Sunday Times Teaser 2859 – Palindromic

by Victor Bryant

Published July 9 2017 (link)

I have assigned to each letter of the alphabet a different number from 0 to 25. Therefore, for example, a three-letter word might stand for a number of three, four, five or six digits. In fact all the letters used in PALINDROMIC have even values. Furthermore, the number represented by this word contains no zeros and is indeed palindromic.

Please send in the number represented by PIN.

2 Comments Leave one →

Brian Gladman permalink


from itertools import permutations

# the values for the letters in PALINDROMIC
dv = {2, 4, 6, 8, 12, 14, 16, 18, 22, 24}

# map P to possible C values based on the fact that PALINDROMIC
# is a palindrome
p2c = {2:(12, 22), 4:(14, 24), 6:(16,), 8:(18,), 22:(2,), 24:(2,)}

# consider possible values for P and C
for P in p2c.keys():
  for C in p2c[P]:
    # permute the remaining numeric values for the remaining letters
    for p in permutations(dv.difference([P, C])):
      # create a dictionary mapping letters to their values
      d = dict(zip('ADILMNORPC', p + (P, C)))
      # find the numeric value of PALINDROMIC as a string
      pal = ''.join(str(d[c]) for c in 'PALINDROMIC')
      # ... and look for a result that is a palindrome
      if pal == pal[::-1]:
        pin = ''.join(str(d[c]) for c in 'PIN')
        t = ', '.join(str(d[c]) for c in 'PALINDROMIC')
        print(f'PIN = {pin} (PALINDROMIC = {t})')

from itertools import permutations

# the values for the letters in PALINDROMIC

dv = {2, 4, 6, 8, 12, 14, 16, 18, 22, 24}

# map P to possible C values based on the fact that PALINDROMIC

# is a palindrome

p2c = {2:(12, 22), 4:(14, 24), 6:(16,), 8:(18,), 22:(2,), 24:(2,)}

# consider possible values for P and C

for P in p2c.keys():

for C in p2c[P]:

# permute the remaining numeric values for the remaining letters

for p in permutations(dv.difference([P, C])):

# create a dictionary mapping letters to their values

d = dict(zip('ADILMNORPC', p + (P, C)))

# find the numeric value of PALINDROMIC as a string

pal = ''.join(str(d[c]) for c in 'PALINDROMIC')

# ... and look for a result that is a palindrome

if pal == pal[::-1]:

pin = ''.join(str(d[c]) for c in 'PIN')

t = ', '.join(str(d[c]) for c in 'PALINDROMIC')

print(f'PIN = {pin} (PALINDROMIC = {t})')

geoffrounce permalink


from itertools import permutations

nums = (24, 22, 18, 16, 14, 12, 8, 6, 4, 2)

for q in permutations(nums):
    p, a, l, i, n, d, r, o, m, c = (str(x) for x in q)
    pal = p + a + l + i + n + d + r + o + m + i + c
    if pal == pal[::-1]:
        print(f'PIN = {p + i + n}, PALINDROMIC = {pal}')

from itertools import permutations

nums = (24, 22, 18, 16, 14, 12, 8, 6, 4, 2)

for q in permutations(nums):

p, a, l, i, n, d, r, o, m, c = (str(x) for x in q)

pal = p + a + l + i + n + d + r + o + m + i + c

if pal == pal[::-1]:

print(f'PIN = {p + i + n}, PALINDROMIC = {pal}')

Sunday Times Teaser 2859 – Palindromic

by Victor Bryant

Published July 9 2017 (link)

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