[Withdrawn: example probably of little use.]

Last edited by daj95376 on Sun Apr 08, 2012 2:46 pm, edited 1 time in total.

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

daj95376 wrote:ronk wrote:The "exocet" definition should be expanded a bit or, depending upon one's POV, the requirements relaxed a bit. Below is the exocet in Golden Nugget and a only-slightly-different pattern in GP-kz0 (morphed).

After reading some recent posts on Exocet puzzles, I'm testing the following Exocet logic.

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`1) Select possible base and target cells (using template solver output).`

2) For each value in the base cells, use coloring to logically perform:

2a) Assume value is true in base cells and perform eliminations.

2b) Perform all resulting Hidden Singles for this value.

2c) Derive an X-Chain (possibly grouped SI's) from one target cell to the other target cell.

(David's two given/solved cells become important here.)

3) Exocet!

as I wrote in the opening posts, the process used to prove the exocet is not important, except for a player.

A player needs something that can easily be seen, and I quite understand david's search for specific patterns and rules.

For a computer, IMO the easiest way (brute force) is the safest way to cover all the field, but any alternative process can be used.

champagne

- champagne
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ronk wrote:champagne wrote:

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`sets`

2789R2 2789R4 2789R5 2789R7

linksets

89C2 79C6 79C8 28C9 r2c4 r2c5 r4c3 r4c5 r5c1 r5c7 r7c1 r7c3

Is there a reason you can't use Allan Barker's format? With his format, readers could simply cut and paste the puzzle (or pencilmarks) and the logic set. To do this, the above would need to look like:

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`16 Truths = {2R2457 7R2457 8R2457 9R2457}`

17 Links = {2c9 7c68 8c29 9c268 57n1 47n3 2n4 24n5 5n7 8b7}

While there is some flexibility in the format, AFAIK the minimum requirements are:You're already mostly doing 3) and 4). If it's OK with you and you start using this Xsudo syntax, I can help out by changing the syntax in your existing posts of this thread. I would simply copy and paste the logic set from Xsudo itself. Let me know.

- Truths and Links lists separately enclosed in braces, "{ ...... }",
- 'n' and 'N' for Truth and Link cells, respectively; for example, 2n4 for link r2c4 above ,
- upper case 'R', 'C, and 'N' in Truths,
- lower case 'r, 'c, and 'n' in Links.

I can explain some of the reasons why I have another writing, but I am not sure it helps.

For sure the format used by Allan will not change, he is not any more active in that field.

I understand from your post that you can prepare a XSUDO diagram just copying the puzzle and the proposed truth links (I have not the knowledge of XSUDO to do that).

In that case, I agree that the active program should meet the specifications of the "frozen" one.

If it's not to much work, I'll change the print in the old program (I think it's not)

I'll surely use AB specifications in the new one.

I have another small problem, I have huge files prepared with the current specification and it's a long task to re run the program.

So even if I change the old program, I'll still produce for a while the old print.

champagne

- champagne
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ronk wrote:An "exocet with a bonus" ???

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`....5...94..1...3..6.7..1....59......8..7.......5.2..73......6...8.9...2.1....4.. eleven-1452 ED=11.2/11.2/9.9`

18 Truths = {1346R18 1346C39 5N46}

27 Links = {1346r5 346c4 1346c6 156n3 18n46 45n9 1b9 3b19 4b37 6b37}

6 Eliminations --> r8c6<>57, r4c9<>8, r6c3<>9, r7c2<>4, r9c1<>6

Please would tell me what rank this solution is, and how it is calculated?

- David P Bird
- 2010 Supporter
**Posts:**1004**Joined:**16 September 2008**Location:**Middle England

David P Bird wrote:Please would tell me what rank this solution is, and how it is calculated?ronk wrote:An "exocet with a bonus" ???

18 Truths = {1346R18 1346C39 5N46}

27 Links = {1346r5 346c4 1346c6 156n3 18n46 45n9 1b9 3b19 4b37 6b37}

I only know that the raw rank is 9 = 27 -18. That's one of the reasons I love 0-rank logic.

- ronk
- 2012 Supporter
**Posts:**4764**Joined:**02 November 2005**Location:**Southeastern USA

champagne wrote:ronk wrote:Is there a reason you can't use Allan Barker's format? With his format, readers could simply cut and paste the puzzle (or pencilmarks) and the logic set. To do this, the above would need to look like:

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`16 Truths = {2R2457 7R2457 8R2457 9R2457}`

17 Links = {2c9 7c68 8c29 9c268 57n1 47n3 2n4 24n5 5n7 8b7}

While there is some flexibility in the format, AFAIK the minimum requirements are:You're already mostly doing 3) and 4). If it's OK with you and you start using this Xsudo syntax, I can help out by changing the syntax in your existing posts of this thread. I would simply copy and paste the logic set from Xsudo itself. Let me know.

- Truths and Links lists separately enclosed in braces, "{ ...... }",
- 'n' and 'N' for Truth and Link cells, respectively; for example, 2n4 for link r2c4 above ,
- upper case 'R', 'C, and 'N' in Truths,
- lower case 'r, 'c, and 'n' in Links.

...

If it's not to much work, I'll change the print in the old program (I think it's not)

I'll surely use AB specifications in the new one.

champagne

May-be slightly more optimistic news.

I have nearly closed the re coding of the old process in my new program.

I currently test the entire file and it should be closed in one day.

If the results are correct, I shall

1) update the data base using these results

2) give a new list of puzzles to be searched for other patterns with a good chance to find one

3) code step by the the other patterns found;

Meantime, I'll continue to enter extended definition for the SK loop and to work on examples describing some possible ways to crack puzzles having these properties.

A lot of work has been done recently in the thread "bi bi pattern in hardest puzzles", I have to digest that work prior to writing on that topic.

champagne

- champagne
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I prepared several examples on the topic

"what to do after an exocet has been seen"

and I used in the corresponding posts the "abi"loop

That loop contributes in a very elegant way to eliminations of solutions in the base of an exocet.

It is explained in that start for "fata morgana"

fata morgana

thanks to "abi" who founds it

"what to do after an exocet has been seen"

and I used in the corresponding posts the "abi"loop

That loop contributes in a very elegant way to eliminations of solutions in the base of an exocet.

It is explained in that start for "fata morgana"

fata morgana

thanks to "abi" who founds it

- champagne
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[edit: withdrawn; gave up trying to understand the unconventional notation]

Last edited by ronk on Tue May 22, 2012 2:56 am, edited 1 time in total.

- ronk
- 2012 Supporter
**Posts:**4764**Joined:**02 November 2005**Location:**Southeastern USA

ronk wrote:champagne wrote:FATA MORGANA V2

So we have 36r5c46

Are you saying it takes all this to show r5c46=36 [and r4c2<>24 due to the exocet]?

My sense of humour is not very good, so I have difficulties to answer.

This wants to be a training demonstration, so I could answer as well

- It's quite perfect on the logical side and I have never seen shorter (in logic)

or as well

- do you have a shorter proof, it will be interesting to study it

or may be

- is there any point needing more explanations

I am prepared to go in any direction as soon as I understand your point

- champagne
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ronk wrote:[edit: withdrawn; gave up trying to understand the unconventional notation]

Hi ronk,

I answer to the deleted posts, another player could have the same question

note: I don't think notation here are not conventional

at most, part of the logic gets out of the trivial one

Now your question "is it a proof linked to the assumption in the base"

the answer is clearly "yes".

I restart with the willingness to prove 13r4c56 false.

As in nearly all cases, we try (and fail) to show 13r4c56 true

The reduced PM looks like that using the exocet properties

Note the 6 in that PM plays no role

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`X 6+ 6+ |16+ 16+ 16+ |X 1+ X`

3+ 3+ X |3+ 3+ X |X X X

3+ X 6+ |136+ X 136+ |X X 1+

13+ 13 6+ |6+ 6+ X |x X 6+

X 6+ 6+ |13 X 13 |x 6+ X

x X 6+ |X 6+ X |3+ 13 16+

1+ X X |136+ X 136+ |3+ X 6+

X X X |X 6+ 6+ |X 6+ 6+

X 1+ X |13+ 13+ 13+ |3+ 3+ X

We have many way to express that UR's r15c46 and r57c46 are not authorized

The useful one in our case is

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`3r3c1|1r3c9`

1r7c1|3r7c7

these are simple logical 'OR'. Both can be true, but this is a strong inference

in columns 2 and 8 we have outside the target of the exocet 1 ^ 3

This is clearly a strong link (exclusive or) since the target must have one of the digits.

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`3r2c2 ^ 1r9c2`

1r1c8 ^ 3r9c8

I add here to more strong links to give an alternative proof

they are properties of the exocet

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`3r2c2 ^ 3r9c8`

1r9c2 ^ 1r1c8

we can now redo the inferred "abi" loop (a 100% standard AIC)

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`1r7c1 - 1r9c2 = 3r2c2 — 3r3c1 = 1r3c9 — 1r1c8 = 3r9c8 - 3r7c7 = 1r7c1`

as in any AIC,

even positions have the same status (true or false in the solution)

odd positions have the same status (true or false in the solution)

must be even or odd true;

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`even true gives {3r2c2;3r9c8 true} => 3r46c5 - 13r5c46`

odd true gives {1r9c2;1r1c8 true} => 1r46c5 - 16r5c46

this is the way abi showed it.

BTW, this is equivalent to show that each sub scenario

13r5c46 + 1r4c2 + 3r6c8

13r5c46 + 3r4c2 + 1r6c8

is false

we have an easy alternative proof that could be of interest in some case.

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`we use here 3r2c2 ^ 3r9c8 ; 1r9c2 ^ 1r1c8 as weak links`

1r1c8 - 1r9c2 = 3r2c2 — 3r3c1 = 1r3c9 — 1r1c8 AIC showing 1r1c8 false

1r9c2 - 1r1c8 = 3r9c8 - 3r7c7 = 1r7c1 - 1r9c2 AIC showing 1r9c2 false

which is not possible, one must be true

I see nothing hard to understand in that and no special notation ( | and ^ are the logical universal notations for a OR condition)

This is an inferred proof, but this is a very common way to establish a fact

All AIC's are pure ones and shown with the nnotation in use in that forum

so where is the problem???

champagne

- champagne
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Here's an alternative proof for this puzzle.

This is the grid after the Exocet deletions with "x" marking the cells where any digit in the base cells must be false:

These are the conjugate chains that would result if each of the digits were true in the base cells:

(1')r6c8 ^ (1")r1c8 ^ (1')r3c8 ^ (1"#2)r35c46 ^ (1'#2)r57c46 ^ (1")r7c1 ^ (1')r9c2 ^ (1")r4c2

(3')r6c8 ^ (3")r9c8 ^ (3')r7c7 ^ (3"#2)r57c46 ^ (3'#2)r35c46 ^ (3")r3c1 ^ (3')r2c2 ^ (3")r4c2

(6')r6c8 ^ (6")r8c8 ^ (6')r7c9 ^ (6"#2)r57c46 ^ (6'#2)r35c46 ^ (6")r3c3 ^ (6')r1c2 ^ (6")r4c2

Note that starting from the same cell in each case, the alternative Xwings in the centre of the chain appear in the opposite order for (1) than they do for (3) and (6). This means that if (1) is true in the base cells one of two UR's will be made with either of (3) or (6) depending on which way round they appear in the target cells. Therefore (1) cannot be a member of the base pair.

A quick way to check is to start from one target cell and note if an odd or even number of links is required to reach the same Xwing cells. Two digits having the same odd/even parity will be compatible, remembering that one will start true and the other false from the selected start point.

I don't know if this proof is specific to this case or if it generalises for other puzzles, but I suspect it will.

This is the grid after the Exocet deletions with "x" marking the cells where any digit in the base cells must be false:

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`*-------------------------*-------------------------*-------------------------*`

| 2458 2467 245678x | 126789x 16789 1678x | 24589 1248x 3 |

| 2348x 2347 1 | 23789x 3789 5 | 6 248 249 |

| 2358 9 2568 | 12368 4 1368 | 258 7 125 |

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

| 12348 136 2468 | 13678x 13678x 9 | 2347x 5 12467x |

| 7 12346x 2469x | 136 5 136 | 2349x 12346x 8 |

| 1389x 5 689x | 4 13678x 2 | 379 136 1679 |

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

| 145 8 457 | 1367 2 13467 | 3457 9 4567 |

| 249 247 3 | 5 6789 4678x | 1 2468 2467x |

| 6 1247 24579 | 13789x 13789 13478x | 234578x 2348 2457 |

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

These are the conjugate chains that would result if each of the digits were true in the base cells:

(1')r6c8 ^ (1")r1c8 ^ (1')r3c8 ^ (1"#2)r35c46 ^ (1'#2)r57c46 ^ (1")r7c1 ^ (1')r9c2 ^ (1")r4c2

(3')r6c8 ^ (3")r9c8 ^ (3')r7c7 ^ (3"#2)r57c46 ^ (3'#2)r35c46 ^ (3")r3c1 ^ (3')r2c2 ^ (3")r4c2

(6')r6c8 ^ (6")r8c8 ^ (6')r7c9 ^ (6"#2)r57c46 ^ (6'#2)r35c46 ^ (6")r3c3 ^ (6')r1c2 ^ (6")r4c2

Note that starting from the same cell in each case, the alternative Xwings in the centre of the chain appear in the opposite order for (1) than they do for (3) and (6). This means that if (1) is true in the base cells one of two UR's will be made with either of (3) or (6) depending on which way round they appear in the target cells. Therefore (1) cannot be a member of the base pair.

A quick way to check is to start from one target cell and note if an odd or even number of links is required to reach the same Xwing cells. Two digits having the same odd/even parity will be compatible, remembering that one will start true and the other false from the selected start point.

I don't know if this proof is specific to this case or if it generalises for other puzzles, but I suspect it will.

- David P Bird
- 2010 Supporter
**Posts:**1004**Joined:**16 September 2008**Location:**Middle England

David P Bird wrote:This is the grid after the Exocet deletions with "x" marking the cells where any digit in the base cells must be false:

Hi David,

This seems a key sentence to justify the chains

can you comment on it

- champagne
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champagne wrote:David P Bird wrote:This is the grid after the Exocet deletions with "x" marking the cells where any digit in the base cells must be false:

Hi David,

This seems a key sentence to justify the chains

can you comment on it

The true base digits are restricted to a diagonal of 3 mini-lines in the Exocet band and must be false in the other 18 cells.

This puzzle satisfies JExocet condition 3 because each digit is restricted to two rows in columns 258 – the most common situation.

In these columns a base digit will occur once in a target cell, and twice outside the Exocet band.

For example if (1) is a base digit, it must occupy 2 cells in r19c258, and so must be false in the other cells in rows 1 & 9.

These two rules determine which cells should be marked with "x"

When the chain for (1) is followed, it can therefore be assumed that (1) is false in all the marked cells which would make all the links conjugate.

DPB

- David P Bird
- 2010 Supporter
**Posts:**1004**Joined:**16 September 2008**Location:**Middle England

For Fata Morgana, it's possible to separately show r5c46=3 and r5c46=6, without depending upon [edit: the exclusions of] an exocet.

It appears this anti-symmetry is exploitable yet again.

________ (clickable thumbnails)

**pastable Xsudo logic sets: **Show

I suspect these logic sets use the same truths as David P Bird's post above but, if so, the POV is very different.

here Allan Barker wrote:A key point is the symmetry, both eliminations have 3 identical layers. Layers 3 and 6 are symmetric, layer 1 is anti-symmetric to 3 and 6.

It appears this anti-symmetry is exploitable yet again.

________ (clickable thumbnails)

I suspect these logic sets use the same truths as David P Bird's post above but, if so, the POV is very different.

Last edited by ronk on Tue May 22, 2012 3:37 pm, edited 1 time in total.

- ronk
- 2012 Supporter
**Posts:**4764**Joined:**02 November 2005**Location:**Southeastern USA

David P Bird wrote:

For example if (1) is a base digit, it must occupy 2 cells in r19c258, and so must be false in the other cells in rows 1 & 9.

DPB

That's ok now.

If I had to explain your 'x' I would show the inferred loops as that one for '1' true in the base

1r9c2 - 1r9c5 = 1r1c5 - 1r1c8 = 1r6c8 - 1r4c2 = 1r9c2 loop

Combining the 3 loops, and applying direct effect of the exocet, I find all your 'x'.

In "abi"s approach, the immediate consequence of a digit true in the base is also used, but we don't need more.

Then, (with a small jump skipped to go from one UR to the other) your chains can be understood.

I agree as well on the conclusion, but it's not so easy to follow.

Regarding the fact that eliminations are working for 13 16 and not for 36, I have a very simple clue in "abi"s approach.

To have a conflict, it's necessary to switch from one row/column holding the target to the other one.

This is possible only if the 2 digits linked to the UR are in 2 separate bands/stack.

here 36 are in stack 1 for the potential UR r35c46 and in stack 3 for the potential UR r57c46

This can not lead to a conflict;

- champagne
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