I was wondering what the "canonical" cyclic lead head should be, and figured it should be either 1345...2 or 1Z2345... where Z is the tenor. As a cycle, either permutation could be considered 2345..., depending on whether you read the permutation left to right or right to left. Since Plain Bob Minimus has the 1342 as the lead head, I decided to focus on 1345...2. Also, it comes if we just sort by the leftmost number. Then I was wondering if there could be a prototypical method associated with cyclic lead heads, just as Plain Bob is associated with PB lead heads. I think this would really just be plain changes ("What were plain changes?" in The Tower Handbook"), with the 1 staying, i.e.

But maybe we can find something more interesting than that.

Double symmetry, etc.

I want to find cyclic bob methods with 134562 lead heads. I don't really know what kind of constraints would be good, but I decided on double symmetry, no long places, the given lead head, and .12. lead end change.

On minor, the only method that satisfies this is Double Cambridge Cyclic Bob, but on higher stages, it doesn't have the desired lead head. If instead I want a .16. lead end change, then that results in at least one long place, and additionally, at least one wrong place.

On major, I tried adding the constraint that there should be no wrong places, but this is unsatisifiable. If I make an exception for a wrong place made at just before the half lead and lead end, the only solution is x145678x1238x38.16.78... (followed by the second half lead, since it's a double method). If instead I make an exception for a wrong place made at just after the HL/LE, the only solution is 36.14x18x1458x78....

A note on double symmetry: I realized how to apply the half lead "twice". Suppose the half lead (i.e. as the treble is finishing lying behind) is 432651. To get the flipped change, you flip it from left to right, and then renumber* the numbers, so you would get 432651 → 156234 → 621543. Then you apply in the usual manner for rows as permutations, so 432651 * 621543 = 134562. In my case, I knew the half lead change and lead end that I wanted, and I tried different half lead ends and half lead heads to see which of them worked. if I wanted to make .56. at the half lead and wanted 135426 as a lead end, the half lead end/head pair had to be 346251 432651. Since I only found one answer, perhaps there is a unique solution and a faster way to calculate the half lead pair other than guess and check.

*Well, actually renumbering and flipping are the same thing. Considering a row as a permutation is like one-line notation, i.e.

is the same as

523416
↓↓↓↓↓↓
123456

so flipping from left to right is equivalent to renumbering the 123456 side.

"Extension" of plain bob minimus

The other thing I considered was, since Plain Bob Minimus has a 1342 lead head, Plain Bob Minimus is technically a cyclic method. So maybe there is a way to extend it? In this case, I'm not interested in double symmetry, because Plain Bob Minimus doesn't have it, and palindromic symmetry precludes cyclic lead ends ("Cyclic Methods" on Phillip Earis's blog)

I decided to go with having a .12. half lead and making the 3 and 6 bells plain hunt, since they end up in the desired location that way.

I think there are only two possibilities for minor (though I don't remember if I had additional constraints): x16x16.34.1236x16x16x12 and x16.12.16.34.16x16x16x12. Note that both have one long place and a wrong place, and I think those are both forced to happen.

For major, I required that 4 had to hunt from 1 to 4, and required no long places. There are only two possibilities: 56.1x23x1x45x1x1x1.56.2 and 56.1x1x1x45x1x67x1.56.2.