Q: In English country dancing a "triple minor" set consists of a line of n couples, numbered 1, 2, 3, 1, 2, 3, ... from the top: each block of three couples 1, 2, 3 is called a "minor set". At each turn of the dance the 1s move down one place and the 2s move up one place, and the 2s and 3s then swap numbers to form new minor sets numbered 1, 2, 3 (omitting the new top couple). During any turn only those in complete minor sets dance (experienced dancers sometimes do a fudged version in an incomplete minor set, but ignore this for now).
The 1s move down the set: when they reach the bottom - 1 position they immediately move to the bottom, and wait two turns to restart as 3s. The others move (more slowly) up the set, alternately dancing as 2s and 3s: when they reach the top - 1 position they wait one turn, dance once as 2s to get to top place, and wait two turns to restart as 1s. Normally the dance has t = 3u + 1 turns, for some u.
Many people believe that couples spend a greater proportion of time
dancing if n is not a multiple of 3. Is this true?
[David Barnert, ECD discussion list, 19 Jun 2001]
A: Yes and no. Call the percentage of everyone's time spent actually dancing as opposed to standing out the "efficiency". For large n it's most efficient to have a multiple of three couples, but for small n adding another couple improves efficiency, because it significantly reduces the proportion of couples involved in end effects. Specifically, with t = 3u + 1 turns:
For example if u = 2 (ie 7 turns) then for various numbers of couples the efficiency is
|3||42.86||4||53.57||5||60.00||n < 6u|
|12||85.71||13||85.71||14||85.71||6u <= n <= 6u + 2|
|15||88.57||16||88.39||17||88.24||n > 6u + 2|
[Adapted from version emailed to ECD discussion list, 19 Jun 2001]
Note that maximising the overall efficiency may produce solutions that are non-optimally equitable among couples for a given dance. However, things should average out over a lifetime's dancing.
Q: Generalise to minor sets of m couples, as follows: after the first turn the 1s have moved down one place and as before become the top of new minor sets, so the couple below them change their number to 2, the next couple to 3, etc. End effects in this case are left as an exercise for the reader. Let n = pm + q, and while we're at it generalise the number of turns to t = um + v. How does the number of couples affect the efficiency?
A: Fix m and the number of turns t = um + v, and let the number of couples n = pm + q vary. Then as before there are three regimes, according to the value of p, although they're a bit more complicated:
Note that in the original problem we assumed v = 1, so we never saw the "rises until" part of regimes 2 and 3.
Whichever regime we're in, all the efficiencies for a given value of p are greater than all those for smaller p.
Notes: There are many "duple minor" English country dances, with m = 2. There are quite a few "triple minors", with m = 3 as above, which were particularly popular in the 17th and 18th centuries because they tend to give the 2s and 3s more flirting time. Many Scottish country dances also have this type of progression, almost always with n = 4, m = 3 and t = 8 (although with such a small n the two sets of end effects meet...). Hampstead Manor is a "quadruple minor", with m = 4, but I don't know details, or any other such dances. A dance with m = 3 or 4 should not be confused with a duple minor that has a triple or quadruple progression. Nor is it the same as a double contra, such as Jim Kitch's Double Chocolate, which is different again.
Back to puzzles
This page is maintained by Thomas Bending,
and was last modified on 22 August 2018.
Comments, criticisms and suggestions are welcome. Copyright © Thomas Bending 2018.