The New Sudoku Players' Forum

by **ronk** » Thu Sep 19, 2013 4:35 pm

daj95376 wrote:When viewed as a Sashimi Swordfish, then [c1] and [c4] are [edit: the logical sectors to be] in the cover set ... leaving C=(8)r8c6 as the only spoiler cell for <8>.

A JExocet "sashimi swordfish" is not just any sashimi. For each base digit 'j' in base set "abcd", there must exist the strong inference (j)Q = (j)R, where Q and R are the targets.

As you have noted, for digit <8> of this puzzle there is a spoiler candidate S at r8c6.

Code: Select all: . . . | . . . | . . . . . . | . . . | . . . 8 / / | 8 / / | / Q8 / ----------+----------+---------- . . . | . . . | . . . / / / | 8 / / | R8 / / . . . | . . . | . . . ----------+----------+---------- . . . | . . . | . . . 8 / / | 8 / S8 | / / / . . . | . . . | . . .

The JExocet sashimi requirement above must be met when the spoiler is false and when the spoiler is true. When the spoiler r8c6 is false we have the sashimi swordfish below; no problem.

Code: Select all: . . . | . . . | . . . . . . | . . . | . . . 8 / / | 8 / / | / Q8 / ----------+----------+---------- . . . | . . . | . . . / / / | 8 / / | R8 / / . . . | . . . | . . . ----------+----------+---------- . . . | . . . | . . . 8 / / | 8 / / | / / / . . . | . . . | . . .

When the spoiler is true, however, a secondary spoiler appears at r3c1. A derived weak inference such that r6c8=8 => r3c1<>8 would solve this issue.

Code: Select all: . . . | . . / | . . . . . . | . . / | . . . S8 / / | 8 / / | / Q8 / ----------+----------+---------- . . . | . . / | . . . / / / | 8 / / | R8 / / . . . | . . / | . . . ----------+----------+---------- . . . | / / / | . . . / / / | / / 8 | / / / . . . | / / / | . . .

At the risk of stating the obvious, Q and R are the two fin cells for the sashimi swordfish in the first case, and for the sashimi x-wing (skyscraper) in the second case.

by **JC Van Hay** » Thu Sep 19, 2013 8:50 pm

Does 2=1+1 ? In the case of Puzzle #5, the answer is yes, as it contains 2 exocets, according to the definition given in page 1 !
Being more explicit than here :-D

:
Exocet : base=AALS (1689)r12c7

Code: Select all: +------------------------+-----------------------+-----------------------------+ | 356789 135679 156789 | 2578 5689 289 | (1689) 14689 146789 | | 456789 45679 56789 | 578 5689 1 | (689) 2 3 | | 7(689) 7(169) 2 | 7(8) 3 4 | 5 (1689) 7(1689) | +------------------------+-----------------------+-----------------------------+ | 24579 124579 1579 | 6 (14589) (89) | 1389 13489 12489 | | 4(69) 4(169) 3 | 4(18) 2 7 | (1689) 5 -4(1689) | | 24569 8 1569 | 145 (1459) 3 | 7 1469 12469 | +------------------------+-----------------------+-----------------------------+ | 23678 2367 678 | 9 (18) 5 | 4 1368 168 | | 3(689) 3(69) 4 | 23(18) 7 2(8) | 3(1689) -3(1689) 5 | | 1 359 589 | 348 (48) 6 | 2 7 89 | +------------------------+-----------------------+-----------------------------+

Rank 5 Logic
19 Truths = {1689R3 1689R5 1689R8 4679N5 4N6 12N7}
24 Links = {1c2457 4c5 6c127 8c147 9c127 8n8 5n9 1689b3 589b5 8b8}
2 Eliminations --> r5c9<>4, r8c8<>3

Code: Select all: r12c7=1->-1r3c89,r58c7;r3c2=1;-1r5c2;1r5c9=1r5c4-1r8c4=1r8c8 :=> 1r5c9=1r8c8 || r12c7=6->-6r3c89,r58c7;6r5c9=XWing(6r35c12)-6r8c12=6r8c8 :=> 6r5c9=6r8c8 || r12c7=8->-8r3c89,r58c7;8r8c1-8r3c1=8r3c4-8r5c4=8r5c9 || || || 8r8c46-(8=14)r79c5-(14=589)r46c5,r4c6-8r5c4=8r5c9 || || || 8r8c8 || || :=> 8r5c9=8r8c8 || r12c7=9->-9r3c89,r58c7;9r5c9=XWing(9r35c12)-9r8c12=9r8c8 :=> 9r5c9=9r8c8

Conclusion : r12c7=x->xr5c9=xr8c8 :=> r5c9,r8c8=1689 or -4r5c9,-3r8c8

And the same, mutatis mutandis, for the exocet : base=(1689)r79c9 [target=r3c8,r5c7].

by **David P Bird** » Thu Sep 19, 2013 11:05 pm

DAJ, I've now run through your grids to prove that any base digit is either in one base pair or the other but not both. We can take it as a theorem that these sashimi fish exist for any digit that only occurs in the S cells in two parallel sectors.

This leaves us with the Leren's 'rogue' digit that occurs in 3 sectors. Now you need to find a supporting AIC to link a spoiler cell (the remote fin) to the cells in the JE band that don't see either pair of the base cells. On the other hand I need to find a supporting AIC linking it to another spoiler to show they can't be true together.

I suspect that our supporting AICs are strongly related, and would be interested in any cases when one approach succeeds when the other fails.

You originally wrote that you didn't like the using the supporting AIC one bit and I tend to concur as it amounts to using a derived inference (a door I don't want to open). My more recent attempt for puzzle 5 is longer but avoids this.

DPB

by **David P Bird** » Thu Sep 19, 2013 11:09 pm

Leren wrote:The DJE properties of puzzles such as Puzzle no 5 can be established from a different POV as follows.

Suppose you have shown that;

1. The puzzle has ADJE properties except that a single "Rogue" digit (in this case 8) has "Spoiler" cells;

and, via a separate process

2. The conjugate properties of the 2 AJE Base cell AAHSs.

The rest of your points follow directly from these two conditions, but what exactly is involved in proving the second one? For my 'aesthetic reasons', I hope it's not splitting off a series of tests, one for each digit.

Leren wrote: I've tested this property against all 2214 puzzles in Champagne's Conjugate AAHS file with no failures, so I am reasonably confident that it is generally true.

Unfortunately Champagne discarded any puzzles where his brute force methods showed the DJE was false, so it would be impossible for you to get any false positives from this collection.

In < this post > I found there were three cases 1) both JEs true, 2) one JE true, 3) both JEs false. So far I've only been looking to prove that both of them are true, but there should be ways to explore if at least one of them must be true.

DPB

by **daj95376** » Fri Sep 20, 2013 2:27 am

ronk wrote:When the spoiler is true, however, a secondary spoiler appears at r3c1. A derived weak inference such that r6c8=8 => r3c1<>8 would solve this issue.

Code: Select all
. . . | . . / | . . . . . . | . . / | . . . S8 / / | 8 / / | / Q8 / ----------+----------+---------- . . . | . . / | . . . / / / | 8 / / | R8 / / . . . | . . / | . . . ----------+----------+---------- . . . | / / / | . . . / / / | / / 8 | / / / . . . | / / / | . . .

At the risk of stating the obvious, Q and R are the two fin cells for the sashimi swordfish in the first case, and for the sashimi x-wing (skyscraper) in the second case.

You missed one other obvious possibility that completely ignores 8r3c1:

(8-2)r8c6 = (2)r1c6 - (2=578)r123c4 - (8)r5c4 = (R8)r5c7

Now, 8r8c6 true is directly linked to R8 true. Specifically, the solution to this grid is equivalent to my solution for DPB's first grid.

Just for completeness, if you had started with the grid after base set r12c7 true for <8>, then 8r8c6 true would be directly linked to (8)r5c9.

by **ronk** » Fri Sep 20, 2013 3:11 am

daj95376 wrote:
ronk wrote:When the spoiler is true, however, a secondary spoiler appears at r3c1. A derived weak inference such that r6c8=8 => r3c1<>8 would solve this issue.

Code: Select all
. . . | . . / | . . . . . . | . . / | . . . S8 / / | 8 / / | / Q8 / ----------+----------+---------- . . . | . . / | . . . / / / | 8 / / | R8 / / . . . | . . / | . . . ----------+----------+---------- . . . | / / / | . . . / / / | / / 8 | / / / . . . | / / / | . . .

You missed one other obvious possibility that completely ignores 8r3c1:

(8-2)r8c6 = (2)r1c6 - (2=578)r123c4 - (8)r5c4 = (R8)r5c7

Now, 8r8c6 true is directly linked to R8 true. Specifically, the solution to this grid is equivalent to my solution for DPB's first grid.

It wasn't obvious to me.

Using 8r5c4 as a spoiler for a hidden single in the target never crossed my mind, but I like it.

This may be as simple as it gets for this AJE in this puzzle.

000000000000001023002034500000600000003027050080003700000905400004070005100006270
20 Truths = {1689R358 2C6 123N4 12N7 79N9}
32 Links = {1c2479 6c1279 8c1479 9c1279 8n6 5n7 38n8 5n9 257b2 1689b3 1689b9}
33 Eliminations --> r1267c1<>6, r1246c1<>9, r1249c2<>9, r1c9<>1689, r3c9<>1689, r8c7<>1689,
r127c2<>6, r14c2<>1, r5c9<>4, r6c4<>1, r8c8<>3, r9c4<>8
Singles-to-the-end (stte).

daj95376 wrote:Just for completeness, if you had started with the grid after base set r12c7 true for <8>, then 8r8c6 true would be directly linked to (8)r5c9.

The grid shown is for base set r79c9, but yes, what you say would be correct for base set r12c7.

by **champagne** » Fri Sep 20, 2013 4:52 am

I coded the extended search to validate a JE digit and applied it to the collection (last public update) of 963 K puzzles of the data base of potential hardest.

In that collection, with my last code, about 100K puzzles had none of the key exotic properties.
V loop, Exocet, Rank 0 logic, symmetry of given, conjugated AAHS.

Applying the extended search for JE, I got 7686 new puzzles having a JE, some of them having a double jexocet not seen as a conjugated AAHS (one surprising point needing more checking)

As with any fresh code, it has to be confirmed, but this is IMO a good result although I was expecting a better yield.
I have still to check whether adding rules in the dynamic expansion can increase the yield.

I am checking the next lot of 75000 puzzles entered in the data base and I have in stand by or nearly ready another lot of puzzles, with at the end a significant amount of puzzles in the 26 clues area;

I intend to publish an update of the data base after these puzzles have been entered in the base and analysed.

by **champagne** » Fri Sep 20, 2013 6:48 am

Here after a list of puzzles where a double exocet can be seen using more than a fish expansion for some digits

These puzzles are not in the group of puzzles where I could establish complementary AAHS, but it could be that they are in the list I published

Hidden Text: Show

EDIT I have to revise the definition of a double exocet in my code. The first puzzle has two exocets in the same band, but with a target belonging to the other base. The 2 JE's are not complementary.

by **Leren** » Fri Sep 20, 2013 12:12 pm

Hi David, The best way I can answer your questions at the moment is to highlight the difficulty of some of the puzzles in Champagnes list.

Here's puzzle no 932 in the list ........1....12.....34...5........64.346..1..16...7.......8.5...5.3....694....31.;748933;dob;12_12_19;32;2789 ;9;2789

Code: Select all: *--------------------------------------------------------------------------------* | 245678 2789 25789 | 5789 35679 35689 | 246789 24789 1 | | 45678 789 5789 | 5789 1 2 | 46789 34789 3789 | | 2678 1 3 | 4 679 689 |t26789 5 T2789 | << |--------------------------+--------------------------+--------------------------| | 2578 2789 25789 | 12589 2359 13589 |B2789 6 4 | | 2578 3 4 | 6 259 589 | 1 t2789 5 | << | 1 6 2589 | 2589 4 7 |B289 2389 35 | |--------------------------+--------------------------+--------------------------| | 3 27 1267 | 1279 8 1469 | 5 2479 b279 | | 278 5 1278 | 3 279 149 | 4 T24789 6 | << | 9 4 2678 | 257 2567 56 | 3 1 b278 | *--------------------------------------------------------------------------------* 278 278 279 89

This PM shows a possible 2789 Double Exocet with 3 Rogue digits 278 and 5 Spoilers: 2r3c1 + r8c3, 7r5c1 + r8c3, 8r8c3 (as best as I can determine). Is this an AAADJE ?

The task to prove the DJE via your method would be to prove that (r3c1 OR r8c3 <> 2) AND (r5c1 OR r8c3 <> 7) AND r8c3 <> 8 - no mean feat I imagine.

The method I outlined uses a consistent way of handling all variations of these puzzles - proving the conjugacy of the 2 Base AAHs enables the underlying JExocets to be
revealed. As is typical in these puzzles once the AAHs eliminations are made all would be Double Exocet eliminations follow via basic follow-on moves.

Of the 2214 puzzles in the file I found 2010 to have a single Rogue digit, others having 2 or 3 Rogue digits and 55 for which I could not prove a complementary AAHS.

Leren

by **ronk** » Fri Sep 20, 2013 1:35 pm

Leren wrote:The best way I can answer your questions at the moment is to highlight the difficulty of some of the puzzles in Champagnes list.

Here's puzzle no 932 in the list ........1....12.....34...5........64.346..1..16...7.......8.5...5.3....694....31.;748933;dob;12_12_19;32;2789 ;9;2789

The most difficult moves required to solve this puzzle are (arguably) w-wings and m-wings. Do you have something a bit tougher?

by **daj95376** » Fri Sep 20, 2013 4:50 pm

Leren wrote:Here's puzzle no 932 in the list

........1....12.....34...5........64.346..1..16...7.......8.5...5.3....694....31.;748933;dob;12_12_19;32;2789 ;9;2789

Code: Select all
*--------------------------------------------------------------------------------* | 245678 2789 25789 | 5789 35679 35689 | 246789 24789 1 | | 45678 789 5789 | 5789 1 2 | 46789 34789 3789 | | 2678 1 3 | 4 679 689 |t26789 5 T2789 | << |--------------------------+--------------------------+--------------------------| | 2578 2789 25789 | 12589 2359 13589 |B2789 6 4 | | 2578 3 4 | 6 259 589 | 1 t2789 5 | << | 1 6 2589 | 2589 4 7 |B289 2389 35 | |--------------------------+--------------------------+--------------------------| | 3 27 1267 | 1279 8 1469 | 5 2479 b279 | | 278 5 1278 | 3 279 149 | 4 T24789 6 | << | 9 4 2678 | 257 2567 56 | 3 1 b278 | *--------------------------------------------------------------------------------* 278 278 279 89

This PM shows a possible 2789 Double Exocet with 3 Rogue digits 278 and 5 Spoilers: 2r3c1 + r8c3, 7r5c1 + r8c3, 8r8c3 (as best as I can determine). Is this an AAADJE ?

I don't know why everyone keeps finding so many spoilers. I see three, in cell r8c3, and they can be resolved using one chain.

Code: Select all: ........1....12.....34...5........64.346..1..16...7.......8.5...5.3....694....31. #748933;dob;12_12_19;32;2789 ;9;2789 after basics +--------------------------------------------------------------------------------+ | 245678 2789 25789 | 5789 35679 35689 | 246789 24789 1 | | 45678 789 5789 | 5789 1 2 | 46789 34789 3789 | | 2678 1 3 | 4 679 689 | 26789 5 2789 | |--------------------------+--------------------------+--------------------------| | 2578 2789 25789 | 12589 2359 13589 | 2789 6 4 | | 2578 3 4 | 6 259 589 | 1 2789 25789 | | 1 6 2589 | 2589 4 7 | 289 2389 23589 | |--------------------------+--------------------------+--------------------------| | 3 27 1267 | 1279 8 1469 | 5 2479 279 | | 278 5 1278 | 3 279 149 | 24789 24789 6 | | 9 4 2678 | 257 2567 56 | 3 1 278 | +--------------------------------------------------------------------------------+ # 165 eliminations remain (1)r8c3 = (1-6)r7c3 = (6-4)r7c6 = (4-1)r8c6 = (1)r8c3 => r8c3<>278 after chain and Naked Single +--------------------------------------------------------------------------------+ | 245678 2789 25789 | 5789 35679 35689 | 246789 24789 1 | | 45678 789 5789 | 5789 1 2 | 46789 34789 3789 | | 2678 1 3 | 4 679 689 | 26789 5 2789 | |--------------------------+--------------------------+--------------------------| | 2578 2789 25789 | 12589 2359 13589 | 2789 6 4 | | 2578 3 4 | 6 259 589 | 1 2789 25789 | | 1 6 2589 | 2589 4 7 | 289 2389 23589 | |--------------------------+--------------------------+--------------------------| | 3 27 267 | 1279 8 1469 | 5 2479 279 | | 278 5 1 | 3 279 49 | 24789 24789 6 | | 9 4 2678 | 257 2567 56 | 3 1 278 | +--------------------------------------------------------------------------------+ # 160 eliminations remain ### -2789- QExocet Base = r46c7 Target = r3c9==r7c8,r8c8==r2c9 ### -2789- QExocet Base = r79c9 Target = r5c8,r3c7==r6c8 *** double QExocet

BTW: I agree with ronk about alternate solutions to this puzzle making ADJE over-kill.

ADDENDUM:

Using the conjugate property of the DJE base cells, we get r56c9,r8c7<>2789. This is followed by a cascade of Singles that stop at a grid containing only <2789> as candidates.

Code: Select all: +--------------------------------------------------------------+ | 6 2789 2789 | 789 5 3 | 2789 4 1 | | 4 789 5 | 789 1 2 | 6 3 789 | | 278 1 3 | 4 79 6 | 2789 5 2789 | |--------------------+--------------------+--------------------| | 5 2789 2789 | 29 3 1 | 2789 6 4 | | 27 3 4 | 6 29 8 | 1 279 5 | | 1 6 289 | 5 4 7 | 289 289 3 | |--------------------+--------------------+--------------------| | 3 27 6 | 1 8 4 | 5 279 279 | | 278 5 1 | 3 27 9 | 4 278 6 | | 9 4 278 | 27 6 5 | 3 1 278 | +--------------------------------------------------------------+ # 61 eliminations remain

by **David P Bird** » Sat Sep 21, 2013 10:14 am

Leren,

Ignoring the more basic eliminations, DAJ's chain suits me admirably.

Off Topic: Show

DPB

by **Leren** » Sat Sep 21, 2013 11:46 am

My apologies to all for that bad example, I had L3 wings turned off and came up with an "unnecessary" ADJE.

I'll do another run with proper settings and see if I can find a decent example.

Leren

by **Leren** » Sun Sep 22, 2013 12:31 am

ronk wrote:
Leren wrote:The best way I can answer your questions at the moment is to highlight the difficulty of some of the puzzles in Champagnes list.

Here's puzzle no 932 in the list ........1....12.....34...5........64.346..1..16...7.......8.5...5.3....694....31.;748933;dob;12_12_19;32;2789 ;9;2789

The most difficult moves required to solve this puzzle are (arguably) w-wings and m-wings. Do you have something a bit tougher?

I think this might be a bit tougher

98.7..6..7...6......6....5.4..3.9.....8.573.....8......2......1..9...5....7.839..;975059;GP;13_03;21;1234 ;3;34

Leren

by **champagne** » Sun Sep 22, 2013 9:35 am

I have to thanks abi who pointed a new kind of pattern having some solving potential.

Here is the first puzzles in the list I published recently

98.7..6..7..5...8...6.....48..3...9..2...........1...85.....9...379....5..8.5..3.;174173;GP;12_11

That list is made of puzzles where several exocet were found by the "extended JE search" and where my code qualified the pattern as a double exocet.
This was a bug in my code, one target of the exocet based in r7c23 belongs to the second exocet based in r8c56
My PM at the start is the following

Code: Select all: 9 8 12345 |7 234 1234 |6 125 123 7 14 1234 |5 2346 12346 |123 8 9 123 15 6 |128 2389 12389 |12357 1257 4 -------------------------------------------------------- 8 14567 145 |3 2467 24567 |12457 9 1267 1346 2 13459 |468 46789 456789 |13457 14567 1367 346 4567 3459 |246 1 245679 |23457 24567 8 -------------------------------------------------------- 5 146 124 |12468 234678 1234678 |9 12467 1267 1246 3 7 |9 246 1246 |8 1246 5 1246 9 8 |1246 5 12467 |1247 3 1267

The potential complementary AAHS are r7c12 and r8c56

It is here very easy to prove that digits 1;4;6 can not share both AAHS.
No question to prove it for digit 2, the solution includes the digit "2" in both bases. (despite the fact that both exocet can be established)

The fact that 3 digits can not share the 2 AAHS is enough to force the last digit to be in at least one of the AAHS, giving a small potential for eliminations.

But here, both AAHS see the cell r8c1 containing only the 4 digits, so the 2 AAHS can not have complementary digits. the digit 2 must be in both AAHS.

This is a very common situation in the list I posted. I suspect that we can extract other examples "not having the 2 exocets established".

I'll try to extract such puzzles from the data base

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almost complementary AAHS