[Withdrawn: abandoning ship as well.]

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[Withdrawn: abandoning ship as well.]

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Last edited by daj95376 on Sat Aug 09, 2014 3:28 am, edited 1 time in total.

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

(withdrawn)

Last edited by blue on Fri Aug 08, 2014 3:54 pm, edited 1 time in total.

- blue
**Posts:**894**Joined:**11 March 2013

Leren wrote: Hi David, I don't like this wording because it seems to me to be a circular argument

On re-reading it I can see the wording could be improved, but I consider the argument to be linear. It produces two conclusions:

1. For the JE to be valid all candidates must be restricted to fewer than three S cells.

2. In a valid JE any base digit candidate restricted to one S cell can be eliminated from the base cells.

From these, one of the things we can conclude is that, only if the JE is invalid can a candidate be true in the base cells and just one S cell.

However, as identified by the introductory sentence, my proof is reverse-engineered (and so is yours). Consequently it may or may not fit in with the rest of a pattern description depending on the presentation order of the other parts. I'll check that out at the appropriate time.

David

- David P Bird
- 2010 Supporter
**Posts:**1043**Joined:**16 September 2008**Location:**Middle England

- Code: Select all
`v v v`

*--------------------------------------------------------------------------------*

| 9 8 124 | 7 125 1235 | 6 345 12345 |

| 7 3 124 | 8 1256 1256 | 9 45 1245 |

| 12 5 6 | 9 4 123 | 1278 378 12378 |

|--------------------------+--------------------------+--------------------------|

| 8 124 127 | 5 126 12467 | 3 9 1467 |

| 1234 6 1237 | 124 8 9 | 147 457 1457 |

| 14 9 5 | 14 3 67 | 78 2 678 |

|--------------------------+--------------------------+--------------------------|

|T2345 7 23-8 | 6 25 245 | 248 1 9 |

| 1246 124 9 | 1234 7 8 | 5 B346 B234 |

|T12456 124 128 |t23-14 9 1245 | 2478 34678 23478 |

*--------------------------------------------------------------------------------*

^ ^ ^

I noticed that Danny and blue have withdrawn their posts regarding the existence of an Exocet in this grid. OK I'll stick my neck out and claim that an Exocet does exist.

A disconcerting aspect of this puzzle is that the companion cell r7c4 to the target cell r9c4 holds the base digit 6, but lets not worry about that too much just yet.

If any combination of base digits 234 were to be in the base cells then since r7c4 does not hold these digits, Twin JExocet conditions would apply and the appropriate base digit would be in two of the target cells with the third target cell containing 5.

If r8c8 is 6 then r8c1 => - 6 r8c1 = 6 r9c1. Also r9c4 <> 6, so if 6 is in a base cell it is in exactly 1 target cell r9c1 with r7c1 = 5.

OK, so if the other base cell is X = 2, 3 or 4 then what is in the other target cell r9c4 ? Well in that case r789c1 <> X, r8c4 <> X, X must appear in the S cells of Column 7. X Must also appear in the S cells of Column 1. If it doesn't, then since r789 <> X Column 1 would be void of X so X could not appear in the base cells with 6. If X does appear in the S cells of Column 1 then it can't appear in the S cells of Column 4 (since it's also in the S cells of Column 7), so r9c4 = X.

So if base cell r8c9 holds X = 234 then either r9c4 holds X = 234 or X can't be in r8c9 if r8c8 = 6.

So summarizing all this, whatever 2 digits are in the base cells, one each of those digits are in 2 of the target cells (with the third target cell r7c1 or r9c1 being 5).

That's an Exocet ! Even though a companion cell holds a base digit! Of course it's not a JEXocet because the proof for digit 6 does not rely on the number of truths in the S cells of the cross lines. Could I coin the term GExocet (ie and Exocet in the more Generalised sense of term, that is not a JExocet) ?

Checking my solver's detection of this GExocet I found that it went through all this logic except for the part in red. However I'm still trying to decide whether the red part is necessary, or does it automatically follow from the truth counts for digits 234.

In other words, is it correct to conclude that if 1. digits X = 234 have at most 2 truths in the S cells => X must appear in at least one target cell if it's in a base cell; and 2. Digit 6 must appear in exactly 1 target cell if it's in a base cell

then whatever combination of digits 2346 appear in the base cells then one each of those digits appears in a Target cell. I think it's OK but I'm open to arguments to the contrary.

Leren

- Leren
**Posts:**3871**Joined:**03 June 2012

David P Bird wrote: On re-reading it I can see the wording could be improved, but I consider the argument to be linear.

Hi David, I now think your argument is in fact not circular, for the following reason.

- Code: Select all
`CL1 CL2 CL3 CL1 CL2 CL3`

. . a | . A . | A . . . . \ | . \ . | \ . .

. . \ | . \ . | \ . . . . \ | . \ . | \ . .

. . B | . b . | b . . . . B | . b . | b . .

------+--------+------- ------+--------+-------

. . \ | . \ . | \ . . . . \ | . \ . | \ . .

. . b | . B . | B . . . . b | . B . | B . .

. . \ | . \ . | \ . . . . \ | . \ . | \ . .

------+--------+------- ------+--------+-------

a b . | . AB . | AB . . a b . | . AB . | AB . .

. . . | . a . | \ . . . . . | . c . | \ . .

. . . | . \ . | a . . . . . | . \ . | d . .

The diagram on the left shows that if a digit a has one truth in the S cells and digit b has 2 truths in the S cells then a is forced into both target cells and a contradiction exists for digit B in CL2 and CL3.

The diagram on the right shows that if digit b has 2 truths in the S cells and if it doesn't appear in either target cell then a contradiction exists for digit B in CL2 and CL3 => b must appear in at least 1 of the target cells.

The 2 statements in red appear to contradict each other, but in fact they don't because the situation on the left is false. In fact the diagram on the left is a particular case of the diagram on the right. c and d could be any digits, as long as they both aren't b - they could both be equal to a.

I haven't thought this through yet but one rider to your statement regarding a base digit with one Truth in the S cells may be that it would have to be a candidate in both Target cells and be forced into both target cells if it were true in a base cell, thus proving it to be false in a base cell. The grid in my previous post is a case in point, although this case was not a JExocet, but a GExocet.

Leren

- Leren
**Posts:**3871**Joined:**03 June 2012

[Withdrawn: abandoning ship as well.]

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Last edited by daj95376 on Sat Aug 09, 2014 3:26 am, edited 1 time in total.

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

Hi Danny, the following is what I think are the full set of eliminations for <2346> GExocet - see if you agree.

r79c1 must both resolve to a base digit or 5 so r9c1 <> 1

r7c3==r9c4 so r7c3 <> 8 and r9c4 <> 14

But also r7c1==r9c6 so r7c1 <> 3 and r9c6 <> 1

Also, I noted with interest your troublesome observation post.

My understanding is that if a base digit has 2 truths in the S cells then what has been proven is that if it is in a base cell it must be in at least one target cell.

The conventional wisdom seems to be that if you have this pattern for all 3 or 4 base digits then this implies that the same pair of digits will appear in the target cells as are in the base cells.

But what happens if a base digit has two S cell truths but can be shown (by some other method) to occupy both target cells if it is in the base. Presumably that means that it can be removed from the base and target cells in a similar way that a digit with 1 truth in the S cells can be - the only difference being that it's presumably quite rare for a 2 truth digit to have this property.

Furthermore, what happens if all, or all but 1 base digits have this property. Then presumably a JExocet does not exist ! So, to be absolutely certain that a JEoxcet has been proven, not only do all digits have to satisfy the 2 Truth rule but at least 2 digits must be shown to occupy at most 1 target cell if they are in the base. No doubt there would be a lot of work required to make this last check and only in extremely rare cases would it disprove the JEoxcet.

I've always been nonplussed by this conventional wisdom argument and I can now see why - it's technically not always true but is true in practice in an overwhelming majority of cases.

Leren

r79c1 must both resolve to a base digit or 5 so r9c1 <> 1

r7c3==r9c4 so r7c3 <> 8 and r9c4 <> 14

But also r7c1==r9c6 so r7c1 <> 3 and r9c6 <> 1

Also, I noted with interest your troublesome observation post.

My understanding is that if a base digit has 2 truths in the S cells then what has been proven is that if it is in a base cell it must be in at least one target cell.

The conventional wisdom seems to be that if you have this pattern for all 3 or 4 base digits then this implies that the same pair of digits will appear in the target cells as are in the base cells.

But what happens if a base digit has two S cell truths but can be shown (by some other method) to occupy both target cells if it is in the base. Presumably that means that it can be removed from the base and target cells in a similar way that a digit with 1 truth in the S cells can be - the only difference being that it's presumably quite rare for a 2 truth digit to have this property.

Furthermore, what happens if all, or all but 1 base digits have this property. Then presumably a JExocet does not exist ! So, to be absolutely certain that a JEoxcet has been proven, not only do all digits have to satisfy the 2 Truth rule but at least 2 digits must be shown to occupy at most 1 target cell if they are in the base. No doubt there would be a lot of work required to make this last check and only in extremely rare cases would it disprove the JEoxcet.

I've always been nonplussed by this conventional wisdom argument and I can now see why - it's technically not always true but is true in practice in an overwhelming majority of cases.

Leren

- Leren
**Posts:**3871**Joined:**03 June 2012

Clearly JExocets are a subset of Exocets, but what defines their envelope?

It was my intention for JExocets to be those Exocets that manual players could recognise as patterns with a bit of perseverance. This then leads us back to the thorny question of defining what a pattern is! I've covered this topic before, but to my mind there are three important aspects to consider in judging if a pattern is acceptable:

a) the number of its identifying elements should be bounded.

b) no logic tracking should be required.

c) It should occur frequently enough to make a search worthwhile.

For me then JExocets just about qualify. In searching for them there are a clear set of conditions that can be tested progressively that will identify the vanilla form, and when these are nearly met there are simple tests is to see if there any locked candidates around that will allow more exotic varieties. Both these steps can be achieved by counting alone. Once one is found it's then simple to identify its sub-type and apply the proven inferences it provides.

Champagne's method of verifying the existence of an Exocet (two cells that must contain different members of a base pair) allows a network of AICs to be used to verify them. Therefore his search routine would also find those that could be found with short unbranched AICs. However I believe the vast majority of the Exocets he found were JExocets. That's one reason that 'single chain' Exocets don't excite me. Another is that the theorems provided by different chain configurations are likely to be all different, and a third is that discussions about them make browsing this thread like wading through treacle.

It was my intention for JExocets to be those Exocets that manual players could recognise as patterns with a bit of perseverance. This then leads us back to the thorny question of defining what a pattern is! I've covered this topic before, but to my mind there are three important aspects to consider in judging if a pattern is acceptable:

a) the number of its identifying elements should be bounded.

b) no logic tracking should be required.

c) It should occur frequently enough to make a search worthwhile.

For me then JExocets just about qualify. In searching for them there are a clear set of conditions that can be tested progressively that will identify the vanilla form, and when these are nearly met there are simple tests is to see if there any locked candidates around that will allow more exotic varieties. Both these steps can be achieved by counting alone. Once one is found it's then simple to identify its sub-type and apply the proven inferences it provides.

Champagne's method of verifying the existence of an Exocet (two cells that must contain different members of a base pair) allows a network of AICs to be used to verify them. Therefore his search routine would also find those that could be found with short unbranched AICs. However I believe the vast majority of the Exocets he found were JExocets. That's one reason that 'single chain' Exocets don't excite me. Another is that the theorems provided by different chain configurations are likely to be all different, and a third is that discussions about them make browsing this thread like wading through treacle.

- David P Bird
- 2010 Supporter
**Posts:**1043**Joined:**16 September 2008**Location:**Middle England

[Withdrawn: abandoning ship as well.]

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Last edited by daj95376 on Sat Aug 09, 2014 3:26 am, edited 1 time in total.

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

(withdrawn: abandoning ship)

Last edited by blue on Sat Aug 09, 2014 6:06 am, edited 2 times in total.

- blue
**Posts:**894**Joined:**11 March 2013

blue wrote:

The last line is interesting. I hadn't spotted the equivalence.

It's true of course, even if r7c1 contains a 5.

You don't need the (secondary) equivalence, though, to show that r7c1<>3.

If 3 was true in a base cell, then it would be true in r9c4, and r79c1 would contain a 5 and the value from the other base cell.

If 3 was false in both base cells, then r79c1 would contain a 5 and one of the other base digits.

Either way r79c1 can't contain a 3.

The r7c1==r9c6 equivalence was new for me and I'll now write it into my solver. I might suggest that you could writer r7c1===r9c6, the extra = sign indicating that an additional check for the S/L digit needs to be made to prove the equivalence.

If 3 was true in a base cell, then it would be true in r9c4, and r79c1 would contain a 5 and the value from the other base cell.

Hmmm .... This sounds like an argument from the conventional wisdom school. Strictly speaking, if all you've done is count the S cell truths, 3 might be true in 2 target cells if it's in the base, so the underlined words are not necessarily true. However if 3 was true in 2 target cells it would be false in both the base cells and target cells, so r7c1 <> 3 in this case also. I only mention this to show how inured we all are to concluding that a digit is true in exactly 1 target cell if its true in the base when all that has been done is check that its S cell count is 2.

Leren

- Leren
**Posts:**3871**Joined:**03 June 2012

Congratulations gentlemen. With the preceding three posts you have completed your hijacking of this thread. You now have control.

- David P Bird
- 2010 Supporter
**Posts:**1043**Joined:**16 September 2008**Location:**Middle England

David,

i agree, that these posts were not about JExocets, but they are related, interesting and keep your thread living (e.g. they brought me to reread your definitions). So i don't quite understand your reaction.

Anyway it should not be a problem for a moderator to put them into an own thread, if you want that.

i agree, that these posts were not about JExocets, but they are related, interesting and keep your thread living (e.g. they brought me to reread your definitions). So i don't quite understand your reaction.

Anyway it should not be a problem for a moderator to put them into an own thread, if you want that.

- eleven
**Posts:**2393**Joined:**10 February 2008

Eleven, thanks for your response, but as you know, I believe in taking a disciplined and well-ordered approach to things, which is neither accepted nor respected by the free spirits that now dominate this forum, so I'm really out of place here.

David

David

- David P Bird
- 2010 Supporter
**Posts:**1043**Joined:**16 September 2008**Location:**Middle England

Accepted, though i regret it.

- eleven
**Posts:**2393**Joined:**10 February 2008