Sunday Times Teaser – Put Your Finger On It

by BRG on July 5, 2024

by Colin Vout

Published Friday July 05 2024 (link)

On a particular stringed instrument, a fingering pattern of 2,2,3,1 means that the pitches of strings 1 to 4 are raised by that many steps respectively from the “open strings” (ie, their pitches when not fingered); this gives a “chord” of pitches C,E,G,A# in some order. The fingering pattern 4,1,0,x (for some whole number x from 0 to 11 inclusive) would give pitches F,A,C,D in some order, if these were all shifted by some fixed amount.

[Pitches range through C,C#,D,D#,E,F,F#,G,G#,A,A#,B, which then repeat.]

What are the pitches of “open strings” 1 to 4, and what is the value of x?

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BRG permalink


from itertools import permutations, product

# the order of the notes  
notes = 'C,C#,D,D#,E,F,F#,G,G#,A,A#,B'.split(',')

# indexes for the specified note sequences and the corresponding shifts
n1, s1 = [notes.index(c) for c in 'C E G A#'.split()], (2, 2, 3, 1)
n2, s2 = sorted((notes.index(c) for c in ('F A C D'.split()))), (4, 1, 0) # + x

# consider rearrangements of the first specified note sequence
for p1 in permutations(n1):
  # shift the four notes down by the first shift amounts
  # to find possible open note values (n4, assumed different)
  n4 = [(p - n) % 12 for p, n in zip(p1, s1)]
    
  # now shift the four open notes up by the second shift amounts 
  # together with an added fixed amount <sh>
  for x, sh in product(range(12), repeat=2):
    r2 = [(p + n + sh) % 12 for p, n in zip(n4, s2 + (x,))]
    # check the second specified order
    if sorted(r2) == n2:
      print(f"Open notes: {' '.join(notes[c] for c in n4)} ({x = }, {sh = }).")

from itertools import permutations, product

# the order of the notes

notes = 'C,C#,D,D#,E,F,F#,G,G#,A,A#,B'.split(',')

# indexes for the specified note sequences and the corresponding shifts

n1, s1 = [notes.index(c) for c in 'C E G A#'.split()], (2, 2, 3, 1)

n2, s2 = sorted((notes.index(c) for c in ('F A C D'.split()))), (4, 1, 0) # + x

# consider rearrangements of the first specified note sequence

for p1 in permutations(n1):

# shift the four notes down by the first shift amounts

# to find possible open note values (n4, assumed different)

n4 = [(p - n) % 12 for p, n in zip(p1, s1)]

# now shift the four open notes up by the second shift amounts

# together with an added fixed amount <sh>

for x, sh in product(range(12), repeat=2):

r2 = [(p + n + sh) % 12 for p, n in zip(n4, s2 + (x,))]

# check the second specified order

if sorted(r2) == n2:

print(f"Open notes: {' '.join(notes[c] for c in n4)} ({x = }, {sh = }).")

Frits permalink

Reusing Jim’s finger routine.


from itertools import permutations

ps = "C C# D D# E F F# G G# A A# B".split()  # pitches
n = len(ps)

# finger strings <ss> with steps <stps>, sgn=-1 means undoing a finger action
def finger(ss, stps, sgn=1):
  return list(ps[(ps.index(x) + sgn * y) % n] for (x, y) in zip(ss, stps))

# second target
tgt = ['C', 'D', 'F', 'A']
inds = sorted(ps.index(x) for x in tgt)
tgt_diffs = sorted(y - x for x, y in zip(inds, inds[1:] + [inds[0] + n]))
tgt.sort()

# choose an order for C E G A#
for perm in permutations(["C", "E", "G", "A#"]):
  # determine 'open strings' by undoing pattern [2, 2, 3, 1] 
  os = finger(perm, [2, 2, 3, 1], -1)
  
  # check pattern for [4, 1, 0, x]
  for x in range(12):
    p = finger(os, [4, 1, 0, x])  
    inds = sorted(ps.index(x) for x in p)
    diffs = sorted(y - x for x, y in zip(inds, inds[1:] + [inds[0] + n]))
    # difference between pitchs must be same as target
    if diffs != tgt_diffs: continue
    # try shifting pattern by some amount
    for shift in range(n):
      if sorted(finger(p, [shift] * 4)) == tgt:
        print(f"answer: 'open strings': {os}, {x = }")
        break # we have found 'some amount'

from itertools import permutations

ps = "C C# D D# E F F# G G# A A# B".split() # pitches

n = len(ps)

# finger strings <ss> with steps <stps>, sgn=-1 means undoing a finger action

def finger(ss, stps, sgn=1):

return list(ps[(ps.index(x) + sgn * y) % n] for (x, y) in zip(ss, stps))

# second target

tgt = ['C', 'D', 'F', 'A']

inds = sorted(ps.index(x) for x in tgt)

tgt_diffs = sorted(y - x for x, y in zip(inds, inds[1:] + [inds[0] + n]))

tgt.sort()

# choose an order for C E G A#

for perm in permutations(["C", "E", "G", "A#"]):

# determine 'open strings' by undoing pattern [2, 2, 3, 1]

os = finger(perm, [2, 2, 3, 1], -1)

# check pattern for [4, 1, 0, x]

for x in range(12):

p = finger(os, [4, 1, 0, x])

inds = sorted(ps.index(x) for x in p)

diffs = sorted(y - x for x, y in zip(inds, inds[1:] + [inds[0] + n]))

# difference between pitchs must be same as target

if diffs != tgt_diffs: continue

# try shifting pattern by some amount

for shift in range(n):

if sorted(finger(p, [shift] * 4)) == tgt:

print(f"answer: 'open strings': {os}, {x = }")

break # we have found 'some amount'

Sunday Times Teaser – Put Your Finger On It

by Colin Vout

Published Friday July 05 2024 (link)

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