## Hint on a puzzle.

Post the puzzle or solving technique that's causing you trouble and someone will help

### Hint on a puzzle.

Hi
Please can somebody give me a *hint*, how to move forward with this puzzle.
So far I've looked for x-wing and swordfish, what next?

The puzzle came from sudoku.org.uk.
Described as "Tough - Should give an experienced solver some entertainment"

Original

Code: Select all
`. . 1 | 9 . . | . . 86 . . | . 8 5 | . 3 .. . 7 | . 6 . | 1 . .------+-------+------. 3 4 | . 9 . | . . .. . . | 5 . 4 | . . .. . . | . 1 . | 4 2 .------+-------+------. . 5 | . 7 . | 9 . .. 1 . | 8 4 . | . . 77 . . | . . 9 | 2 . .`

Stuck here

Code: Select all
`. . 1 | 9 . 7 | . . 86 . . | 1 8 5 | 7 3 .. . 7 | 4 6 . | 1 . .------+-------+------. 3 4 | . 9 . | . . .. . . | 5 . 4 | . . .. . . | . 1 . | 4 2 .------+-------+------. . 5 | . 7 1 | 9 . .. 1 . | 8 4 . | . . 77 . . | . 5 9 | 2 . .`

Solution

Code: Select all
`3 4 1 | 9 2 7 | 5 6 86 9 2 | 1 8 5 | 7 3 48 5 7 | 4 6 3 | 1 9 2------+-------+------1 3 4 | 2 9 6 | 8 7 52 7 8 | 5 3 4 | 6 1 95 6 9 | 7 1 8 | 4 2 3------+-------+------4 2 5 | 3 7 1 | 9 8 69 1 6 | 8 4 2 | 3 5 77 8 3 | 6 5 9 | 2 4 1`

With pencilmarks
Code: Select all
`2345  245   1    | 9   23 7   | 56  456  86     249   29   | 1   8  5   | 7   3    24923589 2589  7    | 4   6  23  | 1   59   259-----------------+------------+--------------1258  3     4    | 267 9  268 | 568 1567 1561289  26789 2689 | 5   23 4   | 368 1679 1369589   56789 689  | 367 1  368 | 4   2    3569-----------------+------------+--------------248   2468  5    | 236 7  1   | 9   468  34629    1     2369 | 8   4  236 | 356 56   77     468   368  | 36  5  9   | 2   1468 1346`
Last edited by bat999 on Tue Sep 16, 2014 2:08 pm, edited 1 time in total.
bat999
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Joined: 15 September 2014
Location: UK

### Re: Hint on a puzzle.

This puzzle is well beyond the X-Wing and Swordfish level of difficulty. There is a Finned X-Wing that isn't too difficult to find, but that doesn't get you very far. To solve the puzzle I needed to move up to ALS-XZ, which is a much more complex technique (as well as a few other things in-between).

JasonLion
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Joined: 25 October 2007
Location: Silver Spring, MD, USA

### Re: Hint on a puzzle.

JasonLion wrote:... There is a Finned X-Wing that isn't too difficult to find...

Hi
I think I've found it...

The x-wing for 4 is at r2c2,r2c9,r9c2,r9c9.
With a fin at r9c8.
So remove the 4 from r7c9.

Is that correct?
bat999
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Posts: 677
Joined: 15 September 2014
Location: UK

### Re: Hint on a puzzle.

Code: Select all
`*--------------------------------------------------------------*| 2345  245   1      | 9     23    7      | 56    456   8      || 6    b249   29     | 1     8     5      | 7     3    a249    || 23589 2589  7      | 4     6     23     | 1     59    259    ||--------------------+--------------------+--------------------|| 1258  3     4      | 267   9     268    | 568   1567  156    || 1289  26789 2689   | 5     23    4      | 368   1679  1369   || 589   56789 689    | 367   1     368    | 4     2     3569   ||--------------------+--------------------+--------------------|| 248   2468  5      | 236   7     1      | 9     468   36-4   || 29    1     2369   | 8     4     236    | 356   56    7      || 7    c468   368    | 36    5     9      | 2    d1468 d1346   |*--------------------------------------------------------------*`

Here's a move from your stuck position - a Grouped Skyscraper.

If r2c9 <> 4 then r2c2 = 4, r9c2 <> 4 so 1 of r9c89 = 4. This => r7c9 <> 4.

This doesn't fully solve the puzzle but does expose a naked triple in Box 9 for further eliminations.

Hope this helps.

Leren
Leren

Posts: 3319
Joined: 03 June 2012

### Re: Hint on a puzzle.

bat999 wrote:So remove the 4 from r7c9.

Is that correct?

Indeed

JasonLion
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Posts: 640
Joined: 25 October 2007
Location: Silver Spring, MD, USA

### Re: Hint on a puzzle.

JasonLion wrote:Indeed

Hi
The finned x-wing has done the same job as Leren's Grouped Skyscraper.

These are the updated pencilmarks.

Code: Select all
`2345  245   1    | 9   23 7   | 56  456  86     249   29   | 1   8  5   | 7   3    24923589 2589  7    | 4   6  23  | 1   59   259-----------------+------------+--------------1258  3     4    | 267 9  268 | 568 1567 1561289  26789 2689 | 5   23 4   | 368 1679 1369589   56789 689  | 367 1  368 | 4   2    3569-----------------+------------+--------------248   2468  5    | 236 7  1   | 9   48   3629    1     2369 | 8   4  236 | 356 56   77     468   368  | 36  5  9   | 2   148  14`
bat999
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Joined: 15 September 2014
Location: UK

### Re: Hint on a puzzle.

OK Things are getting a bit tougher from your current PM, so I'll try and finish off the puzzle for you with a minimum amount of fuss.

Here's your current PM with cells marked a and b and 4 being eliminated in r1c1.

Code: Select all
`*-----------------------------------------------------------------------*| 235-4 b245    1       | 9      23     7       |a56    a456    8       || 6     b249   b29      | 1      8      5       | 7      3      249     || 23589  2589   7       | 4      6      23      | 1      59     259     ||-----------------------+-----------------------+-----------------------|| 1258   3      4       | 267    9      268     | 568    1567   156     || 1289   26789  2689    | 5      23     4       | 368    1679   1369    || 589    56789  689     | 367    1      368     | 4      2      3569    ||-----------------------+-----------------------+-----------------------|| 248    2468   5       | 236    7      1       | 9      48     36      || 29     1      2369    | 8      4      236     | 356    56     7       || 7      468    368     | 36     5      9       | 2      148    14      |*-----------------------------------------------------------------------*`

This move is called ALS ZX Rule. Here Z = 4 and X = 5. The way it works is as follows.

Look at the cells marked a ie r1c78. They are an ALS (Almost Locked Set) because they contain one more different digits (3) than the number of cells (2).

Now suppose r1c8 was not 4. Then r1c78 would be a locked pair (56), so in particular r1c2 would not be a 5.

Now look at the cells marked b. They are also an ALS with 4 digits and 3 cells. Now suppose r1c2 was not 5. These cells would become a locked triple and in particular either r1c2 or r2c2 would have to be 4.

So summing up, we can conclude that at least one of r1c8, r1c2 or r2c2 would have to be 4.

Since r1c1 can see all of these 3 cells we can conclude that r1c1 can't be 4. You can remove the 4 from r1c1.

The way I write this is ALS XZ Rule: X = 5, Z = 4: (4=5) r1c78 - (5=4) r1c2, r2c23 => - 4 r1c1

Z is the elimination digit and X is the digit that links the 2 Almost Locked sets. In general all the X's in ALS1 have to see all of the X's in ALS 2 and you can eliminate all Z's from cells that can see all of the Z's in both ALSs.

As 4 is one of 2 4's in Column 1 there are a few elementary eliminations that should lead you to the following PM:

Code: Select all
`*-----------------------------------------------------------------------*| 235    245    1       | 9      23     7       | 56     456    8       || 6      249    29      | 1      8      5       | 7      3      249     || 23589 a2589   7       | 4      6      3-2     | 1     a59    a259     ||-----------------------+-----------------------+-----------------------|| 1258   3      4       | 267    9      268     | 568    1567   156     || 1289   26789  2689    | 5      23     4       | 368    1679   1369    || 589    56789  689     | 367    1      368     | 4      2      3569    ||-----------------------+-----------------------+-----------------------|| 4     b26     5       | 236    7      1       | 9      8      36      ||b29     1     b2369    | 8      4     c236     |c356   c56     7       || 7     b68     368     | 36     5      9       | 2      14     14      |*-----------------------------------------------------------------------*`

The next move is called ALS XY Wing and is basically an extension of the previous move, except that this time there are 3 ALS's (which I have indicated by a, b and c in the PM).

So the logic of this move is as follows:

Suppose r3c2 and r3c9 were both not 2. Then the cells marked a would be a locked triple and in particular r3c2 would be 8.

In that case r9c2 would not be 8 and the cells marked b would be a locked quad and in particular r8c3 would be 3.

In that case r8c67 would not be 3, r8c678 would be a locked triple and in particular r8c6 would be 2.

So summing up, at least 1 of r3c29 and r8c6 must be 2. Since r3c6 can see all of these cells, r3c6 can't be 2. You can eliminate 2 from r3c6.

I write this move as ALS XY Wing: (2=8) r3c289 - (8=3) r7c2, r8c13, r9c2 - (3=2) r8c678 => - 2 r3c6

Simple ! Well maybe not, but the good news is that you'll find that the puzzle can now be solved by a cascade of singles.

Leren
Leren

Posts: 3319
Joined: 03 June 2012

### Re: Hint on a puzzle.

Leren wrote:... the good news is that you'll find that the puzzle can now be solved by a cascade of singles...

Hi Leren
From your PM I was able to complete the puzzle OK.

Thanks for the clear explanation of ALS XZ rule.
I will be able to refer back to this thread when it's needed in future.
bat999
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Location: UK

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