The New Sudoku Players' Forum

[Kent]

Need help on this.Any particular method I can use? Maybe X-wing or XY-Wing?

Code: Select all

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


[re'born]

Kent,

Here is a (probably) over-complicated solution. Consider the cells (1,[56]), (4, [45]) and (7, [46]). If (1,5)!1 and (4,4)!<39> and (4,5)!8, then we would get a deadly pattern.

Code: Select all

389 389 138 | 2     145* 45* | 6 58 7
7   2   18  | 6     15   3   | 9 4  58
5   6   4   | 8     7    9   | 3 2  1
------------+----------------+---------
2   39  6   | 3459* 458* 7   | 1 38 348
349 1   5   | 349   48   6   | 2 7  348
348 7   38  | 34    2    1   | 5 6  9
------------+----------------+---------
38  38  2   | 45*   9    45* | 7 1  6
6   5   7   | 1     3    8   | 4 9  2
1   4   9   | 7     6    2   | 8 35 35

But if (2,5) = 1, then (1,5)!1. Moreover, (2,5)1 > (2,3)8 > (6,3)3 > (4,2)9 > (4,4)!9. Also, (6,3)3 > (6,4)4 > (5,5)8 > (4,5)!8. Finally, (2,5)1 > (1,[56])[45] > (1,8)8 > (4,8)3 > (4.4)!3. Thus, (2,5)1 would lead to a deadly pattern and we can conclude that (2,5)!1. This solves the puzzle.

[Carcul]

Let's start by noting the Almost Unique Rectangle in cells r4c5/r5c5/r4c9/r5c9, in which we have the link [r4c9|r5c9]=3|5=[r4c5]. Thus, we could write

[r9c9]-3-[r4c9|r5c9]{=3=[r4c8](-3-[r4c4])-3-[r4c2]-9-[r4c4]}=3|5=[r4c5](-5-[r4c4])-5-[r2c5]-1-[r2c3]-8-[r6c3]-3-[r6c4]-4-[r4c4]

which means that, r9c9=3 would make r4c4 an empty cell - a contradiction. So, r9c9<>3 which solve the puzzle.

Carcul

[Neunmalneun]

It also works with the rule "if a cell has the candidates abc and ab leads to a you can eliminate a from this cell".

In my opinion this techinque is easy to apply.

Assuming R4C5=48 => R45C9=3 (uniqueness) => R9C5=5 => R9C8=3 =R4C8=8 => R4C5<>8.

This leads to a single 8 in R5C5 and a blocked 8 in R4C2.

Same technique once again: Assuming R1C3 is 18 => R6C3=3 => R4C2=9 => R5C1=4 => R5C9=3 => R2C9=8 => R2C3=1 which solves the rest.

[tarek]

Code: Select all

*--------------------------------------------------------*
| 389   389   138  | 2     145   45   | 6     58    7    |
| 7     2     18   | 6     15    3    | 9     4     58   |
| 5     6     4    | 8     7     9    | 3     2     1    |
|------------------+------------------+------------------|
| 2     39    6    | 3459  458   7    | 1     38    348  |
| 349   1     5    | 349   48    6    | 2     7     348  |
| 348   7     38   | 34    2     1    | 5     6     9    |
|------------------+------------------+------------------|
| 38    38    2    | 45    9     45   | 7     1     6    |
| 6     5     7    | 1     3     8    | 4     9     2    |
| 1     4     9    | 7     6     2    | 8     35    35   |
*--------------------------------------------------------*
Candidates in r1c8 will force r5c4 to have only 39 as valid Candidates
r1c8=5: r1c8=5 => r1c6=4 => r7c6=5 => r7c4=4 => r5c4<>4 => r5c4=39
r1c8=8: r1c8=8 => r2c9=5 => r2c5=1 => r2c3=8 => r6c3=3 => r6c4=4 => r5c4<>4 => r5c4=39
Threfore r5c4=39
*--------------------------------------------------------*
| 389   389   138  | 2     145   45   | 6     58    7    |
| 7     2     18   | 6     15    3    | 9     4     58   |
| 5     6     4    | 8     7     9    | 3     2     1    |
|------------------+------------------+------------------|
| 2     39    6    | 3459  458   7    | 1     38    348  |
| 349   1     5    | 39    48    6    | 2     7     348  |
| 348   7     38   | 34    2     1    | 5     6     9    |
|------------------+------------------+------------------|
| 38    38    2    | 45    9     45   | 7     1     6    |
| 6     5     7    | 1     3     8    | 4     9     2    |
| 1     4     9    | 7     6     2    | 8     35    35   |
*--------------------------------------------------------*
Candidates in r6c4 will force r1c5 to have only 14 as valid Candidates
r6c4=3: r6c4=3 => r6c3=8 => r2c3=1 => r2c5=5 => r1c5<>5 => r1c5=14
r6c4=4: r6c4=4 => r7c4=5 => r7c6=4 => r1c6=5 => r1c5<>5 => r1c5=14
Threfore r1c5=14
*--------------------------------------------------------*
| 389   389   138  | 2     14    45   | 6     58    7    |
| 7     2     18   | 6     15    3    | 9     4     58   |
| 5     6     4    | 8     7     9    | 3     2     1    |
|------------------+------------------+------------------|
| 2     39    6    | 3459  458   7    | 1     38    348  |
| 349   1     5    | 39    48    6    | 2     7     348  |
| 348   7     38   | 34    2     1    | 5     6     9    |
|------------------+------------------+------------------|
| 38    38    2    | 45    9     45   | 7     1     6    |
| 6     5     7    | 1     3     8    | 4     9     2    |
| 1     4     9    | 7     6     2    | 8     35    35   |
*--------------------------------------------------------*
Candidates in r6c4 will force r5c1 to have only 34 as valid Candidates
r6c4=3: r6c4=3 => r5c4=9 => r5c1<>9 => r5c1=34
r6c4=4: r6c4=4 => r7c4=5 => r7c6=4 => r1c6=5 => r1c8=8 => r4c8=3 => r4c2=9 => r5c1<>9 => r5c1=34
Threfore r5c1=34
*-----------------------------------------------*
| 9    38   138 | 2    14   45  | 6    58   7   |
| 7    2    18  | 6    15   3   | 9    4    58  |
| 5    6    4   | 8    7    9   | 3    2    1   |
|---------------+---------------+---------------|
| 2    9    6   | 345  458  7   | 1    38   348 |
| 34   1    5   | 9    48   6   | 2    7    348 |
| 348  7    38  | 34   2    1   | 5    6    9   |
|---------------+---------------+---------------|
| 38   38   2   | 45   9    45  | 7    1    6   |
| 6    5    7   | 1    3    8   | 4    9    2   |
| 1    4    9   | 7    6    2   | 8    35   35  |
*-----------------------------------------------*
Candidates in r5c9 will force r6c4 to have only 3 as valid Candidates (This would make a simple NICE Loop)
r5c9=3: r5c9=3 => r9c9=5 => r2c9=8 => r1c8=5 => r1c6=4 => r7c6=5 => r7c4=4 => r6c4=3
r5c9=4: r5c9=4 => r5c1=3 => r7c1=8 => r7c2=3 => r1c2=8 => r1c8=5 => r1c6=4 => r7c6=5 => r7c4=4 => r6c4=3
r5c9=8: r5c9=8 => r5c5=4 => r6c4=3
Threfore r6c4=3

Tarek

[tso]

The seven cells in brackets form a continuous forcing loop, eliminating the five candidates marked with 'x':

Code: Select all

 *-----------------------------------------------------------*
 | 389   389   13x8  | 2     14x5 [45]   | 6     58    7     |
 | 7     2    [18]   | 6    [15]   3     | 9     4     58    |
 | 5     6     4     | 8     7     9     | 3     2     1     |
 |-------------------+-------------------+-------------------|
 | 2     39    6     | 3x459 458   7     | 1     38    348   |
 | 349   1     5     | 3x49  48    6     | 2     7     348   |
 |x348   7    [38]   |[34]   2     1     | 5     6     9     |
 |-------------------+-------------------+-------------------|
 | 38    38    2     |[45]   9    [45]   | 7     1     6     |
 | 6     5     7     | 1     3     8     | 4     9     2     |
 | 1     4     9     | 7     6     2     | 8     35    35    |
 *-----------------------------------------------------------*

(Either r6c3 or r6c4 must be 3, so r6c1 must be 3, etc.)


The new grid ...

Code: Select all

 *--------------------------------------------------*
 | 389  389  13   | 2    14   45   | 6    58   7    |
 | 7    2    18   | 6    15   3    | 9    4    58   |
 | 5    6    4    | 8    7    9    | 3    2    1    |
 |----------------+----------------+----------------|
 | 2    39   6    | 359  458  7    | 1    38   348  |
 | 349  1    5    | 39   48   6    | 2    7    348  |
 | 48   7    38   | 34   2    1    | 5    6    9    |
 |----------------+----------------+----------------|
 | 38   38   2    | 45   9    45   | 7    1    6    |
 | 6    5    7    | 1    3    8    | 4    9    2    |
 | 1    4    9    | 7    6    2    | 8    35   35   |
 *--------------------------------------------------*

Now a forcing chain can be found:

r7c4=4 -> r6c4=3 -> r6c3=8 -> r2c3<>8 -> r1c12=8 -> r1c8<>8 -> r4c8=8
r7c4=5 -> r4c4=[39] -> r4c24=[39][39] -> r4c8=8

Therefore, r4c8=8 and the puzzle is solved.

I used the BINARY GROUP method to help find this second step quickly.

[QBasicMac]

Of course if you are in the "Advanced solving techniques" forum, any mention of T&E is rightfully regarded as blaspheme. But this is the "Help with particular puzzles" forum where people come to find out how to solve a puzzle. Anything that solves it is appropriate.

In this particular case, the answer is to simply guess.

Guess r9c9=3. Via singles, you quickly get stopped with no candidates in r4c4. So it has to be 5. Yep. Solves with singles.

Piece of cake!

Hmmm, maybe this is a problem with the forums. We have one whereby fanatics can develop and discuss "Advanced solving techniques." That's fine, but we have only one other forum.

Maybe we need two other forums.

A "Newbie Forum" where ordinary people who are just learning can see how to solve particular puzzles that are bothering them. T&E is as good as any other technique if it leads quickly to the solution.

A "No T&E forum" where people can discuss how to solve puzzles without using T&E.

As it is, you have to announce "I don't want T&E solutions" or the equivalent to limit your responses (and shut me up).

Mac

[vidarino]

Hmm, well, as I see it, if people felt it was okay to use trial and error, they wouldn't need to come here for help, would they? It's not particularly hard to randomly pick a number and a cell.

It just seems to me that most people seeking help want to find the next number or elimination by means of a less arbitrary method.

I'm not trying to disrespect T&E as a solving method. It works, it's quick and easy, and for some, it's fun. I, however, prefer to analyse the grid to spot the next logical step, not just assume one.

Vidar

[ronk]

Assuming R1C3 is 18 => R6C3=3 => R4C2=9 => R5C1=4 => R5C9=3 => R2C9=8 => R2C3=1 which solves the rest.

Starting with the simplest, I have three problems with that expression:

• no mention of R9C9
• no indication of the parallel inference of R6C3=3 => R5C1<>3
• the implication stream starting with R1C3 and ending with R2C3, which doesn't exactly close the loop

Ron

[Ruud]

Assuming R1C3 is 18 => R6C3=3 => R4C2=9 => R5C1=4 => R5C9=3 => R2C9=8 => R2C3=1 which solves the rest.

Starting with the simplest, I have three problems with that expression:

• no mention of R9C9
• no indication of the parallel inference of R6C3=3 => R5C1<>3
• the implication stream starting with R1C3 and ending with R2C3, which doesn't exactly close the loop

And yet, with all these shortcomings, it happens to be one of the few suggested solutions that does not use quantum-chain-notation, but a simple and easy to follow sequence of implications, with an underlying rule that excels in simplicity.

Thanks 9x9, for this simple alternative.

Ruud.

[tso]

In this particular case, the answer is to simply guess.

That's ridiculous, especially since his question was:

Any particular method I can use? Maybe X-wing or XY-Wing?

If *you* had posted a puzzle at a postion, tantilizingly close to the end, but still stuck, asking for help -- and I told you to guess -- you wouldn't feel insulted? You wouldn't feel condescended to? Maybe I should explain to you *how* to take a guess? "Hold the pencil by the eraser over the puzzle, close your eyes and drop it. Which ever candidate it's closest to..."

"Did you try looking in the back of the book? That usually helps. If not, why not try that now?" "Did you try entering it into a solving program? That can help."

I think you agree with the rest of us that simply posting the solution in response to a request for help is a no-no, but I fail to see the difference. If someone posts a simple or complex deduction, Kent can then work out that deduction for himself -- maybe learn something. (He seems to be absorbing this stuff faster than most.) But what exactly is he supposed to do with:

"Guess r9c9=3. Via singles, you quickly get stopped with no candidates in r4c4. So it has to be 5. Yep. Solves with singles."

Can you expect him go go back to his puzzle and "guess r9c9=3"? What would he be learning, other than you know which cell to guess 5?

If you must make this point -- wouldn't you agree that it might be better to say something like: "In a position like this, the quickest way to the solution is most likely to take an educated guess in one of the bivalue cells. If you reach a contradiction, you can eliminate your guess." In fact, there is skill in knowing when, where and how to bifurcate. Maybe that hasn't been spelled out. Possibly you could make a detailed post about it that could be included with the other tactics? I think we all agree that when part of the challenge is to complete the puzzle in the shortest time, given a complex puzzle, guessers will clobber the diehards nearly everytime. It is my opinion that by posting the puzzle here, we can safely assume that the clock is no longer running.

And no, the default is "I'd like help solving without guessing.", not "Would someone please help me take a guess?"

The latter is absurd.

[ronk]

And yet, with all these shortcomings, it happens to be ... a simple and easy to follow sequence of implications, with an underlying rule that excels in simplicity.

Agreed, and I was negligent by not saying so. While lamenting the flaws in the expression, I forgot that I was studying it closely precisely because of the elegance of the deduction. Neunmalneun, my apologies.

Ron

[tso]

It also works with the rule "if a cell has the candidates abc and ab leads to a you can eliminate a from this cell".

Tell me if I've got this rule correct -- Given a cell with the candidates [abc], if 'c' is eliminated and that leads to 'a', then the candidates can be reduced to [bc]? What is this rule called, why does it work and where is it explained?

Assuming R4C5=48 => R45C9=3 (uniqueness) => R9C5=5 => R9C8=3 =R4C8=8 => R4C5<>8.

I assume you mean r9c9 not r9c5. Simple typo, but then ...

This leads to a single 8 in R5C5

How so? I thought the point of this tactic was that if r4c5=[458] and [48] leads to [4] (which it does), then we can exclude [4] leaving [58]. That doesn't give me an 8 in r5c5. Is this a typo or a mistake or is the rule mistated? Double checking -- r4c5 is in fact '4'. Huh?

and a blocked 8 in R4C2.

Huh? What's a "blocked 8"? R4C2 is [39] -- it can't be an 8 by virtue of locked candidates in row 6. Typo? I'm lost -- so I really should pick apart the next one, cause I don't know what state the grid would be, but ...

Same technique once again: Assuming R1C3 is 18 => R6C3=3 => R4C2=9 => R5C1=4 => R5C9=3 => R2C9=8 => R2C3=1 which solves the rest.

Ok, though (R6C3=3 => R4C2=9 => R5C1=4) is not true, ((R6C3=3 AND R4C2=9) => R5C1=4) is. But I suppose that's nitpicking. But then you left out r5c9=3 => r9c9=5. Another nitpick. But THEN ... you close with "r2c3=1 which solve the rest"! What the heck does *that* mean? I According to the rule you gave, that means that since r1c3 was [138] and [18] lead [8], then r1c3 is actually [13] -- which it is. But that hardly solves the puzzle. I see no justification for the leap to r2c3=1 yet.

I wouldn't have listed all these little things -- Iv'e maid plenty of tyops myself -- it's just that I'm not sure were the *big* thing is.

Could you re-read, correct and comment on your post? It looks interesting but I cannot follow it. There appear to by multiple typos, cells refered to that do not have the candidates you claim -- I just can't make it out.

[Edit:]

I found your previous post with this rule. You typed it with an important typo here.

It also works with the rule "if a cell has the candidates abc and ab leads to a you can eliminate a from this cell".

... should be ...

It also works with the rule "if a cell has the candidates abc and ab leads to a you can eliminate b from this cell".

[Kent]

I don't quite understand about the notations

(1,[56]), (4, [45]) and (7, [46]). If (1,5)!1 and (4,4)!<39> and (4,5)!8, then we would get a deadly pattern.

WHat does (4,4)! means and <39> mean?
By the way what's a deadly pattern?

How do you read this??

r9c9]-3-[r4c9|r5c9]{=3=[r4c8](-3-[r4c4])-3-[r4c2]-9-[r4c4]}=3|5=[r4c5](-5-[r4c4])-5-[r2c5]-1-[r2c3]-8-[r6c3]-3-[r6c4]-4-[r4c4]

[re'born]

Kent,

I'll let Carcul answer your last question as he is the real expert. As to the first which is directed at me. This notation (which I have stolen from aeb) should be read as follows:

(i,j)a : row i column j is a
(i,j)!a : row i column j is not a
(i, [jk])<ab> : The correct values for row i column j and row i column k are a and b, in some order (this is how you might right a naked pair).
(i,j)!<ab> or (i,j)![ab] : row i column j is neither a nor b.

Note that I don't make a distinction between using <> and [], whatever looks best to me at the time, I use.

Deadly patterns have to do with unique rectangles. You can either go to the sticky note in the Advanced techniques section to get a link, or I highly recommend aeb's website:

http://homepages.cwi.nl/~aeb/games/sudoku/index.html

The explanations on that site are the best I have found so far.

Hope this helps,
rep'nA