The New Sudoku Players' Forum

[David P Bird]

Withdrawn - response to DAJ

[daj95376]

[Withdrawn: working from an incorrect understanding of the JExocet properties.]

[Leren]

daj95376 wrote : For (9), my solver allows the discontinuous loop: (9)r5c7 - r6c9 = r6c2 - r5c3 = (9)r5c7 ... to force target cell r5c7 true when (9) is true in the base cells.

This looks like a continuous loop to me - which proves nothing. I've been following this thread and I was going to ask you how you proved the Exocet property for 1, 4 and 9. My solver also found this Exocet but for 1, 4 and 9 I had to resort to a multi digit contradiction expansion, which may be unsatisfactory if it's too complicated.

Leren

[daj95376]

[Withdrawn: working from an incorrect understanding of the JExocet properties.]


As for verifying target cells being linked to the base cells, I use templates for each of the base cells values. A template is active for a value if all of the template's cells contain a candidate or solved cell for the value in the puzzle. A template is selected if it is active and one of the base cells is true in it.

There are two sets of concurrent logic for each value.

The first set of logic verifies that at least one target cell is true for every template selected for the value. If only one target cell is always true, then I know there's an alternative to the finned Swordfish pattern. If neither target cell is encountered in a template, then the base cells are rejected and my search continues for base cells.

The second set of logic tracks how may target cells are true for every template selected for the value. If the result is always two in every template, then we have your scenario where that value can be eliminated from the base/target/secondary cells. This happens to (7) in the other QExocet in this puzzle.

Code: Select all

 ***   ux_nono = <7>

 ### -1479- QExocet   Base = r6c89   Target = r4c2,r5c5==r4c3


_

[David P Bird]

Withdrawn - response to DAJ

[blue]

Code: Select all

+---------------------+-----------------------+--------------------+
| 5     13489  123489 | 1236    12346   136   | 2368   7      1246 |
| 6     7      12348  | 1235    9       135   | 2358   12345  1245 |
| 234   134    1234   | 123567  123467  8     | 2356   12345  9    |
+---------------------+-----------------------+--------------------+
| 2349  3459   7      | 1359    8       1359  | 259    6      125  |
| 1     3569   359    | 35679   367     2     | 4      8      57   |
| 29    5689   2589   | 4       167     15679 | 2579   125    3    |
+---------------------+-----------------------+--------------------+
| 379   1359   1359   | 123679  12367   4     | 23567  235    8    |
| 3479  2      3459   | 36789   367     3679  | 1      345    4567 |
| 8     134    6      | 1237    5       137   | 237    9      247  |
+---------------------+-----------------------+--------------------+

Is this what you're looking for ?
[ r3c12,r1c5,r2c8 are the potential base and target cells ]

[David P Bird]

Is this what you're looking for ?

Thanks Blue yes it is - at least it confirms that the situation is possible now, and I can start looking for any new inferences it may present.

Did you find it or compose it?

DPB

[blue]

Did you find it or compose it?

I composed it. The clues except for {8r9c1,8r7c9,8r5c8}, established the conditions, and the 8's gave it a unique solution.

[Leren]

Hi Danny - thanks for the explanation. You don't say when you use your full template analysis instead of a simple chain (as you showed for the 9 in the first Exocet) to prove a digit True in both target cells. ie do you always use your template analysis and then look for a simple chain as a simpler explanation for the template analysis result (to answer questions from simpleton's like me) ?

For example, in the second Exocet, it's obvious that if 7 is in 1 target cell it must be in the other target cell, so can be removed from the base and target cell, but did you actually find it via templates ?

7 has 2 truths in the cross lines so I wouldn't have checked that it must occupy both Target cells - something for me to add to my solver.

Regards, Leren

[Leren]

Here's a revisit of the dobrichev puzzle where I asserted that an Exocet digit (2) could be removed from the base and target cells because it only had 1 Truth in the cross lines (as per David's proposition (theorem ? )).

Here's the position at which I applied this move:

Code: Select all

                   V                  V                  V
*--------------------------------------------------------------------------------*
| 23789   56      2789     | 346789  789     3469     | 29      45      1        |
| 1       56      789      | 46      789     2        | 78      3       45       |
| 23789   237     4        | 3789    5       1        | 6       29      78       |
|--------------------------+--------------------------+--------------------------|
| 4       8       1        | 239     6       39       | 5       7       239      |
| 237     237     5        | 239     1       8        | 4       29      6        |
| 23      9       6        | 5       4       7        | 1       8       23       |
|--------------------------+--------------------------+--------------------------|
|B2789   B27      3        | 1       2789    46       | 2789    456     45       |
| 5       1       2789     | 46789  T78-29   469      | 3       46      29       |
| 6       4       2789     | 789     3       5        |T2789    1       78       |
*--------------------------------------------------------------------------------*

Exocet r7c1 r7c2 r8c5 r9c7 2789 ; r9c9==r8c5

Due to a bug in my solver, I had missed the secondary equivalence r9c9==r8c5, which removes the 2 from r8c5 and then trivially from both base cells. Also examining the Exocet tier it's obvious that if 2 is in one target cell it must be in the other.

So, there now appears to be 4 ways to eliminate an Exocet digit that must occupy both target cells :

1. The 1 truth in the cross lines proposition (theorem ?).

2. It comes about naturally via normal Exocet eliminations.

3. A simple analysis may prove that if the digit occupies one target cell it must occupy the other, and

4. Some more complex analysis, such as Danny's templates method.

Leren

[daj95376]

Hi Danny - thanks for the explanation. You don't say when you use your full template analysis instead of a simple chain (as you showed for the 9 in the first Exocet) to prove a digit True in both target cells. ie do you always use your template analysis and then look for a simple chain as a simpler explanation for the template analysis result (to answer questions from simpleton's like me) ?

I always use template analysis, and then manually look for a simple chain as an explanation.

For example, in the second Exocet, it's obvious that if 7 is in 1 target cell it must be in the other target cell, so can be removed from the base and target cell, but did you actually find it via templates ?

7 has 2 truths in the cross lines so I wouldn't have checked that it must occupy both Target cells - something for me to add to my solver.

It's derived through "second set of logic" from my earlier message.

_

[daj95376]

[Withdrawn: working from an incorrect understanding of the JExocet properties.]

[Leren]

<Edit> - removed reference to typo in previous post - fixed.

I'm starting to see a pattern here : if, for a non-aligned Exocet, a digit d appears in the target cell mini-lines only in one or both target cells, then d can't be part of the Exocet solution.

To illustrate, look at the dobrichev puzzle.

Code: Select all

                   V                  V                  V
*--------------------------------------------------------------------------------*
| 23789   56      2789     | 346789  789     3469     | 29      45      1        |
| 1       56      789      | 46      789     2        | 78      3       45       |
| 23789   237     4        | 3789    5       1        | 6       29      78       |
|--------------------------+--------------------------+--------------------------|
| 4       8       1        | 239     6       39       | 5       7       239      |
| 237     237     5        | 239     1       8        | 4       29      6        |
| 23      9       6        | 5       4       7        | 1       8       23       |
|--------------------------+--------------------------+--------------------------|
|B2789   B27      3        | 1       2789    46       | 2789    456     45       |
| 5       1       2789     | 46789  T2789    469      | 3       46      29       |
| 6       4       2789     | 789     3       5        |T2789    1       78       |
*--------------------------------------------------------------------------------*

2 doesn't appear in r8c46 or r9c89, so if r9c7 is 2 and is in the base cells, r8c5 is 2 and vice versa. So 2 occupies either no target cell or both target cells and consequently can be eliminated in the base and target cells.

Also, suppose target cell r8c5 did not hold a 2 and, say, r9c4 held a 2 (so the 2 was unresolved in box 8) then if if r9c7 is 2 and is in the base cells, there is no 2 in Box 8, so 2 can't be in r9c7 and can be eliminated in the base and target cells.

This pattern also holds for the 7 in your -1479- QExocet Base = r6c89 Target = r4c2,r5c5==r4c3.

Is this new ? At least it's new for me.

Leren

[daj95376]

Is this new ? At least it's new for me.

I don't know why you like to start at the target cells and test for all permutations.

What I have is two streams from the base cells to the target cells. If the base cells are true for "2", then both target cells are true as well.

Code: Select all

 2r7c12 - r7c5                 = 2r8c5
 2r7c12 - r89c3 = r1c3 - r17c7 = 2r9c7


Since the other candidates, when true in the base cells, generate the classic finned Swordfish pattern, then "2" can be deleted in all base/target/secondary cells.


The other Exocets are more interesting!

Code: Select all

 ***   value   = <2> true   locks cells true:   r9c3        must be true
 ***   value   = <4> true   locks cells true:   r7c6
 ***   value   = <6> true   locks cells true:   r7c6
 ***   value   = <9> true   locks cells true:   r7c6,r9c3   eliminate in base/target/secondary cells

 ### -2469- QExocet   Base = r8c89   Target = r7c6,r9c3==r7c5


Code: Select all

 ***   value   = <2> true   locks cells true:   r5c8        must be true
 ***   value   = <7> true   locks cells true:   r9c9
 ***   value   = <8> true   locks cells true:   r9c9
 ***   value   = <9> true   locks cells true:   r5c8,r9c9   eliminate in base/target/secondary cells

 ### -2789- QExocet   Base = r12c7   Target = r5c8==r8c9,r9c9


Code: Select all

 ***   value   = <2> true   locks cells true:   r5c8        must be true
 ***   value   = <7> true   locks cells true:   r2c7
 ***   value   = <8> true   locks cells true:   r2c7
 ***   value   = <9> true   locks cells true:   r2c7,r5c8   eliminate in base/target/secondary cells

 ### -2789- QExocet   Base = r89c9   Target = r5c8==r1c7,r2c7


_

[David P Bird]

Leren, the way I've handled the checks for a JExocet is to say that no base candidate should need more than two 'S' cell cover houses (with nothing about a minimum count). Then when the pattern checks are passed any digit with a single cover house can be eliminated by theorem. Whether the targets are aligned or not is irrelevant as no other digit can pair with the rogue one.

The situation under discussion when the cover house counts are (a) =2, (b) =2, (c) >2, (d) =1
then falls into the Almost JE category
Either
1) JE = true: two digits from (abc)
2) JE = false: one of (abd) true in the base cells and both targets + (c) true in the other base cell and three 'S' cells.
This is a win-win situation because either way we can make significant eliminations.

Despite DAJ's contrary opinion, there are then a number of ways a manual player can use compatibility testing and AICs to try to resolve which of these cases applies. At this point if the targets are aligned (d) won't survive. I presume these are covered by your 'simple analysis' action category.

My computer monitor has died and I've had to resurrect an old one, I also have an appointment today so I haven't been able to spend more than a few minutes on Blues example, but at first sight, if that’s typical, the prospects aren't good.