The New Sudoku Players' Forum

[Jeff]

Although "backtesting" is normally used for to develop a trading strategy, I suggest "Backtesting Eliminations".

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.

[ronk]

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]

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=2

OR

[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

[Carcul]

Hi Ronk.

Thanks for this another good example.

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

[ronk]

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

[Carcul]

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

[ronk]

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

[re'born]

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.

[ronk]

(3,4)8 > (2,9)6 > ......

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

Ron

[re'born]

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.

[Carcul]

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

Completely correct. Good observation.

Regards, Carcul

[aeb]

(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!

[re'born]

Let's see now an example of a type 3 AUR.
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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...

Another Example of a Type 3 AUR
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.
.
(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?

[Carcul]

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

[ronk]

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?