Glyn wrote:I think this would produce the eliminations given by Ronk
(189=6)ALS:r9c235-(6=9)r7c1-(9=168)ALS:r9c235
The closure of the loop allows Carcul's eliminations at the peers of the ends of the weak links (ie the 6's and 9's in Box 7 outside the chain). The ends of the chain give the strong link between the values possible in r9c235 leading to Ronk's additional eliminations.
Although there are no "ends of the chain" for a continuous loop, you have hit upon the very heart -- i.e., the minimum number of cells -- required for this deduction. It is known as the (doubly-linked) ALS xz-rule, aka the Sue de Coq or two-sector disjoint subset (TSDS) technique.
Since it doesn't matter where you start a continuous loop, the shortest NL notation occurs when you duplicate the bivalue:
r7c1 -6- ALS:([r9c23,r9c5] =6|18|9=r9c23) -9- r7c1 - continuous loop
==> r7c2<>69, r8c123<>6, r9c6<>18, r9c8<>8
... although no one else writes "continuous loop" at the end. R9c5 is "split out" in the notation to indicate that digits 6&9 do not occur there. IOW r7c1 must see all occurrences of digits 6&9 in the ALS. I think Carcul would simply write ...
[r7c1]-6-[r9c235]-9-[r7c1], => r7c2<>6,9; r8c123<>6; r9c6<>1,8; r9c8<>8
In AIC notation: (189=186)ALS:r9c235 - (6=9)r7c1 - loop
==> r7c2<>69, r8c123<>6, r9c6<>18, r9c8<>8
... although most people only write the locked digits 18 once in the ALS.
The eliminations are identical if r9c5 is the bivalue and [r7c1,r9c23] is the ALS.
