





XCycles (Part 2) So far, I been looking at XCycles which alternate perfectly all the way round. There are two interesting rules that lead to eliminations when we identify an imperfection in a loop which is called a discontinuity. A discontinuity occurs when we find two strong links next to each other (that is, with no weak link between them) or two weak links next to each other (with no strong link dividing them). These rules work only if there is exactly one discontinuity, and such a loop will always have an odd number of nodes. “Discontinuity” doesn’t mean that the loop is broken or that it’s a chain; it refers only to the imperfection that would otherwise make links alternate strong/weak/strong, and so on. Nice Loops Rule 2
XCYCLE on 1 (Discontinuous Alternating Nice Loop, length 6): 1[J1]+1[G3]1[E3]+1[E8]1[J8]+1[J1]  Contradiction: When 1 is removed from J1 the chain implies it must be 1  other candidates 3/9 can be removed We have two strong links joined at J1; therefore, J1 is 1. One way to make sense of this logically is to trace round the alternative. If J1 was not a 1 G3 and J8 would have to be 1s. That would remove the candidate 1 from E3 and oblige E8 to be a 1. But hang on  that forces two 1s in column 8. A contradiction so the 1 must exist in J1. Nice Loops Rule 3 Our third rule dictates what happens when two weak links form a discontinuity in a loop: If the adjacent links are links with weak inference (broken line), a candidate can be eliminated from the cell at the discontinuity.
XCycles introduced the idea of Weak and Strong links but I want to make a more precise definition of terms since there are subtleties which will be useful in other chaining strategies. The rough and ready distinction between Strong and Weak links is to do with how many candidates are in a unit – namely, Strong links are formed when only two are present, while three or more imply a Weak link. From a strong link we can infer that if not A, then B From a weak link, we can infer only that if A then not B, C, D according to how many candidates there are in a unit This implies that:
However, the following is also true that for a strong link: if A, then not B So, some Strong links can be reversed to give us a "link with weak inference"  if the occasion calls for it. It is perfectly logical to assert on a unit with two candidates of X both:
In Figure 5 we have an array of 6 candidates on a board. A number of strategies can show that the 6 on H9 can be eliminated. I have coloured some cells using Simple Colouring Rule 2 which link up some pairs on the board  either all of the yellow cells will be 6 or all of the cyan cells will be 6. Since H9 can see C9 (yellow) and H5 (cyan) it cannot be a 6 since it can seel cells with both colours.

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