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Jim Randell 9:18 am on 28 February 2023 Permalink Reply
Tags: by: Angela Newing ( 37 )   

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  Jim Randell 9:18 am on 28 February 2023 Permalink | Reply

  I am assuming that each cyclist cycles(!) through their helmets in strict rotation. So each time through they wear the same colours in the same order.

  None of them have duplicate colours, so if we see one of them wearing the same colour on different days, they must have completed a whole number of cycles between those days. And they are all wearing the same colours on 1st September and 1st October, so the length of each of their cycles must be a divisor of 30, less than 12 (i.e. 1, 2, 3, 5, 6, 10).

  For any given cyclist, if we are given a set of days and colours then we can look at days when the same colour is worn. The gap between the days must be a multiple (possibly 1) of the cycle length. And so by considering all possible gaps we can can determine possible cycle lengths. Furthermore, different colours must be worn on different days in the cycle. So

  This Python program calculates possible cycle lengths from the colours given on the specified dates.

  It runs in 55ms. (Internal runtime is 2.2ms).

  Run: [ @replit ]

  from enigma import (
    subsets, group, unpack, union, tuples, mgcd, divisors,
    seq_all_different, delete, update, rev, map2str, printf
  )
  
  # find cycle lengths for values give in ms
  def solve(ms, ns=dict()):
    # are we done?
    if not ms:
      yield ns
    else:
      # choose a key to process (the one with the most entries)
      x = max(ms.keys(), key=(lambda k: len(ms[k].keys())))
      # group days by colour
      g = group(ms[x].items(), by=unpack(lambda k, v: v), f=unpack(lambda k, v: k))
      # collect multiples of cycle length
      s = union((abs(b - a) for (a, b) in tuples(vs, 2)) for vs in g.values())
      # consider possible cycle lengths (must be < 12)
      for n in divisors(mgcd(*s)):
        if n > 11: break
        # is this one already used?
        if n in ns: continue
        # check different colours give different cycle days
        if not seq_all_different(vs[0] % n for vs in g.values()): continue
        # solve for the remaining items
        yield from solve(delete(ms, [x]), update(ns, [(n, x)]))
  
  # consider possible orderings for 11th
  for (a, b, c, d, e) in subsets('RRGMW', size=len, select='mP'):
    # known positions (day 1 = 1st September)
    ms = dict(
      A={ 1: 'M', 11: a, 19: 'R', 31: 'M' },
      B={ 1: 'R', 11: b, 19: 'R', 31: 'R' },
      C={ 1: 'R', 11: c, 19: 'R', 31: 'R' },
      D={ 1: 'O', 11: d, 19: 'O', 31: 'O' },
      E={ 1: 'G', 11: e, 19: 'G', 22: 'O', 23: 'W', 31: 'G' },
    )
    # find possible cycle lengths
    for ns in solve(ms):
      # output solution
      printf("day 11: (A B C D E) = ({a} {b} {c} {d} {e}) {ns}", ns=map2str(rev(ns), sep=" ", enc="[]"))
  

  Solution: On 11th September, the colours were: Alan = mauve; Bill = red; Charles = red; Dave = green; Eric = white.

  Alan has 5 or 10 helmets. Bill has 1 or 2. Charles has 2 or 1. Dave has 3. Eric has 6.

  We can’t work out all the colours, but here is what we do know:

  A: (M ? ? ? ?) or (M ? ? ? ? ? ? ? ? ?)
  B: (R) or (R ?)
  C: (R ?) or (R)
  D: (O ? ?)
  E: (G ? ? O W ?)

  However these cycle lengths are chosen, they all give the same sequence of values for the 11th September.

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