Sunday Times Teaser 2506 – No Title

by Quentin Fowler

Published October 4 2010 (link)

Three brothers share a house in a university city. Each is capable of undertaking two of the four domestic chores required. For instance, only one knows how to vacuum and only one can dust. They have recruited two friends who can do three chores each, so now any chore can be done by three people. Leo does not wash up; in fact, only one person can both wash up and do the laundry. There is one job both Leo and Neil can do, while Keith and Neil can do two jobs together. John can vacuum; Mike cannot. Neil is one of the brothers.

Who are the other two brothers, and who does the dusting?

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Brian Gladman permalink


from itertools import combinations, product
from collections import defaultdict

# dictionary of boy's names indexed on their first letter
boys = {c[0]:c for c in 'John Keith Leo Mike Neil'.split()}
# dictionary of chore names indexed on their first letter
chores = {c[0]:c for c in 'Dust Launder Vacuum Wash'.split()}

# a tuple of pairs of chores
ch_p = tuple(''.join(x) for x in combinations(chores, 2))
# a tuple of triples of chores
ch_t = tuple(''.join(x) for x in combinations(chores, 3))

# consider possible combinations for the brothers
for a, b, c in combinations(boys, 3):
  # the friends
  d, e = sorted(set(boys).difference((a, b, c)))
  # Neil is one of the brothers
  if 'N' not in {a, b, c}:
    continue
  
  # their brothers possible pairs of chores
  for ca, cb, cc in product(ch_p, repeat=3):
    # map the boys to their chores
    b2c = dict(zip((a, b, c), (ca, cb, cc)))

    # only one can vacuum and only one can dust
    t = ''.join(b2c.values())    
    if not (t.count('V') == t.count('D') == 1):
      continue
    
    # consider the possible triples of the friends chores 
    for cd, ce in product(ch_t, repeat=2):
      #  update the map of the boys chores
      b2c.update(zip((d, e), (cd, ce)))
      
      # create a map from chores to the boys who can do them
      c2b = defaultdict(set)
      for boy in b2c.keys():
        for chore in b2c[boy]:
          c2b[chore].add(boy)
  
      # all boys can do three chores
      if not all(len(c2b[ch]) >= 3 for ch in c2b.keys()):
        continue
      # Leo does not wash up
      if 'W' in b2c['L']:
        continue
      # only one person can wash and launder  
      wl_cnt = sum(1 for b in b2c.keys() if {'W', 'L'} <= set(b2c[b]))
      if wl_cnt != 1:
        continue
      # Leo and Neil can share one chore
      if len(set(b2c['L']).intersection(b2c['N'])) != 1:
        continue
      # Keith and Neil can share two chores   
      if len(set(b2c['K']).intersection(b2c['N'])) != 2:
        continue
      # John can vacuum but Mike can't
      if 'V' not in b2c['J'] or 'V' in b2c['M']:
        continue
      for i, boy in enumerate((a, b, c, d, e)):
        t = boys[boy] + (' (brother):' if i < 3 else ' (friend):')
        print(f"{t:15} {', '.join(chores[c] for c in b2c[boy])}")

from itertools import combinations, product

from collections import defaultdict

# dictionary of boy's names indexed on their first letter

boys = {c[0]:c for c in 'John Keith Leo Mike Neil'.split()}

# dictionary of chore names indexed on their first letter

chores = {c[0]:c for c in 'Dust Launder Vacuum Wash'.split()}

# a tuple of pairs of chores

ch_p = tuple(''.join(x) for x in combinations(chores, 2))

# a tuple of triples of chores

ch_t = tuple(''.join(x) for x in combinations(chores, 3))

# consider possible combinations for the brothers

for a, b, c in combinations(boys, 3):

# the friends

d, e = sorted(set(boys).difference((a, b, c)))

# Neil is one of the brothers

if 'N' not in {a, b, c}:

continue

# their brothers possible pairs of chores

for ca, cb, cc in product(ch_p, repeat=3):

# map the boys to their chores

b2c = dict(zip((a, b, c), (ca, cb, cc)))

# only one can vacuum and only one can dust

t = ''.join(b2c.values())

if not (t.count('V') == t.count('D') == 1):

continue

# consider the possible triples of the friends chores

for cd, ce in product(ch_t, repeat=2):

# update the map of the boys chores

b2c.update(zip((d, e), (cd, ce)))

# create a map from chores to the boys who can do them

c2b = defaultdict(set)

for boy in b2c.keys():

for chore in b2c[boy]:

c2b[chore].add(boy)

# all boys can do three chores

if not all(len(c2b[ch]) >= 3 for ch in c2b.keys()):

continue

# Leo does not wash up

if 'W' in b2c['L']:

continue

# only one person can wash and launder

wl_cnt = sum(1 for b in b2c.keys() if {'W', 'L'} <= set(b2c[b]))

if wl_cnt != 1:

continue

# Leo and Neil can share one chore

if len(set(b2c['L']).intersection(b2c['N'])) != 1:

continue

# Keith and Neil can share two chores

if len(set(b2c['K']).intersection(b2c['N'])) != 2:

continue

# John can vacuum but Mike can't

if 'V' not in b2c['J'] or 'V' in b2c['M']:

continue

for i, boy in enumerate((a, b, c, d, e)):

t = boys[boy] + (' (brother):' if i < 3 else ' (friend):')

print(f"{t:15} {', '.join(chores[c] for c in b2c[boy])}")

Sunday Times Teaser 2506 – No Title

by Quentin Fowler

Published October 4 2010 (link)

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