An old pastime.

Size: px

Start display at page:

Download "An old pastime."

Charla Bennett
6 years ago
Views:

1 Ringing the Changes

2 An old pastime

The mechanics of change ringing http://www.

3 The mechanics of change ringing

4 Some Terminology Since you can not rapidly change which bells are rung, it is almost impossible to have church bells create a tune (with rope pulling... you can create tunes with hammers on church bells but that is a totally different thing). The church bells are arranged in tonal order with the highest pitched bell first (the treble) and the lowest pitched bell last (the tenor). The order in which the bells are rung can be modified in a very restricted way. Change ringing involves ringing the bells in all possible orders (with no repeats), and this is called an extent. Going from one order to another is called a change (or sometimes a row).

5 Some terminology n n! Approximate Time Name secs. Singles min. Minimus mins. Doubles mins. Minor 7 5,040 3 hrs. Triples 8 40, hrs.* Major 9 362,880 9 days Caters 10 3,628,800 3 months Royal 11 39,916,800 3 years Cinques ,001, years Maximus * July 27, Loughborough Bell Foundry, England actually took 17 hrs. 58 min.

6 A small example Plain Bob Minimus (with treble plain hunting depicted)

7 A little history Prior to the fourteenth century, church bells in Europe were usually hung on a spindle and chimed by pulling a rope attached to the spindle. The next two centuries saw the development, in England, of a more sophisticated method of hanging a bell, to improve the control that a ringer had over it. In the seventeenth century, the development of the slider and stay, which prevents a bell from going beyond a 360 o turn, gave ringers enough control over the bells to make change ringing possible. During WW II, church bells were needed in civil defense activities, so church bell change ringing was not allowed. This led to the rise of using hand bells for change ringing. This has remained popular even after church bells were again available.

8 The Rules

9 Following the Blue Line There are many rules that change ringers follow and one of them is that no visual aids are allowed while an extent is being rung. Ringers do not memorize thousands of permutations in order to ring extents, rather they memorize the path that their particular bell takes through the various permutations. Visually, if the permutations are listed one below the other, this path of any particular bell is called the blue line for that bell. So, for instance, if the treble bell ringer knows that the treble will be plain hunting (a typical blue line for the treble) throughout the extent, the ringer knows what to do at every change without having to know what the other bells are doing.

10 Following the Blue Line Canterbury Minimus: Treble is plain hunting. Bells 2 and 3 are doing the same work, but Tenor is not.

11 The Mathematics From the mathematical point of view change ringing amounts to generating all permutations in a systematic way according to certain restricting rules. We will approach the mathematical question in stages, getting closer and closer to acceptable solutions. Our first steps will be to satisfy rules 1) 3) and not worry about the others until these are satisfied.

12 Generating Permutations There are several algorithms for generating permutations. In some cases one wants to generate all permutations (without repetition), or generate some special permutations, or random permutations, etc. Algorithms for generating all permutations are only of limited use, since for permutations even on a small number of elements (12-14) the amount of time or storage needed to generate them becomes prohibitive. However, there are some circumstances (ours in particular) in which this is necessary. Since the number of permutations is large, one wants algorithms which can produce them efficiently on a computer.

13 Lexicographical Order One well known algorithm produces permutations in lexicographical order (alphabetical except using numbers). Letting k 1 k 2...k n represent a permutation on n letters. To obtain the next permutation in lexicographical order we: 1) Find the largest i so that k i-1 < k i. 2) For this fixed i find the largest j so that k i-1 < k j. 3) Interchange k i-1 and k j. 4) Reverse the order of the digits k i k i+1...k n.

14 Lexicographical Order Example: With n = 4, the next permutation after in this order is obtained in four steps: 1) i = 2 2) j = 3 3) Interchange k 1 with k 3 to obtain ) Reverse k 2...k 4 to obtain for another example start with : 1) i = 4 2) j = 4 3) Interchange k 3 with k 4 to obtain ) Reverse k 4...k 4 to obtain

15 Lexicographical Order The permutations on 4 symbols in this order are: We can see that this ordering will not satisfy rule 3 of change ringing (look at the second transition), so we must investigate other orderings of these permutations.

16 Factorial Representation The lexicographical ordering of permutations is so common that it is hard to even visualize any other ordering. Any non-negative integer m can be written uniquely in the following factorial form : m = a 1 1! + a 2 2! a n-1 (n-1)! where 0 a i i. The a i 's are called factorial digits and we can write m = (a 1, a 2,..., a n-1 ). Thus, 0 = (0,0,...,0) and 2000 = (0,1,1,3,4,2). For fixed n, the largest integer represented is (1,2,...,n-1) = n! - 1.

17 Factorial Representation The factorial representation using the n-1 digits a 1,a 2,...,a n-1 can represent all the integers from 0 to n!-1, that is, n! integers. Any bijection between these representations and the set of permutations on n symbols will give an ordering of the permutations. There are many ways to set up these bijections, here we give one method due to Marshall Hall Jr. The permutations are written on the symbols 0,1,..., n-1. Given a permutation π, set a i as the number of symbols less than i which actually follow i in π. (Note that we have 0 a i i) (1,0,2) = (0, 2, 3) = 22

18 Inverse The easiest way to show that this is a bijection is to produce the inverse function. To form the permutation on the symbols 0,...,n-1 from the factorial representation (a 1, a 2,..., a n-1 ), we fill the n positions of the permutation in the following way: Work with the symbols starting with the largest (n-1) and going down to 0. Place symbol i in position a i where one counts only unoccupied positions, going from right to left and starting the count at 0. Put 0 in the last empty space. Thus, (1,0,2) would give us the following steps:

19 Johnson-Trotter Algorithm Independently, Johnson and Trotter came up with an algorithm for generating all permutations where each new permutation differs from the last one by only a switch of two adjacent symbols. This algorithm produces a list of permutations which would satisfy rule 3. The algorithm is based on the idea that if you already have a permutation on n-1 symbols, to get a permutation on n symbols you only have to put in the symbol n all come from expanding

20 Johnson-Trotter Algorithm This idea can be used to recursively generate the permutations, but we can actually generate the next permutation in this order knowing only the last one. This algorithm for generating the permutations in the Johnson- Trotter order is due to Even. To each symbol in the permutation we will associate a direction, either left or right, by writing an arrow above the symbol. A symbol is called mobile if its arrow points to a smaller symbol adjacent to it. So, in (no arrow = left arrow) only 3, 5 and 6 are mobile.

21 Johnson-Trotter (Even) The algorithm: Begin with n (all arrows pointing left). While there is a mobile symbol do the following: (1) Find the largest mobile symbol m. (2) Swap m and the adjacent symbol to which it points. (3) Switch the direction of all the arrows above the symbols larger than m. The algorithm stops when there are no mobile symbols.

22 Examples: Johnson-Trotter (Even) is largest mobile swap 3 and change direction on symbols greater than is largest mobile swap 2 and change direction on symbols greater than 2

23 Johnson-Trotter The output for n = 4 is:

24 Fibonacci Sequences How many ways (ordered) can one write a non-negative integer as a sum of 1's and 2's? For the non-negative integer n, let F(n) = F n be the answer to this question. We have: F 2 = 2 since 2 = 2 = 1+1 (2 ways) F 3 = 3 since 3 = 2+1 = 1+2 = (3 ways) F 4 = 5 since 4 = 2+2 = = = = = (5 ways) Clearly, F 1 = 1 and it makes sense to define F 0 = 1, since there is only one way to write 0 this way (use no 1's and no 2's).

25 Fibonacci Sequences Let n 2. Any of these expressions for n must end in either a 1 or a 2. For those that end in a 1, the earlier digits must add up to n-1 (and are just 1's and 2's) and any expression for n-1 consisting of just 1's and 2's with an additional 1 added at the end will give an expression for n (which ends in a 1). Similarly, those expressions that end in a 2 have earlier terms that add up to n-2. So, we must have F n = F n-1 + F n-2 for n 2. This is known as the Fibonacci recursion relation and with the initial conditions that F 0 = F 1 = 1, the sequence of integers {F n } n 0 is known as the Fibonacci sequence.

26 Fibonacci Sequences We wish to obtain an explicit formula for the n th Fibonacci number. Since the recursion relation is linear, any linear combination of solutions to the recursion is also a solution to the recursion. If the number of terms in such a linear combination equals the number of initial conditions, the coefficients of the linear combination can be determined. There are two common methods for solving linear recursion relations and we will look at both of them as applied to our situation.

27 Fibonacci Sequences Method 1: Start with the recursion relation: F n = F n-1 + F n-2 Make the subscripts exponents:(a nonsense move) F n = F n-1 + F n-2 Factor out any common factors and solve the resulting polynomial equation (for F). F 2 F 1 = 0. The solutions of this polynomial raised to the n th power are particular solutions of the recursion relation. f 1 = n and f 2 = n

28 Fibonacci Sequences Method 1 (cont.): Now we form the linear combination of these two solutions to get the general solution: F n = af 1 + bf 2 We solve for the coefficients by using the initial conditions: 1 = F 0 = a + b, and 1 = F 1 = a b To get, a = and b = , F n = n n

29 Fibonacci Sequences Method 2: Form the (ordinary) generating function for the Fibonacci sequence: G t = n 0 F n t n We multiply by powers of t to set up a replacement using the recursion relation: t G t = F n t n 1 n 0 t 2 G t = F n t n 2 n 0 = F n 1 t n n 1 = F n 2 t n n 2

30 Fibonacci Sequences G t t G t t 2 G t = F 0 F 1 t n 2 1 t t 2 G t = F 0 F 1 t F 0 t = 1 so G t = F n t n F 0 t n 2 F n 1 t n n 2 F n 2 t n 1 1 t t 2. Now we need to write G(t) as a power series and then read off the coefficients which will be the Fibonacci numbers. The steps here involve factoring the denominator and then writing this rational function in its partial fraction form: G t = 1 1 t t = t 1 t = A 1 t B 1 t

31 Fibonacci Sequences After finding the constants, α, β, A and B we would have: G t = A n 0 t n so F n = A n B n. B n 0 t n = A n B n t n n 0 Since 1 t t 2 = (1 αt)(1 βt), we must have that αβ = -1 and α + β = 1. Thus, α and β are the roots of the quadratic x 2 -x 1. Finding A and B also involves simple manipulations that we have already seen, so F n = n n

32 Special Involutions The Fibonacci numbers arise in quite a few places where you would not expect them. In change ringing, the transition from one row to the next must either leave a bell fixed or interchange a bell with one that is adjacent to it. When thought of as a permutation, if a transition is repeated, we would fix every bell (that is, get the identity permutation). Such permutations (of order 2) are called involutions. Not all involutions give transitions (consider an involution which interchanges just two non-adjacent bells) and the identity permutation (which is an involution) does not give a proper transition.

33 Special Involutions Let t(n) be the number of change ringing transitions that are permitted with n bells. Theorem: For n 2, t(n) = F n -1. (F n is the n th Fibonacci #) Pf: It is easy to see that t(2) = 1 = F 2-1 and t(3) = 2 = F 3 1. We can then assume the result for n < k and consider t(k) with k 4. There are t(k-1) transitions which fix position k (the last position). Also, there are t(k-2) transitions which interchange positions k and k-1 and at least one other pair. Finally, a single transposition (k-1,k) (not already counted). t(k) = t(k-1) + t(k-2) + 1 = F k F k = F k - 1.

Lecture 16b: Permutations and Bell Ringing

Lecture 16b: Permutations and Bell Ringing Another application of group theory to music is change-ringing, which refers to the process whereby people playing church bells can ring the bells in every possible