The New Sudoku Players' Forum

[sdkNewbie]

Hello out there,
the following puzzle leaves me, well, puzzled. I have tried to put it into a form like I saw on other postings in this forum.
Any help or advice will be greatly appreciated.
Best regards
sdkNewbie
-------------------------------------------

Code: Select all

-8-  -1-  237
7-2  3--  519
-13  --2  684

127  ---  968
-56  9-1  -23
-39  -2-  -51

291  8--  345
378  254  196
---  1--  872


+----------------+----------------+----------------+
| 4569 8    45   | 456  1    569  | 2    3    7    |
| 7    46   2    | 3    468  68   | 5    1    9    |
| 59   1    3    | 57   79   2    | 6    8    4    |
+----------------+----------------+----------------+
| 1    2    7    | 45   34   35   | 9    6    8    |
| 48   5    6    | 9    478  1    | 47   2    3    |
| 48   3    9    | 467  2    678  | 47   5    1    |
+----------------+----------------+----------------+
| 2    9    1    | 8    67   67   | 3    4    5    |
| 3    7    8    | 2    5    4    | 1    9    6    |
| 456  46   45   | 1    369  369  | 8    7    2    |
+----------------+----------------+----------------+


[ronk]

Any help or advice will be greatly appreciated.

Naked pairs in col 1 and box 8
Locked candidaates in row 4
Naked triple in col 6

[Shazbot]

you need to read up here on naked pairs and triples, and locked candidates. There are several eliminations you can make in this grid using those techniques.

Leter there's a technique which I believe is referred to as "forcing chains" (which I USED to believe was the same as colouring but now I'm not so sure) but I'm not up on that. It places a 6 in r1c4 and you can solve the puzzle from there. You can learn about it here

[QBasicMac]

Hello out there

Well, hello to you, too. Tough puzzle you've tackled.

Notice that r5c1 has the candidate "4". So does r6c1.
No other cell in that box has the candidate 4.
Therefore one of those two MUST be a 4, right?

Well, then no other cell in column 1 can have the candidate "4".

But you show "4" in cell r1c1 and r9c1. So erase those two 4's.

When someone here says "Locked candidate x in box y" they are hinting that if you look around in box y you will find such a situation and can benefit. Lets take a new case:

"Box 7 has a locked candidate." That is the kind of help you can expect on this forum. The members believe that nobody should be helped other than to learn a mass of locally-invented jargon. So learn it if you want help. As I showed above, it means to inspect box 7 and note that there are only two 6's and they are in the same row. Hence the other 6's in row 9 (in columns 5 and 6) can be erased. Erase them.

Now it's a bit tougher. Look down Column 6. You will see that r2, r6 and r7 contain the candidates 6,7,and 8: Three cells in the same column which contain only three candidates between them. That means that no matter how the final placement occurs, 6, 7 and 8 MUST be in those three cells and no others. Thus the 6 can be erased from r1c8. Erase it

Unfortunately there are no 7's nor 8's to be erased, but if there were, we would erase them too. Such a pattern is called a "Naked triple". Learn it if you expect to benefit from any help here.

Things are still a bit tough. See cells r5c1, r6c1. Here 4 is a locked candidate in row c1. One of those two cells MUST be 4.

But this is also true for r5c7 and r6c7. AND THEY FORM A BOX! So what? Well convince yourself that r5c5 can never be 4. Erase that candidate. Such a pattern is called an X-Wing here.

Now we have this result:

Code: Select all

-8-  -1-  237
7-2  3--  519
-13  --2  684

127  ---  968
-56  9-1  -23
-39  -2-  -51

291  8--  345
378  254  196
---  1--  872


Code: Select all

+-------------+---------------+----------+
| 569  8   45 | 456  1    59  | 2   3  7 |
| 7    46  2  | 3    468  68  | 5   1  9 |
| 59   1   3  | 57   79   2   | 6   8  4 |
+-------------+---------------+----------+
| 1    2   7  | 45   34   35  | 9   6  8 |
| 48   5   6  | 9    78   1   | 47  2  3 |
| 48   3   9  | 467  2    678 | 47  5  1 |
+-------------+---------------+----------+
| 2    9   1  | 8    67   67  | 3   4  5 |
| 3    7   8  | 2    5    4   | 1   9  6 |
| 56   46  45 | 1    39   39  | 8   7  2 |
+-------------+---------------+----------+


This is where I give up. You can study up on coloring, chains, etc. and maybe find a solution that is ethestically pleasing to you. Personally at such a point, I just use Trial and Error (T&E). Try r9c6=3. That gives a solution. Try r9c6=9. That does not. Therefore the puzzle is valid and r9c6=3 was correct.

Mac

[Shazbot]

"Box 7 has a locked candidate." That is the kind of help you can expect on this forum.

Hmmmm - I DO believe Mac is having a dig at those of us who have learned and make frequent use of terminology known "only to the in crowd".

I must admit when I first started browsing this forum I was overwhelmed by the local jargon and felt there was absolutely NO hope for me to ever learn how to solve any but the easiest puzzles. Then I happened to see a couple of links to "how-to" sites that explained these mysterious terms and actually showed how to use them! (the sites I've just given YOU the links for). So I printed, studied, practiced and voila! I was able to complete puzzles using these advanced techniques and actually help others out of a tight spot as well.

Unfortunately, Mac must have some kind of skill that I do not possess, because even though I vowed to "remember my roots" and try and explain things in plain english for those new to the forum and as equally overwhelmed as I was, I could not resist the pull of the dark side and eventually crossed over to the world of "sudoku-speak" (it's also easier to say and understand "naked triple" than to explain what it means, and AngusJ and SadMan sites provide a better explanation than I could give anyway). Mac, on the other hand, has been a member of the forum for quite a bit longer than me and has managed to stay true to the needs of the rest of the world, and able to explain these things in a way that can be easily understood by the majority. Onya Mac - I must make a point of taking the experience of puzzlers into account in my replies.

However, there WAS a certain satisfaction when I was talking to my mum the other day (a recent Sudoku convert, who believes there are only two tactics required in solving a sudoku - naked single and hidden single, not that she'd call them that), when I listed as alternate techniques: naked and hidden pairs, triples and quads, locked candidates, xwings, swordfish, jellyfish, colouring and forcing chains. The glazed-eye, blank face, "you're pulling my leg, aren't you", "this is too much for me to take in - does not compute, does not compute" expression was the result I was seeking. My job was done.

[SarC]

A little OT but...I get to here with standart techniques

Code: Select all

569    8      45     | 46     1      59     | 2      3      7     
7      46     2      | 3      468    68     | 5      1      9     
59     1      3      | 57     79     2      | 6      8      4     
---------------------+----------------------+----------------------
1      2      7      | 45     34     35     | 9      6      8     
48     5      6      | 9      78     1      | 47     2      3     
48     3      9      | 67     2      678    | 47     5      1     
---------------------+----------------------+----------------------
2      9      1      | 8      67     67     | 3      4      5     
3      7      8      | 2      5      4      | 1      9      6     
56     46     45     | 1      39     39     | 8      7      2     


Then my solver suggest a simple 5-link forcing chain:
[6, 1]=5 => [3, 1]=4 => [4, 1]=6 => [4, 6]=7 => [4, 3]=5 => [6, 1]=9
So [6, 1] can't be a 5 and this solves the puzzle.
But I was looking at the BUG thing the other day, and I'm wondering if this can be applied here, even though there is multiple cells with more than two candidates? Or maybe there is a simpler technique that can be applied?

SarC

[Nick67]

[Edit: argh ... this post is incorrect. I made
a mistake, and then just got lucky. I'll try to
point out the mistake with further edits below.]

Here is Mac's last grid:

Code: Select all

+-------------+---------------+----------+
| 569  8   45 | 456  1    59  | 2   3  7 |
| 7    46  2  | 3    468  68  | 5   1  9 |
| 59   1   3  | 57   79   2   | 6   8  4 |
+-------------+---------------+----------+
| 1    2   7  | 45   34   35  | 9   6  8 |
| 48   5   6  | 9    78   1   | 47  2  3 |
| 48   3   9  | 467  2    678 | 47  5  1 |
+-------------+---------------+----------+
| 2    9   1  | 8    67   67  | 3   4  5 |
| 3    7   8  | 2    5    4   | 1   9  6 |
| 56   46  45 | 1    39   39  | 8   7  2 |
+-------------+---------------+----------+


We can eliminate the 4 in r6c4, because the 4's in
row 4 are confined to box 5. Then, there is a
57-79-59 naked triple in box 2, which lets us
eliminate 5 from from r1c4.

The resulting grid forms a BUG with three "extra candidates",
the 5 in r1c1, the 6 in r2c5, and the 6 in r6c6:

[Edit: actually the resulting grid does not form a BUG,
at least not in the way that I show it below]

Code: Select all

 *--------------------------------------------------*
 | 69+5  8   45   | 46   1    59   | 2    3    7    |
 | 7    46   2    | 3    48+6 68   | 5    1    9    |
 | 59   1    3    | 57   79   2    | 6    8    4    |
 |----------------+----------------+----------------|
 | 1    2    7    | 45   34   35   | 9    6    8    |
 | 48   5    6    | 9    78   1    | 47   2    3    |
 | 48   3    9    | 67   2    78+6 | 47   5    1    |
 |----------------+----------------+----------------|
 | 2    9    1    | 8    67   67   | 3    4    5    |
 | 3    7    8    | 2    5    4    | 1    9    6    |
 | 56   46   45   | 1    39   39   | 8    7    2    |
 *--------------------------------------------------*


By the BUG principle, at least one of those extra candidates
must be the final value of the associated cell.
So r1c1=5 or r2c5=6 or r6c6=6 (and possibly more
than 1 of these statements is true). Each result by itself would
lead to this: r2c6=8, so we can place the 8 in r2c6.

[Edit: so here, placing the 8 in r2c6, I was just lucky.]

Next we can place 8 in r5c5, 4 in r5c1, 8 in r6c1,
7 in r5c7, and 4 in r6c7. These are all singles.

[Edit: I suppose the rest of the post is OK.]

Again we are left with a BUG, this time with just
1 extra candidate, the 5 in r1c1:

Code: Select all

 *--------------------------------------------------*
 | 69+5  8   45   | 46   1    59   | 2    3    7    |
 | 7    46   2    | 3    46   8    | 5    1    9    |
 | 59   1    3    | 57   79   2    | 6    8    4    |
 |----------------+----------------+----------------|
 | 1    2    7    | 45   34   35   | 9    6    8    |
 | 4    5    6    | 9    8    1    | 7    2    3    |
 | 8    3    9    | 67   2    67   | 4    5    1    |
 |----------------+----------------+----------------|
 | 2    9    1    | 8    67   67   | 3    4    5    |
 | 3    7    8    | 2    5    4    | 1    9    6    |
 | 56   46   45   | 1    39   39   | 8    7    2    |
 *--------------------------------------------------*


By the BUG principle, that single extra candidate
must be the final value of that cell. So r1c1=5.

After that, just several singles remain.

[Carcul]

Hi SdkNewbie.

After the "basic steps" that Ronk, Shazbot, and QBasicMac have pointed out above, we can use a more advanced technique, the bilocation-bivalue plot technique, which allow us to find the following nice loop:

[r1c3]=4=[r1c4]=6=[r6c4]=7=[r3c4]-7-[r3c5]-9-[r3c1]-5-[r1c3]
=> r1c3<>5

and that solve the puzzle. Try here for an excelent explanation and some examples of this technique.

Regards, Carcul

[ronk]

Personally at such a point, I just use Trial and Error (T&E). Try r9c6=3. That gives a solution. Try r9c6=9. That does not.

Much like a blind man running into a brick wall ... without his glasses on.

[emm]

Well there we have it - different people see different things, different people want different things.

I’m no longer surprised here to see a newbie respond with enthusiastic gratitude to a reply that I think (privately) is a garbled piece of waffle, while my own (pertinent) response has obviously drawn nothing but a total blank. It’s very levelling the forum - it is.

And it’s good, that element of surprise. Some newbies enthuse over the succinct jargonistic reply, others wallow in the protracted step-by-step explanation. Some will be enamoured of the BUG principle (well maybe) some will run a mile at the very mention. That’s part of the fun – who knows when a newbie posts, what the hell they’re going to want? It adds a certain suspense to what could otherwise get a bit humdrum. The ball’s in the air, anyone can have a shot at it and there don’t seem to be any rules about what’s going to do it for them this time. ‘Viva la difference!’ I say.

Something’s got your goat by the sound of things, Mac, or is it just getting you down being the only one expounding your theory on the forum. There’s someone called elf in this thread who might agree with you.

[ronk]

the following nice loop:

[r1c3]=4=[r1c4]=6=[r6c4]=7=[r3c4]-7-[r3c5]-9-[r3c1]-5-[r1c3]
=> r1c3<>5

and that solve the puzzle.

What are the "implication chains" for that?

I thought one could start anywhere in the loop, and get one implication chain by following the loop forwards, and the second chain by following it backwards. And by choosing a starting point at (or as near as possible to) the middle of the loop, the two chains would be of minimal length. But I don't get your result.

Implication chains:
r3c4=7 => r3c4=9 => r3c1=5 => r1c3<>5 (forward from r3c4)
r3c4<>7 => r6c4=7 => r1c4=6 => r1c3=4 => r1c3<>5 (backward from r3c4)

Oh, never mind, that *is* the same result. Writing it down certainly helps.

BUT ... when looking at the equation, what's the short-cut "rule of thumb" to know the implication is r1c3<>5 (from the right side of the eq'n) ... as opposed to r1c3<>4 (from the left side)?

Can one assume the elimination is from the bivalued end? And is only one end always bivalued?

blbvNewbie

[Jeff]

BUT ... when looking at the equation, what's the short-cut "rule of thumb" to know the implication is r1c3<>5 (from the right side of the eq'n) ... as opposed to r1c3<>4 (from the left side)?

Can one assume the elimination is from the bivalued end? And is only one end always bivalued?

Hi Carcul, could I give it a shot.

Ronk, you might like to take a few minutes reading the description of the bilocation/bivalue plot technique here. The nice thing about this technique is that deductions can be made without having to list out the 2 implications, hence considered non-bifurcational.

The particular deduction was described in Theorem 5.

[ronk]

BUT ... when looking at the equation, what's the short-cut "rule of thumb" to know the implication is r1c3<>5 (from the right side of the eq'n) ... as opposed to r1c3<>4 (from the left side)?

Ronk, you might like to take a few minutes reading the description of the bilocation/bivalue plot technique here. The nice thing about this technique is that deductions can be made without having to list out the 2 implications, hence considered non-bifurcational.

The particular deduction was described in Theorem 5.

No offense intended, but often a simple "yes" or "no" answer is more helpful than referring to a post that reads like a mathematical treatise or a government specification.

I'll assume the answers to my other two questions ...

Can one assume the elimination is from the bivalued end? And is only one end always bivalued?

... are "yes".

[Jeff]

No offense intended, but often a simple "yes" or "no" answer is more helpful than referring to a post that reads like a mathematical treatise or a government specification.

I'll assume the answers to my other two questions ...

Can one assume the elimination is from the bivalued end? And is only one end always bivalued?

... are "yes".

Just tried to be helpful. No hard feeling whatsoever. Cheers

[Carcul]

Hi Ronk.

Let me try to explain how you can follow the deduction in this particular nice loop. R1c3 is part of a strong link in "4". So, the candidate "4" can only be in r1c3 or r1c4, in row 1. If it is in r1c3, then r1c3<>5. If it is in r1c4, then if you follow the implications along the loop you will get r3c1=5, and so r1c3<>5. Hence, wherever the 4 is in row 1, the "5" cannot be in r1c3.
I hope this will help. If not, please make more questions and I (or others) will try to explain better.

Regards, Carcul