S2T2

Jim Randell 8:49 am on 26 December 2019 Permalink Reply
Tags: by: Victor Bryant ( 108 )   

• 

  Jim Randell 8:51 am on 26 December 2019 Permalink | Reply

  This Python program runs in 73ms.

  Run: [ @replit ]

  from enigma import (irange, subsets, group, nconcat, mod, printf)
  
  # available digits
  digits = irange(1, 9)
  
  # choose 3 digits
  for ds in subsets(digits, size=3, select="C"):
    # any permutation must be divisible by 3
    if sum(ds) % 3 != 0: continue
    # compute the residues of the permutations mod 7
    g = group(subsets(ds, size=len, select="P", fn=nconcat), by=mod(7))
    # there can be no permutations exactly divisible by 7
    if g.get(0): continue
    # there can be no non-zero residue that is ambiguous
    if any(len(v) > 1 for (k, v) in g.items() if k != 0): continue
    # output solution
    printf("digits = {ds} [{g}]")
  

  Solution: The three digits were: 4, 8, 9.

  They can be assembled into the following 3-digit numbers (and residues modulo 7):

  489 (6)
  498 (1)
  849 (2)
  894 (5)
  948 (3)
  984 (4)

  Each permutation of the digits is uniquely identified by the residue.

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• 

  GeoffR 4:31 pm on 26 December 2019 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  % Three single digits
  var 1..9:A; var 1..9:B; var 1..9:C;
  constraint all_different([A, B, C]);
  
  % Residues modulus 7 
  var 1..6:a; var 1..6:b; var 1..6:c;
  var 1..6:d; var 1..6:e; var 1..6:f;
  constraint all_different([a, b, c, d, e, f]);
  
  % Three digit numbers 
  var 100..999:ABC = 100*A + 10*B + C;
  var 100..999:ACB = 100*A + 10*C + B;
  var 100..999:BAC = 100*B + 10*A + C;
  var 100..999:BCA = 100*B + 10*C + A;
  var 100..999:CAB = 100*C + 10*A + B;
  var 100..999:CBA = 100*C + 10*B + A;
  
  % All three digit numbers divisible by three
  constraint ABC mod 3 == 0 /\ ACB mod 3 == 0 
  /\ BAC mod 3 == 0 /\ BCA mod 3 == 0 
  /\ CAB mod 3 == 0 /\ CBA mod 3 == 0;
  
  % All three digit numbers were not divisible by seven
  constraint ABC mod 7 == a /\ ACB mod 7 == b
  /\ BAC mod 7 == c /\ BCA mod 7 == d
  /\ CAB mod 7 == e /\ CBA mod 7 == f; 
  
  constraint increasing([A, B, C]);
  
  solve satisfy;
  
  output[ "The three digits were " ++ show(A) ++ ", " 
  ++ show(B) ++ ", " ++ show(C)];
  
  % The three digits were 4, 8, 9
  % % time elapsed: 0.03 s
  % ----------
  % ==========
  
  
  
  

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