Sunday Times Teaser 2649 – Right to Left

1. 
   BRG permalink

   Python

   # len(TO) >= 3 since TO is a palindrome and T wouuld equal O # if T and O both had only one digit # len(CORRECT) = 2 * len(CR) + len(OET) = 9 and len(ETO) > 3 # giving len(C) = len(R) = 1 and len(ETO) = 5 # because REFLECT is a palindrome, if T had only one digit R # would equal T - so len(T) = 2, len(EO) = 3 and R = T % 10 for R in (2, 4, 6, 8): # the first digit of T is 1 or 2 for t1 in (10, 20): T = t1 + R if T > 25: continue # O is even because TO is specified as even for O in set(range(0, 26, 2)) - set((R, T)): # ensure that O and T are different that TO is a palindrome s = str(T) + str(O) if s[::-1] != s: continue # consider the left and right ends of the palindrome REFLECT # if len(E) = len(F) = 1 then ref = tTc -> e = T, f = c ) len(F) # if len(E) = 2 and len(F) = 1 then reEf = tTceE -> f = c = e ) is 2 # if len(E) = 1 then reFf = tTce -> e = T, Ff = ce # if len(E) = 2 then rEeFf = tTceE -> E = Tc, Ff = eE = cT for C in (1, 2): # len(E) = 3 - len(O) if O < 10: E, F = 10 * (T // 10) + C, 10 * C + (T // 10) else: E, F = (T // 10), 10 * C + (T // 10) ss = set((C, E, F, O, R, T)) # ensure that all these values are different if len(ss) != 6: continue # now try all the remaining values for L for L in set(range(26)) - ss: # and skip any for which REFLECT is not a palindrome ref = ''.join(str(x) for x in (R, E, F, L, E, C, T)) if ref[::-1] != ref: continue # now find all permutations of the values for the remaining letters for G, H, I in permutations(set(range(26)) - (ss | set((L,))), 3): # and use only those for which RIGHTTOLEFT is a palindrome r2l = ''.join(str(x) for x in (R, I, G, H, T, T, O, L, E, F, T)) if r2l[::-1] == r2l: print(int(''.join(str(x) for x in (C, O, R, R, E, C, T))))

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   # len(TO) >= 3 since TO is a palindrome and T wouuld equal O # if T and O both had only one digit   # len(CORRECT) = 2 * len(CR) + len(OET) = 9 and len(ETO) > 3 # giving len(C) = len(R) = 1 and len(ETO) = 5   # because REFLECT is a palindrome, if T had only one digit R # would equal T - so len(T) = 2,  len(EO) = 3 and R = T % 10 for R in (2, 4, 6, 8):   # the first digit of T is 1 or 2   for t1 in (10, 20):     T = t1 + R     if T > 25:       continue     # O is even because TO is specified as even     for O in set(range(0, 26, 2)) - set((R, T)):       # ensure that O and T are different that TO is a palindrome       s = str(T) + str(O)       if s[::-1] != s:         continue       # consider the left and right ends of the palindrome REFLECT       # if len(E) = len(F) = 1 then ref = tTc   -> e = T, f = c     ) len(F)       # if len(E) = 2 and len(F) = 1 then reEf = tTceE -> f = c = e )  is 2       # if len(E) = 1 then reFf = tTce   -> e = T,  Ff = ce       # if len(E) = 2 then rEeFf = tTceE -> E = Tc, Ff = eE = cT       for C in (1, 2):         # len(E) = 3 - len(O)         if O < 10:           E, F = 10 * (T // 10) + C, 10 * C + (T // 10)         else:           E, F = (T // 10), 10 * C + (T // 10)           ss = set((C, E, F, O, R, T))         # ensure that all these values are different         if len(ss) != 6:           continue           # now try all the remaining values for L         for L in set(range(26)) - ss:           # and skip any for which REFLECT is not a palindrome           ref = ''.join(str(x) for x in (R, E, F, L, E, C, T))           if ref[::-1] != ref:             continue           # now find all permutations of the values for the remaining letters           for G, H, I in permutations(set(range(26)) - (ss | set((L,))), 3):             # and use only those for which RIGHTTOLEFT is a palindrome             r2l = ''.join(str(x) for x in (R, I, G, H, T, T, O, L, E, F, T))             if r2l[::-1] == r2l:               print(int(''.join(str(x) for x in (C, O, R, R, E, C, T))))

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