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July 8 VH not VH
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Earl



Joined: 30 May 2007
Posts: 546
Location: St Louis MO

PostPosted: Mon Jul 09, 2007 12:51 am    Post subject: July 8 VH not VH Reply with quote

Since todays VH was not VH I found another one,
which was VH. Anyone see a next step?

Thanks

Earl
Code:

+---------------+-----------------+------------+
| 6   3    29   | 5    129   4    | 8  7  12   |
| 1   258  258  | 6    28    7    | 3  4  9    |
| 4   78   789  | 123  89    123  | 5  12 6    |
+---------------+-----------------+------------+
| 9   1    457  | 8    347   35   | 2  6  347  |
| 27  2478 2478 | 1237 6     123  | 9  5  1347 |
| 257 6    3    | 127  12457 9    | 17 8  147  |
+---------------+-----------------+------------+
| 8   247  1    | 237  237   236  | 46 9  5    |
| 3   2457 2457 | 9    127   1256 | 46 12 8    |
| 257 9    6    | 4    1257  8    | 17 3  127  |
+---------------+-----------------+------------+

Play this puzzle online at the Daily Sudoku site
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Mon Jul 09, 2007 12:18 pm    Post subject: Reply with quote

Take a look where <7> goes in Box 9, then in Box 4.

Keith
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jLo



Joined: 30 Apr 2007
Posts: 55

PostPosted: Tue Jul 10, 2007 12:50 am    Post subject: Reply with quote

Don't know if this is the path of least resistance, but here's
how I approached this one.

After doing the box-line reductions on <7>s in box 7 and box 4
you end up here.

Code:

+---------------+-----------------+------------+
|          29   |      129        |       12   |
|     258  258  |      28         |            |
|     78   789  | 123  89    123  |    12      |
+---------------+-----------------+------------+
|          45   |      3457  35   |       347  |
| 27  248  248  | 1237       123  |       1347 |
| 257           | 127  12457      | 17    147  |
+---------------+-----------------+------------+
|     247       | 237  237   236  | 46         |
|     2457 2457 |      1257  1256 | 46 12      |
| 25            |      125        | 17    127  |
+---------------+-----------------+------------+


Coloring <1>s starting at R1C5 eliminates the 1 at R8C5

Code:

+---------------+-----------------+-------------+
|          29   |     +129        |       -12   |
|     258  258  |      28         |             |
|     78   789  | 123  89    123  |    +12      |
+---------------+-----------------+-------------+
|          45   |      347   35   |        347  |
| 27  248  248  | 1237       123  |        1347 |
| 257           | 127  12457      | 17     147  |
+---------------+-----------------+-------------+
|     247       | 237  237   236  | 46          |
|     2457 2457 |     *127   1256 | 46 -12      |
| 25            |      125        | 17     127  |
+---------------+-----------------+-------------+


which leads to a naked triple in box 8 on <237> which solves R7C6,
R8C7, and R7C7. There is now a hidden pair in box 8 on <15>.
Now we are here -

Code:

+---------------+----------------+------------+
|          29   |      129       |       12   |
|     258  258  |      28        |            |
|     78   789  | 123  89    123 |    12      |
+---------------+----------------+------------+
|          45   |      347   35  |       347  |
| 27  248  248  | 1237       123 |       1347 |
| 257           | 127  12457     | 17    .47  |
+---------------+----------------+------------+
|     27        | 237  237       |            |
|     2457 2457 |      27    15  |    12      |
| 25            |      15        | 17    12.  |
+---------------+----------------+------------+


Examining the Unique Rectangle at R6C7, R6C9, R9C7, R9C9 I determined that

If R6C9=1 or R6C9=7, then R9C9=2 to avoid the UR
If R6C9=4, then R4C5=4 and R4C9=7 and R9C9<>7.

That eliminates 7 from R9C9 and solves R9C7 and R6C7.

Now we are here, and things get interesting -

Code:

+---------------+---------------+----------+
|          29   |      129      |      12  |
|     258  258  |      28       |          |
|     78   789  | 123  89   123 |   12     |
+---------------+---------------+----------+
|          45   |      347  35  |      347 |
| 27  248  248  | 1237      123 |      347 |
| 257           | 27   2457     |      47  |
+---------------+---------------+----------+
|     27        | 237  237      |          |
|     2457 2457 |      27   15  |   12     |
| 25            |      15       |      12  |
+---------------+---------------+----------+


If R9C1=5 then R8C2<>5
and
If R9C1=2 then R7C2=7,R5C1=7 and R6C1=5 and R4C3=4 and R8C3=5, so R8C2<>5

That eliminates 5 from R8C2 and solves R2C2
and brings us here.

Code:

+--------------+---------------+----------+
|         29   |      129      |      12  |
|         28   |      28       |          |
|     78  789  | 123  89   123 |   12     |
+--------------+---------------+----------+
|         45   |      347  35  |      347 |
| 27  248 248  | 1237      123 |      347 |
| 257          | 27   2457     |      47  |
+--------------+---------------+----------+
|     27       | 237  237      |          |
|     247 2457 |      27   15  |   12     |
| 25           |      15       |      12  |
+--------------+---------------+----------+


If R1C9=1 then we have an X-Wing on <2>s in Rows 1 and 2.
this eliminates the <2> from R7C5 and R8C5, leaving R7C4=2
If R1C9=2 then R3C8=1 and R8C8=2 and R9C9=1 and R9C1=2

Since R7C2 sees both R9C1 and R7C4, we know R7C2=7

I try to illustrate this below -

Code:

+--------------+---------------+----------+
|        x29   |     x129      |     +12- |
|        x28   |     x28       |          |
|     78  789  | 123  89   123 |  -12+    |
+--------------+---------------+----------+
|         45   |      347  35  |      347 |
| 27  248 248  | 1237      123 |      347 |
| 257          | 27   .457     |      47  |
+--------------+---------------+----------+
|    *27       |+237  .37      |          |
|     247 2457 |      .7   15  |  +12-    |
|-25+          |      15       |     -12+ |
+--------------+---------------+----------+



After R7C2=7, everything else is singles.

There may be a better path. If so, I'd like to see it.
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Earl



Joined: 30 May 2007
Posts: 546
Location: St Louis MO

PostPosted: Tue Jul 10, 2007 4:31 pm    Post subject: VH not VH Reply with quote

jLo,

Your final major step might be somewhat simplified.
Regardless of whether R1C9 is 1 or 2,
2 is eliminated from R8C5 which then must be 7.

Thanks for all your work.

Earl
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jLo



Joined: 30 Apr 2007
Posts: 55

PostPosted: Tue Jul 10, 2007 8:20 pm    Post subject: Reply with quote

Quote:

Regardless of whether R1C9 is 1 or 2,
2 is eliminated from R8C5 which then must be 7.


Why?
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Earl



Joined: 30 May 2007
Posts: 546
Location: St Louis MO

PostPosted: Tue Jul 10, 2007 10:07 pm    Post subject: July 8 VH not VH Reply with quote

jLo,

I saw it only as an alternative to your last major step, i.e. assuming the fifth grid you posted. If R1C9 is 1 the X-wing holds and R8C5 cannot be 2. If R1C9 is 2 R8C8 is 2 by hard pairs and R8C5 cannot be 2.

I might have done something wrong, but that is how I see it. I am still learning these techniques.

Thanks

Earl
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jLo



Joined: 30 Apr 2007
Posts: 55

PostPosted: Tue Jul 10, 2007 10:37 pm    Post subject: Reply with quote

Quote:

I saw it only as an alternative to your last major step


OK, I see that now. I agree that your method is a little cleaner.

I was zoned in on R7C2 because I had figured out it had to be a
7 and was looking for a way to force it. Prior to that I had figured
out that R7C2 and R8C5 were tightly coupled (either both a 2 or
both a 7), but had forgotten that when can up with the "conditional" X-wing.

Thanks.
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Asellus



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

PostPosted: Tue Jul 10, 2007 10:53 pm    Post subject: Reply with quote

Earl,

Technically, it seems to me that that qualifies as a finned X-Wing since the 2 at R8C5 is eliminated either by the X-Wing or by the fin (R1C9) via the coloring on 2.

We normally think of fins only as cells that provide a direct alternate elimination to that of the X-Wing (or other fish). This example shows that the fin elimination can be mediated by an intervening technique. (An Empty Rectangle might provide a similar opportunity, for instance.)

Worth keeping in mind!

PS: I was able to solve this puzzle without recourse to the UR or to what-ifs. However, it required quite a number of steps (some very interesting). I may work on it some more later to see if I can streamline it. If so, maybe I'll post the steps.
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Asellus



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

PostPosted: Wed Jul 11, 2007 8:06 am    Post subject: I'm Back - Part 1 Reply with quote

jLo,

In your 17 UR, I believe it falls into one of the defined "Type" categories. (Type 6 maybe? I'm terrible at remembering the Type numbers!) It is the type where one side of the UR is part of an ALS. In this case, the ALS comprises R169C9, {1247} in 3 cells. The "external" cell(s), R1C9 in this case, does not contain <7>. Thus, <1> can be eliminated from R69C9.

The rule does not allow the elimination of <7> from R9C9. But, I suspect that that follows quickly.
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Asellus



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

PostPosted: Wed Jul 11, 2007 9:17 am    Post subject: I'm Back - Part 2 Reply with quote

Okay... here it is, another way to the solution without using the 17 UR (as tempting as it may be) and without any novel what-ifs. I tried to streamline it, but it didn't cooperate very much. No doubt someone out there will see a shortcut that eludes me. Nevertheless, there is some interesting scenery along my chosen path. I can't argue that it is a "better path." It's just different.

Starting just after jLo's <1> Coloring, but before the UR argument, I extended the Coloring further to eliminate <1> from R6C5. We are now ready for a Color Wing on <2>:
Code:
+---------------+-----------------+--------------+
| 6   3    29   | 5    129   4    | 8   7   12   |
| 1   258  258  | 6    28    7    | 3   4   9    |
| 4   78   789  | 123  89   B12+3 | 5  b12+ 6    |
+---------------+-----------------+--------------+
| 9   1    45   | 8    347   35   | 2   6   347  |
|#27  248  248  | 1237 6    P12-3 | 9   5   1347 |
| 257 6    3    | 127  2457  9    | 17  8   147  |
+---------------+-----------------+--------------+
| 8   27   1    | 237  237   6    | 4   9   5    |
| 3   2457 2457 | 9    27    15   | 6   12- 8    |
|p2-5 9    6    | 4    15    8    | 17  3   12+7 |
+---------------+-----------------+--------------+

The Bridge, Bb, is assigned + polarity. The Pincers, Pp, have - polarity. This eliminates <2> from the target, #, R5C1. (For those unfamiliar with the Color Wing method, consult the solving guide at sudocue.com, for example. The essential point is that the two pincers must derive from opposing ends of the Bridge and have opposite polarity from the Bridge. It is akin to a common form of the Skyscraper, but with additional Color Wrapping. You can see the Skyscraper in C68. The C8 branch is extended via R9 in this case.)

We aren't done with this Color Wing yet. But first...

An ALS elimination:
Code:
+--------------+---------------+-----------+
| 6  3    29   | 5    129  4   | 8  7  12  |
| 1  258  258  | 6    28   7   | 3  4  9   |
| 4  78   789  | 123  89   123 | 5  12 6   |
+--------------+---------------+-----------+
| 9  1   a45   | 8   b347  35  | 2  6  347 |
| 7  248  248  | 123  6    123 | 9  5  134 |
|a25 6    3    | 127  2457 9   | 17 8  147 |
+--------------+---------------+-----------+
| 8  27   1    | 237 b237  6   | 4  9  5   |
| 3  2457 2457 | 9   b27   15  | 6  12 8   |
| 25 9    6    | 4    15   8   | 17 3  127 |
+--------------+---------------+-----------+

Set a: {245} in R4C3/R6C1 and Set b: {2347} in R478C5.
Shared exclusive = <4>
Shared common = <2>
Eliminates <2> from R6C5.

That momentary diversion brings us back to our Color Wing, which we can now extend further:
Code:
+---------------+----------------+-------------+
| 6   3    29   | 5    129  4    | 8  7   12   |
| 1   258  258  | 6    28   7    | 3  4   9    |
| 4   78   789  |@123  89  B12+3 | 5 b12+ 6    |
+---------------+----------------+-------------+
| 9   1    45   | 8    347  35   | 2  6   347  |
| 7   248  248  |#123  6   P12-3 | 9  5   134  |
| 2+5 6    3    |p12-7 457  9    | 17 8   147  |
+---------------+----------------+-------------+
| 8   27   1    | 237  237  6    | 4  9   5    |
| 3   2457 2457 | 9    27   15   | 6  12- 8    |
| 2-5 9    6    | 4    15   8    | 17 3   12+7 |
+---------------+----------------+-------------+

Now... as jLo says... this is where things get interesting! Not only do we have the usual Color Wing elimination, #. But, we have something special: both pincers are in the same Box. This means that they can't both be <2>. And, this means that one of the Bridge cells MUST be <2>. So, the Bridge collapses and becomes a Strong Link, eliminating the <2> in R3C4, @. Maybe I'm the only one, but... I think that that is rather cool!

Now there are some naked pairs (13 in C4 and 27 in R7). After simplification, we end up here:
Code:
+--------------+------------+-----------+
| 6  3    29   | 5  129 4   | 8  7  12  |
| 1  258  258  | 6  28  7   | 3  4  9   |
| 4  78   789  | 13 89  123 | 5  12 6   |
+--------------+------------+-----------+
| 9  1    45   | 8  47  35  | 2  6  347 |
| 7  248  248  | 13 6   123 | 9  5  134 |
| 25 6    3    | 27 457 9   | 17 8  147 |
+--------------+------------+-----------+
| 8  27   1    | 27 3   6   | 4  9  5   |
| 3  2457 2457 | 9  27  15  | 6  12 8   |
| 25 9    6    | 4  15  8   | 17 3  127 |
+--------------+------------+-----------+

First, an XY-Wing pivoting on R4C6 eliminates <1> from R5C6. At the same time, there is a closed XY-Chain loop in R4C35/R6C14 that eliminates <7> from R6C5 and <4> from R4C9. This is the first time that I've used a closed XY-Chain loop! (If I keep this up, I'm going to be dangerous some day.)

Continue to ignore that 17 UR behind the curtain; we're almost there.

A couple of XY-Chains come next:
Code:
+--------------+-----------+-----------+
| 6  3    29   | 5  129 4  | 8  7  12  |
| 1  25-8 25-8 | 6 p28  7  | 3  4  9   |
| 4 p78   789  | 3  89  12 | 5  12 6   |
+--------------+-----------+-----------+
| 9  1    45   | 8  47  35 | 2  6  37  |
| 7  248  248  | 1  6   23 | 9  5  34  |
| 25 6    3    | 27 45  9  | 17 8  147 |
+--------------+-----------+-----------+
| 8 c27   1    |c27 3   6  | 4  9  5   |
| 3  2457 2457 | 9 c27  15 | 6  12 8   |
| 25 9    6    | 4  15  8  | 17 3  127 |
+--------------+-----------+-----------+

The chain cells, c, end at pincers, p, eliminating <8> from R2C23. Next...
Code:
+---------------+-----------+-----------+
| 6  3     9    | 5   12 4  | 8  7  12  |
| 1 p25    25   | 6   8  7  | 3  4  9   |
| 4  78    78   | 3   9  12 | 5  12 6   |
+---------------+-----------+-----------+
| 9  1     45   | 8  c47 35 | 2  6  37  |
| 7  248   248  | 1   6  23 | 9  5  34  |
|c25 6     3    | 27 c45 9  | 17 8  147 |
+---------------+-----------+-----------+
| 8 c27    1    |c27  3  6  | 4  9  5   |
| 3  24-57 2457 | 9  c27 15 | 6  12 8   |
|p25 9     6    | 4   15 8  | 17 3  127 |
+---------------+-----------+-----------+

The pincers eliminate <5> from R8C2. Now, R2C2 = 5, etc.

We end up here for one last step:
Code:
+-------------+----------+-----------+
| 6  3    9   | 5  12 4  | 8  7  12  |
| 1  5    2   | 6  8  7  | 3  4  9   |
| 4  78   78  | 3  9  12 | 5  12 6   |
+-------------+----------+-----------+
| 9  1    45  | 8 c47 35 | 2  6  37  |
| 7  -248 48  | 1  6  23 | 9  5  34  |
|p25 6    3   |c27c45 9  | 17 8  147 |
+-------------+----------+-----------+
| 8 p27   1   |c27 3  6  | 4  9  5   |
| 3  247  457 | 9  27 15 | 6  12 8   |
| 25 9    6   | 4  15 8  | 17 3  127 |
+-------------+----------+-----------+

Another XY-Chain! The pincers eliminate <2> from R5C2.

At this point, if the solution were a snake, it would bite you.
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Mogulmeister



Joined: 03 May 2007
Posts: 695

PostPosted: Wed Jul 11, 2007 11:14 am    Post subject: The easy way out Reply with quote

Did anyone spot this early possibility right at the start of Earl's puzzle ?

Notice that as well as the strong links elimination on <1> that jLo pointed out earlier, we can use the same structure to eliminate <2> as well.

What am I saying ? Simply put, that same structure pictured below eliminates both 1 and 2 from r8c5 which then turns the puzzle to toast.

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jLo



Joined: 30 Apr 2007
Posts: 55

PostPosted: Wed Jul 11, 2007 12:50 pm    Post subject: Reply with quote

Quote:

Notice that as well as the strong links elimination on <1> that jLo pointed out earlier, we can use the same structure to eliminate <2> as well.


I don't think that works because of the 2 at R1C3. Did I miss something?
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Mogulmeister



Joined: 03 May 2007
Posts: 695

PostPosted: Wed Jul 11, 2007 1:10 pm    Post subject: Reply with quote

Yes jLo - this works because of the preponderance of bivalues. If we look at the polarity on 1's we are saying "1" or "not 1". In many cases "not 1" means 2.

Let us assume that green is "not 2" ie 1

If you look in box 2 you will notice that a triple <289> is forced, making column 5 contain 2. Or if you prefer, the 2 (or not 1) in r3c8 eliminates the 2's in r3 in box 2. Smile

Effectively this structure will have an eliminating 2 and an eliminating 1 whatever the polarity.
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jLo



Joined: 30 Apr 2007
Posts: 55

PostPosted: Wed Jul 11, 2007 4:38 pm    Post subject: Reply with quote

Quote:

Effectively this structure will have an eliminating 2 and an eliminating 1 whatever the polarity.


Nice. That's probably the path of least resistance.
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Asellus



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

PostPosted: Thu Jul 12, 2007 1:06 am    Post subject: Reply with quote

I certainly didn't see that <2> elimination. Nice!

It is analogous to the Finned X-Wing discussed above: a technique extended using color wrapping. In this case, it is ALS, {1289}{12}, extended by color wrapping.

In the first case, I wouldn't have recognized the pattern as a possible Finned X-Wing based elimination. And in the second case, I would have neglected the ALS because it doesn't provide any "direct" candidates for elimination.

I think it may take some time before I am able to "see" such possibilities as these, though!
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Mogulmeister



Joined: 03 May 2007
Posts: 695

PostPosted: Thu Jul 12, 2007 5:29 am    Post subject: Reply with quote

I always think it is useful when studying puzzles to see what builds upon an apparently simple strong links pattern. I saw here that there were a good many bivalues which always arouses interest eg think about the xy chains that often have multiple eliminations.
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TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Thu Jul 12, 2007 11:05 pm    Post subject: Reply with quote

That was an excellent catch on the 2 but I disagree with your statement "that same structure pictured below eliminates both 1 and 2..." The 1 is pretty obvious, since it's simple coloring (strong links) but in the case of the 2 the structure shown (highlighted) can in no possible way, by itself, eliminate the 2 at r8c5.
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Mogulmeister



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PostPosted: Fri Jul 13, 2007 8:44 am    Post subject: Reply with quote

We are talking about that structure - I was not setting forth a generalised proof- the statement with respect to that puzzle is absolutely correct.
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Asellus



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Location: Sonoma County, CA, USA

PostPosted: Fri Jul 13, 2007 3:57 pm    Post subject: Reply with quote

I can see Tracy's point in that at least a couple of us couldn't see how that <2> was eliminated until the reply to jLo mentioning the potential locked set. However, I think that's perfectly fine. None of us can be sure something we post will be immediately clear to others. That's the value of a discussion board: things become clear through the back-and-forth dialog.
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TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Fri Jul 13, 2007 11:21 pm    Post subject: Reply with quote

Asellus wrote:
I can see Tracy's point in that at least a couple of us couldn't see how that <2> was eliminated...


I had no problem following the logic as presented by Mogulmeister in his post, which was an excellent deduction, as I said before, and a thorough explanation.

But when that explanation includes the statement
Mogulmeister wrote:
If you look in box 2 you will notice that a triple <289> is forced...

then I don't see how "that structure" (which I took to mean the four pink/green highlighted cells) is the same as the one used to eliminate the 1.
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