[r-t] ringing the Mathieu group
Philip Saddleton
pabs at cantab.net
Tue May 24 21:35:26 UTC 2011
Here's my approach:
We'll try to join together half leads of a plain method using a system
of hunts. For a right place method, there is onl one place within the
half lead where there is a choice of PNs, i.e. when the treble is in
6-7, the two methods being the reverse of each other. Arbitrarily fix
one, and we have PN -8E-8E-25-8E-8E-. For the half-lead and lead-head we
can choose either PN without upsetting the treble. For now, we'll use 1T
for each, and say that we are going to have a call every hl and lh. The
method we finish up with has 0 and E as secondary hunts and a 9 lead
plain course. We can get an 11 lead course by using e.g. PN 25 when the
treble is in 3-4, but that gives four consecutive blows.
Now all we need is to find the Q-sets that will let us join the 440
courses together.
First I try to keep the tenor unaffected as much as possible. We can
then fix most of the Q-sets with a skeleton course
1234567890ET
bb 13T54986E702
ax 186E5T943702
ba 12E65349T708
ab 149T6E320578
xb 1380T92E7654
aa 1E02789T3654
bx 104723T9865E
ab 1T987430526E
bx 19E78503426T
where a=25, b=8E and x=1T
We now try to extend this to a block where 1,E and T occupy each
combination of positions. We must have b at the lh where they are in 0E,
or we will repeat (and they cross in 67 at the hl, so are unaffected).
Using this Q-set and allowing E to be unaffected wherever possible we
join all but two of the leads (underlines denote the course heads or the
extra Q-set)
1234567890ET
------------
bb 13T54986E702
ax 186E5T943702
ba 12E65349T708
ab 149T6E320578
xb 1380T92E7654
aa 1E02789T3654
bx 104723T9865E
ab 1T987430526E
bx 19E78503426T
------------
bb 1ET874356029
ax 13568T47E029
ba 19658E74T023
ab 174T56E92803
xb 1E32T4960587
ab 1960234E8T57
------------
ba 170628E43T59
aa 146E30827T59
xa 19E876230T54
aa 13820E679T54
ba 14280976ET53
aa 1687E2904T53
ab 190478265ET3
------------
ab 12654089T7E3
------------
ab 189T5602E473
xb 103ET9267548
aa 16E2739T0548
bx 1E8720T93546
ab 1T93780E4256
bx 196734E0825T
------------
bb 16T378045E29
ax 10453T876E29
ba 19543678TE20
ab 178T456923E0
xb 1602T895E437
aa 1529E08T6437
bx 127E96T80435
ab 1T80E7623945
bx 185E0326794T
------------
bb 15T0E7634298
ax 16340T7E5298
ba 184305E7T296
ab 1E7T34589026
xb 1569T784230E
aa 1498267T530E
bx 19E285T76304
ab 1T762E590834
bx 17426095E83T
------------
bb 14T62E503987
ax 15036TE24987
bb 1076342T8E95
------------
bx 175368T24E90
ab 1T24358796E0
bx 1203497856ET
------------
The missing two leads can be joined by using the Q-set where E is behind
and T is in 890E, to give a block of 2640 rows wher 1ET occupy all
possible positions once each at hand and back. Using x wherever the hl
or lh is not restricted by the position of E or T gives a nine-part
palindromic block, where we now have 10ET in each position twice in a
23760 (a quarter peal). There are ten places in each part where we now
have an x; for each of these, the PN to use is determined by the
position of 0 in all but one part. If we choose a or b consistently for
each of the remaining ones, or for each symmetrically disposed pair, we
finish up with a round block.
I am sure there must be Q-sets with one element in each quarter (but not
the x's) - I leave this for someone else to find.
regards
Philip
Here is the gsiril code where each x is replaced by a:
12 bells
h=+-8E-8E-25-25-25-
l=+-25-25-25-8E-8E-
ha=h,+25
hb=h,+8E
la=l,+25
lb=l,+8E
aa=ha,la,"aa @"
ab=ha,lb,"ab @"
ax=ha,la,"ax @"
ba=hb,la,"ba @"
bb=hb,lb,"bb @"
bx=hb,la,"bx @"
xa=ha,la,"xa @"
xb=ha,lb,"xb @"
u=" ------------"
uu=" ============"
p1=bb,ab,ba,ab,bb,aa,ba,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,ba,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,aa,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,bb,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,ab,bb,u,ba,ab,ba,
uu
p2=bb,ab,ba,ab,bb,aa,ba,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,aa,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,ba,u,
bb,aa,ba,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,ba,ab,ba,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,ba,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,ba,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,ax,bb,u,ba,ab,ba,
uu
p3=bb,ab,ba,ab,bb,aa,ba,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,aa,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,xb,aa,ba,ab,ba,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,ba,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,ba,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,aa,bb,u,ba,ab,ba,
uu
p4=bb,ab,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,ba,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,ba,u,
bb,aa,ba,ab,xb,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,aa,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,aa,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,bx,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,aa,bb,u,ba,ab,ba,
uu
p5=bb,ab,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,ba,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,ba,u,
bb,aa,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,aa,ba,ab,ba,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,aa,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,aa,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,bb,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,ab,bb,u,ba,ab,ba,
uu
p6=bb,ab,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,aa,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,ba,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,aa,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,bb,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,aa,bb,u,ba,ab,ba,
uu
p7=bb,ab,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,aa,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,ba,u,
bb,aa,ba,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,ba,ab,bx,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,xa,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,aa,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,ba,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,ab,bb,u,ba,ab,ba,
uu
p8=bb,ab,ba,ab,bb,aa,bx,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,ba,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,ba,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,ax,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,bb,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,aa,bb,u,ba,ab,ba,
uu
p9=bb,ab,ba,ab,bb,aa,ba,ab,bb,u,
bb,aa,ba,ab,ab,ab,u,
ba,aa,xa,aa,ba,aa,ab,u,
ab,u,ab,ab,aa,bb,ab,bx,u,
bb,aa,ba,ab,bb,aa,bb,ab,bb,u,
bb,aa,ba,ab,bb,ab,u,
ba,bb,u,ba,bb,u,
ab,bb,aa,ba,ab,bb,u,
bb,ab,ba,ab,ab,aa,ba,ab,ba,u,
bb,ab,ba,ab,ab,ab,u,ab,u,
aa,aa,ba,aa,aa,aa,bb,u,
ab,ab,aa,ba,ab,bb,u,
bb,ab,ba,ab,bb,aa,bb,ab,ba,u,
bb,aa,bb,u,ba,ab,ba,u,bb,ab,bb,u,
bb,ab,ba,u,bb,ab,bb,u,ba,ab,ba,
uu
prove p1,p2,p3,p4,p5,p6,p7,p8,p9
Alexander Holroyd said on 26/04/2011 20:03:
> Here is today's brain teaser.
>
> Consider the three 12-bell place notations
> x 125T 18ET
>
> The group generated by these pns (i.e. the set of all rows you can get
> to from rounds using only these pns) contains 95040 rows. It is a very
> interesting group from a mathematical perspective, called the Mathieu
> Group M_12. (It is the second smallest of the 26 "sporadic groups"). One
> interesting property is that it is "sharply 5-transitive", which means
> that any given 5 bells (e.g. 12345) ring exactly once of each of the
> possible places that 5 bells can occupy (counting different orders of
> 12345 as different), giving 12x11x10x9x8 = 95040 rows.
>
> According to the "Lovasz conjecture", it should be possible to ring a
> true round block of these 95040 rows using only these three pns. Can
> anyone come up with an elegant way of doing this? It would obviously be
> nice to do it right-place, ie without 3 consecutive blows. I don't know
> whether that's possible.
>
> Ander
>
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