Teaser 3096: That strikes a chord
From The Sunday Times, 23rd January 2022 [link] [link]
Skaredahora used a scale with seven notes, labelled J to P, for an idiosyncratic composition. It had a sequence of chords, each comprising exactly three different notes (the order being immaterial). The sequence started and ended with JNP. Each change from one chord to the next involved two notes increasing by one step in the repeating scale, and the other decreasing by two steps (e.g. JNP changing to KLJ).
Every chord (eventually) reachable from JNP was used at least once; every allowable change from one of these chords to another was used exactly once. The number of chord changes between the first pair of JNPs was more than that between the last such pair, but, given that, was the smallest it could be.
With notes in alphabetical order, what was the seventh chord in the sequence?
[teaser3096]
Frits, 
GeoffR, 
Hugh+Casement, and 2 others are discussing. 
NigelR and 
Brian Gladman, and 
Tony Brooke-Taylor, 
S2T2 
Jim Randell 7:20 pm on 21 January 2022 Permalink |
I am assuming notes that are an octave apart are considered identical. So going up three notes from M to P gives the same note as going down 4 notes from M to P. Also that chords are identical if they contain the same three notes, and a chord change requires at least one of the notes to change.
This Python program constructs the graph of transitions between chords, and then finds paths from JNP back to JNP that traverse all the edges (Eulerian circuits). We then apply the remaining conditions, and find the 7th element of the paths.
It runs in 51ms.
Run: [ @replit ]
from enigma import (mod, all_different, ordered, tuples, Accumulator, join, printf) # there are 7 notes in the scale fn = mod(7) # format a sequence, 0..6 represent J..P fmt = lambda s: join("JKLMNOP"[i] for i in s) fmts = lambda ss: join((fmt(s) for s in ss), sep=" ", enc="()") # start node = JNP start = (0, 4, 6) assert fmt(start) == 'JNP' # generate changes for chord <s> def changes(s): (x1, y1, z1) = (fn(n + 1) for n in s) (x2, y2, z2) = (fn(n - 2) for n in s) for t in ((x2, y1, z1), (x1, y2, z1), (x1, y1, z2)): t = ordered(*t) if t != s and all_different(*t): yield t # generate possible transitions, starting with <start> adj = dict() trans = set() ns = {start} while ns: ns_ = set() for s in ns: ts = set(changes(s)) adj[s] = ts trans.update((s, t) for t in ts) ns_.update(ts) ns = ns_.difference(adj.keys()) # find sequences that use every transition in <ts>, starting from <ss> def seqs(ts, ss): # are we done? if not ts: yield ss else: # make a transition s = ss[-1] for x in adj[s]: t = (s, x) if t in ts: yield from seqs(ts.difference([t]), ss + [x]) # find viable sequences def generate(trans, start): for ss in seqs(trans, [start]): # sequence must be a circuit if ss[0] != ss[-1]: continue # find the indices of start js = list(j for (j, x) in enumerate(ss) if x == start) # and distances between them ds = list(j - i for (i, j) in tuples(js, 2)) # first distance is greater than last if not (ds[0] > ds[-1]): continue # return (distances, sequence) yield (ds, ss) # find the smallest first distance r = Accumulator(fn=min, collect=1) for (ds, ss) in generate(trans, start): r.accumulate_data(ds[0], ss) # output solutions for ss in r.data: printf("{x} {ss}", x=fmt(ss[6]), ss=fmts(ss))Solution: The seventh chord in the sequence is: KNO.
The complete sequence is:
[Follow the [@replit] link, click on “Show Files”, and select “Teaser 3096.mp3” to hear the chords played with J–P transposed to notes A–G in a single octave].
There are 4 chords between the first two JNPs, and 3 chords between the last two JNPs.
And this is the only progression of the form (JNP [4] JNP [3] JNP).
There are 4 other viable progressions of the form (JNP [5] JNP [2] JNP), but we are only interested in progressions with the shortest possible distance between the first two occurrences of JNP.
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