by 1to9only » Sat Mar 18, 2023 7:56 am
- r3c7 and r3c8 have candidates 1,2,3 - neither can be 2 as they would then be consecutive cells - so 2 can be removed from both cells (-2r3c78).
- now r3c7 and r3c8 have candidates 1,3 - 2 can be removed from r2c7, r2c8 and r4c8 for same reason as above (would result in consecutive cells).
- at this point blocks 3 and 4 can be solved.
tip for further progress: check r5c5.
when a bit stuck, i usually scan the cells with 2 and 3 candidates for (near-)consecutive candidates (123, 13, 45, etc.) and try apply the logic i've described above.
e.g. r2c6 has 1,2,3 - irrespective of the value in r2c6, 2 cannot be in r1c6 (-2r1c6). 2 can also be removed from r2c5 and r2c7 [in OP image] (-2r2c57).
later, there are a few places where this logic? (technique?) has to be used and the puzzle solves (i also required pointing and 1 naked pair).
Edit: Corrected an error in above!
Last edited by
1to9only on Sat Mar 18, 2023 8:52 pm, edited 2 times in total.