Sunday Times Teaser 2614 – The End of Time

by Victor Bryant

Published: 28 October 2012 (link)

A B C D E
F G H I J
K L M N O
P Q R S T
U V W X Y

In the table the letters represent the numbers 1 to 25 in some order. In each row, each column and each main diagonal the sum of the five numbers is the same. If you list all the letters in increasing numerical order then somewhere in the list you will get … , S, U, N, D, A, Y, T, I , M, … with E coming later.

Which letters are equal to 11, 9, 7, 3, 23 and 22?

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Brian Gladman permalink

Here is my version


from itertools import permutations

# Given the value of S, we can easily find another ten values:
#
#   S+4    B      C     S+3     E
#    F   48-4S    H     S+7     J
#    K     L     S+8    S+2     O
#    P     Q      R      S     S+6
#   S+1    V      W    53-4S   S+5

nbr_set = frozenset(range(1, 26))

# start with the value of S (also gives U, N, D, A, Y, T, I, and M)
# note X = 53 - 4.S < 26 -> S > 6.75 and G = 48 - 4.S > 0 -> S < 12
for s in range(7, 12):
  g, x = 48 - 4 * s, 53 - 4 * s
  # G and X must not be in the range for letters S .. M
  if s <= g <= s + 8 or s <= x <= s + 8:
    continue

  # now find the remaining numeric values
  rest = nbr_set - set(s + z for z in range(9))

  # we know that E is greater than S + 8
  for e in rest:

    # we can now compute q and test that it is valid
    q = 49 - 3 * s - e
    if e < s + 9 or q == e or q not in rest:
      continue

    # allocate values in S .. M
    u, n, d, a, y, t, i, m = (s + z for z in range(1, 9))

    # now treat B, V, R and F as independent variables
    for b, v, r, f in permutations(rest - set((e,)), 4):

      # compute the remaining values
      l, c = 65 - (b + g + q + v), 65 - (a + b + d + e)
      p, w = 65 - (q + r + s + t), 65 - (u + v + x + y)
      h, k = 65 - (c + m + r + w), 65 - (a + f + p + u)
      j, o = 65 - (f + g + h + i), 65 - (k + l + m + n)

      # compile the complete set of values
      t_ltr = ( a, b, c, d, e, f, g, h, i, j, k, l, m,
                n, o, p, q, r, s, t, u, v, w, x, y )
      # and test that it contains the numbers 1 .. 25
      if set(t_ltr) == nbr_set:
        print('Answer =', ''.join(chr(ord('A') + t_ltr.index(z))
                  for z  in (11, 9, 7, 3, 23, 22)))

from itertools import permutations

# Given the value of S, we can easily find another ten values:

# S+4 B C S+3 E

# F 48-4S H S+7 J

# K L S+8 S+2 O

# P Q R S S+6

# S+1 V W 53-4S S+5

nbr_set = frozenset(range(1, 26))

# start with the value of S (also gives U, N, D, A, Y, T, I, and M)

# note X = 53 - 4.S < 26 -> S > 6.75 and G = 48 - 4.S > 0 -> S < 12

for s in range(7, 12):

g, x = 48 - 4 * s, 53 - 4 * s

# G and X must not be in the range for letters S .. M

if s <= g <= s + 8 or s <= x <= s + 8:

continue

# now find the remaining numeric values

rest = nbr_set - set(s + z for z in range(9))

# we know that E is greater than S + 8

for e in rest:

# we can now compute q and test that it is valid

q = 49 - 3 * s - e

if e < s + 9 or q == e or q not in rest:

continue

# allocate values in S .. M

u, n, d, a, y, t, i, m = (s + z for z in range(1, 9))

# now treat B, V, R and F as independent variables

for b, v, r, f in permutations(rest - set((e,)), 4):

# compute the remaining values

l, c = 65 - (b + g + q + v), 65 - (a + b + d + e)

p, w = 65 - (q + r + s + t), 65 - (u + v + x + y)

h, k = 65 - (c + m + r + w), 65 - (a + f + p + u)

j, o = 65 - (f + g + h + i), 65 - (k + l + m + n)

# compile the complete set of values

t_ltr = ( a, b, c, d, e, f, g, h, i, j, k, l, m,

n, o, p, q, r, s, t, u, v, w, x, y )

# and test that it contains the numbers 1 .. 25

if set(t_ltr) == nbr_set:

print('Answer =', ''.join(chr(ord('A') + t_ltr.index(z))

for z in (11, 9, 7, 3, 23, 22)))

Sunday Times Teaser 2614 – The End of Time

by Victor Bryant

Published: 28 October 2012 (link)

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