Enigmatic Code
Programming Enigma Puzzles
Enigma 1180: Anomalies
Posted by on 15 February 2016
From New Scientist #2336, 30th March 2002 [link]
SEVEN is a prime, and as one would expect, SEVEN minus THREE equals FOUR. But perversely, FOUR is a prime, (and is also a prime when the digits are in reverse order), so THREE is not a prime. Another anomaly is that TEN is a perfect square.
Each different capital letter above represents a different digit, which is the same for that letter everywhere.
Find the numerical values of FOUR and TEN.
[enigma1180]
Enigmatic Code 
We can use the [[
SubstitutedSum()]] solver from the enigma.py library to find candidate solutions.This Python program runs in 60ms.
from enigma import SubstitutedSum, is_square, is_prime, printf p = SubstitutedSum(['THREE', 'FOUR'], 'SEVEN') for s in p.solve(): # TEN is a perfect square TEN = int(p.substitute(s, 'TEN')) if not is_square(TEN): continue # FOUR is a prime FOUR = int(p.substitute(s, 'FOUR')) if not is_prime(FOUR): continue # and RUOF is also a prime RUOF = int(p.substitute(s, 'RUOF')) if not is_prime(RUOF): continue # check THREE is not a prime THREE = int(p.substitute(s, 'THREE')) if is_prime(THREE): continue # and SEVEN is also a prime SEVEN = THREE + FOUR if not is_prime(SEVEN): continue printf("FOUR={FOUR} TEN={TEN} [SEVEN={SEVEN} THREE={THREE}]")Solution: FOUR = 3407. TEN = 529.
% A solution in MiniZinc include "globals.mzn"; solve satisfy; var 0..9: E; var 0..9: V; var 0..9: N; var 0..9: T; var 0..9: H; var 0..9: R; var 0..9: F; var 0..9: O; var 0..9: U; var 0..9: S; array[1..10] of var int : fd = [E,V,N,T,H,R,F,O,U,S]; constraint all_different(fd) /\ S>0 /\ T>0 /\ F>0; % is_square predicate is_square(var int: c) = let { var 0..ceil(pow(int2float(ub(c)),(1/2.0))): z } in z * z = c ; % is_prime predicate is_prime(var int: x) = x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x))))) ( (i < x) -> (x mod i > 0)); % SEVEN is prime constraint is_prime(S*10000 + E*1000 + V*100 + E*10 + N); % SEVEN - FOUR = THREE constraint (S*10000 + E*1000 + V*100 + E*10 + N) - (F*1000 + O*100 + U*10 + R) == (T*10000 + H*1000 + R*100 + E*11); % FOUR is a prime, inc its reversed digits constraint is_prime (F*1000 + O*100 + U*10 + R) /\ is_prime(R*1000 + U*100 + O*10 + F); % THREE is not a prime constraint not is_prime (T*10000 + H*1000 + R*100 + E*11); % TEN is a perfect square constraint is_square(T*100 + E*10 + N); output ["SEVEN is " ++ show(S),show(E),show(V),show(E),show(N) ++ "\n" ++ "TEN is " ++ show(T), show(E),show(N) ++ "\n" ++ "FOUR is " ++ show(F),show(O),show(U),show(R) ++ "\n" ++ "THREE is " ++ show(T),show(H),show(R),show(E),show(E)]; % Output % ------- % SEVEN is 62129 % TEN is 529 % FOUR is 3407 % THREE is 58722 % -------------- % Finished in 245msec %# A faster Python 3 solution (125 msec) from itertools import permutations digits = set('1234567890') def is_square(x): return int(x ** 0.5 + 0.5) ** 2 == x def is_prime(n): for x in range(2, int(n**0.5)+1): if n % x == 0: return False return True # construct SEVEN for (s,e,v,n) in permutations(digits,4): if s == '0' : continue seven = int(s + e + v + e + n) if not is_prime(seven): continue #construct TEN and THREE for (t,h,r) in permutations(digits.difference(set((s,e,v,n))),3): if t == '0': continue ten = int(t + e + n) if not is_square(ten): continue three = int(t + h + r + e + e) if is_prime(three) : continue #construct FOUR for (f,o,u) in permutations(digits.difference(set((s,e,v,n,t,h,r))),3): if f == '0' : continue four = int(f + o + u + r) if not is_prime(four): continue ruof = int(r + u + o + f) if not is_prime(ruof): continue #test sum SEVEN - FOUR = THREE if seven - four != three: continue print('FOUR = {}, TEN = {}, SEVEN ={}, THREE = {}' \ .format(four, ten, seven, three)) # FOUR = 3407, TEN = 529, SEVEN =62129, THREE = 58722The following run file gives the arguments to the [[
SubstitutedExpression()]] solver in the enigma.py library necessary to solve this puzzle.It can be run as follows:
Execution time is 81ms.