S2T2

Jim Randell 9:10 am on 26 November 2020 Permalink Reply
Tags: by: Graham Smithers ( 82 )   

• 

  Jim Randell 9:10 am on 26 November 2020 Permalink | Reply

  This puzzle can be solved using the [[ SubstitutedExpression ]] solver from the enigma.py library.

  The following run file executes in 136ms.

  Run: [ @repl.it ]

  #! python -m enigma -rr
  
  SubstitutedExpression
  
  # one pair is even, the other is odd
  "E % 2 == M % 2"
  "N % 2 == K % 2"
  "E % 2 != N % 2"
  
  # each pair has the same sum
  "EWE + RAM == HEN + COCK"
  
  # A, R, K are consecutive (in some order)
  "max(A, R, K) - min(A, R, K) == 2"
  
  # answer
  --answer="NOAH"
  

  Solution: NOAH = 2074.

  The pairs are:

  (EWE, RAM) = (595, 873)
  (HEN, COCK) = (452, 1016)

  Both pairs sum to 1468.

  And (A, R, K) rearranged gives the consecutive numbers (6, 7, 8).

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• 

  Frits 10:13 am on 26 November 2020 Permalink | Reply

  Trying not to copy all Jim’s equations.

  from itertools import permutations
  
  c = 1
  digits = set(range(10)).difference({1})
  
  for Q in permutations(digits):
    e,w,r,a,m,h,n,o,k = Q
    # letter combinations don't start with zero
    if 0 in [e, r, h, n, a]: continue
    # pairs EWE, RAM and HEN, COCK are even or odd
    if (e + m) % 2 == 1 : continue
    if (n + k) % 2 == 1 : continue
    # one pair of the numbers being odd, the other pair being even
    if (e + n) % 2 == 0: continue
    #  both pairs have the same sum
    s1 = (e + r) * 100 + (w + a) * 10 + e + m
    s2 = c * 1000 + (h + o) * 100 + (e + c) * 10 + n + k
    if s1 != s2: continue
    # a,r,k are three consecutive digits in a muddled order
    s3 = sum(abs(x - y) for (x, y) in zip([a, r, k, a], [r, k, a]))
    if s3 != 4: continue
    print(f"NOAH = {n}{o}{a}{h}")
  

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• 

  GeoffR 12:24 pm on 26 November 2020 Permalink | Reply

  
  % A Solution in MiniZinc 
  include "globals.mzn";
  
  var 0..9:E; var 0..9:W; var 0..9:R;
  var 0..9:A; var 0..9:M; var 0..9:C;
  var 0..9:O; var 0..9:K; var 0..9:N;
  var 0..9:H;
  
  constraint E > 0 /\ R > 0 /\ H > 0 /\ C > 0 /\ N > 0;
  constraint all_different ([E, W, R, A, M, C, O, K, N, H]);
  
  var 100..999:EWE = 100*E + 10*W + E;
  var 100..999:RAM = 100*R + 10*A + M;
  var 100..999:HEN = 100*H + 10*E + N;
  var 1000..9999:COCK = 1000*C + 100*O + 10*C + K;
  var 1000..9999:NOAH = 1000*N + 100*O +10*A + H;
  
  % One pair of the numbers being odd, the other pair being even
  constraint 
  ((EWE mod 2 == 0 /\ RAM mod 2 == 0) /\ (HEN mod 2 == 1 /\ COCK mod 2 == 1))
  \/ 
  ((EWE mod 2 == 1 /\ RAM mod 2 == 1) /\ (HEN mod 2 == 0 /\ COCK mod 2 == 0));
  
  % Both pairs of numbers have the same sum
  constraint EWE + RAM == HEN + COCK;
  
  % The three digits of the number ARK are consecutive digits in a muddled order
  % Re-using Jim's equation i.e. "max(A, R, K) - min(A, R, K) == 2"
  constraint max([A,R,K]) - min([A,R,K]) == 2;
  
  solve satisfy;
  
  output ["NOAH = " ++ show(NOAH) ++ "\n"
  ++ "EWE = " ++ show(EWE) ++ " , " ++ "RAM == " ++ show(RAM) ++ "\n"
  ++ "HEN = " ++ show(HEN) ++ " , " ++ "COCK = " ++ show(COCK) ];
  
  % NOAH = 2074
  % EWE = 595 , RAM == 873
  % HEN = 452 , COCK = 1016
  % ----------
  % ==========
  
  
  

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