S2T2

Jim Randell 8:18 am on 7 October 2019 Permalink Reply
Tags: by: Danny Roth ( 84 )   

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  Jim Randell 8:19 am on 7 October 2019 Permalink | Reply

  This program considers when there is a corresponding minute for each hour to make a list of palindromic times, and then looks at pairs of times from this list that satisfy the remaining conditions.

  It runs in 44ms.

  Run: [ @repl.it ]

  from itertools import product
  from enigma import irange, nsplit, concat, printf
  
  # palindrome check
  is_palindrome = lambda s: s[::-1] == s
  
  # record palindromic times
  ts = list()
  for h in irange(0, 23):
    (a, b) = nsplit(h, 2)
    m = 10 * b + a
    # record time (in minutes) and displayed digits
    if m < 60: ts.append((h * 60 + m, (a, b, b, a)))
  
  # now consider two palindromic times (the second time could be the following day)
  for ((t1, ds1), (t2, ds2), t3) in product(ts, ts, (0, 1440)):
  
    # calculate the difference in minutes
    t = t2 + t3 - t1
  
    # it should be a 3-digit palindrome
    if not(99 < t < 1000): continue
    ds = nsplit(t)
    if not is_palindrome(ds): continue
  
    # calculate the sum of the digits, it should also be a palindrome
    s = sum(ds1) + sum(ds2) + sum(ds)
    if not is_palindrome(nsplit(s)): continue
  
    # output a solution
    printf("{ds1} -> {ds2}{x} = {t} minutes, digit sum = {s}", ds1=concat(ds1), ds2=concat(ds2), x=(" (next day)" if t3 else ""))
  

  Solution: The two times were 1001 and 0110 (the following day).

  The first time is 10:01am on one day (displayed as 1001), the second time is 1:10am the following day (displayed as 0110). The difference between these times is 15 hours and 9 minutes = 909 minutes. And the sum of the digits in 1001, 0110, 909 is 22.

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  Frits 11:08 am on 22 December 2020 Permalink | Reply

  Slow solution compared to a similar program with enigma.py’s SubstitutedExpression (resp. 600 ms and 15 ms).

  % A Solution in MiniZinc
  
  % time1 ABBA time2 CDDC
  var 0..2:A; var 0..5:B; var 0..2:C; var 0..5:D; 
  var 1..9:E; var 0..9:F; 
  var 1..4:G;
  var 0..23: AB = 10*A + B; var 0..59: BA = 10*B + A;
  var 0..23: CD = 10*C + D; var 0..59: DC = 10*D + C;
  var 101..999: EFE = 101*E + 10*F;
   
  % difference in minutes time1 ABBA time2 CDDC
  constraint (CD * 60 + DC - (AB * 60 + BA) + 1440) mod 1440 == EFE;              
  
  % sum of the eleven digits in the three palindromes was also palindromic
  constraint 2 * ( A + B + C + D + E) + F == G * 11;
   
  solve satisfy;
   
  output ["Time 1 = " ++ show(A) ++ show(B) ++ ":" ++ show(B) ++ show(A) ++ 
  ", Time 2 = " ++ show(C) ++ show(D) ++ ":" ++ show(D) ++ show(C) ++
  ", Difference  = " ++ show(EFE) ++ ", Sum = " ++ show(G * 11)];
   
  % Time 1 = 10:01, Time 2 = 01:10, Difference  = 909, Sum = 22
  % ----------
  % ==========
  

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