angusj wrote:Solved it with 2 forcing chains.

OK. Here's the way I've navigated through the tricky parts, five in total. Sorry for the heavy use of images - I'm personally finding that images are overused here - but for these really tough puzzles I think an image is a lot easier to understand than wading through lots of text.

In the 5 images below, 3 steps show what I call "multi-color" techniques and 2 show "forcing chains". The multi-color steps also happen to be what

Nick70 calls Turbot fish (which are a very common subset of simple colors and multicolors).

For those not familiar with multi-colors -

Start by filtering on a specific candidate. If 2 conjugate pairs exist (say A, B & Y, Z) where one cell from each conjugate (say A & Y) shares a group, then any cell sharing a group with the other two conjugates (B & Z) mush be 'false'. Reason: at least one of the two cells sharing the first group (A,Y) must be 'false'. Consequently at least one of the other 2 conjugate cells (B &/or Z) must be 'true'. Therefore any cell that contains the specific candidate and shares a group with B and Z can have that candidate removed. (nb: This "multi-color" rule equally applies to conjugate chains.)

Anyhow here are the 5 tricky steps:

1. Multi-color to remove 1 from r7c3.

(Pink and orange are conjugates as are blue and bright green. One or both of the pink and blue cells must be 'false' since they both share a group and can't both be 'true'. Therefore one or both the orange and bright green cells must be 'true'. Cell r7c3 shares a group with both an orange and a bright green cell so must also be a 'false' cell. Remove candidate 1 from that cell.)

2. Multi-color to remove 3 from r7c9.

3. Forcing chain starting at r1c9 to force 7 into r9c7

4. Multi-color to remove 2 from r3c9.

5. Forcing chain starting at r1c1 to force 1 into r3c3