## Almost Unique Rectangles: description and use in nice loops

Advanced methods and approaches for solving Sudoku puzzles
Ronk wrote:Although "backtesting" is normally used for to develop a trading strategy, I suggest "Backtesting Eliminations".

Carcul wrote:I suggest "Logic Test for Networks" or "Logic Test for Forcing Chains".

Hi Ronk, Carcul and all readers, Any objection to use the name "Backtest" with the wording below.

Backtest – a logic test for networks that operates under the principle: If a forcing chain technique that uses a set of cells "A" to eliminate candidate "x" from cell "B" (which can belong or not along to A), then this imply that the back substitution of B=x will lead to some sort of contradiction amongst the cells of set A.
Jeff

Posts: 708
Joined: 01 August 2005

Jeff wrote:Any objection to use the name "Backtest" with the wording below?

Backtest – a logic test for networks that operates under the principle: If a forcing chain technique that uses a set of cells "A" to eliminate candidate "x" from cell "B" (which can belong or not along to A), then this imply that the back substitution of B=x will lead to some sort of contradiction amongst the cells of set A.

Mostly just some improved wording to clarify the purpose. With changes highlighted:

Backtest – a method that validates a deduction technique: If a forcing chain technique uses a set of cells "A" to eliminate candidate "x" from cell "B" (which can belong to set A) -- then this implies that the back substitution of B=x will lead to some sort of contradiction amongst the cells of set A.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

### Almost Unique Rectangles: description and use in nice loops

Hi Carcul,

Although this barely advances puzzle #409 of the top1465, I think you'll be interested in this AUR and its nice loops.
Code: Select all
` 2    47   5    | 8    19   6    | 3    14   79 134  9    134  |^234  7   ^24   | 8    5    6 8    347  6    | 34   5    19   | 2    14   79----------------+----------------+--------------- 147  145  2    | 57   39   349  | 6    8    13 9    8    14   |^24   6   ^234  | 5    7    13 367  356  37   | 1    8    57   | 9    2    4----------------+----------------+--------------- 137  2    137  | 6    13   8    | 4    9    5 5    13   8    | 9    4    13   | 7    6    2 46   46   9    | 57   2    57   | 1    3    8[r2c6]=2=(AUR:[r5c6]=3=[r2c4])=2=[r2c6] => r2c6=2OR[r5c4]=2=(AUR:[r2c4]=3=[r5c6])=2=[r5c4] => r5c4=2`

Most interesting is that there are only three cells in each nice loop ... and all are AUR cells.

Regards, Ron
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Hi Ronk.

Thanks for this another good example.

Ronk wrote:Although this barely advances puzzle #409 of the top1465...

Instead of making the deduction r2c6=2 with the AUR, you can use the AUR in a different loop that will solve the puzzle.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:Instead of making the deduction r2c6=2 with the AUR, you can use the AUR in a different loop that will solve the puzzle.

Is yours different from this?

[r1c5]-9-[r1c9]-7-[r1c2]-4-[r1c8]-1-[r3c8]-4-[r3c4]-3-(AUR:[r2c4]=3=[r5c6])-3-[r4c5]-9-[r1c5] implying r1c5<>9 or r1c5=1

Ron
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Here is another example of a type 3 AUR:

Code: Select all
` *----------------------* |6 . . | . 9 7 | . . . | |. 2 3 | . . . | . . . | |. 1 . | . . . | 4 7 2 | |------+-------+-------| |. . . | . . . | . . 9 | |9 . . | 6 1 . | . . . | |2 6 . | . . 3 | . 8 4 | |------+-------+-------| |3 . 4 | . . . | . . . | |. . . | . 6 . | 3 . . | |. . . | . 7 . | 5 . . | *----------------------*`

(expert puzzle, March 8th, 2006, Koalog Collection)

After the basic steps we arrive at:

Code: Select all
` *--------------------------------------------------------------------* | 6      4      58     | 2      9      7      | 18     135    135    | | 7      2      3      | 158    4      15     | 689    569    56     | | 58     1      9      | 58     3      6      | 4      7      2      | |----------------------+----------------------+----------------------| | 4      35     15     | 7      2      8      | 16     1356   9      | | 9      358    578    | 6      1      4      | 27     235    357    | | 2      6      17     | 9      5      3      | 17     8      4      | |----------------------+----------------------+----------------------| | 3      57     4      | 15     8      159    | 12679  1269   167    | | 158    578    2      | 4      6      159    | 3      19     178    | | 18     9      6      | 3      7      2      | 5      4      18     | *--------------------------------------------------------------------*`

Now, cells r1c89/r5c789 make up a type 3 AUR and we can write

[r1c7]-1-[r1c8|r1c9]=1|2,7=[r5c7|r5c8|r5c9]-7-[r6c7]-1-[r1c7],

which implies r1c7<>1 and it solve the puzzle.

Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:Now, cells r1c89/r5c789 make up a type 3 AUR and we can write

[r1c7]-1-[r1c8|r1c9]=1|2,7=[r5c7|r5c8|r5c9]-7-[r6c7]-1-[r1c7],

which implies r1c7<>1 and it solve the puzzle.

Two simple nice loops also solve the puzzle:

r4c7-1-r4c3-5-r1c3-8-r1c7-1-r4c7 implying r4c7<>1

r7c7-9-r2c7=9=r2c8=6=r7c8=2=r7c7 implying r7c7<>9

Ron
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Carcul,

Or one could use another AUR in r2c46/r7c46 to conclude that (3,4)!8:

(3,4)8 > (2,9)6 > (2,8)9 > (8,8)1 > (8,6)9 > (7,6)!9, a contradiction.

This also solves the puzzle.
re'born

Posts: 551
Joined: 31 May 2007

rep'nA wrote:(3,4)8 > (2,9)6 > ......

Different row, and different col, and different box. Looks like something was left out to me.

Ron
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Ronk,

Sorry. In my mind, when I set (3,4)8, (2,[46]) becomes a naked pair forcing (2,9)6. Occasionally, I forget to write such steps. Of course this is pretty obnoxious and I will try to do better next time.

The correct chain of reasoning should be:

(3,4)8 > (2,[46])[15] > (2,9)6 >> (2,8)9 > (8,8)!9 > (8,6)9 > (7,6)!9, giving our contradiction.

P.S. If you are listening aeb, am I correctly using your notation.
re'born

Posts: 551
Joined: 31 May 2007

rep'nA wrote:Or one could use another AUR in r2c46/r7c46 to conclude that (3,4)!8:

Completely correct. Good observation.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

rep'nA wrote:(3,4)8 > (2,[46])[15] > (2,9)6 >> (2,8)9 > (8,8)!9 > (8,6)9 > (7,6)!9, giving our contradiction.

P.S. If you are listening aeb, am I correctly using your notation.

Yes, you are!
aeb

Posts: 83
Joined: 29 January 2006

Carcul wrote:Let's see now an example of a type 3 AUR.
.
.
.
Code: Select all
` 1   458  45 | 9  6    3  | 7  58 2  6   3    257| 1  2578 257| 9  58 4  78  9    257| 28 2578 4  | 6  1  3    -------------+------------+-----------  47  247  1  | 3  9    257| 48 6  58  5   67   8  | 4  17   167| 2  3  9  9   246  3  | 28 258  256| 14 7  15 -------------+------------+-----------  2   1    9  | 7  3    8  | 5  4  6  478 4578 457| 6  12   12 | 3  9  78  3   78   6  | 5  4    9  | 18 2  178 `

Here we have a type 3 AUR in cells r1c23/r8c23...

Carcul wrote:Another Example of a Type 3 AUR
.
.
.
(this is "unlimited077" from Angus) and let's jump to the point where Simple Sudoku gets stuck:

Code: Select all
` 45  3   8   | 2456 1    7   | 9    56   24     145 7   9   | 2456 8    245 | 1245 356  1234   6   14  2   | 45   3    9   | 8    57   147   -------------+---------------+---------------- 9   168 135 | 7    26   25  | 125  4    1238   48  468 57  | 3    2469 1   | 257  579  278    134 2   1357| 459  49   8   | 157  3579 6     -------------+---------------+---------------- 2   5   4   | 1    7    3   | 6    8    9      138 189 13  | 89   5    6   | 47   2    47     7   89  6   | 2489 249  24  | 3    1    5     `

Here we have a type 3 AUR in cells r8c24/r9c246...

Carcul,

Can you explain what is different in these two examples that allows you to put all 5 cells into your AUR in the second, but only 4 in the first?
re'born

Posts: 551
Joined: 31 May 2007

Hi Rep'nA.

Both examples are Type-3 AURs and both are made up of 5 cells, but it seems that in your first example I forget to mention that r9c2 is also part of the AUR: so, we have a type-3 AUR in cells r1c23/r8c23/r9c2. Thanks (original post edited).

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Carcul wrote:I forget to mention that r9c2 is also part of the AUR: so, we have a type-3 AUR in cells r1c23/r8c23/r9c2.

But why should a rectangle, with four corners, be described with five cells?
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Previous