Sunday Times Teaser 2522

by Victor Bryant

Published: 23 January 2011 (link)

I have numbered the letters of the alphabet from 1 to 26, but not in alphabetical order, giving each letter its own unique value. So, for any word, I can work out its value by adding up the values of its letters. In this way, I have found some words with equal values. For example, the following nine words all have the same value:

I SAY ALL THESE HERE HAVE EQUAL VALUES TOO.

What are the values of Y, O, U, R, S?

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Brian Gladman permalink


# Since all the letters have different values, the minimum 
# for a sum of six is 21. So we must have 21 <= I <= 26.
#
# Since the minimum sum of 5 letters is 15, the value of any
# one of letters in VALUES + 15 equals I.  So the values of
# all these letters lie in the range 1 <= ltr <= I - 15
#
# A bit of elimination show that given I, A, S, E, U we can
# calculate the other values using:
#
#  L = (I - A) // 2
#  H = L + S + U
#  Q = L - E - U
#  Y = 2.L - S
#  V = 2.L - E - H
#  R = I - 2.E - H
#  T = I - 2.E - H - S
#  O = E + (S + H) // 2

from itertools import permutations

# try values for I (see analysis above)
for I in range(21, 27):
  
  # and values for A, E, S and U (see analysis above)
  for A, E, S, U in permutations(range(1, I - 14), 4):
  
    # now compute the dependent values  
    L, r = divmod(I - A, 2)
    if r:
      continue
    H = L + S + U
    O, r = divmod(2 * E + S + H, 2)
    if r:
      continue
    Q, R = L - E - U, I - 2 * E - H
    T, V, Y = R - S, 2 * L - E - H, 2 * L - S
    
    # check that all values are valid and all different
    t = (A, E, H, I, L, O, Q, R, S, T, U, V, Y)
    if all(0 < x < 27 for x in t) and len(set(t)) == 13:
      fs = 'Y = {}, O = {}, U = {}, R = {} and S = {}'
      print(fs.format(Y, O, U, R, S))

# Since all the letters have different values, the minimum

# for a sum of six is 21. So we must have 21 <= I <= 26.

# Since the minimum sum of 5 letters is 15, the value of any

# one of letters in VALUES + 15 equals I. So the values of

# all these letters lie in the range 1 <= ltr <= I - 15

# A bit of elimination show that given I, A, S, E, U we can

# calculate the other values using:

# L = (I - A) // 2

# H = L + S + U

# Q = L - E - U

# Y = 2.L - S

# V = 2.L - E - H

# R = I - 2.E - H

# T = I - 2.E - H - S

# O = E + (S + H) // 2

from itertools import permutations

# try values for I (see analysis above)

for I in range(21, 27):

# and values for A, E, S and U (see analysis above)

for A, E, S, U in permutations(range(1, I - 14), 4):

# now compute the dependent values

L, r = divmod(I - A, 2)

if r:

continue

H = L + S + U

O, r = divmod(2 * E + S + H, 2)

if r:

continue

Q, R = L - E - U, I - 2 * E - H

T, V, Y = R - S, 2 * L - E - H, 2 * L - S

# check that all values are valid and all different

t = (A, E, H, I, L, O, Q, R, S, T, U, V, Y)

if all(0 < x < 27 for x in t) and len(set(t)) == 13:

fs = 'Y = {}, O = {}, U = {}, R = {} and S = {}'

print(fs.format(Y, O, U, R, S))

Sunday Times Teaser 2522

by Victor Bryant

Published: 23 January 2011 (link)

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