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		| David Bryant 
 
 
 Joined: 29 Jul 2005
 Posts: 559
 Location: Denver, Colorado
 
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				|  Posted: Fri Apr 07, 2006 6:49 pm    Post subject: 7 Apr Nightmare -- an "anti-unique rectangle" |   |  
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				| This puzzle is extremely tough, even by "nightmare" standards. 
 After making all the more or less "obvious" moves I arrived at this position.
 
 
  	  | Code: |  	  | 2379    6    249    5    28    248  3789    1    389 259  1279  1259+   3   1268  1268   589   579    4
 35    14=    8     7     9    14~    2    356   356
 4    139    7     2   1568  1368   359  3569  3569
 69     5    169-  146    7    136*  349    8     2
 8    23    26    46    56     9     1   3456    7
 259   279    3    189    4    12     6    579  1589
 1    249  24569  689   236    7   34589 3459  3589
 679    8    469   169   136    5   3479    2    139
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 Here there's a "fork" on the "1"s in row 3 and column 3 that lets us eliminate the possibility of a "1" at r5c6. Then we can use a double-implication chain from r6c4 to show that r5c6 = 3:
 
 A. r6c4 = 6 ==> r5c6 <> 6.
 B. r6c4 = 4 ==> triplet {1, 6, 9} in r5c1, 3 & 4 ==> r5c6 <> 6.
 
 Now the grid looks like this.
 
 
  	  | Code: |  	  | 2379    6    249    5    28    248  3789    1    389 259  1279  1259A   3   1268  1268   589   579    4
 35    14     8     7     9    14B    2    356   356
 4    139    7     2   1568   168   359  3569  3569
 69     5    169a  146    7     3    49     8     2
 8    23    26    46    56     9     1   3456    7
 259   279    3    189b   4    12*    6    579  1589
 1    249  24569  689   236    7   34589 3459  3589
 679    8    469   169b  136    5   3479    2    139
 | 
 
 Next we have a "Nishio" move. Notice that there are only two places for a "1" in column 3.
 
 A. r2c3 = 1 ==> r3c6 = 1 ==> r6c6 <> 1.
 B. r5c3 = 1 ==> r7c4 or r9c4 = 1 ==> r6c6 <> 1.
 
 We conclude that r6c6 = 2; this allows us to eliminate some "2"s in the top left 3x3 box (because the "2" in column 1 is at r1c1 or r2c1), to identify a {5, 7, 9} triplet in row 7 (uncovering a hidden pair {1, 8}), and to eliminate the "5" at r2c8 via a "coloring" argument.
 
 Now the puzzle looks like this.
 
 
  	  | Code: |  	  | 2379    6    49     5    28    48   3789    1    389 259*  179*  159    3   1268   168   589   79     4
 35    14     8     7     9    14     2    356   356
 4    139    7     2   1568   168   359  3569  3569
 69     5    169   146    7     3    49     8     2
 8    23    26    46    56     9     1   3456    7
 59*   79*    3    18     4     2     6    579   18
 1    249  24569  689   36     7   34589 3459  3589
 679    8    469   169   136    5   3479    2    139
 | 
 
 Here there's a very unusual formation in r2c1&2, and r7c1&2. I think it should be called an "anti-unique rectangle." Observe that we cannot have r2c1 = 5 and r2c2 = 7, because that would force two "9"s in row 7.
 
 This is where it gets very cute.
 
 A. r1c1 = 2 ==> r1c5 = 8 ==> r1c6 = 4 ==> r1c3 = 9.
 B. r1c1 = 2 & r1c3 = 9 ==> r2c1 = 5.
 C. r1c1 <> 7 ==> r2c2 = 7.
 
 So we can't have a "2" at r1c1, because of the "anti-unique rectangle." This means that the "2" in the top left 3x3 box must appear at r2c1, and that in turn forces a "2" at r1c5.
 
 
  	  | Code: |  	  | 379    6    49     5     2    48   3789    1    389 2    179*  159    3    168   168   589   79     4
 35*   14     8     7     9    14     2    356   356
 4    139    7     2   1568   168   359  3569  3569
 69     5    169   146    7     3    49     8     2
 8    23    26    46    56     9     1   3456    7
 59*   79*    3    18     4     2     6    579   18
 1    249  24569  689   36     7   34589 3459  3589
 679    8    469   169   136    5   3479    2    139
 | 
 
 Now the "rectangle" is a bit lop-sided (a trapezoid), but it still works:
 
 A. r1c1 = 3 ==> r3c1 = 5.
 B. r1c1 <> 7 ==> r2c2 = 7.
 
 So the "3" in column 1 can't appear at r1c1, and it must go in r3c1. This in turn implies that r2c3 = 5, and the rest of the puzzle is relatively straightforward -- for a little while! After making the obvious moves from here we arrive at one more very tough spot.
 
 
  	  | Code: |  	  | 79     6    49     5     2    48   3789    1    389D 2    179    5     3    18     6    89    79     4
 3    14     8     7     9    14     2    56    56
 4    39A    7     2    18    18    359   356  3569
 69     5     1    46     7     3    49B    8     2
 8    23    26    46     5     9     1    34     7
 5    79     3    18     4     2     6    79C   18
 1    249  2469   89    36     7   34589  345  3589
 679    8    469   19    36     5   3479    2    139
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 Another "Nishio" move -- on the "9"s -- finally breaks through the logjam.
 
 A. r4c2 = 9 ==> r5c7 = 9.
 B. r4c2 = 9 ==> r7c8 = 9.
 C. r5c7 = 9 & r7c8 = 9 ==> r1c9 = 9.
 
 But with "9" at A and D above there's no way left to place a "9" in the top left 3x3 box. So the "9" in the middle left 3x3 box must appear at r5c1.
 
 Oh -- notice that the argument on the "9"s can be operated in the other direction, although it's a bit more complex that way. Start with the two spots for a "9" in row 7.
 
 A. r7c2 = 9 ==> r4c2 <> 9.
 B. r7c2 <> 9 ==> r7c8 = 9 ==> either r4c9 = 9 or else r1c9 = 9.
 B1. r4c9 = 9 ==> r4c2 <> 9.
 B2. r1c9 = 9 ==> r2c2 = 9 ==> r4c2 <> 9.
 
 Has anyone noticed this sort of "anti-unique rectangle" before?  dcb
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		| Steve R 
 
 
 Joined: 24 Oct 2005
 Posts: 289
 Location: Birmingham, England
 
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				|  Posted: Sun Apr 09, 2006 4:31 pm    Post subject: An "anti-unique rectangle" |   |  
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				| I cannot remember seeing the patterns you mention and have been playing with Xs and Ys. 
 The second anti-unique rectangle (trapezoid) elimination is straightforward:
 
  	  | Code: |  	  | -----------------
 | Zetc            |
 |        Zetc     |   Only two places for Z in the box
 | YT              |
 |-----------------|
 |                 |
 |                 |
 | XY      XZ      |
 
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 T can be eliminated from the top left Zetc. Incidentally, the YT cell is not required to be in the box, just in the left hand column.
 
 I have not made much progress with the first anti-unique rectangle: probably not seeing the wood for the trees.
 
 However, the elimination of 1 from r7c6 caught my eye. As far as pattern merchants are concerned it seems to be an extension of the empty rectangle, that is a box in which a candidate X is confined to one row plus one column.
 
 The basic empty rectangle format is
 
  	  | Code: |  	  | -------------------------
 | . . . | . . 1 | . . . |
 | . . 1 | . 1 . | . . . |
 | . . . | . . 1 | . . . |
 -------------------------
 | . . . | . . . | . . . |
 | . . 1 | . . x | . . . |
 | . . . | . . . | . . . |
 -------------------------
 | . . . | . . . | . . . |
 | . . . | . . . | . . . |
 | . . . | . . . | . . . |
 -------------------------
 
 Only two cells for 1 in column 3
 
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 This eliminates 1 from r6c6, which is of no use in the puzzle itself. I do not recall coming across an extension like
 
  	  | Code: |  	  | -------------------------
 | . . . | . . 1 | . . . |
 | . . 1 | . 1 . | . . . |
 | . . . | . . 1 | . . . |
 -------------------------
 | . . . | . . . | . . . |
 | . . 1 | 1 . . | . . . |
 | . . . | . . . | . . . |
 -------------------------
 | . . . | . . . | . . . |
 | . . . | 1 . x | . . . |
 | . . . | . . . | . . . |
 -------------------------
 
 Only two cells for 1 in columns 3 and 4
 
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 eliminating 1 from r8c6, still less an extension of the grouped variety
 
  	  | Code: |  	  | -------------------------
 | . . . | . . 1 | . . . |
 | . . 1 | . 1 . | . . . |
 | . . . | . . 1 | . . . |
 -------------------------
 | . . . | . . . | . . . |
 | . . 1 | 1 . . | . . . |
 | . . . | . . . | . . . |
 -------------------------
 | . . . | 1 . x | . . . |
 | . . . | 1 . x | . . . |
 | . . . | 1 . x | . . . |
 -------------------------
 
 Only two cells for 1 in column 3 and only two places for 1(cell and box) in column 4
 
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 that you actually used in the puzzle.
 
 Nice work.
 
 Steve
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