## Discussion...

Post an idea here...

## ... by: Paul

Oops, hit button too quickly. Anyway I have been thinking about this Trial and Error business. When you think about it, all solutions are basically trial and error. As a programmer myself, if I wrote code to check for solutions, I would start somewhere and apply strategies to each square until I arrived at a solution. The starting square could be any square, but typically I would expect most programmers to start at A1, but it would not matter where you start as long as you check all squares. So is this not 'Trial and Error' ?
## ... by: Paul

Trial and error
## ... by: Manny & Vinny

Well, our spreadsheet's done it again. To work through it with, as it were, just pen and paper, try the combination 41673 in column 9, work that for a bit, till it dead-ends, then try a succession of pairs: 46 col 7, 83 also col 7, 21 also col 7, 85 box 6, and 96 col 6. Each of these entries to be understood as from left to right and higher row to lower. There were 32 permutations for column 9 (the correct one, 41673, came up about half way through, saving a bit of time...:-)).
## ... by: ray

1. Use only basic strategies and the forcing bi-location cell set method.Start

with A5 - bi-location cell for 3,tri-location cell for 8.Progressively form the

3-element bi-location cell set A5G6H1,forcing only the 1-cell to 3-cells run out values until A5G6H1 = (8,6,6) which gives the solution.Other bi-location

cell set solutions found were: A5G6J1 = (8,6,7), A5E2G6 = (8,6,6). N.B: Us-

ing basic to extreme strategies,E2 = 6, G6 = 6, H1 = 6, D8 - 4, E3 = 5 gave 1-cell solutions.

Using this solution,with only basic strategies,the following non bi-location cell set solutions were found: G35J7 = (3,9,6), D48E3 = (6,4,5). N.B: Using basic to extreme strategies,D4 = 6, G3 = 3, G5 = 9, J7 = 6 gave 1-cell solution.

OR

2. Start with the Double Exocet cells H45{4679) and J78{467,9). Forcing H5 =

4, or J7 = 6 and applying only basic and Nishio Forcing Chain strategies, gave

the solution. N.B: H1 = 6, G5 = 9, G6 = 6, C4 = 6, C5 = 6 gave 1-cell solutions.
## ... by: JC Van Hay

0. Basics : J8=9, 1Box8, Skyscraper(9 Col 26), HP(69)C45

1. Exocet (467){H45, Col 269}

1a. aH4|4H5 -> Swordfish(a Col 269) -> aJ2==aG9 => -{aG13, aJ7}

1b. Structure of the 3rd band :

| 35c 8 35c | 159c 159c 59c | 3467 2 b|a |

| 23c 39c 239c | ab ab 8 | 5 1 3c |

| 1567 a|b 145 | 3 2 5c | c 9 8 |

where {a, b, c} = {4, 6, 7}

1c. Inferences :

G9=J2=467

H9=3, J6=5

G13=35

::

| 35 8 35 | 1679 1469 4679 | 467 2 467 |

| 267 4679 249 | 67 46 8 | 5 1 3 |

| 167 467 14 | 3 2 5 | 467 9 8 |

1d. Basics : NP(27)AB6, Swordfish(7 Rows CDJ)

Solving the base of the exocet :

1e. 6H4 -> 0 solution

1f. 6H5 -> 0 solution via HP(67)J12, S-Wing[1F9=1F5-(1=4)G5-4G6=4F6]-4F9 and Skyscraper(8 Rows 36)

:: Basics : H4=7, H5=F6=4, NP(16)EF9, HP(47)D78.G79, J7=6, HP(12)AB7, HP(38)CE7

2. 1J1 -> 0 solution; 16 singles

3. XWing(8 Col 17), EmptyRectangle(8 Col 3, 3 Box 2)

4. 6H2 -> 0 solution; H1=6 and stte