S2T2

Jim Randell 10:09 am on 21 October 2025 Permalink Reply
Tags: by: Danny Roth ( 84 )   

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  Jim Randell 10:10 am on 21 October 2025 Permalink | Reply

  Here is a solution using the [[ SubstitutedExpression() ]] solver from the enigma.py library.

  It runs in 87ms. (Internal runtime of the generated code is 6.8ms).

  from enigma import (SubstitutedExpression, irange, div, sprintf as f, printf)
  
  # calculate (integer) percentage change
  def change(before, after): return div(100 * (after - before), before)
  
  # 2-digit numbers for change percentages
  d2 = irange(10, 99)
  d2o = irange(11, 99, step=2)  # odd numbers
  
  # symbols for each party before (= X1) and after (= X2)
  (L1, C1, D1, L2, C2, D2) = "ABCXYZ"
  
  # constraints for the solver
  exprs = [
    # total number of seats is 40
    f("{L1} + {C1} + {D1} == 40"),
    f("{L2} + {C2} + {D2} == 40"),
  
    # before: Lab had an overall majority, with Con second
    f("{L1} > {C1} + {D1}"),
    f("{C1} > {D1}"),
  
    # after: Con was first, Lab second, Dem third
    f("{C2} > {L2} > {D2}"),
  
    # both Con and Dem increased their numbers of seats
    f("{C2} > {C1}"),
    f("{D2} > {D1}"),
  
    # after: Con formed a coalition with Dem to get a majority
    f("not ({C2} > {L2} + {D2})"),
    f("{C2} + {D2} > {L2}"),
  
    # each party had a 2-digit percentage change in its number of seats
    f("abs(change({L1}, {L2})) in d2"),
    f("abs(change({C1}, {C2})) in d2"),
    f("abs(change({D1}, {D2})) in d2o"),  # Dem's change is odd
  ]
  
  # construct the solver
  p = SubstitutedExpression(
    exprs,
    base=41, # each value is 0 .. 40
    distinct="",
    env=dict(change=change, d2=d2, d2o=d2o),
    answer=f("({L1}, {C1}, {D1}, {L2}, {C2}, {D2})"),
  )
  
  # solve the puzzle
  for (L1, C1, D1, L2, C2, D2) in p.answers(verbose=0):
    # output solution
    printf("before (seats)   -> Lab  {L1:2d},  Con  {C1:2d},  Dem  {D1:2d}")
    printf("after  (seats)   -> Lab  {L2:2d},  Con  {C2:2d},  Dem  {D2:2d}")
    (cL, cC, cD) = (L2 - L1, C2 - C1, D2 - D1)
    printf("change (seats)   -> Lab {cL:+3d},  Con {cC:+3d},  Dem {cD:+3d}")
    (cL, cC, cD) = (change(L1, L2), change(C1, C2), change(D1, D2))
    printf("change (percent) -> Lab {cL:+d}%, Con {cC:+d}%, Dem {cD:+d}%")
    printf()
  

  Solution: The result of the recent election was: Conservatives = 19 seats; Labour = 11 seats; Democrats = 10 seats.

  Before the election the seats were allocated thus:

  Labour = 22 seats
  Conservatives = 10 seats
  Democrats = 8 seats

  So the changes are:

  Labour = −11 seats (−50%)
  Conservatives = +9 seats (+90%)
  Democrats = +2 seats (+25%)

  The percentages don’t add up to 0% because they are the percentage change over the previous number of seats, not the percentage change in the overall share of seats. If we use this measure, then there are many solutions, but the changes will sum to 0%. (This can be seen by making the [[ change() ]] function divide by the total number of seats [[ 40 ]] instead of [[ before ]]).

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• 

  Ruud 8:24 am on 22 October 2025 Permalink | Reply

  As brute force as possible:

  import peek
  import itertools
  
  
  def check(x0, x1, has_to_be_odd=False):
      perc = (abs(x0 - x1) / x0) * 100
      if perc % 1 or perc < 10 or perc >= 100:
          return False
      return not has_to_be_odd or perc % 2 == 1
  
  
  for l0, c0, d0 in itertools.product(range(1, 41), repeat=3):
      if l0 + c0 + d0 == 40 and l0 > c0 + d0 and l0 > c0 > d0:
          for l1, c1, d1 in itertools.product(range(1, 41), repeat=3):
              if l1 + c1 + d1 == 40 and c1 < l1 + d1 and c1 > l1 > d1 and c1 + d1 > 20 and c1 > c0 and d1 > d0:
                  if check(l0, l1) and check(c0, c1) and check(d0, d1, has_to_be_odd=True):
                      peek(l0, l1, c0, c1, d0, d1, l1 - l0, c1 - c0, d1 - d0)
  

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  Frits 4:35 pm on 22 October 2025 Permalink | Reply

  # check for a two-figure number percentage change
  def check(old, new):
    d, r = divmod(100 * abs(new - old), old)
    if r or not (9 < d < 100): return False
    return d
        
  M = 40
  parties = "Labour Conservatives Democrats".split(" ")
  # Labour had a majority, Democrats had at least 2 seats (two-figure %)
  for l1 in range(M // 2 + 1, M - 4):
    for c1 in range((M - l1) // 2 + 1, M - l1 - 1):
      d1 = M - l1 - c1
      # up to 99 percent increase   
      for d2 in range(d1 + 1, min(2 * d1, M // 3)):
        if not (perc := check(d1, d2)): continue
        # the Democrats' percentage increase was an odd number
        if not (perc % 2): continue 
        for c2 in range(max(d1 + 1, (M - d2) // 2 + 1), M // 2 + 1):
          if not check(c1, c2): continue
          l2 = M - d2 - c2
          if l2 <= d2 or not check(l1, l2): continue
          print(list(zip(parties, [l2 - l1, c2 - c1, d2 - d1])))
  

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