under construction

1. Introduction

The history of mathematics records several occasions when failure of the prevailing symbolic language to cope with unwieldy expressions has led to the development of a more compact notation by which complicated ideas could be expressed intelligibly. Such innovations have in turn led to an increase in knowledge, for the chief merit of a compact notation is that the significance of statements can be readily grasped and their implications more easily studied.

With this experience as a guide, a symbolic approach to the theoretical aspects of change ringing promises to be attended by similar advantages.

The present paper records an attempt at applying such a treatment to the properties of false course heads in treble-dominated Major methods. Many of the results obtained are well known, but others have not, as far as the writer is aware, been previously published.

2. Symbols

Before proceeding to applications of the symbolic treatment, the symbols employed must be defined and their properties demonstrated. As there is a superficial resemblance between the behaviour of the symbols and the rules of ordinary algebra, it will be natural to pursue the analogy by borrowing from algebra any technical terms required.

Capital letters of the italic (sloping) alphabet will be used to denote rows. Certain letters will be given a permanent significance, while others will be defined as occasion demands, or will represent unknown rows (in the same way that x proverbially represents the unknown quantity in algebra).

Ordinary numbers will be represented by small (lower case) italic letters.

3. Multiplication

The multiplication sign ‘×’ will be used to mean ‘transposed by.’ For example, 4372856 × 8627534 = 6845237. When rows are represented by letters, however, the multiplication sign will usually be omitted (as in algebra) and U transposed by V will be written ‘UV.’

The multiplication of rows differs in an important respect from ordinary multiplication. 5 × 2 = 2 × 5, and yx = xy; but VU is not generally equal to UV. Symbols must accordingly be written in the correct order.

4. Division

Transposition of one row by another having been regarded as multiplication, we will next consider what operation on rows corresponds to division. Now, in algebra, x ÷ y is often written xy -1 that is ‘x divided by y’ is the same as ‘x multiplied by the reciprocal of y.’ The operation of division has thus been converted into one of multiplication.

Applying the same rule to rows, ‘U divided by V’ will be translated into ‘U multiplied by the reciprocal of V’ and will be written ‘UV -1,’ where the notation ‘V -1 indicates the reciprocal of V, as well be explained in a later section.

The reciprocal of a given row is the row by which the given row must be transposed to produce rounds.

Taking the same values for U and V as in Section 3,

U ÷ V =4372856÷ 8627534
that is, UV -1 =4372856 × (8627534)-1
=4372856 × 4786352

In practice, division can be performed directly by writing the row to be divided over the divisor. The quotient is then obtained by writing down the numerals in the top row in the positions indicated by the numerals under them.

U =4372856As 7 is over 2, write 7 in second place; as 5 is over 3, write 5 in third place; and so on.
V =8627534

UV -1 =7568324

Incidentally, taking the reciprocal of a row is merely a special case of division, with rounds as the numerator.

V =8627534

V -1 =4786352

5. Powers

In algebra, if a term a is multiplied by itself, the result is written a 2 similarly, a × a × a is written a 3. The small numeral is called the ‘index’ and indicates the number of a’s to be multiplied together. a 2 is is second power of a and is read as ‘a squared’; a 3 is the third power and is called ‘a cubed’; the fourth power, a 4, is read as ‘a to the fourth’ and similarly with other high powers.

(The first power, a 1, is a itself, and the index ‘1’ is written only when we wish to emphasise that we are thinking of a as its own first power.)

The same notation will be applied to rows.

Let T =4573862
T 2 =4573862 × 4573862

T 3 =T 2 × T
i.e. T 3 =7365284 × 4573862

If a number of powers of the same term are multiplied together, the index of the result is the sum of the indices of the powers. For instance, T 2 × T 3 = T 2+3 = T 5; and T 4 × T 2 × T = T 4+2+1 = T 7. To realise this, it is simply necessary to remember that T 2 = T × T and T 3 = T × T × T, therefore T 2 × T 3 = T × T × T × T × T = T 5.

We are now in a position to show, as promised in the preceding section, that the index ‘-1’ means ‘the reciprocal of.’

Still taking T = 4573862, its reciprocal is 8523746.

Now T 3 × reciprocal of T
=6583427 × 8523746

But 7365284 =T 2
that is, T 3 × reciprocal of T = T 2

In order that the rule regarding the multiplication of powers shall apply (as it must if it is to be a general rule) the reciprocal of T must be written ‘T -1.’

then T 3 × T -1 = T 3-1 = T 2

A meaning can also be found for a zero index.

For T 1 × T -1 = T 1-1 = T 0

But T -1 is the reciprocal of T and, by definition, any row transposed by its reciprocal produces rounds.

therefore T 0 = rounds

Similarly, the zero power of any row is rounds.

If we have to evaluate a power of a term which is itself a power, such as (T 2)3, the index of the result is the product of the indices, in this case T 6, since (T 2)3 = T 2 × T 2 × T 2.

6. Reciprocal of a Product

If we have an expression such as X = UV, we may wish to express X -1 also in terms of U and V.

Given that X = UV Multiply both sides of the expression by V -1 then XV -1 = UVV -1 But VV -1 = rounds, and any row multiplied by rounds is unaltered. therefore XV -1 = U Next multiply both sides by U -1 giving XV -1U -1 = UU -1 that is XVU -1 = rounds Also XX -1 = rounds, by definition, so that XX -1 = XV -1U -1 therefore X -1 = V -1U -1 By similar manipulation, it can be shown that if X = UVW, then X -1 = W -1V -1U -1

Generally, to take the reciprocal of a product, take the reciprocal of the individual factors, but write them down in reverse order.

7. Lead Heads and Lead Ends

The rows which, occur as the lead heads and lead ends of the plain course have properties not shared by rows in general.

Let H =3527486
I =2345678
J =2436587

H, I, J will always have this significance and will be used in future without further definition. They are strictly eight-bell rows, but as the treble is at home it is not written.

The seven lead heads and the seven lead ends of the plain course can be denoted by expressions involving only H, I, J.

For instance, the first lead head of the plain course of any method is I, which can also be written H 0 (see Section 5). In Plain Bob Major, the second lead head is H, which can be written H 1; the third is H 2. Each lead head is H times the preceding one, hence the lead head after H 6 is H 7. But H 7 is the seventh lead head (4263857) and the next is rounds; therefore H 7 = I = H 0.

Similarly, H 8 = H 1, H 9 = H 2, and, in general, H r = H r+7t where t is any whole number, positive or negative. Thus H r = H r+7 = H r+14 = H r-7 = H r-21 and so on. The equivalent expression for H r that we shall come across most often is H r-7. For example, H 3 = H 3-7 = H -4. Conversely, H -5 = H 7-5 = H 2.

Since the plain course lead heads of all methods are the same, though occurring in different orders, they can always be expressed by H 0, H 1, H 2H 6. If the second lead head of a method is H 4, each lead head will be H 4 times the preceding one. The third lead head will thus be H 4 × H 4 = H 8 = H 1; the fourth will be H 1 × H 4 = H 5, and so on.

The lead ends of the plain course can be expressed in the form JH P. For instance, the first lead end of Plain Bob is 3254768 = JH 6, the second is 5372846 = JH 5; the last is JH 0 = J.

These expressions have the property that each is its own reciprocal; for example, (6847253)-1 = 6847253. In general, JH P = (JH P)-1 = H -PJ -1 = H -PJ (since J -1 = J).

If the first lead end of a method is JH b, the other lead ends of the plain course are each JH b times the preceding lead head. For example, if the second lead head is H a, the second lead end is H a.JH b = JH -aH b = JH b-a. The third lead end will be H 2aJH b = JH b-2a.

Conversely, if a given lead end is JH c, we can find the head of the lead to which it belongs. From above, we have:

A lead end JH b-a follows a lead head H a.

As c takes the place of b-a, we havec =b-a
c-b =-a
a =b-c

Hence a lead end JH c follows a lead head H b-c.

The lead heads and ends of a course other than the plain course are represented by GH 0,GH 1,GH 2GH 6 and GJH 0,GJH 1GJH 6, where G is the course head.

8. False Course Heads of a Method

We are now in a position to derive general expressions for the false course heads of a Treble Bob Major method in respect of courses having the tenors together throughout.

A false course head occurs when two rows in the course have the treble and tenors (1,7,8) in the same places and the rows are of the same nature, both even or both odd. For under these conditions, the same row can appear in different courses.

The plain course of a method is made up of fourteen half leads, seven following the lead heads and seven preceding the lead ends. If the latter seven are inverted, they are identical in structure with the first seven. The course can thus be dissected into fourteen half leads, seven headed by H 0,H 1H 6 and seven headed by JH 0,JH 1JH 6. For brevity, half leads headed by even rows (such as H 4) will be referred to as even half leads, and those headed by odd rows (such as JH 2) as odd half leads.

Different formulæ are obtained for the false course heads, depending on whether the two rows with the treble in the same place in the same half lead are of the same or opposite nature. We will deal with the last-mentioned case first.

Let A be an even row in the first half lead of a method and let its position in the half lead be a. Let B be an odd row with the treble in the same place as in row A and let its position in the first half lead be b. Let F, F2, F3, etc., be false course heads.

The ath rows of even half leads will be even; the bth rows of odd half leads will also be even, and one of them might have 7,8 in the same places as one of the even ath rows. If so, the positions of the working bells (2,3,4,5,6) will be different in the two rows, but it will be possible to select two other courses with the working bells so placed that:

(i) in one such course, the bth row of an odd half lead is identical with the ath row of an even half lead in the plain course;

(ii) in the other such course, the ath row of an even half lead is identical with the bth row of an odd half lead in the plain course.

The two courses will thus be false against the plain course and their heads will be false course heads of the method.

Let H mA (the ath row of an even half lead) and JH nB (the bth row of an odd half lead) be the two rows in the plain course having the tenors in the same positions.

The bth row of an odd half lead in a course F is FJH nB, giving, for case (i):

FJH nB = H mA Multiplying both sides of the equation by B -1 FJH n = H mAB -1 Multiplying both sides by H -n FJ = H mAB -1H -n Multiplying both sides by J -1 F = H mAB -1H -nJ -1 = H mAB -1H -nJ Since J -1 = J

In case (ii), the ath row of an even half lead in a course F is F2H mA, giving:

F2H mA = JH nB F2H m = JH nBA -1 F2 = JH nBA -1H -m

Now the odd half leads (headed JH 0JH 6) differ from the even half leads (headed by H 0H 6), so far as 7,8 are concerned, in that 8 occupies the positions of 7, and 7 occupies the positions of 8. For example, the half leads headed H 0 and JH 0 have the tenors in the same places, row for row, but interchanged, since H 0 = 2345678 and JH 0 = 2436587.

Therefore if, as assumed, the rows H mA and JH nB have the tenors in the same positions, the rows J(H mA) and J(JH nB) will also have the tenors in the same positions. (J.JH nB = H nB since J 2 = I.)

JH mA and H nB are odd rows and give rise to two more false course heads, F3, F4, such that:

(iii) F3H nB = JH mA F3 = JH mAB -1H -n (iv) F4JH mA = H nB F4 = H nBA -1H -mJ

The expressions for the set of false course heads F, F2, F3, F4, will now be examined to see whether a simple relationship exists between them.

Now F = H mAB -1H -nJ Using the rule of Section 6: F -1 = J -1H nBA -1H -m = JH nBA -1H -m But F2 = JH nBA -1H -m therefore F2 = F -1 Also JFJ = J.H mAB -1H -nJ.J = JH mAB -1H -n But F3 = JH mAB -1H -n therefore F3 = JFJ From above, F -1 = JH nBA -1H -m JF -1J = J.JH nBA -1H -m.J = H nBA -1H -mJ But F4 = H nBA -1H -mJ therefore F4 = JF -1J

Any false course head F is thus always accompanied by three other false course heads F -1, JFJ, JF -1J. These false course heads may or may not be all different and, as we shall see later, they may be accompanied by others.

9. False Course Heads Due to Rows of Like Nature

The preceding section dealt with the case where the two rows with the treble in a given place in the first half lead of a method are of opposite nature. The following formulæ are derived for the case where the two rows are of the same nature.

Let C1, C2 be two rows in the first half lead of a method with the treble in the same place, the two rows being either both even or both odd.

As the rows C1, C2 are of the same nature, even half leads can repeat among themselves but cannot repeat with odd half leads. Odd half leads can also repeat with one another, since if C1, C2 are even (or odd), the rows corresponding to them in odd half leads will be odd (or even).

Using the same procedure as in Section 8, but noting that J will appear on both sides of the equations or not at all, the four possibilities for repetition are:

(i) FH nC2 = H mC1 F = H mC1C2-1H -n (ii) F2H mC1 = H nC2 F2 = H nC2C1-1H -m (iii) F3JH nC2 = JH mC1 F3 = JH mC1C2-1H -nJ (iv) F4JH mC1 = JH nC2 F4 = JH nC2C1-1H -mJ

The relationship between these false course heads is the same as for those found in Section 8.

For F -1 = (H mC1C2-1H -n) = H nC2C1-1H -m therefore F2 = F -1 That F3 = JFJ, and F4 = JF -1J, can be seen at a glance.

10. Practical Formulæ for False Course Heads

The formulæ obtained in sections 8 and 9 are not in the form best suited for working out the false course heads of a method in practice. Some more useful expressions will now be derived from them, with the help of the following identity: H μX = Xφμ where X is any row (with the treble omitted) and φμ, is defined as the operation which substitutes for each bell in a row the bell μ places earlier in plain course coursing order. Proof:
If the values of H 0H 1H 6 are written one under the other, as in the annexed list, it is found that each column of figures, read upwards, gives the coursing order of the plain course. For instance, the bells in second place, from bottom to top, are 4687532; those in third place are 2468753; and so on H 0=2345678
H 1=3527486
H 2=5738264
H 3=7856342
H 4=8674523
H 5=6482735
H 6=4263857

In other words, a row H μ can be obtained from rounds by substituting for each bell in rounds the bell μ places earlier in plain course coursing order.

Hence H μ = Iφμ and H μX = IφμX

It now remains to be shown that IφμX = Xφμ.

In the operation of multiplying a row W by another row X, each bell in the row W takes up a new position (which may or may not be different from its original position). The new position is such that the bell in nth place in W now occupies the position of the bell n in X.

Let W = a.b.c.d.e.f.g. Let X = 3482576 then WX = b.c.g.a.d.f.e.

where ag represent the individual numerals constituting the row W.

Suppose a 1 is the bell μ places earlier than a in plain course coursing order, b 1 the bell μ places earlier than b, and so on,

then a 1.b 1.c 1.d 1.e 1.f 1.g 1. = Wφμ and b 1.c 1.g 1.a 1.d 1.f 1.e 1. (Wφμ)X Now suppose that in the original product WX = b.c.g.a.d.f.e. we substitute for each bell the bell μ places earlier in plain course coursing order, we get: (WXμ = b 1.c 1.g 1.a 1.d 1.f 1.e 1. thus WφμX = (WXμ Putting W = I, this becomes IφμX = (IXμ = Xφμ But we showed above that H μX = IφμX therefore H μX = Xφμ Example: Take X = 3482576 and μ = 3, then H μ = H 3 = 7856342 X = 3482576 H 3X = 8527643 Also X = 3482576 Xφμ = 8527643 as before It was shown in Section 8 that when the two rows concerned are of opposite nature, F = H mAB -1H -nJ = H mAB -1JH n since H -nJ = JH n Comparing this expression with the identity just proved (HμX = Xφμ), we see that AB -1JH n takes the place of X, and m takes the place of μ. Therefore F = H m(AB -1JH n) = (AB -1JH nm When the two rows concerned are of like nature (Section 9), F = H m(C1C2-1H -n) = (C1C2-1H -nm

These formulæ are convenient for practical use, as the starting point in the calculation is the pair of rows being tested.

Similar formulæ for the three false course heads associated with F can be obtained by using the same transformation, but they are unnecessary, since the associated false course heads can be more easily obtained from the relationships derived at the end of Sections 8 and 9.

In applying the formulæ, a value of n is chosen to bring behind a pair of bells which course in the plain course. The appropriate value of in is then chosen to convert these bells into 7.8, so producing a row of course head form.

The incidence of falseness is indicated by the values of m and n so obtained. Referring to Section 8 for the case dealing with rows of opposite nature, we see that the false course head F involves repetition between a row in the plain course half lead headed H m and the half lead headed FJH n of the course F. This can be written F.JH n v. I.H m, where ‘v.’ (versus) indicates ‘false against.’

The associated false course head F -1 is produced by repetition between the plain course half lead headed JH n and the half lead headed F -1H m of a course F -1; that is F -1.H m v. I.JH n. The other false course heads JFJ and JF -1J give JFJ.H n v. I.JH m and JF -1J.JH m v. I.H n.

Collecting these expressions, together with corresponding expressions from Section 9, we can tabulate the incidence of falseness in terms of the half lead heads.

Table IA
Table IB
Falseness due to rows
of opposite nature
Falseness due to rows
of same nature.

F.JH nv.I.H mF.H nv.I.H m
F -1.H mv.I.JH nF -1.H mv.I.H n
JFJ.H nv.I.JH mJFJ.JH nv.I.JH m
JF -1J.JH mv.I.H nJF -1J.JH mv.I.JH n

In dealing with a particular method, however, we want to know which complete leads are false against one another, and we want to quote them by their position in the course, as is customary, or (better still, perhaps) by the calling positions of the tenor. The latter alternative will be adopted here, as it seems more logical to designate, for example, the Middle lead by M than to call it 1 in some methods and 3 in others.

Moreover, in a pair of methods differing only in the place made at the treble full lead, identical leads are differently numbered so that the incidence of falseness has to be listed separately for each method. When the leads are designated by the positions of the tenor, one list of falseness applies to both methods.

Now, if the first lead end of the method under consideration is JH b, the head of the lead to which a half lead headed JHC belongs is H b-c, as pointed out in Section 7.

We can thus rewrite Table I. in a form showing the incidence of falseness in terms of lead heads only, giving Table II.

Table IIA
Table IIB
Falseness due to rows
of opposite nature
Falseness due to rows
of same nature

F.H b-nv.I.H mF.Hv.IH m
F -1.H mv.I.H b-nF -1.H mv.IH n
JFJ.H nv.I.H b-mJFJ.H b-nv.I.H b-m
JF -1J.H b-mv.I.H nJF -1J.H b-mv.I.H b-n

Finally, by means of Table III., we can convert the results into calling positions.

Table III.

H 0 = H(Home)
H 1 = W(Wrong)
H 2 = V(Fifths)
H 3 = B(Before)
H 4 = I(In)
H 5 = F(Fourths)
H 6 = M(Middle)

11. Examples

As a demonstration of the use of the formulæ given in the preceding Section, we will work out some of the false course heads of Cambridge Surprise. The first two sections of the method are given here, the even rows being marked + and the odd rows -. Considering the rows with the treble in second place, we see that they are of opposite nature, hence the formula F = AB -1JH nφm applies. A = 2436587 B = 2648375 AB -1 = 2537486 AB -1J = 2354768 Cambridge. 12345678 + A 21436587+ 12463857- B 21648375- C1 26143857+ 62418375 + C2 62148735+ 26417853+ Inspection of the value of AB -1J reveals that 2.3 (which course in the plain course) are in plain course positions, so we transpose by 8674523 to bring them behind; as 8674523 this gives n = 4. then A B -1JH 4 = 8765423 We now substitute for each bell in this row the bell m places earlier in plain course coursing order, choosing m to convert 2.3 into 7.8. Since 8 is three places earlier than 3, the appropriate value is m = 3, giving AB -1JH 4φ3 = 2436578 thus F = 24365. m = 3, n = 4 Referring to Tables IIA and III. in Section 10 and substituting the values of F, m and n, taking b = 5 (since the first lead end of Cambridge is 5372846), the incidence of falseness is: F.H 5-4 v. I.H 3 = 24365 W v. B F -1.H 3 v. I.H 5-4 = 24365 B v. W JFJ.H 4 v. I.H 5-3 = 24365 I v. V JF -1J.H 5-3 v. I.H 4 = 24365 V v. I Before leaving this example, we will show how two of the operations can be condensed into one. As given above, we obtained AB -1 from A and B and then multiplied by J to obtain AB -1J. AB -1 was obtained by taking the numerals of B in the order 2345678 and writing down the corresponding numerals of A. If, instead, we take the numerals of B in the order 2436587, we obtain AB -1J at once. (What we are then doing is to multiply A by (JB)-1, producing A(JB)-1 = AB -1J -1 = A B -1J.) As an example of the case where the two rows involved are of like nature, consider the rows with the treble in third place. The relevant formula is F = C1C2-1H -nφm. C1 = 2643857 C2 = 6248735 C1C2-1 = 6547283 In the row 7.8 are in plain course positions and can be brought behind by multiplying by H 6, giving -n = 6, that is n = -6 = 7-6 = 1. Then C1C2-1H6 = 4625378 and, as this row is already a course head, m = 0. Translating this result by Table IIB and by Table III., we get: F.H 1 v. IH 0 = 46253 W v. H F -1.H 0 v. IH 1 = 46253 H v. W JFJ.H 5-1 v. I.H 5-0 = 32546 I v. F JF -1JH 5-0 v. I.H 5-1 = 32546 F v. I But, referring again to the value of C1C2-1, we see that 5.7 are also in plain course positions; accordingly, the above procedure must be repeated to obtain the falseness resulting from this coursing pair. C1C2-1 = 6547283 C1C2-1H 5 = 2436857 C1C2-1H 5φ1 = 3254678 F = 32546, m = 1, n = 2 giving 32546 V v. W 32546 W v. V 46253 B v. I 46253 I v. B As a further example, consider the rows with the treble in fifth’s in Bristol Surprise. They are of opposite nature and are marked A and B in the annexed section. A = 3254786 B = 2435786 AB -1J = 3256487 Bristol B 24351786- 23457168 - A 32541786+ 35247168- Inspection of the row 3256487 shows that no pair of bells which course in the plain course are in plain course positions. No value for F can therefore be found; hence as far as the rows with the treble in fifth place are concerned, the method has a clean proof scale. One further point should be mentioned in connection with rows of opposite nature, in order to present the arguments used in deriving the formula for F in a more definite manner, A was assumed to be an even row and B an odd row. It is not necessary, when using the formula, to adhere to this convention, since precisely the same final result is obtained if B is even and A is odd.

12. Possible Values of AB -1J and C1C2-1

Having arrived at the stage where, given a pair of rows, we can derive the false course heads produced by them, it will be instructive to reverse the process and see what range of values AB -1J and C1C2-1 can have for a given set of false course heads.

Dealing first with rows C1,C2, of like nature, we can rearrange the formula for F in the following manner:

F = H mC1C2-1H -n H -mF = C1C2-1H -n H -mFH n = C1C2-1 This can be written: C1C2-1 = Fφ-mH n, as shown in Section 10.

Treating the formula pertaining to rows of opposite nature in the same way, we get:

AB -1J = Fφ-mH -n

These expressions for AB -1J and C1C2-1 are similar, the only difference being the minus sign in H -n in the second one, which, as will be seen, has no effect on the range of values which the expressions can take.

For brevity, C1C2-1 will be referred to as a ratio, since it is simply one row divided by another; the term ratio will, in this Section, be regarded as including AB -1J also, since, although this expression is not strictly a ratio (it is in fact a ratio multiplied by J) the possible values of AB -1J are the same as the possible values of the ratio C1C2-1

Now, m and n can each take seven values (0 to 6), giving forty-nine values of the ratio for a given value of F. Both the expressions have the same forty-nine values, as H -n in one expression gives the same result as H n in the other when all seven values of n are substituted. It will thus be sufficient to consider one expression, and Fφ-mH n will be used.

Some practical examples will help to make clear the implications of the expression, and we will consider first the false course head set, 34562, 62345, 46325, 54263.

Table IVa shows the forty-nine values of the ratio calculated from the above expression for the case when F = 34562. All the ratios produced by the same value of in have been placed in the same column, while those produced by the same value of n are in the same row. Thus the first column of the table is obtained by putting m = 0, and giving n the values 0, 1, 2 … 6 in succession; the second column in obtained by putting m = 1, and so on.

Table IVa. Ratios Producing 34562


Now, in a set of false course heads, any course head can be regarded as F, since whichever is so regarded, the associated false course heads F -1, JFJ, JF -1J, are the remaining members of the set. Accordingly, by substituting 62345, 46325, and 54263 in the formula in turn, additional ratios are obtained which lead to the same set of false course heads.

Table IVb. Ratios Producing 62345


Table IVc. Ratios Producing 46325


Table IVd. Ratios Producing 54263


The four tables contain all values of the ratio which lead to the set of false course heads on which the tables are based.

Such a set of tables can be written out for every set of false course heads, but we have not the space to give them all here.

We will therefore confine ourselves to one or two further examples, and will next deal with the frequently occurring false course head set 32546, 46253. Table V gives the forty-nine values of the ratio leading to 32546.

Table V. Ratios Producing 32546


Owing to the way in which the tables are built up, a row of course head form (i.e. with 8 at home) occurs in each column of the tables. In Tables IVa, b, c, d. apart from the course head at the top of column 1 (on which the table is based), the course heads have the tenors parted. When we examine Table V, however, we discover that the course head in column 2 also has 7,8 at home and is, in fact, 46253, the remaining member of the course head set under consideration. It is clear, therefore, that a table based on 46253 would contain the same values of the ratio as Table V.

The presence in the same table of the two course heads 32546, 46253 has an effect on the incidence of falseness in a method having these false course heads. Suppose, for example, that a method has a pair of rows with the treble in the same place such that AB -1J = 3254678. The method then has a false course head F = 32546, with m = 0, n = 0. But it is clear that a method having AB -1J = 3254678 will also have a false course head F = 46253, with m = -1, n = 1, since in Table V, 4625378 is one column to the right of 3254678 and one row lower.

Similarly, whatever the values of m and n for the false course head 32546, the false course head 46253 will also occur and will have values of m and n one less and one greater, respectively, than those for 32546.

When, however, the rows concerned are of the same nature, the values of m and n for 46253 are both one less than for 32546.

The difference between the cases for rows of like and unlike nature is due to the minus sign before the n in the formula F = C1C2-1Hnφm for rows of like nature.

An instance of such a duplication of a set of false course heads is given by the rows with the treble in third place in Cambridge Surprise, worked out in Section 11.

In this example, the additional false course head 46253 is a member of the same set as the first false course head 32546, since when F = 32546, JFJ (and JF -1J) = 46253; conversely, when F = 46253, JFJ = 32546. This turns out to be merely incidental and is not a general rule.

By writing out tables such as Table V. for all the possible false course heads, those which occur in the same table, and which thus occur together as the false course heads of methods, can be discovered. A less laborious way of finding the course heads which belong to one another is, however, available.

The first step in the procedure for working out the false course head for a given value of AB -1J or C1C2-1 consists in bringing behind a pair of bells which course in the plain course. If the ratio has more than one such pair of bells, each such pair produces a false course head, and these false course heads must appear in the same table. We can thus determine which false course heads are inseparable by examining the sixty course heads for coursing pairs.

As an example, consider the false course head 26543. Now F -1, JFJ, and JF -1J are also 26543, so that it seems as if this false course head can occur alone.

But the row 2654378 has 3,5 and 6,4 (in addition to 7,8) in plain course positions, and when we work out the false course heads produced by bringing 35 and 64 behind, we obtain 36245 and 42563 as false course heads which inevitably accompany 26543.

F1 =2654378
F1H 2 =4768235
F1H 2φ2 =3624578 = F2

F1 =2654378
F1H 5 =3582764
F1H 5φ5 =4256378 = F3

We thus see that if we have:
F1 = 26543, m = 0, n = 0)for rows of unlike nature
we also have)
F2 = 36245, m = 2, n = 2)
F1 = 42563, m = 5, n = 5)
F1 = 26543, m = 0, n = 0)for rows of like nature
F2 = 36245, m = 2, n = -2)
F3 = 42563. m = 5, n = -5)

Generally, the complete list of falseness when 26543 occurs as one false course head will be given by substituting in the relevant half of Table II. (Section 10) the following values:

F =26543 mnunlike rows
36245 m+2,n+2
42563 m+5,n+5
26543 mnlike rows
36245 m+2,n-2
42563 m+5,n-5


Suppose F = 26543, m = 2, n = 5, b = 1, and the rows producing this false course head are of opposite nature. Then the incidence of falseness is as follows:

F.H b-n v. I.H m= 26543 H 3 v. I.H 2
= 26543 B v. V
F -1.H m v. I.H b-n= 26543 H 2 v. I.H 3
= 26543 V v. B
JFJ.H n v. I.H b-m= 26543 H 5 v. IH 6
= 26543 F v. M
JF -1J.H b-m v. I.H n= 26543 H 6 v. I.H 5
= 26543 M v. F

and F = 36245, m = 4, n = 7 = 0, b = 1.

F.H b-n v. I.H m= 36245 H 1 v. I.H 4
= 36245 W v. I
F -1.H m v. I.H b-n= 42563 H 4 v. I.H 1
= 42563 I v. W
JFJ.H n v. I.H b-m= 42563 H 0 v. I.H 6
= 42563 H v. M
JF -1J.H b-m v. I.H n= 36245 H 6 v. I.H 0
= 36245 M v. H

and F = 42563, m = 7 = 0, n = 10 = 3, b = 1.

F.H b-n v. I.H m= 42563 H 5 v. I.H 0
= 42563 F v. H
F -1.H m v. I.H b-n= 36245 H 0 v. I.H 5
= 36245 H v. F
JFJ.H n v. I.H b-m= 36245 H 3 v. I.H 1
= 36245 B v. W
JF -1J.H b-m v. I.H n= 42563 H 1 v. I.H 3
= 42563 W v. B

13. Groups of False Course Heads

By carrying out the procedure outlined in the preceding Section, we can marshal the sixty course heads into groups, such that if any member of a group appears as the false course head of a method, the other members of the group also occur.

If we define a set of false course heads as the four represented by F, F -1, JFJ, JF -1J (which may be all alike, all different, or two pairs alike), then we find that some of our groups contain only one set, some two different sets, and one three different sets. In the following list of groups, the sets within the groups are indicated by suffix numbers in the symbols F, JFJ, etc.

Group I. (See Note 3)

F = F -1 = JFJ = JF -1J = 23456

Group II.

F = F -1 = JFJ = JF -1J = 24365

Group III.

F = F -1 = JFJ = JF -1J = 25634

Group IV.

F1 = F1-1 = JF2J = JF2-1J = 32546
F2 = F2-1 = JF1J = JF1-1J = 46253
Group V.

F =F -1= 32465
JFJ =JF -1J= 43265
Group VI.

F1 = F1-1 = JF2J = JF2-1J = 32654
F2 = F2-1 = JF1J = JF1-1J = 45236
Group VII.

F =F -1= 56423
JFJ =JF -1J= 63542
Group VIII.

F =JF -1J= 53462
F -1 =JFJ= 63425
Group IX.

F =JF -1J= 54632
F -1 =JFJ= 65324
Group X.

F =F -1= 53624
JFJ =JF -1J= 65432
Group XI.

F1 =F1-1= JF1J= JF1-1J= 26543
F2 =F3-1= JF2-1J= JF3J= 36245
F2-1 =F3= JF2J= JF3-1J= 42563
Group XII.

F1 =JF2-1J= 23564
F1-1 =JF2J= 23645
F2-1 =JF1J= 26435
F2 =JF1-1J= 25463
Group XIII.

F = 34562
F -1 = 62345
JFJ = 46325
JF -1J = 54263
Group XIV.

F = 52643
F -1 = 36524
JFJ = 65243
JF -1J = 46532
Group XV.

F1 =F1-1= JF2J= JF2-1J= 54326
JF1J =JF1-1J= F2= F2-1= 64352
F3 =JF3-1J= F4-1= JF4J= 64523
F3-1 =JF3J= F4= JF4-1J= 56342
Group XVI.

F1 =JF2J= 45623
F1-1 =JF2-1J= 56234
JF1J =F2= 35642
JF1-1J =F2-1= 62534
Group XVII.

F1 =JF2J= 52364
F1-1 =JF2-1J= 34625
JF1J =F2= 64235
JF2-1J =F2-1= 45362
Group XVIII.

F1 =24536F2 =62453
F1-1 =25346 F2-1 =36452
JF1J =26354 JF2J =53246
JF1-1J =24653 JF2-1J =43526
Group XIX.

F1 =JF1-1J= 34256
F1-1 =JF1-1J= 42356
F2 =JF2-1J= 63254
F2-1 =JF3J= 43652
JF2J =F3-1= 52436
JF2-1J =F3= 35426
F4 =JF4-1J= 42635
F4-1 =JF4J= 35264
Note 1: In the above groups, the marking against the course heads merely indicates the relationship between them, and means nothing more than this. Any course head in a set can be regarded as F, as pointed out in Section 12; and in groups containing more than one set, any course head can be regarded as F1.

Note 2: No attempt has been made to indicate the relationship between the values of m and n for the different sets of false course heads in a group containing more than one set. A more voluminous tabulation would be necessary to show this, and as most of the groups of false course heads are not likely to occur in practical methods, it has not been attempted here.

Note 3: Group I. has been set out as if it consists of a single set; in fact, it consists of seven identical sets, since 2345678 has seven coursing pairs in plain course positions.

© Maurice Hodgson, 1962.
Reprint 1965.