While Robert's explanation may work for the common 2-2-2-2- 2-2-2-2- type, it doesn't translate very well into the other types. I provide a better (in my opinion) proof:

All 16 digits are restricted commons, meaning each digit can only upto 1 time in the loop. Now the 16 cells should have 16 digits, so each of the 16 restricted commons should appear exactly once, which explains the eliminations.

And yeah, as I said, the givens/solved cells don't seem to be important.

I don't get why the givens/solved cells are important. The logic seems perfectly correct even when the hinges have more than one candidate

OK, updating from my previous comment.

I think I have figured out the logic of the SK loop,and how the eliminations take place. But the description has an issue - it is not a matter of all the blue being true or all the greens being true - in fact, in the easter monster puzzle, this is not the case.

Starting at B13, if {3,8} is "true" so that the two cells contain these two values, this propagates all the way around the chain, making all of the blue candidates true. And if {2,7} is "true", this also propagates all the way around (the other direction), and makes all of the green candidates true.

But these are not the only two possibilities. B13 could contain one of {3,8} and also one of {2,7}, and in fact this is what happens in the unique solution. But we still get the eliminations in this case. Suppose there is no "3" in B13 or B79. Then if B13 also contains no "8", it must contain {2,7}, and all the green candidates are "true", including the {3,8} in B79 - this contradicts the assumption that B79 contains no "3". So B13 must contain an "8", meaning B79 cannot. But then B79 does not contain a "3" and does not contain an "8", so it must contain the blue candidates, so that all the blue candidates are true, including {3,8} in B13. So this is also contradictory. So if B13 and B79 do not contain a "3", B13 containing no "8" results in a contradiction, and B13 containing an "8" also results in a contradiction - therefore, B13 or B79 must contain a 3. By the analogous reasoning, B13 and B79 must contain an "8". However, it does not have to be the same group - in the unique solution, it is B13 that contains the "3", and B79 that contains the "8".

So I understand how the loop works now, but the description, which says that all the blue are true or all the green are true, is not correct. It's more subtle than that.

Hello,

It doesn't seem like the comments are monitored closely, but I'll try anyway.

I am really struggling to understand how SK loops work, and the link to pjb's comments isn't helping that much - it seems to presume some knowledge of the technique already.

From the description of the Easter Monster on this page, we have:

"Since we have identified the content of these sixteen cells as a Locked Set we can eliminate along the unit of alignment. So, since B13 contains a blue {3,8} and B79 contains a green {3,8} and the solution will be one or the other, so we can remove 3 and 8 from B5 and B6."

I am interpreting this to mean that either (i) all the blue candidates are "true" and all the green candidates are "false", or (ii) all the blue candidates are "false" and all the green candidates are "true". But according to the solution count" feature in the solver, B1 should be 2 (which is green) and B3 should be 3 (which is blue).

So am I misunderstanding something, or is there something wrong with the description?