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Saturday 25-Jun-2016

... by: ray

Also C7 = 2.

Saturday 25-Jun-2016

... by: JC Van Hay

0. Pointing{9B1} -> -{9r1c45}; HP(89-2347)r78c9
1. Double Exocet (4567){r3c23, r2c89, C167}

1a.
r3c2|3=a -> Swordfish{aC167} -> [ar1c7==ar2c6] -> -{ar1c45, ar2c89}
r2c8|9=a -> Swordfish{aC167} -> [ar1c1==ar3c6] -> -{ar1c45, ar3c23}
|
V
{a,b,c,d}={3,6,7,9} :
r3c23=ab -> r1c7.r2c6=ab -> -{(ab)r1c45,(ab)r2c89}
implies or is implied by
r2c89=cd -> r1c1.r3c6=cd -> -{(cd)r1c45,(cd)r3c23}
|
V
-(abcd)r2c1 or r2c1=1; -(abcd)r3c7 or r3c7=2; -(abcd)r1c45 or r1c45=12

1b. Multifish/Rank 0 Logic : the 12 constraints {(4567)C167} are covered by the 12 constraints {r1c17, r23c6, 4R47, 5R45, 6R58, 7R78}
|
V
-{4r4c28.r7c5, 5r4c234.r5c39, 6r5c35.r8c28, 7r7c3.r8c48}

1c. The solutions of the multifish also exclude {5r4c6, 8r45c1, (12)r5c6}
Proof :
[(45=8)r4c71-(8=74)r7c17-(4=5)r4c7]-5r4c6=HP(45-8)r4c17
HP(56-128)r5c16

2. Updating :
Pointing{8B4} -> NP(13)r7c3.r8c2 -> -{3r9c23}
HT(467-123)r269c8 -> HP(23-47)r8c8.r9c9 -> NP(23)r59c9 -> -{3r6c9}
Swordfish{6R58C8} -> -6r2c6

3. [9r2c5=9r7c5-(9=8)r7c9-8r7c1=(8-6)r8c1=6r8c7-6r9c8=6r2c8]-6r2c5; 2 singles
4. [8r7c1=(8-9)r7c9=9r7c5-(9=4)r2c5-4r9c5=(4-7)r9c8=7r7c7]-7r7c1; 6 singles
5. [4r2c9=4r2c6-4r7c6=(4-7)r7c7=7r1c7]-7r2c9=7r1c79-7r1c3=7r3c3
6. [1r5c8=1r4c8-(1=2)r4c6-2r8c6=2r8c8]-2r5c8=XWing(2C68)-2r4c4
7. [(2=1)r1c5-1r1c4=1r46c4-(1=2)r4c6]-2r5c5; 8 singles
8. [8r4c4=8r4c3-(8=3)r5c3-3r5c8=3r4c8]-3r4c4
stte of a unique solution

Saturday 25-Jun-2016

... by: ray

1. Use only basic strategies and the forcing bi-location cell method.Start with G7 - bi-value and tri-location cell for 4.Progressively form the 3-element
bi-location cell set G7E5B4 by forcing the 1-cel to 3-cell values,only force the run out values,until G7E5B4 = (4,1,7) which gives the solution.Other bi-location cell set solutions found were : B4E5H1 = (7,1,7),E5G37 =(1,1,4), A2G37 = (6,1,4). N.B: using basic to extreme strategies,A2 = 6, B4 = 7 gave1-cell solutions.
2. Using this solution,with basic strategies,the following non bi-location cell set solutions were found: BDF4 = (7,8,3),A5B4C6 = 2,7,6) E5F4J5 =(1,3,4),BEH6 = (4,5,1), BEJ5 = (9,1,4). N.B: using basic to extreme strategies,the following 1-cell solutions were found:B6 = 4,A5 = 2,B1 = 2 .

Saturday 25-Jun-2016

... by: anybody1

1A4 2A5 4G7...

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