Mathematics of Bell-Ringing

Mathematics of bell-ringing

Dr Rob Sturman University of Leeds, School of Mathematics

24th September 2016 Gravity Fields Festival Easy example: on 3 bells, the possible sequences are: 1 2 3   1 3 2   Combinatorics: 2 1 3  3 choices for ﬁrst place, then 2 6 sequences 2 3 1 choices, then 1 choice =⇒  3 1 2  3 × 2 × 1 = 3! (3 factorial)  3 2 1 

Change-ringing Extents

Basic idea: Ring every possible sequence exactly once 4 bells: 4! = 4 × 3 × 2 × 1 = 24 sequences (around 30 seconds) 6 bells: 6! = 720 sequences (takes about 25 minutes) 8 bells: 8! = 40320 sequences (18 hours, in Loughborough, 1963) 12 bells: 12! = 479, 001, 600 sequences (over 30 years!)

Change-ringing Extents

Basic idea: Ring every possible sequence exactly once

Easy example: on 3 bells, the possible sequences are: 1 2 3   1 3 2   Combinatorics: 2 1 3  3 choices for ﬁrst place, then 2 6 sequences 2 3 1 choices, then 1 choice =⇒  3 1 2  3 × 2 × 1 = 3! (3 factorial)  3 2 1  12 bells: 12! = 479, 001, 600 sequences (over 30 years!)

Change-ringing Extents

Basic idea: Ring every possible sequence exactly once

4 bells: 4! = 4 × 3 × 2 × 1 = 24 sequences (around 30 seconds) 6 bells: 6! = 720 sequences (takes about 25 minutes) 8 bells: 8! = 40320 sequences (18 hours, in Loughborough, 1963) Change-ringing Extents

Basic idea: Ring every possible sequence exactly once

4 bells: 4! = 4 × 3 × 2 × 1 = 24 sequences (around 30 seconds) 6 bells: 6! = 720 sequences (takes about 25 minutes) 8 bells: 8! = 40320 sequences (18 hours, in Loughborough, 1963) 12 bells: 12! = 479, 001, 600 sequences (over 30 years!) You can only swap neighbouring bells. Easy example: 3 bells: a = (12), b = (23) [positions, not bells] Alternate these changes: All 6 possible sequences generated, and the ringers only have 1 2 3  to remember two different opera-  a 2 1 3  tions, and to alternate them.   b 2 3 1  a 3 2 1 6 sequences b b 3 1 2   a 1 3 2  1  a b 1 2 3  3 2

Change-ringing A constraint and an opportunity

You can’t just move from one sequence to any other! Easy example: 3 bells: a = (12), b = (23) [positions, not bells] Alternate these changes: All 6 possible sequences generated, and the ringers only have 1 2 3  to remember two different opera-  a 2 1 3  tions, and to alternate them.   b 2 3 1  a 3 2 1 6 sequences b b 3 1 2   a 1 3 2  1  a b 1 2 3  3 2

Change-ringing A constraint and an opportunity

You can’t just move from one sequence to any other! You can only swap neighbouring bells. All 6 possible sequences generated, and the ringers only have to remember two different opera- tions, and to alternate them.

b 1 a

3 2

Change-ringing A constraint and an opportunity

You can’t just move from one sequence to any other! You can only swap neighbouring bells. Easy example: 3 bells: a = (12), b = (23) [positions, not bells] Alternate these changes:

1 2 3   a 2 1 3    b 2 3 1  a 3 2 1 6 sequences b 3 1 2   a 1 3 2   b 1 2 3  Change-ringing A constraint and an opportunity

You can’t just move from one sequence to any other! You can only swap neighbouring bells. Easy example: 3 bells: a = (12), b = (23) [positions, not bells] Alternate these changes: All 6 possible sequences generated, and the ringers only have 1 2 3  to remember two different opera-  a 2 1 3  tions, and to alternate them.   b 2 3 1  a 3 2 1 6 sequences b b 3 1 2   a 1 3 2  1  a b 1 2 3  3 2  The full peal is not generated  by a and b — we need an extra   change, for example    c = (34).  only 8 rows! a   1 2      3 4 b

Change-ringing Plain Bob Minimus

Consider 4 bells and let a = (12)(34) , b = (23)

1 2 3 4 a 2 1 4 3 b 2 4 1 3 a 4 2 3 1 b 4 3 2 1 a 3 4 1 2 b 3 1 4 2 a 1 3 2 4 b 1 2 3 4 The full peal is not generated by a and b — we need an extra change, for example c = (34). a 1 2

3 4 b

Change-ringing Plain Bob Minimus

Consider 4 bells and let a = (12)(34) , b = (23)

1 2 3 4   a 2 1 4 3   b 2 4 1 3    a 4 2 3 1  b 4 3 2 1 only 8 rows! a 3 4 1 2   b 3 1 4 2   a 1 3 2 4   b 1 2 3 4  Change-ringing Plain Bob Minimus

Consider 4 bells and let a = (12)(34) , b = (23)

The full peal is not generated 1 2 3 4   by a and b — we need an extra a 2 1 4 3   change, for example b 2 4 1 3    c = (34). a 4 2 3 1  b 4 3 2 1 only 8 rows! a a 3 4 1 2  1 2  b 3 1 4 2   a 1 3 2 4   b 1 2 3 4  3 4 b Change-ringing Plain Bob Minimus — full extent

a = (12)(34) , b = (23), c = (34)

1 2 3 4 c 1 3 4 2 c 1 4 2 3 a 2 1 4 3 a 3 1 2 4 a 4 1 3 2 b 2 4 1 3 b 3 2 1 4 b 4 3 1 2 a 4 2 3 1 a 2 3 4 1 a 3 4 2 1 b 4 3 2 1 b 2 4 3 1 b 3 2 4 1 a 3 4 1 2 a 4 2 1 3 a 2 3 1 4 b 3 1 4 2 b 4 1 2 3 b 2 1 3 4 a 1 3 2 4 a 1 4 3 2 a 1 2 4 3 c 1 2 3 4 Change-ringing Plain Bob Minor — full peal Change-ringing Extent rules

1 start and end with bells in order musical satisfaction

2 every sequence rung exactly once completeness

3 only swap neighbouring bells physical contraints

4 each bell must move at least every other change keep ringers interested 5 bells do the same work as each other keep ringers equitable

6 the method is palindromic ease of memory? Fabian Stedman 1640–1713

Fabian Stedman: “the father of change- ringing” a contemporary of Isaac Newton (1643–1727) Born in Herefordshire, went to London as apprentice printer in 1655 Learnt bell-ringing at St Mary-le-Bow two publications of bell-ringing a parish clerk at St Bene’t’s Church in Cambridge, 1670 cross-changes Stedman Doubles Stedman Doubles

a = (12)(45), b = (23)(45) 1 2 3 4 5 1 2 3 4 5 a 2 1 3 5 4 b 1 3 2 5 4 b 2 3 1 4 5 a 3 1 2 4 5 a 3 2 1 5 4 b 3 2 1 5 4 b 3 1 2 4 5 a 2 3 1 4 5 a 1 3 2 5 4 b 2 1 3 5 4 b 1 2 3 4 5 a 1 2 3 4 5 Stedman Doubles

a = (12)(45), b = (23)(45), c = (12)(34) (a parting change) 1 2 3 4 5 3 1 5 2 4 a 2 1 3 5 4 b 3 5 1 4 2 b 2 3 1 4 5 a 5 3 1 2 4 a 3 2 1 5 4 b 5 1 3 4 2 b 3 1 2 4 5 a 1 5 3 2 4 a 1 3 2 5 4 b 1 3 5 4 2 c 3 1 5 2 4 c 3 1 4 5 2 Now known as a single, d = (34) will then allow the other sixty combinations to be rung. Grandsire doubles on 5 bells: a = (12)(45), b = (23)(45), c = (12)(34) =⇒ a(bc)4a Always an even number of permutations, so only half the 5! = 120 possible sequences generated. Need a ‘single’, e.g., d = (45) to generate the other ‘odd’ permutations.

Stedman Doubles

a = (12)(45), b = (23)(45), c = (12)(34) (a parting change) So the ‘lead’ (ababac)(bababc) = (ab)2ac(ba)2bc gives the permutations

12345 → 31452 → 43521 → 54213 → 25134 → 12345

“By this method the peal will go to sixty changes, and to carry it farther, extremes must be made”. Grandsire doubles on 5 bells: a = (12)(45), b = (23)(45), c = (12)(34) =⇒ a(bc)4a Always an even number of permutations, so only half the 5! = 120 possible sequences generated. Need a ‘single’, e.g., d = (45) to generate the other ‘odd’ permutations.

Stedman Doubles

a = (12)(45), b = (23)(45), c = (12)(34) (a parting change) So the ‘lead’ (ababac)(bababc) = (ab)2ac(ba)2bc gives the permutations

12345 → 31452 → 43521 → 54213 → 25134 → 12345

a = (12)(45), b = (23)(45), c = (12)(34) (a parting change) So the ‘lead’ (ababac)(bababc) = (ab)2ac(ba)2bc gives the permutations

12345 → 31452 → 43521 → 54213 → 25134 → 12345

“By this method the peal will go to sixty changes, and to carry it farther, extremes must be made”. Now known as a single, d = (34) will then allow the other sixty combinations to be rung. Grandsire doubles on 5 bells: a = (12)(45), b = (23)(45), c = (12)(34) =⇒ a(bc)4a Always an even number of permutations, so only half the 5! = 120 possible sequences generated. Need a ‘single’, e.g., d = (45) to generate the other ‘odd’ permutations. Stedman triples on 7 bells: (ab)2ac(ba)2bc, repeated seven times, produces 84 of the 5040 sequences. Then d = (12)(34)(67) and e = (12)(34) can can give a full extent. Can it be done with only a, b, c and d? These all alternate between odd and even, so it seems possible. In 1994 such an extent was composed. Place notation: Plain Bob Minor (6 bells): [[x16]5x12]5 Bob 14 Single 1234

Change-ringing More bells

Grandsire triples on 7 bells: Bell-ringers could ring a full extent, but could not work out a method to do so using only triple changes, e.g., a = (12)(34)(56), b = (23)(45)(67), c = (12)(34)(56), even though this alternates odd and even permutations. In 1886 a non-bell-ringing mathematician showed (using group theory) that it can’t be done. Stedman triples on 7 bells: (ab)2ac(ba)2bc, repeated seven times, produces 84 of the 5040 sequences. Then d = (12)(34)(67) and e = (12)(34) can can give a full extent. Can it be done with only a, b, c and d? These all alternate between odd and even, so it seems possible. In 1994 such an extent was composed. Change-ringing More bells

R. Duckworth and F. Stedman, Tintinnalogia, or the art of ringing (1668) F. Stedman, Campanologia, or the art of ringing improved (1677) Arthur T. White, Fabian Stedman: the ﬁrst group theorist, American Mathematical Monthly 103 (1996) 771–778. Further reading

Consider a set X of elements, and a binary operation ◦ which combines any pair of elements. We say X is a group under the operation ◦ if the following four axioms hold: 1 Closure: every combination of a pair elements is also a member of X 2 Associativity: order doesn’t matter in this sense: (x ◦ y) ◦ z = x ◦ (y ◦ z) 3 Identity: in X there is an element whose effect is to do nothing when combined with anything else 4 Inverses: every element in X has a partner element — when these are combined, you get the identity element