A "RAINY" SUDOKU ON A rainy day !! 

18 mins. Similar method. 1. Eliminate 2 from R4C7 using uniqueness 2. Using 39 in R2C8, eliminate 4 from R1C5 and R3C5 => 4 in R5C5. Cracked!

Hi again! Used my only single # in all the BVCs strategy again. It was 8 in this puzzle. Using colouring I built out several Houses on 8s. Found r2c3 can not be a 2. Then, realized that r2c7 must be 28 BVC with known colours. IE, that cell can not be a 4. Found multiple eliminations from there on to solve. Regards. Keep up the good work!

Excellent puzzle. Had a lot of fun solving it. Solved it in 29:04. I used the following logic: The first step I noticed from box 9: 2 is either in row 8 or in r7c8. These put a 4 to either r8c1 or r6c8, which put a 2 to either r2c1 or r6c2, thus r1c2, which sees both, cannot be 2, leaving only 34 there. Now earlier I noticed a swordfish on 3s in columns 2, 5, and 8, and in each of these columns there are exactly two possible locations for 3. So in the final solution 3 is either in cells r1c2, r7c5, and r4c8, or 3 is in cells r7c2, r4c5, and r1c8. This means that exactly one of r1c2 and r4c5 will be 3, and since at that point both can only be a 34 pair, exactly one of them will be 4, which eliminates 4 from r4c2 and r1c5. This creates a 29 pair in cells r4c2 and r4c4, eliminating 2 from r4c8. Now both r1c2 and r4c8 can only be 3 or 4, and since in row 6 we have a 24 pair in r6c28, we cannot have r1c2 and r4c8 be both 4, so r1c8 cannot be 3. This places 9 to r1c8, after which the puzzle solves easily.

Easy to Puzzle solving but reading comments it's very hard. 1. 349 XY wing gives 28,28 then puzzle solved. 2. 3s X wing , start with R7C2,R1C2,R2C8,R2C9, R5C9, R5C6, Eliminating 3 in R8C6. 3. The Medusa method... *See the row 7, (39), (238),( 29) connecting with the 3,2 and 9, conjugate pairs. *Col 9, (39) and (29) is 9s conjugate. *Col 6 conjugate with 3 4 and 9. We start with R8C3 (39) 3 is off 9 is on, R8C6 (238) 3 is on, 2 is off, R8C9 (29) 2 is on 9 is off, R2C9 (39) 9 is on 3 is off, R5C9 (23) 3 is on. Now come to col 9.., If R2C9 (39) 9 is on means R2C6 (49) 9 is off and 4 is on. If R5C9 (23) 3is on means R5C6 (34) 3 is off 4 is on. In col 7 two 4s is on, so any other 4 will be spotte so, 4 spotted in R1C6, puzzle solved.

Awesome!