## Discussion...

Post an idea here...

## ... by: ghfick

The solver YZF_Sudoku does include a very thorough JE implementation [including aligned targets] as well as MSLS.

In fact, this solver will search out all the various MSLS versions at a particular step.

This solver can be found in the forum under Software.

The developer has built his solver from the ground up but has used many facets of Bernhard Hobiger's HoDoKu including a 'Solution path' tab and an 'All Possible Steps' tab.
## ... by: hehe

*****************************

The correct answer is:

3 6 9 |5 2 1 |7 4 8

5 4 7 |3 6 8 |1 9 2

2 8 1 |7 9 4 |5 3 6

------ ------ ------

7 1 5 |6 4 2 |9 8 3

6 2 3 |9 8 7 |4 1 5

4 9 8 |1 5 3 |6 2 7

------ ------ ------

1 3 6 |2 7 9 |8 5 4

8 5 2 |4 1 6 |3 7 9

9 7 4 |8 3 5 |2 6 1

*****************************

26

(6,8:7),(8,8:8),(8,6:7),(8,3:8),(8,2:9),(8,0:6),(7,6:7),(7,5:9),(7,3:8),(6,7:9),(6,5:9),(5,6:7),(5,2:9),(4,8:8),(4,7:8),(3,6:7),(3,3:8),(2,7:7),(2,1:3),(1,5:6),(8,1:9),(7,7:9),(7,2:9),(7,0:6),(6,4:9),(6,0:4),(5,3:8),(5,1:9),(5,0:8),(4,5:8),(4,4:8),(4,1:6),(3,4:9),(3,2:9),(2,0:4),(1,8:6),(1,7:6),(1,4:6),(1,1:5),(0,5:7),(6,1:9),(5,7:8),(4,0:8),(3,8:8),(3,5:9),(3,0:8),(2,6:7),(2,4:8),(1,2:5),(0,8:7),(0,7:7),(0,4:7),(0,3:7),(0,2:5),(7,1:9),(5,4:9),(2,3:8),(0,6:7),(100,100:1000)

3293

81,58
## ... by: Algo

23> Hidden single in rE | Cand=5 / pv removed 7 : G9=5,H9=5,I9=5,E9=4789

> Box line pairs in c9 rG,H | Cand=9 / pv removed 3 : G8=9,H7=9,H8=9

> Contradiction: H9 pv [1, 6, 7, 9] canot be 1 (vio) / pv removed 1

24> Guess1x in E5 = 8 from pv [3, 4, 6, 8] / pv removed 13 : E3=8,E4=8,E6=8,A5=8,B5=8,D5=8,I5=8,D4=8,F4=8,F6=8,E5=346

25> Hidden single in rD | Cand=8 / pv removed 5 : B8=8,F8=8,F9=8,D8=79

26> Hidden single in rB | Cand=8 / pv removed 6 : I6=8,A4=8,B6=4567

27> Hidden single in rA | Cand=8 / pv removed 2 : A9=67

28> Hidden single in rF | Cand=8 / pv removed 5 : F3=12347

29> Hidden single in rI | Cand=8 / pv removed 4 : I4=1256

30> Hidden single in c8 | Cand=9 / pv removed 3 : B7=9,B8=67

31> Hidden single in blkA2 | Cand=5 / pv removed 6 : A1=5,A2=5,G4=5,A4=267

> Pointing pairs in blkA2 rC | Cand=7 / pv removed 4 : C1=7,C3=7,C8=7,C9=7

> Pointing pairs in blkA3 rC | Cand=6 / pv removed 3 : C1=6,C4=6,C6=6

> Pointing pairs in blkA3 c7 | Cand=7 / pv removed 3 : D7=7,E7=7,H7=7

> Pointing pairs in blkB2 c6 | Cand=3 / pv removed 3 : G6=3,H6=3,I6=3

> Pointing pairs in blkB3 rF | Cand=7 / pv removed 3 : F1=7,F4=7,F6=7

32> Naked single in rF | Cand=1 / pv removed 4 : F1=1,D4=1,G4=1,D5=1

> Naked pairs in c5 rB,D | Cand=46 / pv removed 3 : A5=6,H5=6,I5=6

...to soln
## ... by: Cenoman

@BobW,

Your solver is not the only one missing the JE2A. Andrew's and Phil's do not implement this pattern either. Have a look to David's JE Compendium, but don't be afraid of subtleties. JE with aligned targets work exactly the same as JE with diagonal targets. You have just to keep in mind that in target boxes, the mirror nodes are also aligned and located in the band third line. Mirror nodes are made of two cells each (mini-line but the companion cell) To me, that's all.

I have no exocet at all in my solver. JE is a pattern easy to spot manually.
## ... by: BobW

@Cenoman,

When I implemented JExocets in my solver, I didn't fully understand how the ones with aligned target cells worked. And so, I only implemented the version with diagonal target cells. I found that David P. Bird's write up is very accurate, but also very concise, and has to be read extremely carefully, in order not to miss subtle but important details. I'll have to go back and read his material again.
## ... by: Cenoman

WU#376 is rated SE 11.1

JE2A (1234)HJ3, F1, C1 (JE2A: Junior Exocet with Aligned target cells F1, C1)

The pattern is a Junior Exocet because:

- companion cells F2, C2 contain no base digit,

- in 'S' cells CFG456789, each base digit has two cover houses (1: col 49, 2: col 48, 3: col 68, 4: col 69)

Eliminations from the pattern: -67C1, -7F1 (non base digits in target cells)

Using David P. Bird's criteria for base pair compatibility, it can be concluded that pairs (1,2) and (2,3) are incompatible base pairs ('S' cells not in diagonal boxes)

For the pair (1,3) the same criterion is not fulfilled but the pair can easily be shown incompatible by T&E:

HJ3=13 & F1=1 & C1=3 =>contradiction (with basics)

HJ3=13 & F1=3 & C1=1 =>contradiction (with basics)

The three pairs containing 1, 2, 3 are incompatible => 4 is a true base digit; +4J3

The puzzle is now rated SE 9.2, not yet easy to solve.

A further directed T&E can be processed to get more information on the second true base digit:

- J3=4 & H3=1 =>contradiction (with basics)

1 is a false base digit, then H3=23, C1=F1=234

Another directed T&E can be processed to guess which target cell houses base digit 4:

- J3=4 & H3=F1=23 & C1=4 =>contradiction (with basics)

So, eventually J3=4 & H3=C1=23 & F1=4 =>solution (with a sequence of AICs and krakens, puzzle solvable by Andrew's solver, puzzle now rated SE 8.9)
## ... by: Frans Goosens

With trial and error

Combination A9=678 and H3=123

************************************************************

A9=6 H3=1 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=6 H3=2 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=6 H3=3 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=7 H3=1 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=7 H3=2 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=7 H3=3 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=8 H3=1 No solution, Fixed

Combination 2-digits cells

A5=2 No solution, Undo calculation

A5=6 No solution, Undo calculation

A7=3 No solution, Undo calculation

A7=7 No solution, Undo calculation

B3=4 No solution, Undo calculation

B3=7 No solution, Undo calculation

F9=4 Wrong, Undo calculation

F9=7 No solution, Fixed

G8=3 Wrong, Undo calculation

G8=5 Wrong, Undo calculation

All reset to initial position

************************************************************

A9=8 H3=2 No solution, Fixed

Combination 2-digits cells

A5=2 ( D7=9 ) Solved,

#376--------------------------Solution

009 001 040-------------369 521 748

000 300 002-------------547 368 192

080 090 500-------------281 794 536

005 002 003-------------715 642 983

000 000 010-------------623 987 415

090 050 600-------------498 153 627

006 070 800-------------136 279 854

800 400 000-------------852 416 379

970 000 000-------------974 835 261

Total solving time is : 318 sec.

Number of logical steps is : 27923
## ... by: SuDokuFan

I suspect there may be some form of SK loop structure here, but no MSLS or Junior Exocet came out

Phil’s solver confirms the above, giving an Almost SK +1 leading to the following: +8F3, -67C1, -7C3, -7F1, -7F3, -6H5, -68J5, -7H7, -2A4, -4B6, -1D4, -34E6. Anyone have any tips on how to spot these?
## ... by: James Havard

My solver couldn't get only 2 for a basic solve. Going box to box manually, I found two together for a basics solve in box 4. F1=4 and F3=8.
## ... by: Neil

E5=8 to completion.