S2T2

Jim Randell 10:33 am on 12 December 2021 Permalink Reply
Tags: by: Mr. W. Blackman   

• 

  Jim Randell 10:35 am on 12 December 2021 Permalink | Reply

  The following program steps through multiples of the largest price, and looks for multiples of the other prices which come in close to the same price, but with differing “pence” values.

  We find that there are many sets of 4 sales figures that differ only in the pence column, with values of (1, 2, 3, 4).

  If the sales target had been met (or exceeded) we would not know which of these to choose (if we find one set of figures that works, any larger set of figures would also work).

  However, if the sales target had not been met, and that is enough information to work out what the figures are, then we must be interested in the lowest set of possible sales figures, and the target must be higher than this, but less than the second lowest possible set of figures.

  The program runs in 166ms, so is fast enough, but this isn’t a suitable approach for a manual solution.

  Run: [ @replit ]

  from enigma import irange, inf, div, singleton, unpack, printf
  
  # pence -> (pounds, shillings, pence)
  def d2psd(n):
    (n, d) = divmod(n, 12)
    (p, s) = divmod(n, 20)
    return (p, s, d)
  
  # (pounds, shillings, pence) -> pence
  def psd2d(p, s, d):
    return 240 * p + 12 * s + d
  
  # solve for the specified prices, (A, B, C, D)
  def solve(A, B, C, D):
    # look for multiples of the largest price
    for n in irange(0, inf, step=D):
      (p, s, d) = d2psd(n)
      if 0 < d < 5:
        # so the other amounts were are interested in are...
        r = dict()
        for d_ in (1, 2, 3, 4):
          if d_ == d: continue
          k = singleton(x for x in (A, B, C) if div(psd2d(p, s, d_), x))
          if k:
            r[k] = (p, s, d_)
          else:
            break
        # have we found a value for each d?
        if len(r.keys()) == 3:
          r[D] = (p, s, d)
          yield r
  
  # prices (in ascending order)
  prices = list(psd2d(0, s, 11) for s in (3, 4, 5, 6))
  
  # look for the smallest solution
  for r in solve(*prices):
    for (k, (p, s, d)) in sorted(r.items(), key=unpack(lambda k, v: v)):
      n = div(psd2d(p, s, d), k)
      (_, ks, kd) = d2psd(k)    
      printf("({p}, {s}s, {d}d) = {n} pkts @ ({ks}s, {kd}d) each")
    printf()
    break
  

  Solution: The sales figures are:

  £53878, 1s, 1d = 182123 pkts @ 5s, 11d each
  £53878, 1s, 2d = 275122 pkts @ 3s, 11d each
  £53878, 1s, 3d = 219165 pkts @ 4s, 11d each
  £53878, 1s, 4d = 155792 pkts @ 6s, 11d each

  And the target must be above this.

  The next set of figures starts at £77098, 0s, 1d, so the target must be below this (otherwise there would be a choice of two sets of figures).

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  Hugh+Casement 2:10 pm on 12 December 2021 Permalink | Reply

  I wouldn’t like to have had to work that out in 1954, with no computer to hand.
  It would have been easier if the sales had been all the same: £68088 14s. 1d. is the LCM of the four packet prices, each a prime number of pence.

  In the days before the ubiquitous teabags, tea was normally sold in quarter-pound packets.
  I seem to remember a price of about 1s. 6d., but probably less in the ’50s.

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• 

  GeoffR 9:00 am on 14 December 2021 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  % Four standard sale prices (3s, 11d, 4s, 11d, 5s, 11d, 6s, 11d)
  int: P1 == 47; int: P2 == 59; int: P3 == 71; int: P4 == 83;
  
  % Amounts sold for each packet price
  var 100000..300000:pkt1; var 100000..300000:pkt2;
  var 100000..300000:pkt3; var 100000..300000:pkt4;
  
  constraint all_different([pkt1,pkt2,pkt3,pkt4]);
  
  % testing range for sales totals is £50,000 to £60,000
  % range values in old pence (240d = £1)
  var 12000000..14400000: T1;  
  var 12000000..14400000: T2;
  var 12000000..14400000: T3;
  var 12000000..14400000: T4;
  
  constraint T1 == pkt1 * P1 /\ T2 == pkt2 * P2
  /\ T3 == pkt3 * P3 /\ T4 == pkt4 * P4;
  
  constraint all_different([T1, T2, T3, T4]);
  
  % the four total sales values are 1d, 2d, 3d and 4d different 
  constraint max([T1, T2, T3, T4]) - min([T1, T2, T3, T4]) == 3;
  
  solve satisfy;
  
  output ["Totals(old pence) [T1, T2, T3, T4] = " ++ show([T1, T2, T3, T4]) 
  ++ "\nPackets (No.) = " ++ show([pkt1, pkt2, pkt3, pkt4])];
  
  % Totals(old pence) [T1, T2, T3, T4] = [12126799, 12126801, 12126800, 12126798]
  % Packets (No.) = [258017, 205539, 170800, 146106]
  % ----------
  % Totals(old pence) [T1, T2, T3, T4] = [12193116, 12193117, 12193114, 12193115]
  % Packets (No.) = [259428, 206663, 171734, 146905]
  % ----------
  % Totals(old pence) [T1, T2, T3, T4] = [12930734, 12930735, 12930733, 12930736]
  % Packets (No.) = [275122, 219165, 182123, 155792]  *** Published answer ***
  % ----------
  % Totals(old pence) [T1, T2, T3, T4] = [13664874, 13664872, 13664873, 13664871]
  % Packets (No.) = [290742, 231608, 192463, 164637]
  % ----------
  % ==========
  
  
  

  I worked in old decimal pence (240p = £1), looking for 1d differences in total sales values between 12,000,000 and 14,400,000 sales totals, in old pence. This search range is £50,000 to £60,000 in current money terms and was used to reduce search time. The Chuffed solver was used.

  Using Python’ divmod function for converting old money totals to current monetary values:

  T1 = 12930734 in old pence is:
  divmod(12930734,240) = (53878, 14) = £53878 1s 2d

  T2 = divmod(12930735,240)
  = (53878, 15) = £53878 1s 3d

  T3 = divmod(12930733,240)
  = (53878, 13) = £53878 1s 1d

  T4 = divmod(12930736,240)
  = (53878, 16) = £53878 1s 4d

  There were several solutions, including the published solution.

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