This post is a somewhat technical discussion of some mathematical aspects of change ringing. I’ll assume you have familiarity with changes, leadheads, Plain Bob leadheads, palindromic methods, etc.

A question that I’ve always had about change ringing is: Why Plain Bob leadheads? Is there some sense in which this particular set of leadheads is mathematically or musically unique, or is their predominance largely due to historical accident? Until recently, my money would have been on the latter: Many of the earliest treble-dominated methods (Plain Bob, Double Court / DNCB, Kent, Oxford, Cambridge) all had PB leadheads, and it’s compositionally easier if new methods use the familiar group. I’ve never seen another explanation, although it’s entirely possible that one is out there and I’ve missed it — please let me know if so!

To my (mathematically-amateur) mind, this seems to suggest some deeper connection between the Plain Bob leadheads and palindromic methods than I previously suspected. I doubt I’m the first person to notice this, but I couldn’t find discussion of it anywhere so I thought I’d write it up and see what others had to say.

Nearly all methods that we ring have ordinary palindromic symmetry. It’s fairly well-known that there are some leadheads that simply aren’t reachable in a palindromic method; the canonical example is the cyclic leadheads. Let’s take the Minor leadhead 162345 as an example — there’s no way for a method that is palindromic about two changes to reach that leadhead. Let’s look at why that is.

When we say that a method is palindromic, we mean that we can write it as follows:

$$l = g_1 \cdot g_2 … g_n \cdot h \cdot g_n … g_2 \cdot g_1 \cdot c$$

where $l$ is the leadhead, $h$ is the half-lead change, $c$ is the lead-end change, and $g_i$ are arbitrary changes. For notational convenience, let’s define

$$g = g_1 \cdot g_2 … g_n$$

It should be easy to confirm then that, because all changes are their own inverse,

$$g^{-1} = g_n … g_2 \cdot g_1$$

We can thus rewrite our method above as

$$l = g \cdot h \cdot g^{-1} \cdot c$$

Our goal now is to find a test that we can apply that will let us know whether a given leadhead can be reached in a palindromic method; for convenience, I’m going to call any leadhead with this property palindromizable. To that end, let’s multiply both sides of the above equation by $c$ — remember that changes are their own inverses, so this cancels out the $c$ on the right-hand side.

$$l \cdot c = g \cdot h \cdot g^{-1}$$

Now, square both sides:

$$(l \cdot c)^2 = (g \cdot h \cdot g^{-1})^2$$

Written out in full, the right-hand side expression is $g h g^{-1} g h g^{-1}$; remembering again that changes are their own inverses, it’s hopefully easy to see that this all cancels down to the identity $e$. This gives us our test:2

A leadhead $l$ is palindromizable if there exists some change $c$ s.t.

$$(l \cdot c)^2 = e$$

Let’s return to the cyclic leadhead 162345. It’s relatively easy to confirm by exhaustive check that there is no valid leadend change such that this propety holds:3

$$(l \cdot \left<12\right>)^2 = (2,4,6)$$ $$(l \cdot \left<14\right>)^2 = (3,4,5)$$ $$(l \cdot \left<16\right>)^2 = (3,5,6)$$ $$(l \cdot \left<1234\right>)^2 = (2,4)(3,6)$$ $$(l \cdot \left<1236\right>)^2 = (1,4)(2,6)$$ $$(l \cdot \left<1256\right>)^2 = (2,5)(4,6)$$ $$(l \cdot \left<1456\right>)^2 = (3,5)(4,6)$$

So the cyclic leadhead 162345 is not palindromizable. But what about the other cyclic leadheads? It’s fairly easy to exhaustively check them, but actually we can do one better: Results about palindromizability hold for leadhead classes.

The cyclic leadhead 162345 doesn’t come alone — the powers of this change form the rest of the cyclic leadhead class4: 156234, 145623, & 134562. These leadheads together with rounds form a subgroup of $S_6$ that’s isomorphic to $Z_5$. All non-differential leadheads on Minor will similarly belong to leadhead classes of 4 distinct non-rounds leadheads, each of which can be expressed as powers of each other. More formally, two non-differential leadheads $l, m$ belong to the same class if $l^i = m$ for some integer $i$.

$$l \cdot c \cdot l \cdot c = e$$

If we multiply both sides on the right by $c$, we get a useful equivalence:

$$l \cdot c \cdot l = c$$

Let’s say that there is some $m$ such that $l^i = m$ (i.e. $l$ & $m$ are in the same class). We want to prove that $l$ being palindromizable implies that $m$ is palindromizable — i.e. that the expression $(m \cdot c)^2$ is equal to the identity. We can rewrite this expression in terms of $l$:

$$(l \cdot l … l) \cdot c \cdot (l \cdot l … l )\cdot c$$

Now from our equality above, if $l$ is palindromizable then $l \cdot c \cdot l = c$, so we can replace the middle sequence with just c:

$$l \cdot l … (l \cdot c \cdot l) \cdot l … l \cdot c$$

Hopefully it’s clear that we can do this repeatedly until we wind up with the expression $l \cdot c \cdot l \cdot c$ — which, because we’re assuming $l$ is palindromizable, we know is just $e$. Therefore we know that $m$ is also palindromizable.

(You can take a second to confirm that the inverse is true: if $l$ not palindromizable, then neither is $m$. You can see this by remembering that if $l^i = m$, then there’s some $j$ such that $m^j = l$; as such, if we assume $l$ is not palindromizable and $m$ is, we run into a contradiction.)

## Palindromizers

You may have noticed that we actually proved something stronger than what we wanted. Let’s say that $c$ palindromizes $l$ if $(l \cdot c)^2 = e$. You could imagine a situation in which two members of a leadhead class $l$ and $m$ are both palindromizable, but $l$ is palindromized by $c$ and $m$ is palindromized by a different change $c^\prime$. But in fact we proved above that this is imppossible: what we proved above is that if $c$ palindromizes one member of a leadhead class, it palindromizes all members of that class.

This is where I’m going to switch from proof to just inspection: Everything from here on is based on exhaustive search of the non-differential leadheads on all stages from Minimus to Royal. Brute-force search is a bit too difficult on Cinques and above, so I don’t have results there. The results I do have imply a pattern, but any generalization to higher stages is conjecture.

Here’s what I did: On all stages from Minimus to Royal, I picked one representative from each leadhead class. I then exhaustively searched all of the possible changes on that stage to check which palindromized each leadhead class. The results on odd stages (/even numbers of working bells) are complex — there’s no clear-cut pattern. On even stages, however, we get the following results:5

2. Some leadhead classes are palindromized by exactly one change — and always a change with at most one internal place. (So on Minor there are methods palindromized by only $\left<14\right>$ but not by $\left<1234\right>$.)
3. Exacly one leadhead class is palindromized by more than one change: The Plain Bob leadheads are palindromized by $\left<12\right>$ and $\left<1n\right>$.

## Final thoughts

A lot of this is conjecture: It’s still entirely possible that this is only true for the even stages that I’ve tested, and will turn out to be false on higher stages. But I think even if that turns out to be the case, it’s still true that on two of the most commonly rung stages the Plain Bob leadheads have this unique property; given that we mostly ring palindromic methods, it seems highly likely that this has something to do with the preponderance of the Plain Bob leadhead class.

2. I’ve left out an important detail here: $c$ needs to be a change that fixes the treble. In practice, we can simplify this discussion by assuming that leadheads on $n$ bells are drawn from the set of permutations on $n-1$ bells — e.g. if for minor, $l$ and $c$ above are elements of $S_5$. ↩︎
3. In what follows, I’ll use the notation $\left<…\right>$ to indicate place notation — so $\left<16\right> = (23)(45)$ ↩︎