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20 JULY VH

 
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Thu Jul 19, 2007 11:37 pm    Post subject: 20 JULY VH Reply with quote

I suppose this one was not so VH, one xy-chain solves the puzzle i think.
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Captain Pete



Joined: 09 Jun 2007
Posts: 55
Location: Oley, PA

PostPosted: Fri Jul 20, 2007 1:51 pm    Post subject: Reply with quote

Could you describe the XY-Chain in this puzzle?
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sdq_pete



Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL

PostPosted: Fri Jul 20, 2007 2:30 pm    Post subject: Reply with quote

I found myself unable to get past the following point:

Code:

+------------+----------+----------+
| 8  36  369 | 5 1  29  | 23 4   7 |
| 17 167 469 | 3 69 249 | 28 68  5 |
| 2  5   346 | 8 67 47  | 1  36  9 |
+------------+----------+----------+
| 4  9   2   | 1 5  8   | 37 37  6 |
| 37 37  8   | 9 2  6   | 4  5   1 |
| 15 16  56  | 7 4  3   | 9  2   8 |
+------------+----------+----------+
| 9  4   7   | 2 8  5   | 6  1   3 |
| 35 2   35  | 6 79 1   | 78 789 4 |
| 6  8   1   | 4 3  79  | 5  79  2 |
+------------+----------+----------+

Play this puzzle online at the Daily Sudoku site

Peter
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Fri Jul 20, 2007 3:33 pm    Post subject: XY-CHAIN Reply with quote

XY-chain length 4.

Starting with number 2 in [R1C6] = 3 in [R1C7] = 6 in [R3C8] = 7 in [R3C5(chain end)].

Starting with number 9 in [R1C6] = 7 in [R9C6].
This means either 7 resides in [R3C5] or in [R9C6] or in both [R3C5,R9C6]
Both cells [R3C5,R9C6] can see the bi-value cell [47] in R3C6, which eliminates the 7 from R3C6, that solves the puzzle.

Both 7's in R2C5 and R2C6 must be erased in the grid, I couldn't delete them before copying and pasting.

[/code]
Code:

+------------+-------------+----------+
| 8  136 369 | 5 169  29   | 23 4   7 |
| 17 167 469 | 3 1679 2479 | 28 68  5 |
| 2  5   346 | 8 67   47   | 1  36  9 |
+------------+-------------+----------+
| 4  9   2   | 1 5    8    | 37 37  6 |
| 37 37  8   | 9 2    6    | 4  5   1 |
| 15 16  56  | 7 4    3    | 9  2   8 |
+------------+-------------+----------+
| 9  4   7   | 2 8    5    | 6  1   3 |
| 35 2   35  | 6 79   1    | 78 789 4 |
| 6  8   1   | 4 3    79   | 5  79  2 |
+------------+-------------+----------+

[/quote]
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jLo



Joined: 30 Apr 2007
Posts: 55

PostPosted: Fri Jul 20, 2007 3:57 pm    Post subject: Reply with quote

sdq_pete wrote:
I found myself unable to get past the following point:

Code:

+------------+----------+----------+
| 8  36  369 | 5 1  29  | 23 4   7 |
| 17 167 469 | 3 69 249 | 28 68  5 |
| 2  5   346 | 8 67 47  | 1  36  9 |
+------------+----------+----------+
| 4  9   2   | 1 5  8   | 37 37  6 |
| 37 37  8   | 9 2  6   | 4  5   1 |
| 15 16  56  | 7 4  3   | 9  2   8 |
+------------+----------+----------+
| 9  4   7   | 2 8  5   | 6  1   3 |
| 35 2   35  | 6 79 1   | 78 789 4 |
| 6  8   1   | 4 3  79  | 5  79  2 |
+------------+----------+----------+

Play this puzzle online at the Daily Sudoku site

Peter


Notice that the <6> in row 1 will be from box 1. That eliminates the <6> entries
in box 1 that are not in row 1 (R2C2, R2C3 and R3C3). This opens up an XYZ-wing
with a pivot in box 2.
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Fri Jul 20, 2007 6:20 pm    Post subject: July 20 VH Reply with quote

Peter,

On your grid there is a four-step xy chain beginning at R9C6, ending at R4C8 which removes the 7 from R9C8, and opens the puzzle.

Earl
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sdq_pete



Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL

PostPosted: Fri Jul 20, 2007 6:52 pm    Post subject: Reply with quote

jLo wrote:

Notice that the <6> in row 1 will be from box 1. That eliminates the <6> entries
in box 1 that are not in row 1 (R2C2, R2C3 and R3C3).


I can't believe I missed that!

Earl wrote:
there is a four-step xy chain beginning at R9C6, ending at R4C8 which removes the 7 from R9C8


I'm not sure I understand this. I presume you are referring to the strong link chain of 7's but I can't follow it to R4C8 (there are 3 candidate 7's on row 8 ).

Incidentally, this is probably more trial and error, but in box 3, setting either R3C8 to 3 or R2C8 to 6 quickly leads in both cases to duplicate values (illogical state) showing that the values must be 6 and 8 respectively. Perhaps not the most satsfactory solution though.

Peter
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Fri Jul 20, 2007 9:11 pm    Post subject: July 20 VH Reply with quote

Peter,

It is an xy chain, not a strong link of 7's.
If R9C6 is 7, R9C8 cannot be 7.
If R9C6 is 9, R9C8 cannot be 7 because R4C8 is 7.

(If R9C6 is 9, R8C5 is 7, R3C5 is 6, R3C8 is 3, and R4C8 is 7).

Earl
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sdq_pete



Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL

PostPosted: Sat Jul 21, 2007 11:53 am    Post subject: Reply with quote

Hi Earl

Ah, is that an XY chain? It actually seems a similar argument to what I mentioned myself - a presumed value leading to a logical impasse. Is there anything special about that route? I mean, I could just as easily have said, if R9C6 = 9, then R1C6 = 2, R1C7 = 3 and R4C7 = 7... So, is there something special about your route to R4C8 = 7 or is it trial and error?

Peter
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Sat Jul 21, 2007 6:21 pm    Post subject: July 20 VH Reply with quote

Peter,

I suppose any chain can be thought of as trial and error, but an xy chain does have a pincer effect which is logic. One might think of an xy chain as an xy wing with one arm extended.

Earl
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keith



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

PostPosted: Sat Jul 21, 2007 7:33 pm    Post subject: Reply with quote

I know I'm coming in late on this discussion, but the puzzle can be solved with a couple of XYZ-wings.

Keith
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