Withdrawn - response to DAJ

Last edited by David P Bird on Wed Jan 28, 2015 9:25 pm, edited 1 time in total.

- David P Bird
- 2010 Supporter
**Posts:**1007**Joined:**16 September 2008**Location:**Middle England

[Withdrawn: working from an incorrect understanding of the JExocet properties.]

Last edited by daj95376 on Wed Jan 28, 2015 1:16 pm, edited 3 times in total.

- daj95376
- 2014 Supporter
**Posts:**2624**Joined:**15 May 2006

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

- Leren
**Posts:**3217**Joined:**03 June 2012

[Withdrawn: working from an incorrect understanding of the JExocet properties.]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

_

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

_

Last edited by daj95376 on Wed Jan 28, 2015 1:18 pm, edited 2 times in total.

- daj95376
- 2014 Supporter
**Posts:**2624**Joined:**15 May 2006

Withdrawn - response to DAJ

Last edited by David P Bird on Wed Jan 28, 2015 9:26 pm, edited 1 time in total.

- David P Bird
- 2010 Supporter
**Posts:**1007**Joined:**16 September 2008**Location:**Middle England

- 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 ]

- blue
**Posts:**601**Joined:**11 March 2013

Blue wrote: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

- David P Bird
- 2010 Supporter
**Posts:**1007**Joined:**16 September 2008**Location:**Middle England

David P Bird wrote: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.

- blue
**Posts:**601**Joined:**11 March 2013

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

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
**Posts:**3217**Joined:**03 June 2012

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:

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

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

Last edited by Leren on Wed Jan 28, 2015 3:00 am, edited 3 times in total.

- Leren
**Posts:**3217**Joined:**03 June 2012

Leren wrote: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
- 2014 Supporter
**Posts:**2624**Joined:**15 May 2006

[Withdrawn: working from an incorrect understanding of the JExocet properties.]

Last edited by daj95376 on Wed Jan 28, 2015 1:18 pm, edited 2 times in total.

- daj95376
- 2014 Supporter
**Posts:**2624**Joined:**15 May 2006

<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.

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

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

Last edited by Leren on Wed Jan 28, 2015 4:38 am, edited 1 time in total.

- Leren
**Posts:**3217**Joined:**03 June 2012

Leren wrote: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

_

- daj95376
- 2014 Supporter
**Posts:**2624**Joined:**15 May 2006

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.

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.

- David P Bird
- 2010 Supporter
**Posts:**1007**Joined:**16 September 2008**Location:**Middle England