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7 Apr Nightmare -- an "anti-unique rectangle"

 
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Fri Apr 07, 2006 6:49 pm    Post subject: 7 Apr Nightmare -- an "anti-unique rectangle" Reply with quote

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


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


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

PostPosted: Sun Apr 09, 2006 4:31 pm    Post subject: An "anti-unique rectangle" Reply with quote

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      |


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

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


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


that you actually used in the puzzle.

Nice work.

Steve
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