little Zivvy wrote:Would be grateful if someone could let me know if this is correct coloring.

I couldn't quite understand your chain in the 7's, but you do appear to be eliminating the correct candidate 7's. Here is my chain of conjugate pairs of 7's, cell-by-cell. (R=red, Y=yellow)

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`r2c3 (R) r3c1 (Y)`

r3c1 (Y) r8c1 (R)

r8c1 (R) r7c2 (Y)

r7c2 (Y) r7c9 (R)

r7c9 (R) r8c7 (Y)

r8c7 (Y) r6c7 (R)

r6c7 (R) r4c9 (Y)

r2c3 (R) and r4c9 (Y) allow you to eliminate the candidate 7 from r4c3.

r6c7 (R) and r7c2 (Y) allow you to eliminate the candidate 7 from r6c2.

However, I found that forming this long chain in the 7's was useless because the eliminations didn't actually let me fill in any cells. By comparison, I found colouring the 2's much easier. Here is my chain of conjugate pairs of 2's.

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`r2c3 (R) r2c7 (Y)`

r2c7 (Y) r3c8 (R)

r3c8 (R) r9c8 (Y)

r2c3 (R) and r9c8 (Y) allow you to eliminate the candidate 2 from r9c3. At this point I could proceed filling in cells.

I hope this explanation helped. (And I hope I got all my coordinates right, so as not to confuse anyone.)

Update: It took me a while, but I was able to verify that your elimination of the candidate 2 from r6c2 is correct.

So, continuing from where I was above, I have eliminated the candidate 2 from r9c3. The chain of conjugate pairs of 2's can be extended.

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`r2c3 (R) r3c1 (Y)`

r3c1 (Y) r8c1 (R)

r8c1 (R) r7c2 (Y)

At this point we can use r2c7 (Y) and r8c1 (R) to eliminate the candidate 2 from r8c7. Extend the chain again.

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`r9c8 (Y) r7c9 (R)`

r7c9 (R) r4c9 (Y)

r4c9 (Y) r6c7 (R)

Finally, we can use r6c7 (R) and r7c2 (Y) to eliminate the candidate 2 from r6c2.

So, in the end, your method is absolutely correct; it allows you to place a number in the cell r6c2. However, note that it is much easier to form a short chain of 2's to eliminate a candidate 2 in r9c3 and thereby fill in that cell.