Programming Enigma Puzzles
From New Scientist #2758, 1st May 2010 [link] [link]
On three identical square cards I wrote 3 × 3 magic squares (where the sums of each row, column and major diagonal are equal) using the numbers 1 to 9 on the first, 10 to 18 on the second and 19 to 27 on the third. I formed them into a sandwich in some order. I then found that I could push a pin through the cards so that it passed through three multiples of three. Furthermore, writing the three numbers in order along the pin gave a five-figure perfect square. One number on the middle card was sandwiched between two perfect squares on the other cards.
Which number?
[enigma1593]
 | Frits on Enigma 773: Duodecimal |
 | Jim Randell on Enigma 773: Duodecimal |
 | Brian Gladman on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Frits on Enigma 773: Duodecimal |
| Jim Randell on Enigma 773: Duodecimal |
Enigmatic Code 
The following Python program runs in 39ms.
import itertools from enigma import irange, concat, is_square # generate all rotations/reflections of 3x3 magic square indices def magic(): s = [(7, 0, 5, 2, 4, 6, 3, 8, 1)] # initial magic square # rotations for i in irange(0, 2): s.append(tuple(s[-1][i] for i in (6, 3, 0, 7, 4, 1, 8, 5, 2))) # reflections for j in irange(0, 3): s.append(tuple(s[j][i] for i in (2, 1, 0, 5, 4, 3, 8, 7, 6))) return s squares = magic() # generate possible magic squares from a list of 9 consecutive numbers def generate(l): for s in squares: yield tuple(l[i] for i in s) # generate the possible cards for C1 in generate(irange(1, 9)): for C2 in generate(irange(10, 18)): for C3 in generate(irange(19, 27)): # generate the possible stacking order for the cards for (T, M, B) in itertools.permutations((C1, C2, C3)): # try the pin in each position for i in irange(0, 8): pin = (T[i], M[i], B[i]) # are they all multiples of 3? if any(n for n in pin if n % 3 > 0): continue # and does it make a 5-digit square? n = int(concat(*pin)) if not is_square(n): continue # find where the squares are on the top and bottom cards for j in irange(0, 8): if not(is_square(T[j]) and is_square(B[j])): continue printf("middle={m} [pin={pin} squares={s}]", m=M[j], s=(T[j], M[j], B[j])) # we only need to consider one orientation for one of the cards breakSolution: The number on the middle card is 7.