Sunday Times Teaser 2847 – Easter Parade

by Andrew Skidmore

Published April 16 2017 (link)

Following the success of last year’s Easter parade our village is going to make it an annual event. To celebrate today’s parade I have taken three numbers and I have consistently replaced digits by letters to give

ANNUAL EASTER PARADE

In fact the third number is the sum of the other two.

What is the number PARADE?

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Brian Gladman permalink


from itertools import permutations
from functools import reduce

dgts2nbr = lambda seq: reduce(lambda x, y: 10 * x + y, seq)

# choose digits for the leading letters
for A, E, P in permutations(range(1, 10), 3):
  # from the most significant column of the sum
  if A + E <= P:
    # choose digits for the other letters in the addends
    s1 = set(range(10)).difference([A, E, P])
    for D, R, S, T in permutations(s1, 4):
      easter = dgts2nbr((E, A, S, T, E, R))
      parade = dgts2nbr((P, A, R, A, D, E))
      annual = parade - easter
      # check that any already assigned letters are correct
      # and that all the letters represent different digits
      try:
        a, N, n, U, _a, L = (int(x) for x in str(annual))
      except ValueError:
        continue
      s2 = s1.difference([D, L, N, R, S, T, U])
      if _a == a == A and n == N and not s2:
        print(f'{annual} + {easter} == {parade}')

from itertools import permutations

from functools import reduce

dgts2nbr = lambda seq: reduce(lambda x, y: 10 * x + y, seq)

# choose digits for the leading letters

for A, E, P in permutations(range(1, 10), 3):

# from the most significant column of the sum

if A + E <= P:

# choose digits for the other letters in the addends

s1 = set(range(10)).difference([A, E, P])

for D, R, S, T in permutations(s1, 4):

easter = dgts2nbr((E, A, S, T, E, R))

parade = dgts2nbr((P, A, R, A, D, E))

annual = parade - easter

# check that any already assigned letters are correct

# and that all the letters represent different digits

try:

a, N, n, U, _a, L = (int(x) for x in str(annual))

except ValueError:

continue

s2 = s1.difference([D, L, N, R, S, T, U])

if _a == a == A and n == N and not s2:

print(f'{annual} + {easter} == {parade}')

The solution can also be found using my AlphaSum alphametic solver:


from alphasum import AlphaSum

AlphaSum(['ANNUAL', 'EASTER'], 'PARADE').go()

from alphasum import AlphaSum

AlphaSum(['ANNUAL', 'EASTER'], 'PARADE').go()

or one of Jim Randell’s solvers in his enigma.py library.

Sunday Times Teaser 2847 – Easter Parade

by Andrew Skidmore

Published April 16 2017 (link)

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