S2T2

Jim Randell 9:31 am on 20 June 2023 Permalink Reply
Tags: by: Adrian Somerfield ( 13 )   

• 

  Jim Randell 9:32 am on 20 June 2023 Permalink | Reply

  This Python program uses the [[ SubstitutedExpression() ]] solver from the enigma.py library to find candidate values that are each individually within 5 of a cube number. We then check the answers for cases where three of the distances are 0.

  It runs in 68ms. (Internal runtime is 11ms).

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, iroot, printf)
  
  # distance to nearest kth root
  def dist(n, k=3):
    x = iroot(n, k)
    return min(n - x**k, (x + 1)**k - n)
  
  # make an alphametic puzzle
  p = SubstitutedExpression(
    [
      # each must be less than 5 away from a cube
      "dist(THREE) < 5",
      "dist(CAN) < 5",
      "dist(BE) < 5",
      "dist(CUBES) < 5",
    ],
    answer="(THREE, CAN, BE, CUBES)",
    env=dict(dist=dist),
    verbose=0
  )
  # solve the puzzle
  for ans in p.answers():
    # count the distances
    ds = list(dist(x) for x in ans)
    # three must have distance 0
    if ds.count(0) != 3: continue
    # output solution
    CUBE = ans[-1] // 10
    printf("CUBE = {CUBE} {ans}")
  

  Solution: CUBE = 1968.

  The numbers are:

  THREE = 74088 = 42³
  CAN = 125 = 5³
  BE = 68 = 4³ + 4
  CUBES = 19683 = 27³

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• 

  Frits 1:57 pm on 20 June 2023 Permalink | Reply

  More messy (the reorder=0 isn’t even necessary).

     
  #!/usr/bin/env pypy -m enigma -r
   
  SubstitutedExpression
  
  # each must be a cube or less than 5 away from a cube
  "(c1 := is_cube_p(BE)) or \
    BE - (r := int(BE**(1/3)))**3 < 5        or (r + 1)**3 - BE < 5"
  "(c2 := is_cube_p(CUBES)) or (c1 and \
    (CUBES - (r := int(CUBES**(1/3)))**3 < 5 or (r + 1)**3 - CUBES < 5))"
  "(c3 := is_cube_p(CAN)) or (c1 and c2 and \
    (CAN - (r := int(CAN**(1/3)))**3 < 5     or (r + 1)**3 - CAN < 5))"
  "is_cube_p(THREE) or (c1 and c2 and c3 and \
    (THREE - (r := int(THREE**(1/3)))**3 < 5 or (r + 1)**3 - THREE < 5))"
  
  --answer="(THREE, CAN, BE, CUBES)"
  
  # [optional] less verbose output
  --template=""
  
  --reorder=0
  

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  • 

    Frits 7:02 pm on 20 June 2023 Permalink | Reply

    The 2 missing ” and ” don’t cause a syntax error.

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  • 

    Jim Randell 8:26 am on 21 June 2023 Permalink | Reply

    @Frits: I think I see what this is doing. (I removed the extraneous trailing commas, and fixed up some missing and keywords).

    But I think it would allow the case where all four of the words were cubes.

    ---

    This is what my [[ SubstitutedExpression ]] run file looks like:

    Run: [ @replit ]

    #! python3 -m enigma -rr
    
    SubstitutedExpression
    
    # distance to nearest kth root
    --code="def _dist(n, k=3): x = iroot(n, k); return min(n - x**k, (x + 1)**k - n)"
    --code="dist = cache(_dist)"
    
    # check cube distance is less than 5 (if flag) or 0
    --code="def check(n, f): x = dist(n); return (0 < x < 5 if f else x == 0)"
    
    # n = 1..4 denotes which statement is not a cube
    --invalid="0|5|6|7|8|9,n"
    
    # each is less than 5 away from a cube
    "check({THREE}, {n} ==  1)"
    "check({CAN}, {n} == 2)"
    "check({BE}, {n} == 3)"
    "check({CUBES}, {n} == 4)"
    
    --distinct="ABCEHNRSTU"
    --answer="CUBE"
    --template="(THREE, CAN, BE, CUBES)"
    --solution="n"
    

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• 

  GeoffR 5:43 pm on 20 June 2023 Permalink | Reply

  Using Frits method of solution i.e.each number must be a cube or less than 5 away from a cube.

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:T; var 0..9:H; var 0..9:R; var 0..9:E;
  var 1..9:C; var 0..9:A; var 0..9:N; var 1..9:B;
  var 0..9:U; var 0..9:S;
  
  constraint all_different ([T, H, R, E, C, A, N, B, U, S]);
  
  % Four numbers are BE, CAN, CUBES, THREE
  var 10..99:BE == 10*B + E;
  var 100..999:CAN == 100*C + 10*A + N;
  var 10000..99999:CUBES == 10000*C + 1000*U + 100*B + 10*E + S;
  var 10000..99999:THREE == 10000*T + 1000*H + 100*R + 11*E;
  
  % Required answer is CUBE
  var 1000..9999:CUBE == 1000*C + 100*U + 10*B + E; 
  
  % Sets of cubes for 2-digit, 3-digit and 5-digit numbers
  set of int: cb2 = {27, 64};
  set of int: cb3 = {n * n * n | n in 5..9};
  set of int: cb5 = {n * n * n | n in 22..46};
  
  % Three of the numbers are cubes
  constraint sum([THREE in cb5, CAN in cb3, BE in cb2, CUBES in cb5]) == 3;
  
  % Allow for the fourth number to be less than five away from a cube
  % Number BE
  constraint BE in cb2 \/ (BE - 1) in cb2  \/ (BE - 2) in cb2 
  \/ (BE - 3) in cb2   \/ (BE - 4) in cb2  \/ (BE + 1) in cb2 
  \/ (BE + 2) in cb2   \/ (BE + 3) in cb2  \/ (BE + 4) in cb2; 
  % Number CAN
  constraint CAN in cb3 \/ (CAN - 1) in cb3 \/ (CAN - 2) in cb3
   \/ (CAN - 3) in cb3  \/ (CAN - 4) in cb3 \/ (CAN + 1) in cb3 
   \/ (CAN + 2) in cb3  \/ (CAN + 3) in cb3 \/ (CAN + 4) in cb3;
  % Number THREE
  constraint THREE in cb5 \/ (THREE - 1) in cb5  \/ (THREE - 2) in cb5
   \/ (THREE - 3) in cb5  \/ (THREE - 4) in cb5  \/ (THREE + 1) in cb5 
   \/ (THREE + 2) in cb5  \/ (THREE + 3) in cb5  \/ (THREE + 4) in cb5;
  % Number CUBES
  constraint CUBES in cb5 \/ (CUBES - 1) in cb5 \/ (CUBES - 2) in cb5
   \/ (CUBES - 3) in cb5  \/ (CUBES - 4) in cb5 \/ (CUBES + 1) in cb5 
   \/ (CUBES + 2) in cb5  \/ (CUBES + 3) in cb5 \/ (CUBES + 4) in cb5;
                                                                                    
  solve satisfy;
  
  output["CUBE = " ++ show(CUBE)  ++ "\n" ++   
  "[THREE, CAN, BE, CUBES] = " ++ show ([THREE, CAN, BE, CUBES]) ];
  
  % CUBE = 1968
  % [THREE, CAN, BE, CUBES] = [74088, 125, 68, 19683]
  % ----------
  % ==========
  
  
  

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