 7      2      35     
 15      8       37     | 6      17     2      | 4      9      35     
 49      24      279    |*57    *9-57   3      | 1      6      8     


BUG-lite (type 1) at r359c45 => -57 r9c5; stte

Phil

[SteveG48]

Yarg. But here it is.

Code: Select all

 *--------------------------------------------------*
 | 135 i135  4    | 9    8    7-5  | 2    17   6    |
 | 7    6    8    | 4    2    1    | 5    3    9    |
 | 159  125  29   | 357  357  6    | 8    17   4    |
 *----------------+----------------+----------------|
 | 38  b37   6    | 2    4   a57   | 9    58   1    |
 | 48  c47   1    | 357  357  9    | 6    58   2    |
 | 2    9    5    | 1    6    8    | 3    4    7    |
 *----------------+----------------+----------------|
 | 6   h15   39   | 8    19   4    | 7    2    35   |
 |g15   8    37   | 6   f17   2    | 4    9    35   |
 |e49  d24   279  |f57  f579  3    | 1    6    8    |
 *--------------------------------------------------*

(5=7)r4c6 - r4c2 = (7-4)r5c2 = r9c2 - (4=9)r9c1 - (9=157)b8p578 - 1r8c1 = r7c2 - (1|3=5)r1c2 => -5 r1c6 ; stte
                 = 3r4c2 --------------------------------------------------------/



[SteveG48]

Code: Select all

 135     135     4      | 9      8      57     | 2      17     6     
 7       6       8      | 4      2      1      | 5      3      9     
 159     125     29     |*357   *357    6      | 8      17     4     
------------------------+----------------------+---------------------
 38      37      6      | 2      4      57     | 9      58     1     
 48      47      1      |*357   *357    9      | 6      58     2     
 2       9       5      | 1      6      8      | 3      4      7     
------------------------+----------------------+---------------------
 6       15      39     | 8      19     4      | 7      2      35     
 15      8       37     | 6      17     2      | 4      9      35     
 49      24      279    |*57    *9-57   3      | 1      6      8     


BUG-lite (type 1) at r359c45 => -57 r9c5; stte

Phil

Phil, I don't see your BUG-lite. It seems to me that one of r3c45 and one of r5c45 has to be a 3, so there's no possibility of a 5/7 UR. But if r9c5 is not a 9, then one of r9c45 has to be a 5 and one of r9c45 has to be a 7, eliminating 3/5 and 3/7 URs as well. So it's OK for r9c5 to be a 5 or 7. Am I missing something?

[omerosler]

I used the same method, the explanation is because for an assignment of R14C6 there are two solutions that can be interleaved:

Code: Select all

.....Y...
.........
...X3...
.....X...
...3Y...
.........
.........
...YX....

Code: Select all

.....Y...
.........
...3X...
.....X...
...Y3...
.........
.........
...XY....


EDIT:
And one assignment must bes true => two solutions

[pjb]

Code: Select all

35    53
73    37
57    75


Steve, these two arrangements are possible. There are thus 2 solutions if r9c5 <> 9.
Phil

[Leren]

Code: Select all

35   53  | 73   37
73   37  | 35   53
57   75  | 57   75

It seems to me that four arrangements are possible. For the first two arrangements r1c6 = 7 and r4c6 = 5. For the second two arrangements r1c6 = 5 and r4c6 = 7.

Arrangements 1 and 2 are thus impermeable within the rows, columns and boxes in which they are situated, as are Arrangements 3 and 4.

This means that the solution status of Arrangements 1 and 2 are the same : they are both either part of a valid solution or part of a non-solution. Similarly for Arrangements 3 and 4.

Thus, if r9c5 <> 9 the number of valid solutions of the puzzle could be 0, 2 or 4. If r9c5 = 9 the number of valid solutions of the puzzle could be 1, 3 or 5.

Since we were not warned that the puzzle may have multiple solutions then we are entitled to assume that the number of solutions is 1. The only way this can happen is if r9c5 = 9 and none of the four arrangements is part of a valid solution.

Leren

[JC Van Hay]

Code: Select all

+-------------------+----------------+-------------+
| 135  (135)   4    | 9    8    (57) | 2  17  6    |
| 7    6       8    | 4    2    1    | 5  3   9    |
| 159  (125)   (29) | 357  357  6    | 8  17  4    |
+-------------------+----------------+-------------+
| 38   (37)    6    | 2    4    5-7  | 9  58  1    |
| 48   47      1    | 357  357  9    | 6  58  2    |
| 2    9       5    | 1    6    8    | 3  4   7    |
+-------------------+----------------+-------------+
| 6    (15)    (39) | 8    19   4    | 7  2   (35) |
| 15   8       37   | 6    17   2    | 4  9   35   |
| 49   24      279  | 57   579  3    | 1  6   8    |
+-------------------+----------------+-------------+


[(7=3)r4c2-3r1c2==2r3c2-2r3c2==3r7c3-3r7c9==1r7c2-1r1c2=*XYWing(357)r1c16.r4c2]-(7=5)r4c6; stte

[Sudtyro2]

Code: Select all

 abc .   .   | abc  .  .  |  ab9 .   .
 abc .   .   | abc  .  .  |  ab  .   .
 .   .   .   | .    .  .  |  .   .   .

Without the 9, wouldn't this be a basic MUG pattern (including morphs)?
Hence, the 9 must be true.

SteveC

[Leren]

Code: Select all

 abc .   .   | abc  .  .  |  ab9 .   .
 abc .   .   | abc  .  .  |  ab  .   .
 .   .   .   | .    .  .  |  .   .   .

Without the 9, wouldn't this be a basic MUG pattern (including morphs)?
Hence, the 9 must be true.

SteveC

At the risk of being labelled a pain in the neck I'd say 1. Yes this would be a MUG pattern and 2. The 9 must be True IF THE PUZZLE HAS EXACTLY ONE SOLUTION !

My understanding is that MUGs are an extension of the BUG principle, for which I have found an old proof here.

The proof states at the end (my underlining) :

In summary, we have shown that a BUG either has no solution, or has at least two solutions if it has any solution. We can conclude that no BUG has a unique solution.

If the MUG had more than one solution, would the 9 have to be True ? Not necessarily as far as I can see.

The reason that we can say that the 9 must be True in this puzzle is that a convention is being followed : if a puzzle maker doesn't say anything about the uniqueness of the puzzle, we are entitled to assume that the puzzle has a unique solution, so the MUG state must have no solutions for the puzzle in question.

Leren

[SteveG48]

Code: Select all

35    53
73    37
57    75


Steve, these two arrangements are possible. There are thus 2 solutions if r9c5 <> 9.
Phil

Thanks, Phil. Senior moment. I see it now.

[bat999]

Code: Select all

.--------------------.-----------------.------------.
|  15-3  d135b   4   | 9     8     57B | 2  17   6  |
|  7      6      8   | 4     2     1   | 5  3    9  |
| a159   d125a   29  | 357A  357A  6   | 8  17   4  |
:--------------------+-----------------+------------:
|  38D    7-3    6   | 2     4     57C | 9  58D  1  |
|  48     47     1   | 357   357   9   | 6  58   2  |
|  2      9      5   | 1     6     8   | 3  4    7  |
:--------------------+-----------------+------------:
|  6     d15b   c39  | 8     19    4   | 7  2    35 |
| b15     8     b37  | 6    b17    2   | 4  9    35 |
| c49    c24    279  | 57    579   3   | 1  6    8  |
'-------------------'-----------------'------------'

(5)r3c1 - (5=3)r8c135 - (3=2)r7c3,r9c12 - (2=3)r137c2 - (3)r1c1,r4c2
(5)r3c2 - (5=3)r17c2 - (3)r1c1,r4c2
(5)r3c45 - r1c6 = r4c6 - (5=3)r4c18 - (3)r1c1,r4c2
=> -3 r1c1,r4c2; stte

[Sudtyro2]

Code: Select all

 abc .   .   | abc  .  .  |  ab9 .   .
 abc .   .   | abc  .  .  |  ab  .   .
 .   .   .   | .    .  .  |  .   .   .

Without the 9, wouldn't this be a basic MUG pattern (including morphs)?
Hence, the 9 must be true.
SteveC

At the risk of being labelled a pain in the neck ...
The reason that we can say that the 9 must be True in this puzzle is that a convention is being followed : if a puzzle maker doesn't say anything about the uniqueness of the puzzle, we are entitled to assume that the puzzle has a unique solution, so the MUG state must have no solutions for the puzzle in question.
Leren

No danger of being a PITN...please continue to keep us honest!
FWIW, to those interested, here's a link to a nice MUG discussion:
http://forum.enjoysudoku.com/forming-mugs-from-bug-lite-composites-t3210.html

SteveC

[Leren]

Code: Select all

35    53
73    37
57    75


Steve, these two arrangements are possible. There are thus 2 solutions if r9c5 <> 9.
Phil

Thanks, Phil. Senior moment. I see it now.

I can't resist having one more PNIN moment. Hi Steve, as you can apparently see that there are 2 solutions if r9c5 <> 9, perhaps you can show them to us.

I gave it a bit of a go myself. I manually removed 9 from r9c5 and forced the puzzle into the first of the UR cell arrangements.

This was my solver's "solution".

Code: Select all

*--------------------------*
| 1 3 4  | 9 8 7  | 2 0 6  |
| 7 6 8  | 4 2 1  | 5 3 9  |
| 9 0 2  | 3 5 6  | 8 7 4  |
|--------+--------+--------|
| 3 7 6  | 2 4 5  | 9 8 1  |
| 8 4 1  | 7 3 9  | 6 5 2  |
| 2 9 5  | 1 6 8  | 3 4 7  |
|--------+--------+--------|
| 6 1 3  | 8 9 4  | 7 2 5  |
| 5 8 7  | 6 1 2  | 4 9 3  |
| 4 2 9  | 5 7 3  | 1 6 8  |
*--------------------------*

You'll note that r1c8 and r3c2 have 0 in them - that means there are no candidates left to put into these cells. Maybe you'll have better luck.

Leren

[SteveG48]

Code: Select all

35    53
73    37
57    75


Steve, these two arrangements are possible. There are thus 2 solutions if r9c5 <> 9.
Phil

Thanks, Phil. Senior moment. I see it now.

I can't resist having one more PNIN moment. Hi Steve, as you can apparently see that there are 2 solutions if r9c5 <> 9, perhaps you can show them to us.

I gave it a bit of a go myself. I manually removed 9 from r9c5 and forced the puzzle into the first of the UR cell arrangements.

This was my solver's "solution".

Code: Select all

*--------------------------*
| 1 3 4  | 9 8 7  | 2 0 6  |
| 7 6 8  | 4 2 1  | 5 3 9  |
| 9 0 2  | 3 5 6  | 8 7 4  |
|--------+--------+--------|
| 3 7 6  | 2 4 5  | 9 8 1  |
| 8 4 1  | 7 3 9  | 6 5 2  |
| 2 9 5  | 1 6 8  | 3 4 7  |
|--------+--------+--------|
| 6 1 3  | 8 9 4  | 7 2 5  |
| 5 8 7  | 6 1 2  | 4 9 3  |
| 4 2 9  | 5 7 3  | 1 6 8  |
*--------------------------*

You'll note that r1c8 and r3c2 have 0 in them - that means there are no candidates left to put into these cells. Maybe you'll have better luck.

Leren

Not quite. I don't see that there are 2 solutions if r9c5 <> 9. I see that if there is at least one solution with r9c5 <> 9, then there must be at least 2 solutions. A subtle distinction, but crucial. As you've shown, there are no solutions if r9c5 <> 9. This is what we expect, since we've assumed from the beginning that it's a proper puzzle with a unique solution. That's the basis of Phil's elimination, and all eliminations involving uniqueness arguments.