S2T2

Jim Randell 9:39 am on 7 March 2021 Permalink Reply
Tags: by: D. G. Beech   

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  Jim Randell 9:40 am on 7 March 2021 Permalink | Reply

  For each of the three subjects a student can either pass or fail. So there are 8 regions in the corresponding Venn diagram.

  And if we had been given the square numbers for those that passed in all three subject and failed in all three subjects we would have 8 equations in 8 variables, so we would be able to work out the values for each of the variables.

  And the missing squares must be less than 15².

  The following Python program runs in 79ms.

  from enigma import (matrix, as_int, subsets, powers, printf)
  
  # solve the equations given values for MPC and X
  def solve(MPC, X):
  
    # set up the equations
    eqs = [
      # X  M  P  C  MP MC PC MPC = k
      ((1, 1, 1, 1, 1, 1, 1, 1), 440),
      ((0, 0, 0, 1, 0, 1, 1, 1), 200),
      ((1, 1, 0, 1, 0, 1, 0, 0), 210),
      ((0, 1, 0, 0, 1, 1, 0, 1), 220),
      ((0, 0, 0, 1, 0, 0, 1, 0),  66),
      ((0, 1, 0, 0, 0, 1, 0, 0),  22),
      ((0, 0, 0, 0, 0, 0, 0, 1), MPC),
      ((1, 0, 0, 0, 0, 0, 0, 0),   X),
    ]
  
    # solve for non-negative integers
    try:
      (X, M, P, C, MP, MC, PC, MPC) = (as_int(x, "0+") for x in matrix.linear(eqs))
    except ValueError:
      return
  
    printf("X={X} M={M} P={P} C={C} MP={MP} MC={MC} PC={PC} MPC={MPC}")
  
  # consider possible squares for X and MPC
  for (MPC, X) in subsets(powers(0, 14, 2), size=2, select="M"):
    solve(MPC, X)
  

  Solution: (1) 144 pupils failed all three subjects; (2) 143 passed in chemistry and also physics.

  We can make a table of the regions of the Venn diagram

  M P C = 121
  M P - =  77
  M - C =  13
  M - - =   9
  - P C =  22
  - P - =  10
  - - C =  44
  - - - = 144
  total = 440
  

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  Frits 12:25 pm on 8 March 2021 Permalink | Reply

  Based on Jim’s solution, with some analysis so no “==” operators are used (to avoid loops).

  @Jim, the generated code is not efficient as possible as there is a loop for O and Q when CF is already known.

  from enigma import SubstitutedExpression, is_square
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
    # X    M     P    C    MP   MC   PC   MPC
    
    #XYZ + MA + PDE + CF + HI + KL + OQ + STU = 440
    #"352 - XYZ - HI - STU = PDE",   # substitute MA + KL = 22 and CF + OQ = 66
    "352 - XYZ - HI - STU >= 0",
   
    #                 CF +      KL + OQ + STU = 200
    "112 + MA = STU",          # substitute CF + KL = 88 - MA - OQ
    
    #XYZ + MA +       CF +      KL            = 210
    "188 - CF = XYZ", 
    
    #      MA +            HI + KL +      STU = 220
    "198 - STU = HI",          # substitute KL = 22 - MA
    
    #                 CF +           OQ       = 66
    "66 - OQ = CF",
    
    #      MA +                 KL            = 22             
    "22 - MA = KL",
    
    "is_square(STU)",
    "is_square(XYZ)",
    ],
    answer="STU, HI, KL, MA, OQ, 352 - XYZ - HI - STU, CF, XYZ",
    d2i=dict([(0, "SX")] +
             [(k, "M") for k in range(3, 7)] +
             [(k, "MCO") for k in range(7, 10)]),
    distinct="",
    verbose=0
    )
  
  for (_, ans) in p.solve():
    print("pupils failed in all three subjects =", ans[0])
    print("Of those who passed in chemistry", ans[0] + ans[4], 
          "also passed in physics")
  

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  • 

    Frits 12:57 pm on 8 March 2021 Permalink | Reply

    It can easily be seen that MA has to be 9 (square 121) and CF has to be 19 or 44 (squares 169 and 144).

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  John Crabtree 5:15 pm on 10 March 2021 Permalink | Reply

  C = 66 – CP and M = 22 – CM.
  From chemistry, CM + CMP = 200 – 66 = 134, with CM <= 22
  And so CMP = 121 and CM = 13.
  From maths, MP = 200 – 22 – CMP = 77
  From physics, P = 230 – MP – CP – CMP = 32 – CP, and so CP <= 32
  From all eight variables, 318 – CP + X = 440, ie X = 122 + CP
  And so X = 144 and CP = 22
  The answers follow directly.

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