S2T2

Jim Randell 9:21 am on 19 October 2023 Permalink Reply
Tags: by: Angela Newing ( 37 )   

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  Jim Randell 9:21 am on 19 October 2023 Permalink | Reply

  If we use variables A, B, C, D, E to correspond to the points allocated (according to mothers name), then we can allocate the first names and surnames of the cubs to the same set of variables to give us a set of simultaneous linear equations, which we can attempt to solve to give the points values.

  I used the [[ Matrix.linear() ]] solver from the enigma.py library to find assignments that give non-negative points values. (Although there are no further solutions if negative values are permitted).

  I also ensure that the pairings mentioned in the text all consist of distinct cubs. (Although, again, there are no further solutions if a pairing can refer to the same cub twice).

  The following Python program runs in 375ms.

  Run: [ @replit ]

  from enigma import (Matrix, subsets, catch, printf)
  
  # suppose the points are: A, B, C, D, E (according to mothers name)
  # then we can map the first and surnames to these values
  
  # variables
  vs = "ABCDE"
  
  # a function to construct equations
  def eq(vs, k, fn=Matrix.equation(vs)):
    r = fn(vs, k)
    assert {0, 1}.issuperset(r[0])  # reject coefficients that aren't 0 or 1
    return r
  
  # validate point values
  def valid(x):
    assert not x < 0
    return x
  
  # for firstnames I, J, K, M, N
  for (I, J, K, M, N) in subsets(vs, size=len, select='P'):
  
    # for surnames P, R, S, T (= Brown), U = (Jones)
    for (P, R, S, T, U) in subsets(vs, size=len, select='P'):
  
      # try to solve the equations
      r = None
      try:
        # construct equations
        eqs = [
          eq(['E', S], 17),  # Enid + Smith = 17
          eq([U, R], 16),  # Jones + Robinson = 16
          eq([I, R], 14),  # Ivan + Robinson = 14
          eq([J, T], 12),  # John + Brown = 12
          eq(['B', M], 11),  # Brenda + Mike = 11
          eq(['A', 'D'], 9),  # Ann + Debbie = 9
          eq(['C', N], 8),  # Christine + Nigel = 8
          eq([P, 'D'], 6),  # Perkins + Debbie = 6
        ]
  
        # solve the equations
        r = Matrix.linear(eqs, valid=valid)
      except (ValueError, AssertionError):
        pass
      if r is None: continue
  
      # check there is a pair = 13 (two youngest), and a pair = 10 (Ken + friend)
      v = dict(zip(vs, r))
      if not any(v[x] + v[K] == 10 for x in vs if x != K): continue
      if not any(v[x] + v[y] == 13 for (x, y) in subsets(vs, size=2)): continue
  
      # output solution
      mn = { 'A': 'Anne', 'B': 'Brenda', 'C': 'Christine', 'D': 'Debbie', 'E': 'Enid' }
      fn = { I: 'Ivan', J: 'John', K: 'Ken', M: 'Mike', N: 'Nigel' }
      sn = { P: 'Perkins', R: 'Robinson', S: 'Smith', T: 'Brown', U: 'Jones' }
      for k in vs:
        printf("{fn} {sn} = {v} points [{mn}]", mn=mn[k], fn=fn[k], sn=sn[k], v=v[k])
      printf()
  

  Solution: The names of the cubs are:

  Mike Smith = 7 points [Anne]
  Ivan Perkins = 4 points [Brenda]
  Ken Jones = 6 points [Christine]
  Nigel Brown = 2 points [Debbie]
  John Robinson = 10 points [Enid]

  So the youngest two are Mike and Ken (= 13 points in total), and Ivan must be Ken’s best friend (= 10 points in total).

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