This article aims to look at ways of seeing the possible connections between the rows in methods or between the lead heads in touches, by drawing diagrams. We can use these graphs to explore the patterns in methods or touches and then help to construct new methods or touches.


Let's start with just three bells. We know there are 6 possible rows using three bells (as 3×2×1=6).

But how can we make a method out of these 6 possible rows, within the rules of ringing? And can we draw a diagram to help us navigate between the rows?

We start, conventionally, with rounds; the row 123. Then we can only swap pairs of adjacent bells. At first, we can either swap 1 and 2 over, or swap 2 and 3 over. In the first case (shown below in red), we reach the row 213; in the second case (shown in blue), we reach the row 132.

Next, where can we go from the row 213? If we swap over the first pair, we just get back to rounds. However, if we swap over the second pair (bells 1 and 3), we get a row we haven't already had; 231.

Similarly, from the row 132, if we swap the second pair over, we just get back to rounds, but if we swap the first pair, we reach the row 312, which is another new one.

I've kept the colour-coding; a red line for swapping the first pair, and a blue line for swapping the second pair.

The one remaining row is 321, which can be reached from either 312 (by swapping the last pair, 1 and 2) or from 231 ( by swapping the first pair, 2 and 3).

This is the complete diagram for Singles. We've got all the rows and every legal change.

So, which methods can we have? Going round the diagram from rounds in a clockwise direction gives Plain Hunt (Original) Singles, while going anti-clockwise gives Reverse Hunt (Reverse Original) Singles. These are the only methods possible that have all the possible rows, once each. There are other methods, which have the rows more than once, or which use the Identity Change (where no bells change places).


So far, so easy, but what can we do with four bells? There are 24 possible rows (as 4×3×2×1=24) and more possible changes between them.

Let's use our extent of Singles as a starting point and just put the 4th at the end of each of those rows. Again, the red lines on the diagram correspond to swapping the first pair of bells, while the blue lines show the swaps between the 2nd and 3rd bells in each row.

To get any other rows, we are going to need a change that affects the bell in 4ths place. We shall try swapping the final pair of bells. Doing that from each of the rows we have already will give six new rows, each of these shown by the yellow lines.

That's twelve rows reached, so we're half-way there now. But before we move on, we can draw in some more changes. You can get from 1243 to 2143, by swapping the first pair, so that needs a red line. Similarly, 2341 and 3241 are linked by swapping the first pair, as are 3142 and 1342.

You can see that some points on the diagram (i.e. the rows) have three lines from them, corresponding to each of the three possible changes we've looked at so far (all the single changes, given by place notations 12, 14 and 34). And these lines are all bidirectional, as repeating the same change gets you back to the one you were at before. We can see the newest 6 rows don't have any blue lines leading from them. So we know that we need to apply to each of them the change that swaps over the middle pair of bells.

Now we need to add some more lines. We can get from 1432 to 1423 by swapping the last pair, so that will be a yellow line. And so will the changes from 2413 to 2431 and from 3412 to 3421.

Before we try to find the remaining six rows, let's look at some of the patterns that are emerging in the diagram.

Remember that we started with a hexagon, which gave all the permutations of three bells. Our first hexagon kept the bell in 4ths place fixed, while swapping the first three bells. The changes were alternately red and blue.

But now we have some other hexagons. Looking at the top left of the diagram, we have a hexagon with blue and yellow edges. All the rows on its perimeter have the treble leading, while we get all possible permutations of the bells in the last three places. There's another similar hexagon at the right, consisting of rows in which the 2nd is leading, again with blue and yellow lines. And then at the bottom left, there's another hexagon of rows with the 3rd leading.

There are some other loops too - can you see them? They have a set of red and yellow edges, each colour alternately, giving a group of four rows.

There's a quadrilateral at the middle of the left side (1342, 1324, 3124, 3142), in which the first and last pairs have swapped amongst themselves, but always kept the two pairs separate. Again, this is the full set of possibilities for this scenario, having 1 and 3 in the first two places and 2 and 4 in the last two places. There are similar quadrilaterals at the top right (1234, 1243, 2143 and 2134) and at the bottom right (2314, 2341, 3241, 3214).

Which rows haven't we had yet? We said the blue and yellow hexagons had treble, 2nd and 3rd in the lead, but we haven't had any rows where the 4th is leading. And we can predict that there will be another hexagon linking them in the diagram. And while we're thinking about hexagons, we might expect there will be other ones like the original (blue and red) one, keeping the same bell at the back all the time.

We can see that last 6 rows that we added have only got two changes from them. That's because those points only have yellow and blue edges connected to them. So we need to add some red lines, i.e. the changes swapping the first pair of bells.

Hey Presto! We've got those missing six rows, as we needed, all with the 4 leading. And we can add in the last few changes:

This is the complete graph for Minimus using single changes (i.e. only one pair of bells swapping at each change). We've got 24 points, one for each of the possible rows. Each point has three lines connecting to it, one for each of the possible changes to get to another row.

What about those other properties we were expecting? There is a large hexagon round the edge, linking those rows with the 4 leading and we've now got three more quadrilaterals with red and yellow edges (top left, bottom left and right). But the last three hexagons aren't quite so obvious, unless you look at the colours. We were looking for alternate blue and red edges, each connecting rows with the same bell in 4ths place. They appear quite elongated in the diagram; at the left hand side we have all the rows ending with 2 (4132, 1432, 1342, 3142, 3412, 4312), then at the top right we have the rows with the 3 in 4ths place, and finally all the rows with treble at the back are in a hexagon at the bottom right.

To get a method which is an extent of Minimus, we need at route that starts at rounds and visits every other point once and only once, before returning to rounds. Do have a go. But here is one example, which is the map of the method Double Canterbury Minimus (shown in green).

Other possibilities are Double Court and Landbeach Minimus. But why can't we do Plain Hunt or Plain Bob? That's because we haven't yet allowed both pairs to swap at the same time.

Let's see where those changes are on the diagram. For instance, in Plain Hunt we want to go from 1234 to 2143. This involves crossing to the opposite corner of a quadrilateral. In the same quadrilateral, we can see that the other corners have the same property; going from row 1243 to row 2134 involves both pairs swapping over. And it doesn't take long to see that this works for every quadrilateral in the diagram, as the rows around each quadrilateral are all linked to each other in the same way.

We can add these changes onto our diagram:

Can you see where Plain Hunt is on the diagram? You should find that it consists of just blue and purple edges, as we know it has alternate changes with place notation X and 14.

It's pleasingly symmetrical, isn't it? And so was the Double Cantebury.

Here's Plain Bob Minimus, which you will see has three of those D-shapes, each of which is a lead of Plain Hunt, just with a gap for the lead end, where you jump from one course of Plain Hunt to another one.

Now for the best bit. Drawing this graph in 3-D gives a polyhedron, where the hexagons are all regular and the quadrilaterals are all squares. The outer heaxagon in the diagram above is the same size as all the others; it's just round the back of the shape. This is a truncated octahedron. You can make your own, with a net available from

The lines for Plain Hunt, which were a D-shape in the 2-D diagram above become circumferences of the truncated octahedron, which are (irregular) octagons, each in a single plane. The three cosets of Plain Hunt lie in three orthogonal planes.


Don't worry, I'm not going to draw a graph of all 120 rows (5×4×3×2×1=120).

In the last episode of Lockdown Learning, about Lead Heads, we mentioned in passing that all the rows where the treble was leading in a 120 of Plain Bob Doubles were equivalent to the rows in an extent of Plain Bob Minimus, if we ignored the treble and just compared the positions of the four working bells with the four bells in Minimus. And here we're going to do a similar thing.

We're just going to look at the Lead Heads of Plain Bob Doubles, that is the rows at the treble's backstroke lead.

At a Plain, the first lead head of Plain Bob Doubles after rounds is 13524. And using this same permutation (taking the 3rd bell, the 5th bell, the 2nd bell, then the 4th bell) can be applied to that row to give subsequent lead heads. In the row 13524, the 3rd bell is the 5, the 5th bell is the 4, the 2nd bell is the 3 and the 4th bell is the 2, so we arrive next at the row 15432. Applying the permutation again gives 14253, and then once more gives 12345. We have generated all the lead heads of a plain course of Plain Bob Doubles. We shall ignore the treble, as that is always in the lead, to give this diagram:

Note that the lines now have arrows. They don't represent changes, with adjacent pairs swapping over, which could happen in either direction, but permutations which, when repeated, don't take you back to where you were before.

And that's all the lead heads we can get in a plain course. To get the other lead heads, we need some Bobs.

Starting from rounds again, with a Bob at the end of the first lead, we reach the lead head 12354. As we know, the 2 has run in becomes 2nds place bell again, while the 3 runs out and becomes 3rds place bell. If we repeat this permutation, we get back to rounds, as 4 and 5 swap places again. So the lines for these permutations will have double-headed arrows. In a true touch, we won't want two Bobs in a row, but this just shows that when looking for paths around our diagram (which will correspond to touches, rather than methods this time), we can go either way along the blue arrows.

So let's see which lead heads we can reach with a Bob from each of the four lead heads that we have already.

As before, we can then add Plain leads from each of those lead heads, adding more squares to the diagram, before adding Bobbed leads to those new points, until we reach the final diagram. Again, there are 24 points, each a different lead head, each with a red arrow and a blue arrow leading away from them, leading to the lead heads obtained respectively by a Plain lead or a Bobbed lead.

You can see there are now six squares in total (including the big outer one), each corresponding to a plain course, starting from a different row.

But there are also some hexagons, although they are slightly disguised by some of their points coming on straight lines. Each of the hexagons has alternate red and blue edges, corresponding to the touch PBPBPB, which gives a 60 of Plain Bob Doubles. There are eight of these hexagons; four are long and thin, on the top, bottom, left and right edges of the diagram. The other four are between the squares at top middle, bottom middle, left middle and right middle.

Squares and hexagons, you say? This diagram can also be drawn on a truncated octahedron. Here's a net with the rows already written on, from Hugh C Pumphrey.

Now, this diagram is slightly different, as if you have a touch that visits all the points on it, you would get a 240, with each row once at handstroke and once at backstroke. A 120 only has 12 lead heads, but you mustn't include two lead heads which each generate a lead containing the same rows, once forwards and once backwards, or the touch will be false. For instance the leads starting (1)2345 and (1)3254 will contain the same rows. (Note that these point are diametrically opposite each other on the truncated octahedron.)

If we tried to make a 120 using the touch PPBPPBPPBPPB, we'd get to rounds OK:

But we've passed through both lead heads (1)2345 and (1)3254, so it will be false. In fact, we've also hit (1)4523 and (1)5432, which will generate false leads too.

Here's the usual 120 of Plain Bob Doubles, calling the 5 unaffected. It's an even more symmetric pattern on the truncated octahedron.

Another way of showing these lead heads in 3-D was developed by Revd. John Morgan and published in the Ringing World in 1995 (p. 930), with a further mention in 1996 (p.365). He used cubes, where each face represented a plain course, like the squares in our diagram. Bobs were represented by transitions onto neighbouring faces. He showed that a 240 using just Bobs, with each row once at handstroke and once at backstroke, was not possible, but did use three cubes, to develop a 720, with each row three times at backstroke and three times at handstroke.

This method will work for any method with the same lead ends as Plain Bob (such as St Nicholas), but for something like St Simon's, it gets very messy, as lines have to cross, partly because a Bobbed course of St Simon's is four leads in length.

For more information see:

Hugh Pumphrey's "Bell-ringing methods as polyhedra"

John Harrison's "Ringing Shapes"

Plus Magazine's "Ringing the Changes"

I also recommend Sarah Hart's recent Gresham College lecture on "The Mathematics of Bell Ringing"