Sunday Times Teaser 2534 – Fiftieth Anniversary

by Victor Bryant

Published: 17 April 2011 (link)

We have recently celebrated the 50th anniversary of the Teaser column. At the party for the setters they gave each letter of the alphabet a different number from 1 to 26 (eg they made A=7). Appropriately, this was done in such a way that, for each setter present, the values of the letters of their surname added up to 50. Angela Newing was there (so N+E+W+I+N+G=50), as were Nick MacKinnon and Hugh Bradley. Only two of Graham Smithers, Danny Roth, Andrew Skidmore, John Owen and Victor Bryant could make it.

Which two?

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Brian Gladman permalink


# This teaser will be too slow without some prior analysis.
#
# We have:
#
#    B + R + A + D + L + E + Y = 50
#    M + A + C + K + I + N + N + O + N = 50
#
# Adding these gives:
#
#    A + 2.N + sum(A, B, C, D, E, I, K, L, M, N, O, R, Y) = 100
#
# The minimum value of 13 different numbers is 91 so A + 2.N is
# at most 9. And we are told that A = 7 so N must be 1 and all
# the letters in the sum are in the range 1..13 inclusive.

from itertools import permutations
from functools import reduce
from sys import exit

# possible attendees at the anniversary
guests = ('Victor Bryant', 'John Owen', 'Andrew Skidmore', 
          'Graham Smithers', 'Danny Roth')

# values from the preliminary analysis
l2d ={ 'A':7, 'N':1 }

# permute the digits needed for MacKinnon and check its digit sum
for p1 in permutations(set(range(2, 14)).difference([7]), 5):
  l2d.update(zip('CIKMO', p1))
  if sum(l2d[c] for c in 'MACKINNON') != 50:
    continue
  
  # permute the new digits needed for Bradley and check its digit sum  
  for p2 in permutations(set(range(2, 14)).difference(p1 + (7,))):
    l2d.update(zip('BRDLEY', p2))
    if sum(l2d[c] for c in 'BRADLEY') != 50:
      continue
    
    # now find the digit values needed to make the digit sum 
    # for Newing equal to 50
    for g in range(14, 27):
      w = 50 - g - sum(l2d[c] for c in 'NEIN')
      if not 14 <= w <= 27:
        continue
      l2d.update(zip('GW', (g, w)))
      
      # at this point we can sum the letter values for some
      # possible guests and hence tell if they are present;
      # for those with surnames with unallocated letters we
      # we don't necessarily need to know their letter sums
      # since unallocated letters are at least 14 and this
      # may already mean that their sums must be above 50
      possible = list(guests)
      for g in guests:
        surname = g.split()[1].upper()
        
        # find the sum of already assigned letters
        val = 50 - sum(l2d[c] for c in surname if c in l2d)
        
        # find the unassigned letters
        ltrs = ''.join(c for c in surname if c not in l2d)
        
        # remove guests if their digit sums cannot be fifty  
        if not ltrs and val != 50 or 14 * len(ltrs) > val:
          possible.remove(g)
      
      # if only two guests are possible they must provide 
      # the solution
      if len(possible) == 2:
        print('{} and {}.'.format(*possible))
        exit(0)

# This teaser will be too slow without some prior analysis.

# We have:

# B + R + A + D + L + E + Y = 50

# M + A + C + K + I + N + N + O + N = 50

# Adding these gives:

# A + 2.N + sum(A, B, C, D, E, I, K, L, M, N, O, R, Y) = 100

# The minimum value of 13 different numbers is 91 so A + 2.N is

# at most 9. And we are told that A = 7 so N must be 1 and all

# the letters in the sum are in the range 1..13 inclusive.

from itertools import permutations

from functools import reduce

from sys import exit

# possible attendees at the anniversary

guests = ('Victor Bryant', 'John Owen', 'Andrew Skidmore',

'Graham Smithers', 'Danny Roth')

# values from the preliminary analysis

l2d ={ 'A':7, 'N':1 }

# permute the digits needed for MacKinnon and check its digit sum

for p1 in permutations(set(range(2, 14)).difference([7]), 5):

l2d.update(zip('CIKMO', p1))

if sum(l2d[c] for c in 'MACKINNON') != 50:

continue

# permute the new digits needed for Bradley and check its digit sum

for p2 in permutations(set(range(2, 14)).difference(p1 + (7,))):

l2d.update(zip('BRDLEY', p2))

if sum(l2d[c] for c in 'BRADLEY') != 50:

continue

# now find the digit values needed to make the digit sum

# for Newing equal to 50

for g in range(14, 27):

w = 50 - g - sum(l2d[c] for c in 'NEIN')

if not 14 <= w <= 27:

continue

l2d.update(zip('GW', (g, w)))

# at this point we can sum the letter values for some

# possible guests and hence tell if they are present;

# for those with surnames with unallocated letters we

# we don't necessarily need to know their letter sums

# since unallocated letters are at least 14 and this

# may already mean that their sums must be above 50

possible = list(guests)

for g in guests:

surname = g.split()[1].upper()

# find the sum of already assigned letters

val = 50 - sum(l2d[c] for c in surname if c in l2d)

# find the unassigned letters

ltrs = ''.join(c for c in surname if c not in l2d)

# remove guests if their digit sums cannot be fifty

if not ltrs and val != 50 or 14 * len(ltrs) > val:

possible.remove(g)

# if only two guests are possible they must provide

# the solution

if len(possible) == 2:

print('{} and {}.'.format(*possible))

exit(0)

Sunday Times Teaser 2534 – Fiftieth Anniversary

by Victor Bryant

Published: 17 April 2011 (link)

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