Stuck again

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Stuck again

Postby wychwood » Wed Nov 14, 2007 1:30 pm

Hi folks

Here I am, stuck again! Just when I was having a good run of solving puzzles, this one comes along and gets me flummoxed.

Starting problem is this:
Code: Select all
*-----------*
 |...|.8.|3..|
 |.2.|...|..4|
 |.7.|5..|...|
 |---+---+---|
 |.5.|...|.7.|
 |8..|.3.|..6|
 |.9.|...|.2.|
 |---+---+---|
 |...|..9|.5.|
 |6..|...|.4.|
 |..1|.4.|...|
 *-----------*


I get to this, after using one ALS in house 1 (is that the correct term for the top left block?):

Code: Select all
 
 
 *-----------------------------------------------------------------------------*
 | 49      6       459     | 2479    8       247     | 3       1       57      |
 | 1359    2       358     | 1679    79      1367    | 57      68      4       |
 | 134     7       348     | 5       16      1346    | 29      68      29      |
 |-------------------------+-------------------------+-------------------------|
 | 234     5       2346    | 124689  29      12468   | 148     7       138     |
 | 8       1       247     | 247     3       2457    | 45      9       6       |
 | 347     9       3467    | 14678   57      145678  | 1458    2       1358    |
 |-------------------------+-------------------------+-------------------------|
 | 27      4       27      | 3       16      9       | 168     5       18      |
 | 6       3       59      | 1278    257     12578   | 1279    4       1279    |
 | 59      8       1       | 267     4       2567    | 2679    3       279     |
 *-----------------------------------------------------------------------------*
[/code]

Where do we go from here please? (please give an explanation of the move if anything that is different from nakeds, X wing, Y wing, XY wing and ALSs).

By the way, I guess it is obvious that the fewer starting cells are given, then generally (but not always, I realise) the harder the problem (depending on the positions and 'shapes' of the given cells).
Does anyone have a view as to when the extra difficulty clicks in? I find that when 19 or less are given (and especially 17 or less) than I am pretty certain that I am going to run into difficulties later. Any other views on that?

Thanks in advance for your suggestions and help.

Wychwood
Code: Select all
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Re: Stuck again

Postby ronk » Wed Nov 14, 2007 1:51 pm

wychwood wrote:Where do we go from here please? (please give an explanation of the move if anything that is different from nakeds, X wing, Y wing, XY wing and ALSs).


Look for a y-wing (aka w-wing, aka semi-remote naked pair) using a <57> pair and empty rectangle (aka hinge).
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Postby re'born » Wed Nov 14, 2007 2:17 pm

While I prefer ronk's solution, here is an ALS way to proceed:

A={2,7,9} on r24c5
B={2,4,5,7} on r5c467
x=2
z=7, r6c5<>7

With either move, it seems that you reduce down to needing to use either uniqueness or an xy-chain.
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Postby wychwood » Wed Nov 14, 2007 3:59 pm

Thanks Ronk - but alas, I faer that I do not undertsand what you are saying here. Obviously I do not understand Y wings after all (I thought that was one of the techniques I understood but it obviously is not).

Could you translate your suggestion into plain english please....!!!!

Thanks
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Re: Stuck again

Postby RW » Wed Nov 14, 2007 4:10 pm

wychwood wrote:By the way, I guess it is obvious that the fewer starting cells are given, then generally (but not always, I realise) the harder the problem (depending on the positions and 'shapes' of the given cells).
Does anyone have a view as to when the extra difficulty clicks in? I find that when 19 or less are given (and especially 17 or less) than I am pretty certain that I am going to run into difficulties later. Any other views on that?

Actually, from a statistical point of view, no. The known puzzles with 17 clues are in general a lot easier than the known minimal puzzles with 30+ clues. Most of the hardest known puzzles have 21-24 clues. However, these are the results when looking at randomly generated puzzles. The puzzles you see in newspapers, books and on the web have usually went through a selection process to make sure that they are solvable with basic techniques. As a puzzle with fewer clues in the beginning might have fewer possible simple moves, a human solver might find it harder to solve, even though it doesn't require any more advanced techniques than a puzzle with more initial clues.

Btw. All valid puzzles have at least 17 clues. Feel free to prove me wrong on this point...

[Edit:
A little hint to ronks move:
Code: Select all
 *-----------------------------------------------------------------------------*
 | 49      6       459     | 2479    8       247     | 3       1       57      |
 | 1359    2       358     | 1679   -79      1367    |*57      68      4       |
 | 134     7       348     | 5       16      1346    | 29      68      29      |
 |-------------------------+-------------------------+-------------------------|
 | 234     5       2346    | 124689  29      12468   | 148     7       138     |
 | 8       1       247     | 247     3       2457    | 45      9       6       |
 | 347     9       3467    | 14678  *57      145678  | 1458    2       1358    |
 |-------------------------+-------------------------+-------------------------|
 | 27      4       27      | 3       16      9       | 168     5       18      |
 | 6       3       59      | 1278    257     12578   | 1279    4       1279    |
 | 59      8       1       | 267     4       2567    | 2679    3       279     |
 *-----------------------------------------------------------------------------*

The pairs he mentioned are marked with '*'. Look at those pairs and box 6. Either r2c7=7 or r2c7=5 => ... => r6c5=?? Leaving some blank spots there for you to fill in. What is the implication by this on r2c5? ]

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Postby wychwood » Wed Nov 14, 2007 7:19 pm

OK, I followed the first two tips and made progress - but got to the point below and am stuck again.
Code: Select all
 
 *--------------------------------------------------*
 | 49   6    49   | 7    8    2    | 3    1    5    |
 | 15   2    58   | 16   9    3    | 7    68   4    |
 | 13   7    38   | 5    16   4    | 29   68   29   |
 |----------------+----------------+----------------|
 | 34   5    346  | 9    2    168  | 148  7    138  |
 | 8    1    2    | 4    3    7    | 5    9    6    |
 | 7    9    346  | 168  5    168  | 148  2    138  |
 |----------------+----------------+----------------|
 | 2    4    7    | 3    16   9    | 168  5    18   |
 | 6    3    59   | 128  7    158  | 129  4    29   |
 | 59   8    1    | 26   4    56   | 269  3    7    |
 *--------------------------------------------------*

I note that in his initial response, re'born said this reduces to an XY chain - but I am not sure what one of those is. But, if it is some form of forcing chain, then it is a technique that I will try to avoid using (because I don't regard it as a logic-based technique, but rather an up-market form of 'trial and eror' or even guessing).

So, is there another solving technique that can solve this problem?
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Postby RW » Wed Nov 14, 2007 8:09 pm

Code: Select all
 *--------------------------------------------------*
 | 49   6    49   | 7    8    2    | 3    1    5    |
 |A15   2    58   |B16   9    3    | 7    68   4    |
 | 13   7    38   | 5   C16   4    |*29   68  *29   |
 |----------------+----------------+----------------|
 | 34   5    346  | 9    2    168  | 148  7    138  |
 | 8    1    2    | 4    3    7    | 5    9    6    |
 | 7    9    346  | 168  5    168  | 148  2    138  |
 |----------------+----------------+----------------|
 | 2    4    7    | 3   D16   9    | 168  5    18   |
 | 6    3    59   | 128  7    158  |-129  4   *29   |
 |-59   8    1    | 26   4   E56   | 269  3    7    |
 *--------------------------------------------------*

An XY-chain is a chain of bivalue cells only, such that if one end of the chain doesn't have a certain value, the other end must have that value. Above an XY-chain is marked A-E. If A<>5 => E=5 (and if E<>5 => A=5). Any cell that can see both A and E must not be 5.

There's also a UR type 1 in r38c79 that may be used to solve the puzzle.

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Postby wychwood » Thu Nov 15, 2007 1:34 pm

Thanks RW, I think I feel that the UR is slightly more logic-based solution than a forcing chain (or XY chain), only because the former expolits the key facet of a puzzle and the desire for a 'unique' solution, whereas a chain still ahs to start with a hypothesis (i.e. IF cell is X, then what happens?) to find the other cell at the end.

Thanks for the tips anyway.
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help understanding ALS?

Postby stumble » Sat Nov 24, 2007 8:51 pm

Wychwood’s starting problem

Code: Select all
.------------------------.------------------------.------------------------.
| 1459    146     4569   | 124679  8       12467  | 3       169     12579  |
| 1359    2       35689  | 13679   1679    1367   | 156789  1689    4      |
| 1349    7       34689  | 5       1269    12346  | 12689   1689    1289   |
:------------------------+------------------------+------------------------:
| 1234    5       2346   | 124689  1269    12468  | 1489    7       1389   |
| 8       14      247    | 12479   3       12457  | 1459    19      6      |
| 1347    9       3467   | 14678   1567    145678 | 1458    2       1358   |
:------------------------+------------------------+------------------------:
| 2347    348     23478  | 123678  1267    9      | 12678   5       12378  |
| 6       38      235789 | 12378   1257    123578 | 12789   4       123789 |
| 23579   38      1      | 23678   4       235678 | 26789   3689    23789  |
'------------------------'------------------------'------------------------'


After he applied an ALS to box 1
Code: Select all
.------------------------.------------------------.------------------------.
| 49      6       459    | 2479    8       247    | 3       1       57     |
| 1359    2       358    | 1679    79      1367   | 57      68      4      |
| 134     7       348    | 5       16      1346   | 29      68      29     |
:------------------------+------------------------+------------------------:
| 234     5       2346   | 124689  29      12468  | 148     7       138    |
| 8       1       247    | 247     3       2457   | 45      9       6      |
| 347     9       3467   | 14678   57      145678 | 1458    2       1358   |
:------------------------+------------------------+------------------------:
| 27      4       27     | 3       16      9      | 168     5       18     |
| 6       3       59     | 1278    257     12578  | 1279    4       1279   |
| 59      8       1      | 267     4       2567   | 2679    3       279    |
'------------------------'------------------------'------------------------'


I’m trying to duplicate his ALS, is this right?:
Set A R1C3 {4,5,9}
Set B R2C1 {1,3,5,9}
X=9 (restricted common)
Z=5 (restricted common)
R1C1<>5
R3C1<>9
R2C3<>9
R3C3<>9

If this isn’t correct, where am I wrong? Could you write out the correct way to get his result? I don’t find ALS-XZ very intuitive, even after looking at the excellent example
http://www.sudopedia.org/wiki/ALS-XZ
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Postby Ruud » Sat Nov 24, 2007 9:34 pm

Here is the ALS-XZ move suggested by re'born:

Code: Select all
.------------------------.------------------------.------------------------.
| 49      6       459    | 2479    8       247    | 3       1       57     |
| 1359    2       358    | 1679   A79      1367   | 57      68      4      |
| 134     7       348    | 5       16      1346   | 29      68      29     |
:------------------------+------------------------+------------------------:
| 234     5       2346   | 124689 A29      12468  | 148     7       138    |
| 8       1       247    |B247     3      B2457   |B45      9       6      |
| 347     9       3467   | 14678   5-7     145678 | 1458    2       1358   |
:------------------------+------------------------+------------------------:
| 27      4       27     | 3       16      9      | 168     5       18     |
| 6       3       59     | 1278    257     12578  | 1279    4       1279   |
| 59      8       1      | 267     4       2567   | 2679    3       279    |
'------------------------'------------------------'------------------------'

set A: r24c5 with digits 279 (2 cells, 3 digits)
set B: r5c467 with digits 2457 (3 cells, 4 digits)
restricted common: 2 (each 2 in set A can see each 2 in set B)
r6c5<>7 (it can see all 7's in sets A and B)

The reason your ALS is not correct:

Set A R1C3 {4,5,9} (1 cell, 3 digits, not an ALS)
Set B R2C1 {1,3,5,9} (1 cell, 4 digits, not an ALS)

An ALS is defined as a group of N cells with candidates for N+1 digits.
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Re: help understanding ALS?

Postby Sudtyro » Sat Nov 24, 2007 11:29 pm

stumble wrote:I’m trying to duplicate his ALS, is this right?:
Set A R1C3 {4,5,9}
Set B R2C1 {1,3,5,9}


Except for a bivalue cell, an ALS must have more than one cell, so I’m not sure about your ALS notation. As near as I can tell...

Starting grid after basics (no wings or fish):
Code: Select all
+----------------+--------------------+----------------+
|  4-59  6  459  |    2479   8    247 |     3   1   57 |
|  1359  2 358-9 |    1679  79   1367 |    57  68    4 |
|   134  7  348  |       5  16   1346 |    29  68   29 |
+----------------+--------------------+----------------+
|   234  5 2346  |  124689  29  12468 |   148   7  138 |
|     8  1  247  |     247   3   2457 |    45   9    6 |
|   347  9 3467  |   14678  57 145678 |  1458   2 1358 |
+----------------+--------------------+----------------+
|    27  4   27  |       3  16      9 |   168   5   18 |
|     6  3   59  |    1278 257  12578 |  1279   4 1279 |
|    59  8    1  |     267   4   2567 |  2679   3  279 |
+----------------+--------------------+----------------+


There is an X-chain, a single-digit method using an alternating inference chain (AIC), to first remove the 5 in r1c1:
(5): r1c9 = r6c9 – r6c5 = r8c5 – r8c3 = r9c1 => r1c1 <> 5.

After that elimination, there is an XYZ-wing: (459)r1c13|r8c3 => r2c3 <> 9.

The XYZ-wing is equivalent to either of two ALS-XZ rule moves:

Option 1:
Set A: r1c1 {4,9}
Set B: r18c3 {4,5,9}
X=4 (restricted common)
Z=9
r2c3 <> 9
AIC equivalent: (9=4)r1c1 – (4=59)r18c3 => r2c3 <> 9

Option 2:
Set A: r1c13 {4,5,9}
Set B: r8c3 {5,9}
X=5 (restricted common)
Z=9
r2c3 <> 9
AIC equivalent: (9=45)r1c13 – (5=9)r8c3 => r2c3 <> 9

[Edit to add: At this point the grid position matches Wychwood's, but I don't know what he used for his "one ALS in house 1."
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Postby Carcul » Mon Nov 26, 2007 1:26 pm

After the basic steps we can have

[r2c5]-7-[r6c5]-5-[r6c9]=5=[r1c9]=7=[r2c7]-7-[r2c5], => r2c5<>7;

then the UR solves the puzzle.
Last edited by Carcul on Wed Nov 28, 2007 5:46 am, edited 1 time in total.
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Postby ArkieTech » Mon Nov 26, 2007 4:57 pm

carcul said:
[r2c5]-7-[r6c5]-5-[r6c9]=5=[r1c6]=7=[r2c7]-7-[r2c5], => r2c5<>7;



Welcome back! I am looking forward to some nice lessons.:D

shouldn't it be r1c9?

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Re: help understanding ALS?

Postby stumble » Mon Nov 26, 2007 4:58 pm

Sudtyro wrote:Except for a bivalue cell, an ALS must have more than one cell, so I’m not sure about your ALS notation. As near as I can tell...

Starting grid after basics (no wings or fish):
Code: Select all
+----------------+--------------------+----------------+
|  4-59  6  459  |    2479   8    247 |     3   1   57 |
|  1359  2 358-9 |    1679  79   1367 |    57  68    4 |
|   134  7  348  |       5  16   1346 |    29  68   29 |
+----------------+--------------------+----------------+
|   234  5 2346  |  124689  29  12468 |   148   7  138 |
|     8  1  247  |     247   3   2457 |    45   9    6 |
|   347  9 3467  |   14678  57 145678 |  1458   2 1358 |
+----------------+--------------------+----------------+
|    27  4   27  |       3  16      9 |   168   5   18 |
|     6  3   59  |    1278 257  12578 |  1279   4 1279 |
|    59  8    1  |     267   4   2567 |  2679   3  279 |
+----------------+--------------------+----------------+


There is an X-chain, a single-digit method using an alternating inference chain (AIC), to first remove the 5 in r1c1:
(5): r1c9 = r6c9 – r6c5 = r8c5 – r8c3 = r9c1 => r1c1 <> 5.

I am amazed that I can intuitively understand AIC, using
http://www.sudopedia.org/wiki/Alternating_Inference_Chain
as a helping tool. (If A is false then B is true, if B is true then C is false, etc) helped me to understand.


After that elimination, there is an XYZ-wing: (459)r1c13|r8c3 => r2c3 <> 9.

The XYZ-wing is equivalent to either of two ALS-XZ rule moves:

Option 1:
Set A: r1c1 {4,9}
Set B: r18c3 {4,5,9}
X=4 (restricted common)
Z=9
r2c3 <> 9

AIC equivalent: (9=4)r1c1 – (4=59)r18c3 => r2c3 <> 9

But that AIC baffles me. I don't understand the 9=4 notation, to start off with, could you clarify what it says, in English? and also the (4=59)?


Option 2:
Set A: r1c13 {4,5,9}
Set B: r8c3 {5,9}
X=5 (restricted common)
Z=9
r2c3 <> 9
AIC equivalent: (9=45)r1c13 – (5=9)r8c3 => r2c3 <> 9

[Edit to add: At this point the grid position matches Wychwood's, but I don't know what he used for his "one ALS in house 1."
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Re: help understanding ALS?

Postby Sudtyro » Mon Nov 26, 2007 6:38 pm

stumble wrote:AIC equivalent: (9=4)r1c1 – (4=59)r18c3 => r2c3 <> 9

But that AIC baffles me. I don't understand the 9=4 notation, to start off with, could you clarify what it says, in English? and also the (4=59)?


The (9=4)r1c1 is just shorthand notation for the strong-inference link between those two digits. You could just as well write (9)r1c1 = (4)r1c1.

(4=59)r18c3 is the Eureka notation for an ALS embedded in a chain, where the first and last digits are the ones that weakly link to other digits on either side of the ALS. Note also that the weak links to (9)r2c3 at the two ends of the AIC are implied and therefore not explicitly shown.

Many definitions and examples of AIC's, ALS's, etc. can be found in:
http://www.sudopedia.org/wiki/Main_Page.
See also http://www.sudopedia.org/wiki/Eureka for several examples of the Eureka notation used above.
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