The New Sudoku Players' Forum

[mmoo9154]

Here's a puzzle from the 10/27 Contra Costa Times:

Code: Select all

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


I can get it to this point:

Code: Select all

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


And here it is with pencil marks:

Code: Select all

*-----------------------------------------------*
| 58   56   7   | 68   1    3   | 9    4    2   |
| 2    3    89  | 78   79   4   | 1    6    5   |
| 1    69   4   | 2    69   5   | 7    8    3   |
|---------------+---------------+---------------|
| 3    17   6   | 5    8    9   | 2    17   4   |
| 4    8    2   | 167  67   17  | 5    3    9   |
| 57   59   19  | 4    3    2   | 8    17   6   |
|---------------+---------------+---------------|
| 9    2    3   | 17   4    8   | 6    5    17  |
| 6    17   5   | 3    2    17  | 4    9    8   |
| 78   4    18  | 9    5    6   | 3    2    17  |
*-----------------------------------------------*


My question is how do you move this forward? Any help at all will be appreciated. But, I'm a newbie, so go ahead and use your acronyms (AIM FISH etc.), but any links or "plain english" descriptions that explain the logic you use would be very deeply appreciated.

-Mark

[keith]

Mark,

This is a BUG, also known as a BUG+1.

Code: Select all

*-----------------------------------------------*
| 58   56   7   | 68   1    3   | 9    4    2   |
| 2    3    89  | 78   79   4   | 1    6    5   |
| 1    69   4   | 2    69   5   | 7    8    3   |
|---------------+---------------+---------------|
| 3    17   6   | 5    8    9   | 2    17   4   |
| 4    8    2   | 167  67   17  | 5    3    9   |
| 57   59   19  | 4    3    2   | 8    17   6   |
|---------------+---------------+---------------|
| 9    2    3   | 17   4    8   | 6    5    17  |
| 6    17   5   | 3    2    17  | 4    9    8   |
| 78   4    18  | 9    5    6   | 3    2    17  |
*-----------------------------------------------*

R5C4 must be <7>.

You can search Google for SUDOKU BUG, etc., but if R5C4 is not <7>, the puzzle does not have a unique solution.

Keith

[re'born]

Mark,

In situations such as these, I think Keith's BUG solution is the best solution. However, if you're not comfortable with using uniqueness, the next best alternative is an xy-chain.

In this puzzle, one such xy-chain is

Code: Select all

  *--------------------------------------------------*
 | 58   56   7    | 68   1    3    | 9    4    2    |
 | 2    3    89B  | 78-  79A  4    | 1    6    5    |
 | 1    69   4    | 2    69   5    | 7    8    3    |
 |----------------+----------------+----------------|
 | 3    17   6    | 5    8    9    | 2    17   4    |
 | 4    8    2    | 167  67   17   | 5    3    9    |
 | 57   59   19   | 4    3    2    | 8    17   6    |
 |----------------+----------------+----------------|
 | 9    2    3    | 17F  4    8    | 6    5    17   |
 | 6    17D  5    | 3    2    17E  | 4    9    8    |
 | 78   4    18C  | 9    5    6    | 3    2    17   |
 *--------------------------------------------------*


If you follow the chain of cells above from A to F (or F to A) you will see that either A is a 7 or F is a 7. Hence, any cell seeing both A and F is not a 7. In particular, r2c4 <> 7.

[udosuk]

As always, I look for ways without uniqueness assumption and not in the chain representation. But I'm surprised I couldn't find any simple y-wing etc or ALS-xz. Not even ALS-xy-wing. Which is hard to believe considering the number of unsolved cells left. Probably I haven't look hard enough.

The "simplest" way I can solve this is through a complex y-wing:

Code: Select all

 *--------------------------------------------------*
 | 58   56   7    | 68   1    3    | 9    4    2    |
 | 2    3   *89   |*78   79   4    | 1    6    5    |
 | 1    69   4    | 2    69   5    | 7    8    3    |
 |----------------+----------------+----------------|
 | 3   @17   6    | 5    8    9    | 2    17   4    |
 | 4    8    2    | 167  67   17   | 5    3    9    |
 | 57   59  #19   | 4    3    2    | 8    17   6    |
 |----------------+----------------+----------------|
 | 9    2    3    |#17   4    8    | 6    5    17   |
 | 6   -17   5    | 3    2   @17   | 4    9    8    |
 | 78   4    18   | 9    5    6    | 3    2    17   |
 *--------------------------------------------------*

r2c34 can't be both 8
=> r2c3=9 or r2c4=7
=> At least one of r6c3 & r7c4 must be 1
=> At least one of r4c2 & r8c6 must be 7
=> r8c2 can't be 7

In other words, we first established a "strong link" of 1 in r6c3+r7c4, then use y-wing via r4c2+r8c6 to eliminate 7 from r8c2.

But the tricky thing is, we can't even transform this move to an ALS-xy-wing (the {17} in b8 is not an ALS).

Added later:

Found it. A very simple ALS-xy-wing :

Code: Select all

 *--------------------------------------------------*
 | 58   56   7    | 68   1    3    | 9    4    2    |
 | 2    3   -89   |@78   79   4    | 1    6    5    |
 | 1    69   4    | 2    69   5    | 7    8    3    |
 |----------------+----------------+----------------|
 | 3    17   6    | 5    8    9    | 2    17   4    |
 | 4    8    2    | 167  67   17   | 5    3    9    |
 | 57   59   19   | 4    3    2    | 8    17   6    |
 |----------------+----------------+----------------|
 | 9    2    3    |@17   4    8    | 6    5    17   |
 | 6   #17   5    | 3    2   *17   | 4    9    8    |
 | 78   4   #18   | 9    5    6    | 3    2    17   |
 *--------------------------------------------------*

ALS A: r27c4={178}
ALS B: r8c2+r9c3={178}
ALS C: r8c6={17}
restricted common between A & C: 1
restricted common between B & C: 7
common between A & B: 8

Therefore r2c3 can't be 8.

Logic:

r8c6 must be 1 or 7.
If r8c6=1 => r7c4=7 => r2c4=8.
If r8c6=7 => r8c2=1 => r9c3=8.
Therefore at least one of r2c4 & r9c3 must be 8.
Hence r2c3, seeing both these 2 cells, can't be 8.

Note these cells are almost the same as re'born's xy-chain, but I like this representation better.

[re'born]

Note these cells are almost the same as re'born's xy-chain, but I like this representation better.

Perhaps I'm overlooking something, but can't you always write an xy6-chain deduction as an ALS xy-wing, or more generally any xy-chain is an ALS xy-chain?

[udosuk]

Perhaps I'm overlooking something, but can't you always write an xy6-chain deduction as an ALS xy-wing, ...

Can you rewrite your xy6-chain into an ALS-xy-wing? I don't think I can. Nor can I rewrite my "complex y-wing" (which in fact is an xy6-chain) into ALS-xy-wing. The core of the problem lies within the {17} in b8 being not an ALS.

... or more generally any xy-chain is an ALS xy-chain?

I don't like the ALS-xy-chain or any move with the word "chain" attached.

BTW my ALS-xy-wing is a 5-cell move, so should be better than your 6-cell chain no matter what.

[ronk]

The core of the problem lies within the {17} in b8 being not an ALS.

With 2 candidates in 1 cell, a bivalue sure fits the definition of an ALS -- N+1 candidates in N cells that see each other. It's the limit case (N=1), but IMO it's still an ALS.

[udosuk]

With 2 candidates in 1 cell, a bivalue sure fits the definition of an ALS -- N+1 candidates in N cells that see each other. It's the limit case (N=1), but IMO it's still an ALS.

I was talking about 2 candidates in 2 cells there (r7c4+r8c6={17}).

It would be a locked set, not an almost locked set.

If you consider them as 2 separate ALS's, then you will have 4 ALS's in the whole move, so you can't form any ALS-xy-wing which can't have more than 3 ALS's.

[ronk]

I was talking about 2 candidates in 2 cells there (r7c4+r8c6={17}).

It would be a locked set, not an almost locked set.

Aha, had you mentioned either r7c4 or a naked pair in b8 earlier then ...

[Para]

The "simplest" way I can solve this is through a complex y-wing:

Code: Select all

  *--------------------------------------------------*
 | 58   56   7    | 68   1    3    | 9    4    2    |
 | 2    3   *89   |*78   79   4    | 1    6    5    |
 | 1    69   4    | 2    69   5    | 7    8    3    |
 |----------------+----------------+----------------|
 | 3   @17   6    | 5    8    9    | 2    17   4    |
 | 4    8    2    | 167  67   17   | 5    3    9    |
 | 57   59  #19   | 4    3    2    | 8    17   6    |
 |----------------+----------------+----------------|
 | 9    2    3    |#17   4    8    | 6    5    17   |
 | 6   -17   5    | 3    2   @17   | 4    9    8    |
 | 78   4    18   | 9    5    6    | 3    2    17   |
 *--------------------------------------------------*

r2c34 can't be both 8
=> r2c3=9 or r2c4=7
=> At least one of r6c3 & r7c4 must be 1
=> At least one of r4c2 & r8c6 must be 7
=> r8c2 can't be 7

In other words, we first established a "strong link" of 1 in r6c3+r7c4, then use y-wing via r4c2+r8c6 to eliminate 7 from r8c2.

But the tricky thing is, we can't even transform this move to an ALS-xy-wing (the {17} in b8 is not an ALS).

There is a simpler complex y-wing, that i find easier to spot because it uses common bivalue cells.

Code: Select all

 *--------------------------------------------------*
 |*58   56   7    | 68   1    3    | 9    4    2    |
 | 2    3   *89   |#78   79   4    | 1    6    5    |
 | 1    69   4    | 2    69   5    | 7    8    3    |
 |----------------+----------------+----------------|
 | 3    17   6    | 5    8    9    | 2    17   4    |
 | 4    8    2    | 167  67   17   | 5    3    9    |
 | 57   59   19   | 4    3    2    | 8    17   6    |
 |----------------+----------------+----------------|
 | 9    2    3    |@17   4    8    | 6    5    17   |
 | 6   @17   5    | 3    2    7-1  | 4    9    8    |
 |#78   4    18   | 9    5    6    | 3    2    17   |
 *--------------------------------------------------*

Both R2C4 and R9C1 are {78} and together sees all 8's in N1(so a y-wing). So they can't both be 8. They both see a cell with {17} (R7C4 and R8C2). R7C4 and R8C2 can't both be 7 as this would force both R2C4 and R9C1 to 8, so at least one has to be a 1. So we can eliminate 1 from R8C6.
So basically this is a y-wing with 2 ALS's or 2 y-wings linked together.
And yes it can also be represented as a xy-chain. But i found this step while searching for a y-wing and for me this way it is easier to spot.

greetings

Para

[udosuk]

Aha, had you mentioned either r7c4 or a naked pair in b8 earlier then ...

Probably my bad for being not clear enough. But if I was referring to a single cell I would say something like the 1|7 for r8c6 instead of the {17} in b8. Anyway if you'd worked through my moves you should have realised that I do know single bivalue cells are valid ALS's.

BTW nice move Para. It's a prettier complex y-wing than mine. But I think my ALS-xy-wing (with 5 cells) is still the simpler move.

[mmoo9154]

Well, thanks to everyone for the replies!

I'd not heard about the BUG technique. That definitely solved the puzzle I posted. I can also see how to apply most of the other "Uniquness" tests.

I gotta say I don't quite get how to apply the BUG+2 (or more) technique, but it sounds like forcing chains are really the next thing to start to recognize. BUG's nice, but it's a boat-load of work to get all the pencil marks but one down to two alternatives.