S2T2

Jim Randell 8:19 am on 18 June 2024 Permalink Reply
Tags: by: J S Rowley ( 7 )   

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  Jim Randell 8:20 am on 18 June 2024 Permalink | Reply

  Let the finishing positions be (best to worst) (a, b, c), and the runner numbers be (smallest to largest) (x, y, z).

  These six numbers are all different, odd, and greater than 1.

  Taking the cartesian product of these sets we get:

  {a, b, c} × {x, y, z} = {ax, ay, az, bx, by, bz, cx, cy, cz}

  And for each boy in each race the product of his runner number and his finishing position gives a unique value, so they correspond directly with the 9 values produced by this cartesian product.

  The following Python program considers divisor pairs of 841, and then splits each divisor into 3 odd numbers meeting the requirements. We then use the [[ SubstitutedExpression() ]] solver from the enigma.py library to assign runner numbers and finishing positions in each race.

  The program runs in 116ms. (Internal runtime is 38ms).

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, divisors_pairs, decompose, div, cproduct, sprintf as f, join, printf)
  
  # find 3 odd numbers that give the required total
  def numbers(t):
    r = div(t - 3, 2)
    if r:
      for (a, b, c) in decompose(r, 3, min_v=1, sep=1):
        yield (2 * a + 1, 2 * b + 1, 2 * c + 1)
  
  # solve for positions <pos>, numbers <num>
  def solve(pos, num):
    # let the positions and numbers in the races be:
    #           pos   | num
    #           T D H | T D H
    #  junior:  A B C | D E F
    #  colt:    G H I | J K L
    #  senior:  M N P | Q R S
  
    # an expression to calculate the sum of products
    def fn(ps, ns):
      ps = join((f("pos[{x}]") for x in ps), sep=", ", enc="[]")
      ns = join((f("num[{x}]") for x in ns), sep=", ", enc="[]")
      return f("dot({ps}, {ns})")
    # construct an alphametic puzzle
    exprs = [
      # "D beat T and H in the junior race ..."
      "pos[B] < pos[A]",
      "pos[B] < pos[C]",
      # "... but his runner number was less than his position number"
      "num[E] < pos[B]",
      # "D has the largest runner number in the senior race ..."
      "num[R] > num[Q]",
      "num[R] > num[S]",
      # "... but his runner number was less than his position number"
      "num[R] < pos[N]",
      # "the sum of H's products were smaller than T's or D's"
      f("{H} < min({T}, {D})", T=fn("AGM", "DJQ"), D=fn("BHN", "EKR"), H=fn("CIP", "FLS")),
    ]
    p = SubstitutedExpression(
      exprs,
      base=3, # values are (0, 1, 2)
      distinct="ABC DEF GHI JKL MNP QRS AGM BHN CIP DJQ EKR FLS".split(),
      env=dict(pos=pos, num=num),
    )
    # solve the alphametic
    for s in p.solve(verbose=""):
      # output solution
      for (t, ps, ns) in zip(["junior", "colt  ", "senior"], ["ABC", "GHI", "MNP"], ["DEF", "JKL", "QRS"]):
        ((pT, pD, pH), (nT, nD, nH)) = ((pos[s[k]] for k in ps), (num[s[k]] for k in ns))
        printf("{t}: T (#{nT}) = {pT}; D (#{nD}) = {pD}; H (#{nH}) = {pH}")
  
  # consider the values of (a + b + c) * (x + y + z)
  for (p, q) in divisors_pairs(841, every=1):
    if p < 15 or q < 15: continue
    for (abc, xyz) in cproduct([numbers(p), numbers(q)]):
      # all numbers and products are different
      if len(set(abc + xyz)) != 6: continue
      if len(set(i * j for (i, j) in cproduct([abc, xyz]))) != 9: continue
      # solve the puzzle
      solve(abc, xyz)
  

  Solution: In the Colts race Tom (#15) finished 5th; Dick (#11) finished 7th; Harry (#3) finished 17th.

  The full results are:

  Junior: Tom (#11) = 17th; Dick (#3) = 5th; Harry (#15) = 7th
  Colts: Tom (#15) = 5th; Dick (#11) = 7th; Harry (#3) = 17th
  Senior: Tom (#3) = 7th; Dick (#15) = 17th; Harry (#11) = 5th

  In each race the runner numbers are #3, #11, #15 and the positions are 5th, 7th, 17th.

  And the sum of the nine different products of these numbers is:

  (3 + 11 + 15) × (5 + 7 + 17) = 29 × 29 = 841

  In fact the only viable factorisation of 841 is 29 × 29.

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