Most methods have palindromic symmetry. That is, the second half of a lead has the same changes as the first half, but in reverse order.

For instance, here's a palindromic Major method. The first half of the lead had place notation X18X16.34.58.12, so the second half must have place notation 12.58.34.16X18X:

There is a line of symmetry at the half-lead (the change where the treble lies behind). I could put a mirror on that line and the bottom half would look exactly like a reflection of the top half. Here the mirror is shown by a dotted line:

Some people find it easier to see the symmetry if the diagram of the grid is turned this way, then the right side is a reflection of the left:

There is another line of symmetry at the lead-end (the change where the treble leads), again shown below both vertically and horizontally. Whatever happens as the treble goes down to the front will happen in reverse as the treble goes back out again.

Note that we haven't yet worried about what happens at the half-lead or the lead-end. But whatever the half-lead and lead-end changes might be, the fact that one lead has palindromic symmetry means that the blue line will have palindromic symmetry, as long as there is another place made at one (or both) of those positions. This must always happen with even bell single-hunt methods.

The blue line will also have two lines of top-to-bottom symmetry (or left-to-right symmetry when written out as below). The blue line of a method is generally written out starting at a line of symmetry, for instance, starting with 2nds place bell for a 2nds-place method. So the second half of the blue line is a reflection of the first half of the blue line. Not that this is an excuse for not learning the second half of the blue line as thoroughly as the first - it's just a help. Here is the blue line for Chester Bob Major, which is the method shown in the diagrams above, using a 78 half-lead and a 12 lead-end. The 12 lead end means we start the blue line with 2nds place bell. Then, for instance, we have a double-dodge in 3-4 up near the start, with places either side of it, with a corresponding double-dodge down right near the end.

The other line of symmetry comes at the half-lead, as with all single-hunt even-bell methods. This is because having an even number of bells means that you'll have an odd number of working bells. So your blue line will have an odd number of leads, and therefore its mid-point must be half-way through a lead. With single-hunt odd-bell methods, the points of symmetry are either both at a lead end (like St Simon's Triples or Reverse Canterbury Doubles) or both at the half-lead (like Croydon Bob Triples or Dunston Place Doubles).

Looking at which pairs of bells swap over at the half-lead in a palindromic method tells us immediately what row we will get at the next handstroke lead of the treble.

Let's look at a different method this time. Here's the first half-lead:

The method is palindromic, so we know the second half of the lead is the reverse of the first (i.e. place notation 56.14.16X14X), but we're not going to look at the detail of the blue line at the moment. We just want to know where each bell will be when the treble gets back to the lead.

The easiest bell to think about is the 5th. It makes a place at the half-lead, so it retraces its path in the second half of the lead. A place bell that does this is sometimes known as a "pivot bell".

Next we will look at the 2nd and 4th, which swap in 1-2 at the half-lead. Thinking about where that "mirror" is, we see that in the second half of the lead, the 2nd will follow the reverse of the line of the 4th in the first half:

And, correspondingly, the 4th's path in the second half of the lead will be the reverse of that of the 2nd in the first half:

So, the 2nd and 4th end up swapped over at the lead-end row, the handstroke lead of the treble. They swapped at the half-lead and so they take each other's positions at the end of the lead.

The other pair of bells that has swapped is the 3rd and the 6th. They must therefore have changed place with each other in the lead-end row, compared with rounds. We don't actually need to draw in their lines to know that we are going to reach the row 146253:

From here, we just have to choose which lead-end change to use and we'll have all the information we need to draw the blue line. (For more on lead-end variants, see here.)

We could make it a 2nds place method, so the lead-end change has place notation 12, giving the last row of the lead as 142635. For this one, we draw out the blue line starting from the 2, to show the symmetry. This is Crick Bob Minor.

Or we could have made a 6ths place method, by having place notation 16 for the lead-end change, giving the row 164523. To show the symmetry, we draw the blue line starting from 6ths place bell. This is Watson Bob Minor.

Alternatively, there could be a 14 place notation at the lead-end, giving a 4ths place method with the final row of the lead being 164235. We write this out starting from 4ths place bell, giving Wilkins Bob Minor.

(You might be forgiven for wondering whether there is a Franklin Bob Minor, named after Rosamund Franklin, but it appears there is not. And in fact Crick Bob Minor was named after the place in Northamptonshire, not after Francis Crick.)

Here's another example, but without drawing any blue lines. The first half of a lead of Fork Handles Bob Triples is here:

At the half lead, the pairs swapping are 4 & 6, 2 & 7 and 3 & 5.
So, given that this is a palindromic method, we can say that the lead end row
will be **1756342**, as the 4 and 6 have swapped places from rounds, so have
the 2 and 7, as also have the 3 and 5. The lead-end place notation in this method
is 125, bringing up the row **1765324**.

We can use these properties in reverse. To get a particular lead-end row in a palindromic method, we can see which bells must have swapped at the half-lead.

For instance, if we want to reach the row 15372846 in a palindromic Major method, we compare that row with rounds, to see which pairs of bells have changed places:

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

↓ | |||||||

1 | 5 | 3 | 7 | 2 | 8 | 4 | 6 |

We can see that 2 and 5 have swapped over, 4 and 7 have changed places, and 6 and 8 are in opposite positions. The 3 is still in the same place.

So at the half lead, this is what has to happen; the 3 must make a place, then the pairs swapping must be 2 & 5, 4 & 7 and 6 & 8. This will happen in different positions and different orders in different methods. And these don't have to have the treble plain-hunting, or even getting right up to the back, as long as those pairs of bells swap over and the 3rd makes a place. Here is a variety of examples, which all give the treble handstroke row 15372846.

Yuletide Bob Major:

Little London Little Bob Major:

Cambridge Surprise Major (only part of this one is shown here):

We also realise why there are some lead-ends which can't be obtained from a palindromic method, as they don't have a set of pairs of bells swapping.

For instance, we can't get the Doubles handstroke row **15324**. The 3rd
is back in its home place, so it would have to be a pivot bell, making a place
at the half-lead. But there are no pairs of bells swapping - 2, 4 and 5 cycle
in a group of three, each going to the place of the next one.

In his recent webinar on creating and solving Methodoku puzzles, Mark Davies referred to cyclic lead heads (i.e. the backstroke treble lead). These are ones where, apart from the treble, the bells are in rotation of rounds, starting with a bell other than the 2, e.g. 15678234, 134562 or 1723456. He said that they couldn't be obtained from palindromic methods.

Let's try to get a lead *head* of **156234**, for example. Working
backwards from there, the lead *end* (i.e. the row of the handstroke treble
lead) could be **152634**, by using a change with place notation 1256.

Comparing 152634 with rounds, we see that 4 and 6 have swapped, which is promising. But the remaining bells, 2, 3 and 5, have changed places in a cycle of three bells, rather than a straight swap. We can't have three bells swapping like that in a single change at the half-lead, so this option doesn't work.

However, although not possible on 6 bells or more, cyclic lead heads can be obtained from palindromic Doubles methods, such as Stedman Bob Doubles. And all three-lead palindromic Minimus methods have cyclic lead ends.

Mark himself created the method **Spinning Jennie Delight Royal**, which
is palindromic, but does produce cyclic permutations when started from the snap
lead. The first lead end is shown here - NB '0' is used for bell 10.

Comparing the handstroke row **1843590267** with rounds, **1234567890**,
we see that the pivot bell is the the 5th. Then the pairs which must swap at
the half-lead are 2 & 8, 3 & 4, 6 & 9 and 7 & 10. We can see
that is just what happens at the half-lead:

Then look at the start of the first two leads:

We can see that the working bells in the row **12345386079** in the first
lead have rotated to give the row **1453860792** in the second lead. And
so the corresponding rows in subsequent leads are all further rotations of that
same row. So the composition has the method starting with rounds at that row,
to exploit the run-based music that is thereby produced.