Need help with this diabolical puzzle!

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

Need help with this diabolical puzzle!

Postby NeilB » Sun Jan 22, 2006 9:56 pm

I'm stuck on this puzzle. I can solve it using Sudoku Cue but it won't give me a hint on how to solve it using logic. Any help? This is as far as I've gotten:
Code: Select all
7 4 2|. 9 1|6 5 .
8 . .|4 . .|1 . .
. 3 1|6 . .|. 4 .
-----+-----+-----
. . .|1 . 5|. 3 6
. . .|9 . .|. . 1
3 1 .|7 . 2|5 9 4
-----+-----+-----
2 7 9|. . 6|4 1 5
. . 3|2 1 4|9 8 7
1 8 4|5 7 9|3 6 2

.---------------.---------------.---------------.
| 7    4    2   | 38   9    1   | 6    5    38  |
| 8    569  56  | 4    25   37  | 1    27   39  |
| 59   3    1   | 6    25   78  | 27   4    89  |
:---------------+---------------+---------------:
| 49   29   78  | 1    48   5   | 278  3    6   |
| 456  256  5678| 9    3468 38  | 278  27   1   |
| 3    1    68  | 7    68   2   | 5    9    4   |
:---------------+---------------+---------------:
| 2    7    9   | 38   38   6   | 4    1    5   |
| 56   56   3   | 2    1    4   | 9    8    7   |
| 1    8    4   | 5    7    9   | 3    6    2   |
'---------------'---------------'---------------'

-Neil
NeilB
 
Posts: 8
Joined: 22 January 2006

Postby tarek » Sun Jan 22, 2006 10:43 pm

Hi there NeilB,

This is one way of solving it:
Code: Select all
*--------------------------------------------------------*
| 7     4     2    | 38    9     1    | 6     5     38   |
| 8     569   56   | 4     25    37   | 1     27    39   |
| 59    3     1    | 6     25    78   | 27    4     89   |
|------------------+------------------+------------------|
| 49    29    78   | 1     48    5    | 278   3     6    |
| 456   256   5678 | 9     3468  38   | 278   27    1    |
| 3     1     68   | 7     68    2    | 5     9     4    |
|------------------+------------------+------------------|
| 2     7     9    | 38    38    6    | 4     1     5    |
| 56    56    3    | 2     1     4    | 9     8     7    |
| 1     8     4    | 5     7     9    | 3     6     2    |
*--------------------------------------------------------*
Candidates in r4c1 will force r4c3 to have only 7 as valid Candidates
r4c1=4: r4c1=4 => r4c5=8 => r4c3=7
r4c1=9: r4c1=9 => r3c1=5 => r2c3=6 => r6c3=8 => r4c3=7
Threfore r4c3=7
*--------------------------------------------------------*
| 7     4     2    | 38    9     1    | 6     5     38   |
| 8     569   56   | 4     25    37   | 1     27    39   |
| 59    3     1    | 6     25    78   | 27    4     89   |
|------------------+------------------+------------------|
| 49    29    7    | 1     48    5    | 28    3     6    |
| 456   256   568  | 9     3468  38   | 278   27    1    |
| 3     1     68   | 7     68    2    | 5     9     4    |
|------------------+------------------+------------------|
| 2     7     9    | 38    38    6    | 4     1     5    |
| 56    56    3    | 2     1     4    | 9     8     7    |
| 1     8     4    | 5     7     9    | 3     6     2    |
*--------------------------------------------------------*
Candidates in r4c2 will force r2c5 to have only 5 as valid Candidates
r4c2=2: r4c2=2 => r4c7=8 => r4c5=4 => r4c1=9 => r3c1=5 => r3c5=2 => r2c5=5
r4c2=9: r4c2=9 => r4c1=4 => r4c5=8 => r4c7=2 => r5c8=7 => r2c8=2 => r2c5=5
Threfore r2c5=5


The second step looks like one of those discontiuous xy chains

Tarek
User avatar
tarek
 
Posts: 2622
Joined: 05 January 2006

Postby NeilB » Sun Jan 22, 2006 10:50 pm

Thanks Tarek. I also realized that I have an X-Wing at Rows 4 and 6, Columns 3 and 5. Eliminating the 8 at R7C5, leads to an easy solve. Interesting in that the solvers I used missed this X-Wing.
NeilB
 
Posts: 8
Joined: 22 January 2006

Postby tarek » Sun Jan 22, 2006 10:57 pm

NeilB wrote:I also realized that I have an X-Wing at Rows 4 and 6, Columns 3 and 5


one of those columns has more than 2 8s & one of the rows has more than 2 8s, it doesn't look like an x-wing to me.

Tarek
User avatar
tarek
 
Posts: 2622
Joined: 05 January 2006

Postby CathyW » Sun Jan 22, 2006 11:42 pm

An alternative next few steps:

xy-wing in box 1 - (eliminate 5 from r5c1 - wrong!) - the elimination is from r2c2.
Another xy-wing [r3c1][r2c2][r8c2] - r8c1=6

Then you need to start looking at those cells which only have 2 numbers in them (bivalue cells) - I'm just learning about these myself but I think the following are valid:

[r5c1] -4- [r4c1] -9- [r3c1] -5- [r2c3] -6- [r6c3] -6- [r5c1] - forms a closed loop that is linked by one of the numbers of each cell in turn. You can eliminate 6 from r5c3.

[r4c3] -8- [r4c5] -4- [r4c1] -4- [r5c1] -6- [r6c3] -8- [r4c3] - you can eliminate 8 from r5c3.

[r5c3] -7- [r5c8] -2- [r2c8] -2- [r2c5] -5- [r2c3] -5- [r5c3] - you can eliminate 7 from r5c7.

xy-wing in box 6 - exclude 2 from r4c7.

From there a hidden single in r4c2 and the rest is singles.:)
Last edited by CathyW on Sun Jan 22, 2006 8:56 pm, edited 1 time in total.
CathyW
 
Posts: 316
Joined: 20 June 2005

Postby Crazy Girl » Mon Jan 23, 2006 12:24 am

CathyW wrote:An alternative next few steps:

xy-wing in box 1 - eliminate 5 from r5c1. - How:?:
Another xy-wing [r3c1][r2c2][r8c2] - r8c1=6



surely if r2c2=5 then r3c1=9 and r8c2=6 so r8c1<>6.

I don't think this is an xy-wing and one can't make these eliminations.:!:
Crazy Girl
 
Posts: 189
Joined: 08 November 2005

Postby CathyW » Mon Jan 23, 2006 12:55 am

Oops - that first elimination of 5 should have been from r2c2. Sorry:!:
CathyW
 
Posts: 316
Joined: 20 June 2005

Postby NeilB » Mon Jan 23, 2006 1:06 am

tarek wrote:one of those columns has more than 2 8s & one of the rows has more than 2 8s, it doesn't look like an x-wing to me.
Ooops. I got lucky then.
NeilB
 
Posts: 8
Joined: 22 January 2006


Postby tarek » Mon Jan 23, 2006 1:31 am

Well, after analising the original puzzle with my solver, I'm not sure how you reached the position you posted

Starting position:
Code: Select all
 7 . . | . 9 . | 6 5 . 
 8 . . | . . . | 1 . . 
 . 3 1 | 6 . . | . 4 . 
-------+-------+------
 . . . | 1 . 5 | . 3 6 
 . . . | . . . | . . . 
 3 1 . | 7 . 2 | . . . 
-------+-------+------
 . 7 . | . . 6 | 4 1 . 
 . . 3 | . . . | . . 7 
 . 8 4 | . 7 . | . . 2

Everything until this point seems to go with what you have done
Code: Select all
*--------------------------------------------------------*
| 7     4     2    | 38    9     1    | 6     5     38   |
| 8     569   56   | 4     25    37   | 1     27    39   |
| 59    3     1    | 6     25    78   | 27    4     89   |
|------------------+------------------+------------------|
| 49    29    78   | 1     48    5    | 278   3     6    |
| 456   256   5678 | 9     3468  38   | 2578  27    1    |
| 3     1     568  | 7     68    2    | 589   89    4    |
|------------------+------------------+------------------|
| 2     7     9    | 38    38    6    | 4     1     5    |
| 56    56    3    | 2     1     4    | 89    89    7    |
| 1     8     4    | 5     7     9    | 3     6     2    |
*--------------------------------------------------------*

you made some eliminations in Box 6, How?????

a possible next step would be (which is even better than the chains I posted above)
Code: Select all
Candidates in r4c1 will force r5c8 to have only 7 as valid Candidates
r4c1=4: r4c1=4 => r4c5=8 => r5c6=3 => r2c6=7 => r2c8=2 => r5c8=7
r4c1=9: r4c1=9 => r3c1=5 => r3c5=2 => r3c7=7 => r2c8=2 => r5c8=7
Therefore r5c8=7


& that solves it
User avatar
tarek
 
Posts: 2622
Joined: 05 January 2006

Postby bennys » Mon Jan 23, 2006 5:36 am

Code: Select all
+----------------+----------------+----------------+
| 7    4    2    | 38   9    1    | 6    5    38   |
| 8    569  56   | 4    25   37   | 1    27   39   |
|%59   3    1    | 6   %25  %78   |%27   4    89   |
+----------------+----------------+----------------+
|^49   29   78   | 1   ^48   5    | 278  3    6    |
| 456  256  5678 | 9    3468 38   | 278  27   1    |
| 3    1    68   | 7    68   2    | 5    9    4    |
+----------------+----------------+----------------+
| 2    7    9    | 38   38   6    | 4    1    5    |
| 56   56   3    | 2    1    4    | 9    8    7    |
| 1    8    4    | 5    7    9    | 3    6    2    |
+----------------+----------------+----------------+
Using the Almost Locked sets xz rule

A=%
B=^
x=9
z=8

and we get R5C6<>8 and that solve the puzzle.
bennys
 
Posts: 156
Joined: 28 September 2005

Postby tso » Mon Jan 23, 2006 6:49 am

bennys wrote:
Code: Select all
+----------------+----------------+----------------+
| 7    4    2    | 38   9    1    | 6    5    38   |
| 8    569  56   | 4    25   37   | 1    27   39   |
|%59   3    1    | 6   %25  %78   |%27   4    89   |
+----------------+----------------+----------------+
|^49   29   78   | 1   ^48   5    | 278  3    6    |
| 456  256  5678 | 9    3468 38   | 278  27   1    |
| 3    1    68   | 7    68   2    | 5    9    4    |
+----------------+----------------+----------------+
| 2    7    9    | 38   38   6    | 4    1    5    |
| 56   56   3    | 2    1    4    | 9    8    7    |
| 1    8    4    | 5    7    9    | 3    6    2    |
+----------------+----------------+----------------+
Using the Almost Locked sets xz rule

A=%
B=^
x=9
z=8

and we get R5C6<>8 and that solve the puzzle.



I just don't get the ALS method. It seems to me that it's the "emperor's new tactic". I can't see how identifying these "almost locked sets" helps find a solution -- instead, once an exclusion is found -- in this case, by a simply discontinuous xy-chain --


Code: Select all
+----------------+----------------+----------------+
| .    .    .    | .    .    .    | .    .    .    |
| .    .    .    | .    .    .    | .    .    .    |
| 59   .    .    | .    25   78   | 27   .    .    |
+----------------+----------------+----------------+
| 49   .    .    | .    48   .    | .    .    .    |
| .    .    .    | .    .    38   | .    .    .    |
| .    .    .    | .    .    .    | .    .    .    |
+----------------+----------------+----------------+
| .    .    .    | .    .    .    | .    .    .    |
| .    .    .    | .    .    .    | .    .    .    |
| .    .    .    | .    .    .    | .    .    .    |
+----------------+----------------+----------------+


r4c5=8 => r5c6=3
r4c5=4 => r4c1=9 => r3c1=5 => r3c5=2 => r3c7=7 => r3c6=8 => r5c6=3
Therefore, r5v6=3

-- or the same thing by nice loop --

[r4c5]-4-[r4c1]-9-[r3c1]-5-[r3c5]-2-[r3c7]-7-[r3c6]-8-[r5c6]-8-[r4c5]

-- which can routinely be found by humans with pencil and paper -- we can *then* decide which sets of cells I want to identify as *almost locked sets* -- after the fact.

How did you identify r3c1567 as an ALS? Why not r3c1569, r3c1589, r3c1689 or r3c5689? Aren't the all equally valid ALS? for that matter, isn't each and every single bivalue cell an ALS? This seems to be more of a forensic technique rather than solving method. r34c1 is an ALS. r4c5+r5c6 is an ALS. Everywhere I look are ALS. Does this actually help a human solver find the answer, or is this a carbon based tactic?
tso
 
Posts: 798
Joined: 22 June 2005

Postby bennys » Mon Jan 23, 2006 7:54 am

The only part that I agree is that you don't get the ALS method.
bennys
 
Posts: 156
Joined: 28 September 2005

Postby Carcul » Mon Jan 23, 2006 9:28 am

Hi Tso.

Tso wrote:I can't see how identifying these "almost locked sets" helps find a solution


In the case of this particular puzzle, it is very easy to find a solution without the need for ALSs, and is probably "harder" to find the "correct" combination of AlSs, because there are so many to choose from. However, in extremely difficult puzzles, we will not get very far with nice loops, and, in my opinion, it is in these cases where the ALSs can be very helpful.

Tso wrote:Does this actually help a human solver find the answer, or is this a carbon based tactic?


Oh yes it helps. I guess one of the "secrets" is, when solving a puzzle, to keep track of some ALSs and then try to relate them with cells or/and with each other.

Regards, Carcul
Carcul
 
Posts: 724
Joined: 04 November 2005

Postby tso » Mon Jan 23, 2006 6:05 pm

bennys wrote:The only part that I agree is that you don't get the ALS method.


You disagree with everything else I suggested? The forcing chains I gave? The nice loop? That any bivalue cell is an Almost Locked Set? That there are dozens of Almost Locked Sets to choose from?

Carcul wrote:Hi Tso.

Tso wrote:I can't see how identifying these "almost locked sets" helps find a solution


In the case of this particular puzzle, it is very easy to find a solution without the need for ALSs, and is probably "harder" to find the "correct" combination of AlSs, because there are so many to choose from.


If this is the case, how is it different from describing this...

[12][23][31]|[..][..][..]|[..][..][..]

... as a continuous nice loop --> [r1c1]-2-[r1c2]-3-[r1c3]-1-[r1c1] --> that eliminates 1s, 2s and 3s from the rest of the row? Sure, its true, but all human solvers will find it easier to identify this as a simple naked triple.

Carcul wrote:However, in extremely difficult puzzles, we will not get very far with nice loops, and, in my opinion, it is in these cases where the ALSs can be very helpful. ... I guess one of the "secrets" is, when solving a puzzle, to keep track of some ALSs and then try to relate them with cells or/and with each other.

Regards, Carcul


Ok, but where are the algorithms? There are several methods of tracking down finding forcing chains that have been described in these forums. The ALS tactics do not seem to be a simple pattern, nor have any methods of *finding* them been described. And unlike forcing chains in which I must only hold two pieces of information in my head at once, ALS seems to require me to idenify and hold in my head, SET A, SET B, maybe SET C, D, x, y and z ... oh, wait, I see, it's so *obvious* to me now. How are you finding x, y and z?

Look at one of bennys original posts, Almost locked rules (for now)

bennys wrote:
Code: Select all
Almost locked rules
-------------------

ALS - almost locked set
For two sets A,B that have common candidate x we will say that x is restricted common if it can't be
in A and B in the same time.(being in A exclude it from B)

Almost locked sets xz rule
--------------------------
If
1)A,B ALS
2) x is restricted common
Then Any other common candidate (lets say z) cant appear outside of A and B if it can see all the z candidates in both A and B.
The reason?
If z appear then both A and B are locked but only one can get the x.

 +-------------+-------------+-------------+
 | 24  7   8   | 24  6   5   | 1   9   3   |
 | 9   3   24  | 248 1   48  | 56  7   56  |
 | 5   1   6   | 7   3   9   | 8   4   2   |
 +-------------+-------------+-------------+
 | 28  9   23  | 458 7   6   |#35  1  #45  |
 |*17  6   5   | 3   9   14  | 2   8  #47  |
 | 178 4  *13  | 58  2   18  | 357 6   9   |
 +-------------+-------------+-------------+
 | 6   5   7   | 1   4   2   | 9   3   8   |
 | 14  2   14  | 9   8   3   | 67  5   67  |
 | 3   8   9   | 6   5   7   | 4   2   1   |
 +-------------+-------------+-------------+
A={R5C1,R6C3}
B={R4C7,R4C9,R5C9}
x=7
z=3




This seems like we're arbitrarily creating more specific patterns and adding complications where a general pattern already exists that is sufficient to explain these exclusions more easily. The 5 cells marked form 5/6 of a forcing chain that will exclude a 3 from both r4c3 and r6c7.

r6c3=1 => r5c1=7 => r5c9=4 => r4c9=5 => r4c7=3 => (r4c3<>3 AND r6c7<>3)
r6c3=3 => (r4c3<>3 AND r6c7<>3)
Therefore, (r4c3<>3 AND r6c7<>3)

Or, written as nice loops (I've marked the strong links, though, as this is a pure bivalue chain, no strong links are required):

[r6c3]-1-[r5c1]=7=[r5c9]=4=[r4c9]-5-[r4c7]=3=[r4c3]=3=[r6c3]
[r6c3]-1-[r5c1]=7=[r5c9]=4=[r4c9]-5-[r4c7]=3=[r6c7]=3=[r6c3]


Here's the partial grid with the extraneous removed:

Code: Select all
 +-------------+-------------+-------------+
 | .   .   2x3 | .   .   .   |#35  .  #45  |
 |*17  .   .   | .   .   .   | .   .  #47  |
 | .   .  *13  | .   .   .   |x357 .   .   |
 +-------------+-------------+-------------+


Solving by ALS requires that I identify the cells marked # as an ALS and those with an * as another -- exclusions marked with 'x'. How these were chosen from among the other combinations using these same cells, I don't know. (Just in box 6, I count three 2-cell ALS and four 3-cell ALS -- and 3 one-cell ALS.)

But:

Code: Select all
 +-------------+-------------+-------------+
 | .   .   2x3 | .   .   35  | .   .   45  |
 |*17  .   .   | .   .   .   | .   .   47  |
 | .   .  *13  | .   . x357  | .   .   .   |
 +-------------+-------------+-------------+


The exact same forcing chains / nice loops exist making the same exclusions for the same reasons -- but the arbitrary # ALS is gone! ... implying that the ALS tactic was coincidental.

Actually, each consecutive pair of cells in the chain, taken together form an ALS.

The ALS method of solving this puzzle requires me keep four bits of information in my head at once the sets A and B and the value and location of x and z.
tso
 
Posts: 798
Joined: 22 June 2005

Next

Return to Help with puzzles and solving techniques