S2T2

Jim Randell 9:10 am on 27 August 2020 Permalink Reply
Tags: by: Victor Bryant ( 139 )   

• 

  Jim Randell 9:10 am on 27 August 2020 Permalink | Reply

  The common difference is (TWO − ONE), and there are 8 instances of the common difference between ONE and NINE.

  So, we can solve this puzzle straightforwardly using the [[ SubstitutedExpression() ]] solver from the enigma.py library.

  The following run file executes in 110ms.

  Run: [ @replit ]

  #! python -m enigma -rr
  
  SubstitutedExpression
  
  "div(NINE - ONE, TWO - ONE) = 8"
  "div(NINE, 9) > 0"
  
  --answer="WIN"
  

  Solution: WIN = 523.

  The progression is: ONE = 631, TWO = 956, …, NINE = 3231.

  The common difference is: TWO − ONE = 325.

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  GeoffR 10:39 am on 27 August 2020 Permalink | Reply

  % A Solution in MiniZinc 
  include "globals.mzn";
  
  var 1..9: O; var 1..9: N; var 1..9: E;
  var 1..9: T; var 1..9: W; var 1..9: I;
  
  constraint all_different ([O, N, E, T ,W, I]);
  
  var 100..999: ONE = 100*O + 10*N + E;
  var 100..999: TWO = 100*T + 10*W + O;
  var 100..999: WIN = 100*W + 10*I + N;
  var 1000..9999: NINE = 1000*N + 100*I + 10*N + E;
  
  % Define 9th term of series
  constraint NINE = ONE + 8 * (TWO - ONE)
   /\ NINE mod 9 == 0;
  
  solve satisfy;
  
  output [ "Series is " ++ show(ONE) ++ "," ++ show(TWO) ++ 
  "..." ++ show(NINE) ++ "\n" ++ "WIN = " ++ show(WIN)]
  
  % Series is 631,956...3231
  % WIN = 523
  % ----------
  % ==========
  
  
  

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  GeoffR 11:02 am on 27 August 2020 Permalink | Reply

  
  from itertools import permutations
  
  for Q in permutations(range(1,10), 6):
    O, N, E, T, W, I = Q
    ONE, TWO = 100*O + 10*N + E, 100*T + 10*W + O
    WIN = 100*W + 10*I + N
    NINE = 1010*N + 100*I + E
    if NINE == ONE + (TWO - ONE) * 8 and NINE % 9 == 0:
      print(f"ONE = {ONE}, TWO = {TWO}, NINE = {NINE}, WIN = {WIN}")
  
  # ONE = 631, TWO = 956, NINE = 3231, WIN = 523
  
  

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  • 

    Frits 1:26 pm on 27 August 2020 Permalink | Reply

    @GeoffR, shouldn’t you allow for E, I and W to be zero as well?

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  Frits 1:20 pm on 27 August 2020 Permalink | Reply

   
  # elements in range <r> unequal to values of D[i]
  def domain(i, r):
    vals = set()
    for s in D[i]:
      # use globals() iso eval()
      vals.add(globals()[s])
    return [x for x in r if x not in vals]
  
  # set up dictionary of for-loop iterator names
  def init(li):
    for i in range(1,len(li)):
      D[li[i]] = li[0:i]
      
  # concatenate list of integers 
  to_num = lambda *args: int("".join(map(str, args)))    
  
  
  D = {}
  init(['O','N','E','T','W'])
  
  
  for O in range(1,10):
    for N in domain('N', range(1,10)):
      for E in domain('E', range(0,10)):
        ONE = to_num(O,N,E)
        for T in domain('T', range(1,10)):
          if T > O:  
            for W in domain('W', range(0,10)):
              TWO = to_num(T,W,O)
              unit = TWO - ONE
              units8 = 8*unit + ONE
              if units8 % 9 != 0 or units8 < 1000: continue
              
              s = str(units8)
              if s[3] != str(E): continue 
              if s[0] != str(N) or s[2] != str(N): continue 
              # check letter I
              if int(s[1]) in {O,N,E,T,W}: continue
        
              WIN = str(W)+s[1]+str(N)
              print("WIN ONE TWO NINE")
              print(WIN, ONE, TWO, units8)
  
  # WIN ONE TWO NINE
  # 523 631 956 3231
  

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  GeoffR 1:47 pm on 27 August 2020 Permalink | Reply

  @Frits: In theory maybe, but it would not have made any difference in practice.
  As the title of the teaser says – “Inconsequential” ?

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