Hi Yogi,

From what you wrote, I think your understanding of chains, group nodes, and the Eureka notation has taken tremendous steps forward! Great job!

Yogi wrote:The two chains do it in a much more straightforward way to show that r9c8 can’t be 8 or 9. I don’t quite see it as a kite but I was mainly interested in the Eureka chaining language.

Well, it's not a simple Kite so it's not as easy to see, but it's a (Dual) Grouped Kite. In a simple Kite you only have simple strong links, i.e. only two candidates of the digit in the relevant row and column, connected by a box. In a Grouped Kite you can have three candidates in one or both lines, as long as two of them are in the connecting box forming a group node (or two).

In this case the row has a group node in box 2 (forming a grouped strong link in that row), but the column strong link remains simple because it has only two candidates. The same logic would work if r3c4 had 8 and 9 as candidates, so you could have a group node (and a grouped strong link) in the column as well.

In a Kite the end-points of the chain must still be single candidates so that they can both see the elimination. On the other hand, in a Grouped Skyscraper one of the end-points is a group node (and it can have up to three candidates, thus a total of four in that line). It works because the elimination(s) is/are in the same box as the group node.

Here’s how I see it. Given that you have identified by logic or otherwise that the eliminations are in r9c8, you want to write the chain to explain it in such a way that the endpoints both see this cell. So we go from r9c4 to r2c8, and we will have an AIC if we can start and end with a strong link and alternate strong and weak links throughout the chain.

(8)r9c4=r1c4-r2c56=r2c8 => -8r9c8 and

(9)r9c4=r1c4-r2c56=r2c8 => -9r9c8 => 5r9c8 stte

Exactly. That's two Grouped Kites, both with a single group node (r2c56) in box 2/row 2. The

r1c4-r2c56 is a

grouped weak link and the

r2c56=r2c8 is a

grouped strong link.

The tricky part was seeing r1c4-r2c56 as a single weak link. I would not have thought of that as a single link but clearly it is valid. Are we seeing r2c56 as a single entity in the chain, a node, perhaps? Is the weak link here saying “When r1c4 is 8, NEITHER r2c5 nor r2c6 can be 8 and therefore r2c8 must be 8,” proving that r9c8 cannot be 8 because it can see both ends of the chain, at least one of which must be 8?

Exactly, on all accounts. The two candidates r2c56 form a

group node because they're in the same box and row. Thus they can be used as a single entity within that box and row, but not in their columns (obviously, because they're in different ones).

Finally, if everything works the way it should, does the AIC work in the opposite direction?

Yes, as it should.

Seems so. If r2c8 is NOT 8, one of r2c5 or r2c6 must be 8, so r1c4 cannot be 8 and r9c4=8 => -8r9c8.

Works fine if r2c56 is a single point in the chain.

Exactly, again. When approaching from that direction, you can think of the three candidates in row 2 as an almost-claiming. If r2c8 is not 8, then the two remaining 8s in row 2 are both in box 2. That's a claiming, which would eliminate 8r1c4 (and any other 8 in that box that is not in row 2). A Grouped Skyscraper has an almost-claiming as well, but there the group node is an end-point. The same logic works for almost-pointing situations, which is how ERs (i.e. Grouped Cranes) work.

In general, you can think of simple bilocation strong links as almost-hidden-singles, while grouped strong links are either almost-claimings or almost-pointings.