Sunday Times Teaser 2892 – Political Pairings

by Graham Smithers

Published February 25 2018 (link)

A debating society arranged for a series of eight political debates between different Labour and Conservative politicians, each time with a different presenter.

The presenters were: MARR, NEIL, NEWMAN, PARKINSON, PAXMAN, POPPLEWELL, ROBINSON and SNOW.

The Labour politicians were: ABBOTT, ABRAHAMS, CHAKRABARTI, CORBYN, EAGLE, LEWIS, STARMER and WATSON.

The Conservative politicians were: BRADLEY, GRAYLING, GREEN, HAMMOND, LEADSOM, LIDINGTON, MCLOUGHLIN and MUNDELL.

For each debate there were 2 letters in common for any pairing from presenter, Labour politician and Conservative politician. The Labour politicians are listed alphabetically. What is the corresponding order for the Conservative politicians?

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Brian Gladman permalink

A solution using my Python solver for this type of puzzle (available here: grouping.py).


from grouping import groups, output_groups, shared_letters

pre = ('MARR',  'NEIL',  'NEWMAN',  'PARKINSON',  'PAXMAN', 
       'POPPLEWELL', 'ROBINSON', 'SNOW')

lab = ('ABBOTT', 'ABRAHAMS', 'CHAKRABARTI',  'CORBYN',  'EAGLE',  
       'LEWIS',  'STARMER', 'WATSON')

con = ('BRADLEY',  'GRAYLING',  'GREEN',  'HAMMOND',  'LEADSOM', 
       'LIDINGTON', 'MCLOUGHLIN', 'MUNDELL')

for s in groups([lab, con, pre], shared_letters(2)):
  f, sp, l = (', '.join(g[1].title() for g in s)).rpartition(', ')
  print(' and '.join((f, l)))
  print()
  for g in sorted(s, key=lambda s:s[2]):
    print(f'({g[2].title()}, {g[1].title()} and {g[0].title()})')

from grouping import groups, output_groups, shared_letters

pre = ('MARR', 'NEIL', 'NEWMAN', 'PARKINSON', 'PAXMAN',

'POPPLEWELL', 'ROBINSON', 'SNOW')

lab = ('ABBOTT', 'ABRAHAMS', 'CHAKRABARTI', 'CORBYN', 'EAGLE',

'LEWIS', 'STARMER', 'WATSON')

con = ('BRADLEY', 'GRAYLING', 'GREEN', 'HAMMOND', 'LEADSOM',

'LIDINGTON', 'MCLOUGHLIN', 'MUNDELL')

for s in groups([lab, con, pre], shared_letters(2)):

f, sp, l = (', '.join(g[1].title() for g in s)).rpartition(', ')

print(' and '.join((f, l)))

print()

for g in sorted(s, key=lambda s:s[2]):

print(f'({g[2].title()}, {g[1].title()} and {g[0].title()})')

Sunday Times Teaser 2892 – Political Pairings

by Graham Smithers

Published February 25 2018 (link)

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