## Unique rectangle type 3/XY-wing hybrid?

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### Unique rectangle type 3/XY-wing hybrid?

Unique rectangle type 3 in blue (3/9) r5c7,r6c7,r5c2,r6c2
Making a virtual cell (1/7) r56c2 in box 4 and column 2 (pincer-yellow) that can form an XY-wing r4c1 (pivot-green) and r8c1 (pincer-yellow)
eliminating 1 from r8c2 (pink)
Is this a valid move?

+--------------+--------------+--------------+
| 79 6 2 | 39 8 37 | 4 5 1 |
| 4 79 15 | 6 19 57 | 8 3 2 |
| 15 8 3 | 2 15 4 | 7 6 9 |
+--------------+--------------+--------------+
| 37 2 8 | 17 347 9 | 6 14 5 |
| 579 379 45 | 137 2 6 | 39 14 8 |
| 6 139 14 | 5 34 8 | 39 2 7 |
+--------------+--------------+--------------+
| 2 4 7 | 8 35 35 | 1 9 6 |
| 13 13 6 | 79 79 2 | 5 8 4 |
| 8 5 9 | 4 6 1 | 2 7 3 |
+--------------+--------------+--------------+
Source: Fiendish, Enjoy Sudoku app
bd=002080051400600000083004760020009005000020000600500020007800190000002004850060200
Attachments
20191109_145631.jpg (58.33 KiB) Viewed 278 times
Houdini

Posts: 3
Joined: 15 October 2019

### Re: Unique rectangle type 3/XY-wing hybrid?

What I can say from the image is that, if the puzzle is valid then:

• r4c1 = 7 would give r1c1=9, r5c1 = 5, r5c3 = 4, r6c3 = 1

That leaves r56c2 = {39} which gives a "uniqueness" contradiction, so r4c1 must be 3

Mathimagics
2017 Supporter

Posts: 1583
Joined: 27 May 2015
Location: Canberra

### Re: Unique rectangle type 3/XY-wing hybrid?

Hi Houdini,

Houdini wrote:Unique rectangle type 3 in blue (3/9) r5c7,r6c7,r5c2,r6c2
Making a virtual cell (1/7) r56c2 in box 4 and column 2 (pincer-yellow) that can form an XY-wing r4c1 (pivot-green) and r8c1 (pincer-yellow)
eliminating 1 from r8c2 (pink)
Is this a valid move?

Yes, it's a valid move, and a very nice one too. There's another similar virtual XY-Wing there as well, using the 3s (r4c1,r8c2) as the eliminating endpoints. The end result is the same.

First, yours:

Code: Select all
`.--------------------.--------------.-------------.|   79    6       2  | 39   8    37 |  4    5   1 ||   4     79      15 | 6    159  57 |  8    3   2 ||   15    8       3  | 2    15   4  |  7    6   9 |:--------------------+--------------+-------------:|  b37    2       8  | 17   347  9  |  6    14  5 ||   579  c39+7    45 | 137  2    6  | c39+  14  8 ||   6    c39(+1)  14 | 5    34   8  | c39+  2   7 |:--------------------+--------------+-------------:|   2     4       7  | 8    35   35 |  1    9   6 || a[1]3   3-1     6  | 79   79   2  |  5    8   4 ||   8     5       9  | 4    6    1  |  2    7   3 |'--------------------'--------------'-------------'`

(1=3)r8c1 - (3=7)r4c1 - (7=UR=1)r56c27 => -1 r8c2

The other one:

Code: Select all
`.-------------------.--------------.-------------.|   79    6      2  | 39   8    37 |  4    5   1 ||   4     79     15 | 6    159  57 |  8    3   2 ||   15    8      3  | 2    15   4  |  7    6   9 |:-------------------+--------------+-------------:| a[3]7   2      8  | 17   347  9  |  6    14  5 ||   579  b9+7-3  45 | 137  2    6  | b39+  14  8 ||   6    b9+1-3  14 | 5    34   8  | b39+  2   7 |:-------------------+--------------+-------------:|   2     4      7  | 8    35   35 |  1    9   6 ||   1-3  c1(3)   6  | 79   79   2  |  5    8   4 ||   8     5      9  | 4    6    1  |  2    7   3 |'-------------------'--------------'-------------'`

(3=7)r4c1 - (7=UR=1)r56c27 - (1=3)r8c2 => -3 r56c2,r8c1

SpAce

Posts: 2583
Joined: 22 May 2017

### Re: Unique rectangle type 3/XY-wing hybrid?

Hi Mathimagics,

Mathimagics wrote:What I can say from the image is that, if the puzzle is valid then:

• r4c1 = 7 would give r1c1=9, r5c1 = 5, r5c3 = 4, r6c3 = 1

That leaves r56c2 = {39} which gives a "uniqueness" contradiction, so r4c1 must be 3

That's valid logic too, though a bit more complicated and perhaps less elegant because it's presented as a contradiction move. Of course it's easy to turn into a pincering AIC as well:

Code: Select all
`.----------------------.--------------.-------------.| d(79)    6        2  | 39   8    37 |  4    5   1 ||   4      79       15 | 6    159  57 |  8    3   2 ||   15     8        3  | 2    15   4  |  7    6   9 |:----------------------+--------------+-------------:|   3-7    2        8  | 17   347  9  |  6    14  5 || d(79)5  a39[+7]  c45 | 137  2    6  | a39+  14  8 ||   6     a39+1    b14 | 5    34   8  | a39+  2   7 |:----------------------+--------------+-------------:|   2      4        7  | 8    35   35 |  1    9   6 ||   13     13       6  | 79   79   2  |  5    8   4 ||   8      5        9  | 4    6    1  |  2    7   3 |'----------------------'--------------'-------------'`

(7=UR=1)r56c27 - (1=4)r6c3 - (4=5)r5c3 - (5=97)r51c1 => -7 r4c1

Or in packed notation:

(7)r5c2 =UR= (145)b4p896 - (5=97)r51c1 => -7 r4c1

The same elimination (+ another one) is available a bit more simply with a virtual H-Wing:

Code: Select all
`.---------------------.--------------.-------------.|  79     6        2  | 39   8    37 |  4    5   1 ||  4      79       15 | 6    159  57 |  8    3   2 ||  15     8        3  | 2    15   4  |  7    6   9 |:---------------------+--------------+-------------:| (3)-7   2        8  | 17   347  9  |  6    14  5 ||  579   a9[+7]-3  45 | 137  2    6  | a39+  14  8 ||  6     a39+1     14 | 5    34   8  | a39+  2   7 |:---------------------+--------------+-------------:|  2      4        7  | 8    35   35 |  1    9   6 || c13    b13       6  | 79   79   2  |  5    8   4 ||  8      5        9  | 4    6    1  |  2    7   3 |'---------------------'--------------'-------------'`

(7=UR=1)r56c27 - (1=3)r8c2 - r8c1 = (3)r4c1 => -3 r5c2, -7 r4c1

SpAce

Posts: 2583
Joined: 22 May 2017