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Wednesday 22-Jan-2020

... by: Cenoman

@SuDokuFan,
'Any flaws in the logic?' [quote]

None.
I have read again DPB's compendium. In his maps, companion cells are present for the JE2 pattern (The '2' referencing the presence of two target cells) and they are used to differentiate the JE2 pattern from the JE+ pattern, where the targets are replaced by two object pairs (containing a non-base locked digit). In JE+ pattern, the mirror nodes are unknown until the location of the base digits in their respective object pair are known. But in WU#383, there is a "GE2" pattern and your logic is right.
I apologize for my spurious comment.

Wednesday 22-Jan-2020

... by: numpl_npm

[123]J7 -> [123]G2 7G3 6G9
| 1J7 -> 1C13 1[F13|F6]=[1F13-G2|1F6-1H4-G2]=1G2
| 2J7 -> 2[F1|F56]=[2F1-G2|2F56-H4-G2]=2G2
| 3J7 -> 3D89 3[C1|C56]=[3C1-G2|3C56-H4-G2]=3G2

[14|24|34]GH8 -> [23|13|12]GJ2 4H3
| 14GH8 -> 23HJ7 2C56 3D89
| _ _ _ _ . 2[F1|F56]=[2F1-GJ2|2F56-HJ4-G2]=2GJ2
| _ _ _ _ . 3[C1|C56]=[3C1-GJ2|3C56-HJ4-G2]=3GJ2
| _ _ _ _ . 23GJ2
| 24GH8 -> 13HJ7 1A89 3D89
| _ _ _ _ . 1[F13|F6]=[1F13-GJ2|1F6-HJ4-G2]=1GJ2
| _ _ _ _ . 3[C1|C56]=[3C1-GJ2|3C56-HJ4-G2]=3GJ2
| _ _ _ _ . 13GJ2
| 34GH8 -> 12HJ7 2C56
| _ _ _ _ . 1[F13|F6]=[1F13-GJ2|1F6-HJ4-G2]=1GJ2
| _ _ _ _ . 2[F1|F56]=[2F1-GJ2|2F56-HJ4-G2]=2GJ2
| _ _ _ _ . 12GJ2

Amazing.

Tuesday 21-Jan-2020

... by: numpl_npm

With basics and DFC(Digit Forcing Chains),

6G9
| 6H7 -> [123]J7=...==contra

7G3 4H3
| [14|24|34]GH8=...=[7G3 4H3]

Then to the end.

Tuesday 21-Jan-2020

... by: SuDokuFan

@Cenoman
If a certain digit, say A, is locked into the base of an Exocet, then it must be in one target cell and one mirror node because if it were in both targets, then it means there would be no contradictions on the second true base digit, and as such there would not even be an Exocet. Therefore, if a digit is not mirrored, then that digit is invalid because it would either make that digit true in both targets (therefore no contradiction and no Exocet) or true in no targets, making the assumption that the digit is true in the base invalid.
If A is in the companion cell, then placing A in the base cell would not cause a contradiction when removing the digit from the target because IT ISN'T IN THE TARGET, therefore the Exocet wouldn't prove
Any flaws in the logic?

Tuesday 21-Jan-2020

... by: Cenoman

@SuDokuFan
As I said in my preamble, I just reminded the solution given by abi, totuan and champagne on The New Sudoku Players' Forum. I didn't try to find another one.

As regards the output "4 is a false base digit", missing 4s in the mirror node is an inference of the Junior Exocet pattern, and you stated (as I do) that the pattern is a GE, not a JE. So, to invoke the "missing 4s in the mirror node", the JE requirements should also be invoked, namely "no base digit in companion cells", requirement complied with in WU#383.

Now, once this has been seen, the other JE inference "If a mirror node contains a locked digit, any other digits it contains of the same type (known base or non-base are false" (rule nr 9 in DPB's compendium), could be used to eliminate 6H7, leading to +6G9 more easily than in the old solution.

BTW, I doubt that the pattern could be treated as an almost JE; there are too many spoiler candidates 1 and 3.

Monday 20-Jan-2020

... by: SuDokuFan

NVM below, forgot the extra 3 instance in A4, no strong link exists

Monday 20-Jan-2020

... by: SuDokuFan

Also, is there potential for elimination using (13)A28=JE2(1234)G56, J2, H8?

Monday 20-Jan-2020

... by: SuDokuFan

@Cenoman there is an easier way to spot the elimination of the 4, it is in the targets but is not mirrored anywhere.

Monday 20-Jan-2020

... by: Cenoman

@jkl Thanks for spotting the clone !
Reminder of the solution given months ago, to WU#299, transposed to #383 (see Gordon's (ghfick) link)
1. Identification of an exocet.
(1234)G56, J2, H8 is a good candidate. Unfortunately, three cover houses in 'S' cells for digits 1 & 3 do not comply with Junior Exocet requirements (no more than two cover houses).
Checking the fundamental exocet property for 1 & 3 (by T&E)
+1G6 & J2=234=H8 =>-1G9 AND swordfish (1)ADE248 =>-1ADE9; column 9 void of 1 (contradiction)
+3G56 & J2=124=H8 =>-3G9 AND swordfish (3)ADE248 =>-3ADE9; column 9 void of 3 (contradiction)
Digits 2 & 4 comply with JE requirements, therefore GE (1234)G56, J2, H8

2. Directed T&E on possible true base digits in the above exocet:
4 is a false base digit. +4G56 =>+4H8=>+4J2 (contradiction with the GE pattern, a true base digit can't be true in both target cells) Therefore -4G56, -4H8, -4J2; four placements AND GE (123)G56, J2, H8
G56=13 =>G9=6
G56=12 =>G9=36 AND (J2, H8)=12 =>J7=3; G9=6
G56=23 =>G9=16 AND (J2, H8)=23 =>J7=1; G9=6
For the three possible base pairs, G9=6; nine placements and basics. End of solution with a few AICs

Sunday 19-Jan-2020

... by: Algo

A1 3
B3 6

Sunday 19-Jan-2020

... by: yzfwsf

Simply put, 1,3 has four crosslines covered by three houses.

Saturday 18-Jan-2020

... by: SuDokuFan

@ghfick
Both 1 and 3 are proven in the GE definition by basic moves (that is, Andrew's solver steps 1-6)

Saturday 18-Jan-2020

... by: ghfick

As jkl has pointed out. #383 is a morph of #299. Back in June 2018, #299 was discussed on the forum:

forum.enjoysudoku.com/help-to-solve-hardest-puzzles-t34876.html#p267762

A while later, David P Bird, considered and expanded on the Almost Junior Exocet. [AJE]
If we insist on the 2 cover house rule, then, in #299 , the Exocet is not a JE since digits 1 & 3 have 3 cover houses. David might call this Exocet an AJE if the moves required establish 1 & 3 are non-assumptive.
SuDokuFan is calling the move a GE while yzfwsf is calling it a JE.

Maybe the names are not that crucial but it would be helpful to see why 1 & 3 are OK.

Saturday 18-Jan-2020

... by: Frans Goosens

With trial and error

Combination A9=138 and F1=127

************************************************************
A9=1 F1=1 No solution, Fixed

Combination 2-digits cells

A1=6 Wrong, Undo calculation
A1=7 No solution, Fixed
A2=3 No solution, Undo calculation
A2=6 Wrong, Undo calculation
A2=3 No solution, Fixed
A4=4 Wrong, Undo calculation
A4=8 Wrong, Undo calculation

All reset to initial position

************************************************************
A9=1 F1=2 No solution, Fixed

Combination 2-digits cells

E9=6 Wrong, Undo calculation
E9=8 No solution, Fixed
G9=3 Wrong, Undo calculation
G9=6 No solution, Fixed
I1=1 Wrong, Undo calculation
I1=3 No solution, Fixed
I2=2 Wrong, Undo calculation
I2=4 Wrong, Undo calculation

All reset to initial position

************************************************************
A9=1 F1=7 No solution, Fixed

Combination 2-digits cells

E9=6 No solution, Undo calculation
E9=8 No solution, Undo calculation
F3=1 Wrong, Undo calculation
F3=4 No solution, Fixed
F5=2 Wrong, Undo calculation
F5=3 Wrong, Undo calculation

All reset to initial position

************************************************************
A9=3 F1=1 No solution, Fixed

Combination 2-digits cells

B8=5 Wrong, Undo calculation
B8=7 Wrong, Undo calculation

All reset to initial position

************************************************************
A9=3 F1=2 No solution, Fixed

Combination 2-digits cells

B8=5 No solution, Undo calculation
B8=7 Wrong, Undo calculation
B8=5 No solution, Fixed
A8=1 ( B4=8 ) Solved,


#383--------------------------Solution
002 000 000-------------972 546 813
400 010 002-------------436 819 752
050 700 064-------------158 723 964

500 000 000-------------549 382 671
003 000 400-------------763 195 428
080 600 095-------------281 674 395

800 900 500-------------827 931 546
090 008 007-------------694 258 137
005 067 089-------------315 467 289

Total solving time is : 394 sec.
Number of logical steps is : 39806

Saturday 18-Jan-2020

... by: SuDokuFan

GE2(1234)G56, J2, H8, eliminations -4J2, -4H8, -4G56, -6H7, solvable.

Saturday 18-Jan-2020

... by: numpl_npm

yzfwsf, nice finding!

| _ _ 2 | _ _ _ | _ _ _ | A
| 4 _ _ | _ 1 _ | _ _ 2 | B
| _ 5 _ | 7 _ _ | _ 6 4 | C

| 5 _ _ | _ _ _ | _ _ _ | D
| _ _ 3 | _ _ _ | 4 _ _ | E
| _ 8 _ | 6 _ _ | _ 9 5 | F

| 8 _ _ | 9 _ _ | 5 _ _ | G
| _ 9 _ | _ _ 8 | _ _ 7 | H
| _ _ 5 | _ 6 7 | _ 8 9 | J

(u)G56 -> (u)[H8|J2] where (u in 1234)
| 1G6 -> 1[C7-H8|F7-H8|F13&C13-J2]
| 2G56 -> 2C56 2[F7-H8|F1-J2]
| 3G56 -> 3[C7-H8|F7-H8|F56&C1-J2]
| 4G56 -> 4H8

So Exocet [1234]G56 H8 J2
(uv)G56 -> (uv)H8.J2 where (uv in 1234)

-4H8 -4J2 -4G56
| 4H8 -> 4J2 contradiction.
| 4J2 -> 4H8 contradiction.

-6H7
| (uv)G56 -> (uv)J12 (uv)H78 where (uv in 1234)

+4J4 +4G8 +6G9

Then

| _ _ 2 | _ 4 6 | _ _ _ | A
| 4 _ 6 | _ 1 _ | _ _ 2 | B
| _ 5 8 | 7 _ _ | _ 6 4 | C

| 5 4 9 | _ _ _ | 6 _ _ | D
| _ 6 3 | _ _ _ | 4 _ _ | E
| _ 8 _ | 6 _ 4 | _ 9 5 | F

| 8 _ _ | 9 _ _ | 5 4 6 | G
| 6 9 4 | _ _ 8 | _ _ 7 | H
| _ _ 5 | 4 6 7 | _ 8 9 | J

+1F3
| 1[HJ7-A89-C1-F3|H8-J1-F3]

[13]A9 -> 8E9
| 1A9 -> 8E9
| 3A9 -> 18DE9 237DE8.F7 27E18 [15]E4 5[A4|A8]=[5A4|5A8 7B8 3B2 58AB4 5AB4]=5AB4 1E4 8E9

Then to the end with basics.

Saturday 18-Jan-2020

... by: yzfwsf

Junior Exocet:Base Cells-r7c5,r7c6;Target Cells-r8c2,r9c2,r8c8,r9c8;Cross Cells-r1c248,r2c248,r3c248,r4c248,r5c248,r6c248 Locked Member in T2: 4
Mirror Check:r9c2<>4,r8c7<>6,r8c8<>4
"S" Cells Need Include:1c9,3c9
By my solver

Saturday 18-Jan-2020

... by: jkl

#299

Saturday 18-Jan-2020

... by: James Havard

Two subs together for a not too easy basics solution, D4=3, E8=2.

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