S2T2

Jim Randell 8:21 am on 22 May 2026 Permalink Reply
Tags: by: Nick MacKinnon ( 58 )   

• 

  Jim Randell 8:23 am on 22 May 2026 Permalink | Reply

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

  The following run-file executes in 78ms. (Internal runtime of the generated code is 211µs).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  # suppose the prices (in pence) are:
  #
  #  ABC = cheapest
  #  DEF
  #  GHI = most expensive
  #
  --digits="1-9"
  
  # the prices are in order
  "A < D < G"
  
  # the middle price is 1p less than 3 times the cheapest
  "3 * ABC - 1 = DEF"
  
  # the total came to a whole number of pounds (= multiple of 100p)
  "(BC + EF + HI) % 100 = 0"
  
  # answer is the 3 prices
  --answer="(ABC, DEF, GHI)"
  

  Solution: The prices were: 239p, 716p, 845p.

  The total amount spent being 1800p = £18.

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• 

  ruudvanderham 9:52 am on 22 May 2026 Permalink | Reply

  import peek
  import istr
  
  set123456789=set(istr.range(1, 10))
  for price2 in istr.range(1, 1000):
      price1 = 3 * price2 - 1
      if price2 < 1000 and (price1 | price2).all_distinct():
          for price0 in istr.range(price1 + 1, 1000):
              if set(price0 | price1 | price2) == set123456789 and (price0 + price1 + price2)[-2:]=="00":
                  peek(price0, price1, price2, price0 + price1 + price2)
  

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• 

  ruudvanderham 1:37 pm on 22 May 2026 Permalink | Reply

  Here’s a different (significantly faster) version:

  import peek
  import istr
  
  set123456789 = set(istr.range(1, 10))
  for price2 in istr.range(1, 1000):
      price1 = 3 * price2 - 1
      if price2 < 1000:
          for total in istr.range((price2 + price1 * 2 + 1).ceil(100), 3000, 100):
              price0 = total - price1 - price2
              if price0 >= 1000:
                  break
              if set(price0 | price1 | price2) == set123456789:
                  peek(price0, price1, price2, price0 + price1 + price2)
  

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