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Puzzle 10/10/12: B XY

 
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daj95376



Joined: 23 Aug 2008
Posts: 3854
PostPosted: Tue Oct 12, 2010 5:28 am    Post subject: Puzzle 10/10/12: B XY Reply with quote
Code:
 +-----------------------+
 | 8 1 . | 5 . . | 2 9 . |
 | 2 5 . | 8 . . | . 3 . |
 | . . 3 | . . . | . 8 . |
 |-------+-------+-------|
 | 1 8 . | 9 5 7 | . 2 3 |
 | . . . | 3 . 4 | . . . |
 | . . . | 2 1 . | . 4 . |
 |-------+-------+-------|
 | 5 . . | . . . | . 7 2 |
 | 3 4 8 | 7 . 5 | 1 6 . |
 | . . . | 4 . . | 3 . 8 |
 +-----------------------+

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peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK
PostPosted: Tue Oct 12, 2010 12:42 pm    Post subject: Reply with quote
Sure someone will make something elegant out of all those URs - for me another wing..
Quote:
w-wing(19) ; (9=1)r2c6 - r9c6=r7c4 - (1=9)r7c3 ; r2c3<>9
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
PostPosted: Tue Oct 12, 2010 5:10 pm    Post subject: Reply with quote
Did someone mentions URs? How about a 10-cell BUG-Lite+2 (3689)r1567c1567.

Danny and others, is this is a valid pattern? It only has four candidate DP digits, 3689, in four rows and four columns but the staggered (68) bivalues in box5 shares both digits in stack2 and band 2 for a total of five boxes. Each candidate digit in every ADP cell "sees" only one other occurrence of that digit so it is possible to "walk" all DP digits through all ADP cells in the fashion of a deadly pattern.

Assuming it is valid, we get r1c5=4,r1c5=7,r5c1=7 for internal SIS; r1c6<>6=3.
(4)r1c5-(4=6)r1c9;
||
(7)r1c5-(7=6)r1c3;
||
(7)r5c1-(9)r5c1=r5c7-(9=8)r6c7-(8=6)r6c6;
As fun as it was (given it is a valid ADP) this move does not solve the puzzle.

So we try again with the same SIS: r1c5=4,r1c5=7,r5c1=7; r5c5<>36
(4)r1c5-(4=6)r1c9-r3c9=r3c4-r7c4=(6)r9c5;
||
(7)r1c5-(7=9)r3c5-(9=6)r9c5;
||
(7)r5c1-(9)r5c1=r5c7-(9=8)r6c7-(8=6)r6c6;
But this move is equivalent to the first attempt.

Looking at the external SIS, we find that the only inferences are various combinations of the digit 6, all of which seem to result in r1c6<>3 as noted in the first attempt again. Thus, a one step move like the one by Peter is still needed to complete the puzzle.


The real issue, for me, is the question of the validity of the possible BUG-Lite. Is it valid??

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3854
PostPosted: Tue Oct 12, 2010 5:44 pm    Post subject: Reply with quote
After r1c5<>47 and r5c1<>7:

Code:
  r1c5=3
 *----------------------------------------------------*
 |  8    1    7    | 5   *3   *6    |  2    9    4    |
 |  2    5    69   | 8    49   19   |  7    3    146  |
 |  4    79   3    | 1    79   2    |  5    8    16   |
 |-----------------+----------------+-----------------|
 |  1    8    4    | 9    5    7    |  6    2    3    |
 | *9    27   257  | 3   *6    4    | *8    1    57   |
 | *6    3    57   | 2    1   *8    | *9    4    57   |
 |-----------------+----------------+-----------------|
 |  5    69   19   | 16  *8   *3    |  4    7    2    |
 |  3    4    8    | 7    2    5    |  1    6    9    |
 |  7    267  127  | 4    9    19   |  3    5    8    |
 *----------------------------------------------------*

Code:
  r1c5=6
 *----------------------------------------------------*
 |  8    1    7    | 5   *6   *3    |  2    9    4    |
 |  2    5    69   | 8    49   19   |  7    3    146  |
 |  4    79   3    | 1    79   2    |  5    8    16   |
 |-----------------+----------------+-----------------|
 |  1    8    4    | 9    5    7    |  6    2    3    |
 | *6    27   257  | 3   *8    4    | *9    1    57   |
 | *9    3    57   | 2    1   *6    | *8    4    57   |
 |-----------------+----------------+-----------------|
 |  5    69   19   | 16  *3   *8    |  4    7    2    |
 |  3    4    8    | 7    2    5    |  1    6    9    |
 |  7    267  127  | 4    9    19   |  3    5    8    |
 *----------------------------------------------------*

Your DP seems valid to me. I just don't know its name.

At this point, your following logic is sufficient to declare r1c6<>6.

Code:
 +--------------------------------------------------------------+
 |  8     1     67    |  5     3467  3-6   |  2     9     46    |
 |  2     5     69    |  8     469   19    |  7     3     146   |
 |  4     79    3     |  16    79    2     |  5     8     16    |
 |--------------------+--------------------+--------------------|
 |  1     8     4     |  9     5     7     |  6     2     3     |
 |  679   267   257   |  3     68    4     |  89    1     57    |
 |  69    3     57    |  2     1     68    |  89    4     57    |
 |--------------------+--------------------+--------------------|
 |  5     69    19    |  16    38    38    |  4     7     2     |
 |  3     4     8     |  7     2     5     |  1     6     9     |
 |  67    267   127   |  4     69    19    |  3     5     8     |
 +--------------------------------------------------------------+
 # 42 eliminations remain

 (4)r1c5-(4=6)r1c9;
 ||
 (7)r1c5-(7=6)r1c3;
 ||
 (7)r5c1-(9)r5c1=r5c7-(9=8)r6c7-(8=6)r6c6;
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
PostPosted: Tue Oct 12, 2010 7:06 pm    Post subject: Reply with quote
daj95376 wrote:

Your DP seems valid to me. I just don't know its name.

At this point, your following logic is sufficient to declare r1c6<>6.

Code:
 +--------------------------------------------------------------+
 |  8     1     67    |  5     3467  3-6   |  2     9     46    |
 |  2     5     69    |  8     469   19    |  7     3     146   |
 |  4     79    3     |  16    79    2     |  5     8     16    |
 |--------------------+--------------------+--------------------|
 |  1     8     4     |  9     5     7     |  6     2     3     |
 |  679   267   257   |  3     68    4     |  89    1     57    |
 |  69    3     57    |  2     1     68    |  89    4     57    |
 |--------------------+--------------------+--------------------|
 |  5     69    19    |  16    38    38    |  4     7     2     |
 |  3     4     8     |  7     2     5     |  1     6     9     |
 |  67    267   127   |  4     69    19    |  3     5     8     |
 +--------------------------------------------------------------+
 # 42 eliminations remain

 (4)r1c5-(4=6)r1c9;
 ||
 (7)r1c5-(7=6)r1c3;
 ||
 (7)r5c1-(9)r5c1=r5c7-(9=8)r6c7-(8=6)r6c6;

Thanks for the feedback and codes Danny. All of my posted examples essentially set r1c6<>6=3.

Ted
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Wed Oct 13, 2010 4:24 am    Post subject: Reply with quote
peterj wrote:
Sure someone will make something elegant out of all those URs - for me another wing..
Quote:
w-wing(19) ; (9=1)r2c6 - r9c6=r7c4 - (1=9)r7c3 ; r2c3<>9


I've done it twice and get the same result both times; W-Wing on 69; r7c4, r9c12<>6. I can't duplicate Peter's W-Wing on 19 because I'm stuck with the 6 in r2c6.

Code:

+-------------+-------------+----------+
| 8   1   67  | 5  3467 36  | 2  9 46  |
| 2   5   69  | 8  469  169 | 7  3 146 |
| 4   79  3   | 16 79   2   | 5  8 16  |
+-------------+-------------+----------+
| 1   8   4   | 9  5    7   | 6  2 3   |
| 679 267 257 | 3  68   4   | 89 1 57  |
| 69  3   57  | 2  1    68  | 89 4 57  |
+-------------+-------------+----------+
| 5   69  19  | 16 38   38  | 4  7 2   |
| 3   4   8   | 7  2    5   | 1  6 9   |
| 67  267 127 | 4  69   169 | 3  5 8   |
+-------------+-------------+----------+

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peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK
PostPosted: Wed Oct 13, 2010 7:47 am    Post subject: Reply with quote
Marty there is a 368 triple in r167c6 that eliminates it - or a 19 hidden pair whichever you see first (I find HP's hard to see personally)
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Wed Oct 13, 2010 3:44 pm    Post subject: Reply with quote
peterj wrote:
Marty there is a 368 triple in r167c6 that eliminates it - or a 19 hidden pair whichever you see first (I find HP's hard to see personally)

Arrrgh, I'm missing basics every puzzle, or so it seems. Embarassed
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