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tough one

 
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garytorborg



Joined: 19 Jan 2011
Posts: 28

PostPosted: Tue Feb 22, 2011 7:32 pm    Post subject: tough one Reply with quote

I tried the techniques that I know so far, but I still have WAY too many candidates in this messy puzzle. After "basics," I am at this position:

Code:

*-----*-----*-----*-----*-----*-----*-----*-----*-----*
* 1236*  4  * 1567*12378*  9  *12358* 5678* 567 * 678 *
* 39  * 3579* 57  * 378 * 345 *  6  *  1  *  2  * 478 *
* 126 * 157 *  8  * 127 * 245 * 1245* 4567*  9  *  3  *
*-----*-----*-----*-----*-----*-----*-----*-----*-----*
* 1469* 179 * 1467* 39  *  8  * 34  *  2  * 1467*  5  *
* 149 * 189 *  2  *  5  *  6  *  7  *  3  * 14  * 1489*
*  5  * 789 *  3  * 29  *  1  * 24  * 4678* 467 *46789*
*-----*-----*-----*-----*-----*-----*-----*-----*-----*
*  7  *  6  * 145 * 1238* 235 *12358*  9  * 1345* 14  *
*  8  *  2  *  9  *  4  * 35  * 135 * 567 *13567* 167 *
* 134 * 135 * 145 *  6  *  7  *  9  * 45  *  8  *  2  *
*-----*-----*-----*-----*-----*-----*-----*-----*-----*



Some notes:

I have never before wound up, after basics, with this many 5-candidate cells. In fact, it's very very rare for me to have more than one cell in the entire grid with 5 candidates.

My "basics" include the usual search for hidden singles (I use "squeezing"), naked doubles, triples, and quads. No, I'm not worth a crap at finding hidden doubles and triples, and that may be my problem here. With the basics, I was only able to solve 3 cells.

In this puzzle, I note the shortage of bivalue cells. I count 8 of them; usually there are between 13 and 18 such cells. And in this puzzle, unless I'm just not seeing things today, not a single one of them can be used as pivot or pincer for an xy-wing or an xyz-wing.

In fact, that's another notable about this puzzle: I have found no wings that result in any eliminations at all. There are 2 finned x-wing patterns in the puzzle but neither result in any eliminations whatsoever.

I have also searched for URs and W-wings, which I suspected of being there because of the pair of bivalues: 39 in r2c1 and r4c4, and 14 in r5c8 and r7c9. I'm certain there's something I can use those pairs for, but I haven't yet found it.

Any ideas?


Last edited by garytorborg on Tue Feb 22, 2011 10:27 pm; edited 1 time in total
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Feb 22, 2011 10:18 pm    Post subject: Reply with quote

There are a few basics remaining, none of which do any good that I can see.

R7c3<>1 due to locked candidates
R4c2<>8 due to 8 being solved in that row
R3c6<>4 due to locked candidates
R4c8<>6 due to locked candidates

There's an XYZ Wing, r5c2<>1
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garytorborg



Joined: 19 Jan 2011
Posts: 28

PostPosted: Tue Feb 22, 2011 10:26 pm    Post subject: Reply with quote

Marty R. wrote:
There are a few basics remaining, none of which do any good that I can see.

R7c3<>1 due to locked candidates
R4c2<>8 due to 8 being solved in that row
R3c6<>4 due to locked candidates
R4c8<>6 due to locked candidates

There's an XYZ Wing, r5c2<>1


Actually, thanks for this help. I have not yet completed the puzzle, but each of the above points did accomplish SOME stuff:

The first, third, and fourth eliminations created two pairs and a triple that I can use to expand my search for wings of various kinds.

The second one is actually an error. R4C2 should have "179" not 189 as previously noted. I will edit my post above to reflect that change. Hopefully that will further help you all help me.

One last question on the above: you posted the elimination from the XYZ wing but I cannot see the wing itself. Where is the pivot and pincers?

Thanks!
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Wed Feb 23, 2011 12:14 am    Post subject: Reply with quote

After basics:
Code:
+-------------------+-------------------+-------------------+
| 1236  4     1567  | 12378 9     12358 | 5678  567   678   |
| 39    3579  57    | 378   345   6     | 1     2     478   |
| 126   157   8     | 127   245   125   | 4567  9     3     |
+-------------------+-------------------+-------------------+
| 1469  179   1467  | 39    8     34    | 2     147   5     |
| 149   189   2     | 5     6     7     | 3     14    1489  |
| 5     789   3     | 29    1     24    | 4678  467   46789 |
+-------------------+-------------------+-------------------+
| 7     6     45    | 1238  235   12358 | 9     1345  14    |
| 8     2     9     | 4     35    135   | 567   13567 167   |
| 134   135   145   | 6     7     9     | 45    8     2     |
+-------------------+-------------------+-------------------+
There is nothing that I can see.

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Wed Feb 23, 2011 1:19 am    Post subject: Reply with quote

Quote:
One last question on the above: you posted the elimination from the XYZ wing but I cannot see the wing itself. Where is the pivot and pincers?

Gary, when I first saw this post, r4c2 contained 189 and the 8 got removed because 8 was already solved in that row. There is no wing that I can see in the current grid.
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Wed Feb 23, 2011 1:20 am    Post subject: Reply with quote

You've tackled a puzzle that's far outside the techniques you use.

It's possible that a deadly pattern exists that'll crack the puzzle. That's beyond what my solver can handle. However, my solver found two chains that open it up for many steps.

Code:
(7=5)r2c3 - (5=4)r7c3 - r9c13 = r9c7 - r3c7 = (4-8)r2c9 = (8)r2c4   =>  r2c4<>7
(8)r2c4 = (8-4)r2c9 = (4-6)r3c7 = (6-2)r3c1 = (2-3)r1c1 = (3)r1c46  =>  r2c4<>3

If you'd like a faster solution, assume r2c5=4:

Code:
 +-----------------------------------------------------------------------+
 |  1236   4      1567   |  12378  9      12358  |  5678   567    678    |
 |  39     3579   57     |  378    345    6      |  1      2      478    |
 |  126    157    8      |  127    245    125    |  4567   9      3      |
 |-----------------------+-----------------------+-----------------------|
 |  1469   179    1467   |  39     8      34     |  2      147    5      |
 |  149    189    2      |  5      6      7      |  3      14     1489   |
 |  5      789    3      |  29     1      24     |  4678   467    46789  |
 |-----------------------+-----------------------+-----------------------|
 |  7      6      45     |  1238   235    12358  |  9      1345   14     |
 |  8      2      9      |  4      35     135    |  567    13567  167    |
 |  134    135    145    |  6      7      9      |  45     8      2      |
 +-----------------------------------------------------------------------+
 # 110 eliminations remain

(4)r2c5 - r3c56 = r3c7 - (4=5)r9c7 - r9c23 = r7c3 - (5=7)r2c3 ...

this forces r2c2=5, r3c2=1, and r1c3=6 ...

At this point, you've eliminated all 6s in [box 3] and have a contradiction. Thus, you can deduce that r2c5<>4 ... and the puzzle cracks with Singles.
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garytorborg



Joined: 19 Jan 2011
Posts: 28

PostPosted: Wed Feb 23, 2011 5:33 pm    Post subject: Reply with quote

Thanks, everyone. Whew! I'm going to have to brush up on my chains as a technique for solving tough puzzles like this one.
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Sun Mar 27, 2011 3:46 am    Post subject: Reply with quote

Gary Torborg
I totally agree with your point stated in the other discussion “I disagree that any puzzles have no solution other than guessing”. At this discussion, I don’t think daj95376’s technique will satisfy you. To me it is a kind of forcing net . I think a method I used in another puzzle posted in sudokuxtra forum to solve this puzzle perhaps can satisfy you better. I am looking forward to hearing comments from all of you, especially from Keith and Marty.
The puzzle :
Code:
+-----------------------------------------------------------------------+
 |  1236   4      1567   |  12378  9      12358  |  5678   567    678    |
 |  39     3579   57     |  378    345    6      |  1      2      478    |
 |  126    157    8      |  127    245    125    | -4567   9      3      |
 |-----------------------+-----------------------+-----------------------|
 |  1469   179    1467   |  39     8      34     |  2      147    5      |
 |  149    189    2      |  5      6      7      |  3      14     1489   |
 |  5      789    3      |  29     1      24     |  4678   467    46789  |
 |-----------------------+-----------------------+-----------------------|
 |  7      6      45     |  1238   235    12358  |  9      1345   14     |
 |  8      2      9      |  4      35     135    |  567    13567  167    |
 |  134Y   135Y   145Y   |  6      7      9      |  45Y    8      2      |
 +-----------------------------------------------------------------------+

Set a Yellow Zone : r9c1=134, r9c2=135, r9c3=145, r9c7=45(marked with Y)
[Note that if r9c1<>4, r9c3<>4 and r9c7<>4 then - contradiction]
The step not allowed : r3c7=4 .(due to technical reasons some details will be followed in the next post)
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Sun Mar 27, 2011 3:50 am    Post subject: Reply with quote

Gary Torborg
(Continued)

The logic : r3c7=4 => r9c7<4>r2c9=78,
r3c7=4 =>r3c1=6=>r1c3=157=>(a naked triple 157 in box 1)=>r2c2=39=>(a naked pair 39 in row2)=>r2c4=78=>(a naked pair 78 in row2)=>r2c3=5=>r7c3=4
=>r9c1<>4 and r9c2<>4.
The result: r3c7<>4 and the puzzle can then be solved by singles.
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Sun Mar 27, 2011 4:12 am    Post subject: Reply with quote

Gary Torborg

Quote:
The logic : r3c7=4 => r9c7<4>r2c9=78,


Should be replaced as "The logic : r3c7=4 => r9c7<>4;"(and r3c7=4=>r2c9=78.)
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Sun Mar 27, 2011 4:58 am    Post subject: Reply with quote

It took a bit of deciphering, but I did manage to create a close clone to Koala's solution.

Code:
 +-----------------------------------------------------------------------+
 |  1236   4      1567   |  12378  9      12358  |  5678   567    678    |
 |  39     3579   57     |  378    345    6      |  1      2      478    |
 |  126    157    8      |  127    245    125    |  4567   9      3      |
 |-----------------------+-----------------------+-----------------------|
 |  1469   179    1467   |  39     8      34     |  2      147    5      |
 |  149    189    2      |  5      6      7      |  3      14     1489   |
 |  5      789    3      |  29     1      24     |  4678   467    46789  |
 |-----------------------+-----------------------+-----------------------|
 |  7      6      45     |  1238   235    12358  |  9      1345   14     |
 |  8      2      9      |  4      35     135    |  567    13567  167    |
 |  134    135    145    |  6      7      9      |  45     8      2      |
 +-----------------------------------------------------------------------+
 # 110 eliminations remain

Consider this initial chain segment that leaves [band 1] as shown:

Code:
(4)r9c7 = r9c13 - (4=5)r7c3 - (5=7)r2c3

 +-----------------------------------------------------------------------+
 |  1236   4      156    |  12378  9      12358  |  5678   567    678    |
 |  39     359    7      |  38     345    6      |  1      2      48     |
 |  126    15     8      |  127    245    125    |  4567   9      3      |
 |-----------------------+-----------------------+-----------------------|

Now, consider this AIC in the resulting [band 1]:

Code:
(6)r3c7 = r3c1 - (6=15)r1c3,r3c2 - (5=39)r2c21 - (3=8)r2c4 - (8=4)r2c9  =>  r3c7<>4

If r9c7<>4 and r3c7<>6, then we get r3c7<>4. If r9c7=4 or r3c7=6, then we get r3c7<>4.

Whew!!!
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Sun Mar 27, 2011 5:14 am    Post subject: Reply with quote

Very nice.
Quote:
r3c7<>4

A backdoor solution. With, I suspect, an "a posteriori" (after the fact) justification.

Can anyone explain how or why a human using pencil and paper would find this?

Keith

Personal to Marty: I apologize for my current Italian phase.
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Sun Mar 27, 2011 8:45 am    Post subject: Reply with quote

Daj95376 and Keith
Thank you so much for giving me an immediate and helpful response. First, Daj95376 is so nice and clever to let me know a better way to get r3c7<>4. Please take a look at my story behind the work and give me your comments again.
Secondly, Keith has enough reasons to suspect my work being a justification of a backdoor found. I must confess that this is not my first version of the work, it did come afterwards, but not the way you think. My story is that I did this in 3 steps (shown in the following) and then I found it can be condensed in one step.

YZ-method (1)
Code:
+-----------------------------------------------------------------------+
 |  1236   4      156    |  12378  9      12358  |  5678   567    678    |
 |  39     357*9  57Y    |  37*8    345    6     |  1      2      47*8   |
 |  126    157*   8      |  127    245    125    |  4^567  9      3      |
 |-----------------------+-----------------------+-----------------------|
 |  1469   179    14^67* |  39     8      34     |  2      147    5      |
 |  149    189    2      |  5      6      7      |  3      14     1489   |
 |  5      789    3      |  29     1      24     |  4678   467    46789  |
 |-----------------------+-----------------------+-----------------------|
 |  7      6      45Y    |  1238   235    12358  |  9      1345   14     |
 |  8      2      9      |  4      35     135    |  567    13567  167    |
 |  134^   135    14^5   |  6      7      9      |  45^    8      2      |
 +-----------------------------------------------------------------------+
Set a Yellow Zone : r2c3=57, r7c3=45(marked with Y)
[Note that if r2c3<>7, r7c3<>4 then - contradiction]
Note that the presence of any candidate with ^ will cause r7c3<>4 and also the presence of any candidate with * will cause r2c3<>7 .
The step not allowed: r2c9=7. As we see that if 7 in r2c9 then 4 in r3c7.
YZ-method (2)
Code:
+-----------------------------------------------------------------------+
 |  1236   4      156    |  12378  9      12358  |  5678   567    678    |
 |  39     357*9  57Y    |  37*8    345    6     |  1      2      48     |
 |  126    157*   8      |  127    245    125    |  4^567  9      3      |
 |-----------------------+-----------------------+-----------------------|
 |  1469   179    14^67* |  39     8      34     |  2      147    5      |
 |  149    189    2      |  5      6      7      |  3      14     1489   |
 |  5      789    3      |  29     1      24     |  4678   467    46789  |
 |-----------------------+-----------------------+-----------------------|
 |  7      6      45Y    |  1238   235    12358  |  9      1345   14     |
 |  8      2      9      |  4      35     135    |  567    13567  167    |
 |  134^   135    14^5   |  6      7      9      |  45^    8      2      |
 +-----------------------------------------------------------------------+
Set a Yellow Zone : r2c3=57, r7c3=45(marked with Y)
[Note that if r2c3<>7, r7c3<>4 then - contradiction]
Note that the presence of any candidate with * will cause r2c3<>7 and also
the presence of any candidate with ^ will cause r7c3<>4.
The step not allowed: r2c4=7. As we see that if 7 in r2c4 then 8 in r2c8 and 4 in r3c7..
YZ-method (3)
Code:
 +-----------------------------------------------------------------------+
 |  1236   4      156    |  12378  9      12358  |  5678   567    678    |
 |  3*9    3*579  57     |  38Y     3*4^5  6     |  1      2      48Y    |
 |  126    15     8      |  127    245    125    | 4^567   9      3      |
 |-----------------------+-----------------------+-----------------------|
 |  1469   179    1467   |  39     8      34     |  2      147    5      |
 |  149    189    2      |  5      6      7      |  3      14     14^89  |
 |  5      789    3      |  29     1      24     |  4678   467    4^6789 |
 |-----------------------+-----------------------+-----------------------|
 |  7      6      45     |  1238   235    12358  |  9      1345   14^    |
 |  8      2      9      |  4      35     135    |  567    13567  167    |
 |  134    135    145    |  6      7      9      |  45     8      2      |
 +-----------------------------------------------------------------------+
Set a Yellow Zone : r2c4=38, r2c9=48(marked with Y)
[Note that if r2c4<>3, r2c9<>4 then - contradiction]
Note that the presence of any candidate with * will cause r2c4<>3 and also
the presence of any candidate with ^ will cause r2c9<>4.
If r3c7=4 then 4 not in r2c9 .
Also if r3c7=4 then r3c1=6-(naked pair 15 in box 1)-r2c3=7-(naked pair 39 in row2)-r2c4<>3.
The step not allowed: r3c7=4.

Keith, are you satisfied with my story. By the way, I think I have solved the puzzle you mentioned in the discussion under the topic “Necessary and sufficient set of solving techniques” using YZ method. I will post it soon and I am looking forward to your comments then.
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Sun Mar 27, 2011 6:45 pm    Post subject: Reply with quote

There is a bit too much "if this is true then that happens" in this YZ-method concept for my taste. It seems to me that this is replicating (multi)coloring with transports. If so, I prefer the coloring because it is more mechanical without requiring strings of if-then statements. Let's look at this same thing using coloring, starting with the weak inference <5>s in those c3 bivalues, which I will color "a" and "b":
Code:
+-----------------------------------------------------------------------+
|  1236   4      156    |  12378  9      12358  |  5678   567    678    |
|  39     3579   5a7A   | @3-78*  345    6      |  1      2     #4*-78  |
|  126    157    8      |  127    245    125    |  4567   9      3      |
|-----------------------+-----------------------+-----------------------|
|  1469   179    1467   |  39     8      34     |  2      147    5      |
|  149    189    2      |  5      6      7      |  3      14     1489   |
|  5      789    3      |  29     1      24     |  4678   467    46789  |
|-----------------------+-----------------------+-----------------------|
|  7      6      4B5b   |  1238   235    12358  |  9      1345   14     |
|  8      2      9      |  4      35     135    |  567    13567  167    |
|  134    135    145    |  6      7      9      |  4B5b   8      2      |
+-----------------------------------------------------------------------+

I have filled in the small Aa and Bb clusters, using the grouped links in b7 to color the candidates in r9c7. AB is the strong inference pair.

Now, 4B in r9c7 can be transported to 4B in r2c9 (#) eliminating <7> there. I would not actually mark a color transport so have used * to make the transport more apparent in the grid above. Due to the conjugate <8>s in r2, I can continue and transport 4B in r2c9 to 8B in r2c4 (@) eliminating <7> there as well. This has quickly replicated the first two YZ-method cases.

Now we base a new cluster on the conjugate <8>s in r2:
Code:
+-----------------------------------------------------------------------+
|  123c6  4      156    |  12378  9      12358  |  5678   567    678    |
|  39     3579   5a7A   |  3c8C   34c5   6      |  1      2      4C8c   |
|  12*6   15     8      |  127    24C5   125    | $4c56*7 9      3      |
|-----------------------+-----------------------+-----------------------|
|  1469   179    1467   |  39     8      34     |  2      147    5      |
|  149    189    2      |  5      6      7      |  3      14     1489   |
|  5      789    3      |  29     1      24     |  4678   467    46789  |
|-----------------------+-----------------------+-----------------------|
|  7      6      4B5b   |  1238   235    12358  |  9      1345   14     |
|  8      2      9      |  4      35     135    |  567    13567  167    |
|  134    135    145    |  6      7      9      |  4B5b   8      2      |
+-----------------------------------------------------------------------+

I have left the Aa and Bb markings in place since these were actual color markings. (The weak inference <4>s in c7 provide us with a bC strong pair and, by inference with the AB strong pair, an AC strong pair as well. However, these are not of immediate use.) I will focus only on the Cc cluster.

Due to the grouped link in b1, I can mark 3c in r1c1. Due to the conjugate <2>s in c1, 3c in r1c1 is transported to 2c in r3c1. Then, the conjugate <6>s in r3 continue the transport to 6c in r3c7 ($). But there cannot be two "c" values in r3c7 so all "c" values are false. This replicates the third YZ-method case. (It also avoided all those messy ALS sequences!)
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Mon Mar 28, 2011 7:05 am    Post subject: Reply with quote

Hi Asellus

Quote:
Asellus wrote :
There is a bit too much "if this is true then that happens" in this YZ-method concept for my taste.


Thanks for your opinion. Yet I think in any logical reasoning, you can do nothing without using “if ... then …” no matter you mention explicitly or just in mind. I am sure you know very well your coloring transport method is supported by many “if … then …”.
May be you think it is too obvious to mention, but just do something good for the beginners, any objection?
To me, your way in doing the puzzle can be expressed in the following 3 lines:

(7=5)r2c3-(5=4)r7c3-(4=135)r9c123-(5=4)r9c7-(4)r3c7=(4-7)r2c9
(7=5)r2c3-(5=4)r7c3-(4=135)r9c123-(5=4)r9c7-(4)r3c7=(4-8)r2c9=(8-7)r2c4
(4)r3c7-(4=8)r2c9-(8=3)r2c4-(3)r2c12=(3-2)r1c1=(2-6)r3c1=(6-4)r3c7

People call these "Forcing Nets".

Koala
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Mar 28, 2011 11:24 pm    Post subject: Reply with quote

Hi Koala,

These are old and endless debates. Where one comes down in them regarding solving practices is pretty much a personal choice.

Yes, "if-then" reasoning underlies all logical techniques. But that doesn't mean that I am performing "if-then"s in my head when I use a solving technique. I can use an XY-Wing or X-Wing or the like by recognizing the pattern. It is true that to understand how it works (why it is valid) I have to have done some "if-then" thinking initially. But I don't redo that reasoning any longer after I have been convinced. I apply the method mechanically. It is the same with coloring and multi-coloring: I apply rules mechanically without making assumptions or doing any "if-then" thinking.

All coloring (and most anything else in sudoku) can be expressed as alternate implication chains (AICs), which are what I call the things you have notated. "Forcing Net" is a more general term. I do not consider AICs to be forcing because, again, they can be constructed and used without making any truth assumptions or performing any "if-then" thinking. I merely need to know how to recognize weak and strong implication instances within the grid and note occasions where they link up end-to-end in alternating fashion that leads to a loop discontinuity (contradiction).

Not everyone thinks of AICs in this way and so it creates confusion with forcing methods, in my opinion. If people are making true/false assumptions and then doing "if-thens" around the grid in order to discover their AICs, that's fine. But it's not how I do it.

As I said, it's an old debate and a matter of personal preference. However, I do believe that there is a more or less clear divide between basic forcing approaches and those that involve the application of mechanical rules. I prefer to stay on the latter side of the divide, myself.

PS: Years ago I raised objections similar to yours in response to Myth Jellies. Clearly my thinking has evolved since because I now agree with him.
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Tue Mar 29, 2011 4:52 am    Post subject: Reply with quote

Hi Asellus,

Thank you so much for your helpful response! First, I agree that once a device is set up with logic we need not mention “if-then” anymore. Secondly, this is the first time I learn the transports skill from you (it is great). I didn’t understand at the first reading so I mistook it as some kind of forcing. Sorry! In the past, I took (multi)coloring as a pair of conjugate groups with two different colors. You may get some candidates eliminated, or one group disappeared, or nothing happen at all. I feel that the transporting to coloring is far more fruitful than the fins to the fishes. Anyway, thanks a lot! Now I wish to let you know about the YZ method. When I did the VH, I began by looking for X-Wing, fishes, XY- wing and so on. Someday I found myself just like a child looking for the hidden rabbits in a picture. It is no fun anymore! I started with considering naked pairs. Everybody loves it I suppose. It is so natural to eliminate some candidates just because they make a conflict in these two cells. Now, if instead of a naked pair we have, for example, 23 and 25 in different cells in one unit, is it possible to eliminate some candidates just because they make a conflict in these two cells? YZ method is to find candidates that eliminate 3 and 5 separately first and then to see if we are able to find the one which can do both. I did this for quite some time and enjoyed myself. I applied YZ method to do the “tough one” and posted. Someone took it as a justification of a backdoor stuff, someone took it as a replication of something. So I am not going to post controversial things anymore. BACK TO VH FAMILY! If you would be so kind to have a look at the work I did for the 2011-3-27 VH posted I should appreciate very much!

Best regards,

Koala
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Tue Mar 29, 2011 9:41 am    Post subject: Reply with quote

Koala,

Don't worry about posting new ideas that may be "controversial." That's how one learns. Plus, you never know if some new idea may turn out to lead others to discover something useful.

I looked at the 2011-3-27 posting. There are 3 different ways apparent to me for how that same elimination can be derived with standard mechanical rules.

First, an XY-Chain: (5=6)r1c2-(6=9)r1c8-(9=4)r3c8-(4=5)r2c9

Second, as simple coloring. The weak inferences in the XY-Chain above are all conjugate pairs so coloring would propagate along the chain and reveal those pincer <5>s as well. (Multi-coloring and transports are not necessary here.)

Third, as an ALS or XZ technique. The r1c2 bivalue is one ALS. The other is the 3-cell 4569 ALS in box 3. The <6>s in row 1 are shared exclusive and the <5>s are shared common:
(5=6)r1c2 - ALSr13c8|r2c9[(6)r1c8=(5)r2c9]

In general, your "Yellow Zone" method is exploiting the strong inference that exists within the YX and XZ 2-cell ALS: Y=Z. The <X>s have a weak inference and if they are a conjugate pair (as in the 2011-3-27 puzzle) then you can form a single color cluster. Otherwise, you can form two color clusters, one from Y and one from Z with the X-X weak inference serving to "bridge" the two clusters (as in the example above in this thread). Note that any candidate thus "trapped" and eliminated by the coloring produces a contradiction that would result in both Y and Z being false, the desired result of your method but without any truth assumptions or "if-then"s. Seen in this way, there is no reason to limit oneself to a 2-cell ALS composed of two bivalues. Any strong inference in any ALS is a potential germ of such a method.

As for those fish fins, they can be transported too, you know. (I believe that is an example of what some call Kraken Fish... but don't quote me on that!)
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Koala



Joined: 15 Mar 2011
Posts: 8

PostPosted: Tue Mar 29, 2011 3:09 pm    Post subject: Reply with quote

Asellus

I am so happy to learn so many things from you. I just wonder our posts may be off the topic “tough one” and will be banned anytime. I don’t know in what way I can get in touch with you again. I believe I understand all (you are marvelous!) you stated (took quite some time I must confess) but the following :

“the Otherwise, you can form two color clusters, one from Y and one from Z with the X-X weak inference serving to "bridge" the two clusters (as in the example above in this thread). Note that any candidate thus "trapped" and eliminated by the coloring produces a contradiction that would result in both Y and Z being false,”

If you don’t mind could you let me have more details on what you said above under the topic 27March VH?

Thank you

Koala
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