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Saturday 28-May-2016

... by: ray

1. Use only basic strategies and the forcing bi-location set method.Start
with F3{23568} - bi-location cell for 6,tri-location cell for 5 and 6.Progress-
ively form the 3-element bi-location cell set F3C5A7 by forcing only the run out cell values until F3C5A7 = (3,7,7) which gives the solution.Other bi-location cell set solutions found were: B1C5F9 = (2,7,4). N.B: using basic to extreme strategies, A7 =7,B1 = 2,C5 = 7,F9 = 4 give the solution.
2. Using this solution,the following non bi-location set solutions were found: BE4C5 = (9,4,7), AB1C5 = (8,2,7), B46C5 = (9,4,6). N.B: using basic to extreme strategies,B4 = 9, A1 = 8, D1 = 4, E4 = 4, D5 = 2, H5 = 9, G6 = 9 give the solution.

Saturday 28-May-2016

... by: David P Bird

Same JExocet as JCVH
(1)Simple Colouring r7c4 = r7c7 - r12c7 = r2c9 => r2c4 <> 1
(1)BoxLine:b2c5 => r89c5 <> 1
(478)JE2:r89c5,r1c6,r4c4
=> r1c6 <> 9, r4c4 <> 2 (non-base digits in targets)
=> r23c6 <> 9, r56c4 <> 2 (6 & 5 locked as the non-base digits in these mirror cells)
Singles (2)r4c5, (9)r7c6
Pairs (58)r7c13, (29)r23c4, (56)r9c78, (39)r9c19

Saturday 28-May-2016

... by: JC Van Hay

0a. D3=1; Pointing{5EF4} -> -{5BC4}; HP(17)HJ2
0b. Skyscraper{1A57,1G47} -> -{1B4,1HJ5}

1. Exocet (478){HJ5,RowsADG} + (478)F6 :

4H5 -> Swordfish{4A16,4D147,4G8} -> [4A6==4D4]-4B4.F6
7H5/7J5 -> Swordfish{7A678,7D478,7G78} -> [7A6==7D4]-4C4.F6
8J5 -> Swordfish{8A135,8D14,8G13} -> [8A6==8D4]-8C4.F6

Implications : HJ5=ab -> A6.D4=ab and -(ab)F6 -> D5=2, G6=9, BC4=29

The puzzle is now "solvable". Or, for example :

Updating :
2a. NP(58)G13; HP(56)J78; HP(39)J19
2b. Swordfish{8CEF2, 8ACF6, 8EF9} -> -8C5

3. [1A5=(1-5)B5=5B7-(5=671)JDG7]-1A7; 2 singles
4. Swordfish{7CH5,7ACF6,7CFH9} -> -7H28; 5 singles
5. Kraken Row 4D148 -> [5B5==7C5==5C8]-5C5; 13 singles
||4D1-4A1=4A6-(4=5)B5
||4D4-(4=7)G4-7H5=7C5
||4D8-6D8=(6-5)J8=5C8
6. [5G1=(5-6)F1=6B1-(6=4)B6-4A6=(48)AD1]-(48)FG1; 2 singles
7. XWing{3C239,3J19}-3B1
8. 6C3=6F3-6F1=*[3B2=3B7-(3=2)F7-(2=3)F1-3HJ1=3H3] -> -3C3
9. 3C2=3C9-3J9=3J1-3H3=HP(35-6)EF3=6C3-(6=8)C6 -> -3EF3, -8C2; stte of a unique solution

Saturday 28-May-2016

... by: anybody1

9B4 2C4

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