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JC Van Hay · Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium

0. Pointing{4B3}
1a. The valid Continuous All Cell Loop A={r28c1379,r1379c28} of 16 cells is a subset of the 16 constraints B={(27)R2, (12)C8, (19)R8, (47)C2, (36)B1, (58)B3, (36)B9, (58)B7}.
1b. The set of 13 candidates B-A={(27)r2c5, 1r5c8, 2r56c8, 1r8c5, 4r45c2, 7r5c2, 3r3c3, 6r1c1, 5r1c7, 6r9c7} is excluded.
2a. The valid Continuous All Cell Loop C={r5c2468, r2468c5} of 8 cells is a subset of the 8 constraints D={(356)R5C5, (47)B5}.
2b. The set of 11 candidates D-C={(35)r5c37, 6r5c17, 3r3c5, 5r1c5, 6r19c5, 4r6c4} is excluded.

Solving A :

3. r79c2=58 -> r8c139=19! => 0 solution
4. r8c13=58 -> r79c2=47 -> r13c2=36 -> r2c13=27 -> r2c79=58 -> r13c8=12 -> r79c8=36 -> r8c79=19 -> r8c13=58 [Loop] => 0 solution because meanwhile r5c28=5!

5. 5r7c2 : NP(19)r8c39; 4 singles
5a. + 6r1c2 -> 11 singles; XYWing(127)r1c8.r27c9 - 1r79c8 => 0 solution
5b. + 6r2c1 -> 2 singles; Pointing(2B1) and NP(14)r4c19, XWing(1R48), HT(129)r5c179, XWing(4L57) => 0 solution
5c. Conclusion : r8c3=5, r9c2=r8c6=8

6a. 6r1c2 + 5r1c8 -> NP(27)r2c17, 40 singles, XWing(3L67) => stte of a unique solution
6b. 6r1c2 + 5r2c7 -> NP(35)r45c2, 10 singles, XWing(1C58) and XWing(7C29) => 0 solution

7a. 6r2c1 + 5r1c8 -> NP(47)r17c2, NP(27)r2c37 => 0 solution
7b. 6r2c1 + 5r2c7 -> 11 singles, NP(14)r9c45, 2 singles, Pointing(2B4) => 0 solution

Ajò Dimonios · Joined: 01 May 2017 Posts: 339 Location: Sassari Italy

If 2 is true in R1C1 => contradiction, in block 7 there are two 5 sure, then 2 is false and can be eliminated.
If 6 is true in R1C1=> contradiction, in block 7 there are two 5 sure, then 6 is false and can be eliminated.
If 7 is true in R1C1=> contradiction, in block 7 there are two 5 sure, then 7 is false and can be eliminated.
If 1 is true in R3C4 => contradiction, R2C9 is empty, then 1 is false and can be eliminated
If 3 is true in R3C4 => contradiction, R2C9 is empty, then 3 is false and can be eliminated
If 2 is true in R3C8 => contradiction, in the second column there are two 4 sure,then 2 is false and can be eliminated.
If 1 is true in R8C5 => contradiction, in line 9 there are two 6 sure, then 1 is false and can be eliminated
If 1 is true in R8C7=> contradiction, R9C6 is empty, then 1 is false and can be eliminated.
If 9 is true in R8C7=> contradiction, R9C6 is empty ,then 9 is false and can be eliminated.
If 4 is true in R9C3=> contradiction, R8C5 is empty, then 4 is false and can be eliminated.
If 2 is true in R9C7=> contradiction, R2C7 is empty ,then 2 is false and can be eliminated.
If 6 is true in R9C7=> contradiction, R2C1 is empty ,then 6 is false and can be eliminated.
If 4 is true in R4C2=> contradiction, in block 7 there are two 5 sure, then 4 is false and can be eliminated.
If 2 is true in R5C7=> contradiction, R2C3 is empty ,then 2 is false and can be eliminated.
If 2 is true in R5C8=> contradiction, R8C1 is empty, then 2 is false and can be eliminated.
If 2 is true in R6C8=> contradiction, R8C1 is empty then 2 is false and can be eliminated.
If 1 is true in R9C8=> contradiction, in the column 5 there are two 4 sure, then 1 is false and can be eliminated.
If 6 is true in R9C5=> contradiction, in block 5 there are two 3 sure, then 6 is false and can be eliminated.
If 2 is true in R3C6=> contradiction, R6C1 is empty ,then 2 is false and can be eliminated.
If 1 is true in R7C8=> contradiction, in block 5 there are two 3 sure, then 1 is false and can be eliminated
If 2 is true in R7C9=> contradiction, R8C1 is empty ,then 2 is false and can be eliminated.
If 7 is true in R7C2=> contradiction, R5C6 is empty ,then 7 is false and can be eliminated.
If 1 is true in R8C1=> contradiction, R2C1 is empty ,then 1 is false and can be eliminated.
If 4 is true in R9C2=> contradiction, in column 1 You can not enter 1 in any cell,
then 4 is false and can be eliminated.
If 7 is true in R9C7=> contradiction in block 7 there are two 8 sure, then 7 is false and can be eliminated.
If 1 is true in R9C5=> contradiction, R9C2 is empty ,then 1 is false and can be eliminated.
If 4 is true in R9C4=> contradiction, in block 4 there are two 6 sure ,then 4 is false and can be eliminated.
If 6 is true in R9C4=> contradiction R2C5 is empty, then 6 is false and can be eliminated.
If 9 is true in R9C3=> contradiction in block 6 there are two 4 sure, then 9 is false and can be eliminated.
If 8 is true in R9C4=> contradiction R6C1 is empty ,then 8 is false and can be eliminated.

Ajò Dimonios · Joined: 01 May 2017 Posts: 339 Location: Sassari Italy

Hi everyone

It is also possible to start with a multifish that produces 17 eliminations

4s at r456c9 only ones in row/column => -4 r456c7.

Multi-Fish pattern with base set of 3,5,6 and 8
16 Truths = {3568R2, 3568R4, 3568R6, 3568R8}
16 Links = {68c1, 35c3, 356c5, 356c7, 8c9,R2C4,R4C2,R6C4,R6C8,R8C6}
17 Eliminations: R4C2<>4, R6C4<>4, R6C8<>2, R1C1<>6, R1C5<>5, R1C5<>6, R1C7<>5, R3C3<>3, R3C5<>3, R5C1<>6, R5C3<>3, R5C3<>5, R5C7<>3, R5C7<>5, R5C7<>6, R9C5<>6,R9C7<>6.