Hanson and Marans’ four slink loops were derived by the “double nasty” trial loops of the previous post. Here we follow up on two rounds of removals. The identical “opening volley” is obtained by the corresponding SK loop, also displayed as a checkpoint for readers in this post.

By determining that aligned 3 and 7 candidates in SW and C boxes, and aligned 4 and 6 candidates in W and S boxes are not both true and not both false, the previous post established slinks between them. Combining the new slinks with the natural slink between two sole candidates of the same number in a box, the four loops of Hanson and Marans appear.

Like Lucy in the Sky, they come with diamonds. First the extra 3, 4, 6 and 7 candidates not on the loops see both ends of the new slinks, and go quietly (purple). Then it dawns how to take advantage of those runs pairs, 18 in r6, 15 in c6, 29 in c8 and 28 in c1. Each line must contain one of 3 and7, one of 4 and 6, and two copies a pair, one for each box. Additional candidates of these numbers are excused (maroon).

The immediate consequence of these removals is a naked triple in c1, spawning NE1 and a naked triple in r3, with more cloud dispersal.

The next step in Hanson and Marans’ attack plan is to “synchronize the loops”. In sysudokie speak, that means to establish slinks between the 37 loops and the 46 loops. It’s to be done in the same way determining the slink partners as the pairs neither both true or both false, by trial. When I see the slink net expanding, I get out my crayons.

But before we embark on deep coloring, let’s review the implications of the alignment of the key numbers 3467 in this puzzle. It’s the same as the SK signpost of Steve Kurzhal’s EM “opening volley “.

The four free lines of the EM SK loop are c1, r6, c6, and r8. Here, as in the EM, each free line has four cells available for 3467 true candidates. Compared to the Hanson and Marans “both true” and “both false” trials, it is much easier to verify that a free line will supply only one of its available “inside” 3467 candidates to each intersecting box, not two to one box and none to the other.

That means that the box can spare no inside candidates for the two boxes outside of the free lines. Inside candidates are removed for those eight cells.

Also, the two free line cells outside of the intersecting boxes are reserved for the two inside candidates its intersecting boxes cannot supply. Their outside candidates are therefore removed.

If you’re feeling dizzy, just place your own diamonds on the above, dealing with one of these two restrictions at a time. Or if not ready for that, do your own tests that a line cannot supply two inside candidates to one of its free lines. Then do the diamonds. They should match the checkpoint grids above.

Next, we continue to follow Hanson and Marans’ post SK loop strategy of “synchonizing the loops”on the HMEM. I found my crayons and I’m ready to go.