Sunday Times Teaser 2606 – Cue for a Queue

by Victor Bryant

Published: 2 September 2012 (link)

Alan, Brian, Colin, Dave and Ed have surnames Smith, Jones, Rogers, Mason and Hall, but not in that order. They form themselves into a queue. Brian is somewhere ahead of Mr Smith who is somewhere ahead of Ed. Similarly, Mr Jones is ahead of Colin who is ahead of Dave who is ahead of Mr Hall. Mr Mason’s two neighbours in the queue are Alan and Mr Rogers.

If I told you Alan’s surname it would now be possible to work our all their surnames and their positions in the queue.

What is Alan’s surname and what is his position in the queue?

2 Comments Leave one →

Brian Gladman permalink

Here is my solution:


from itertools import permutations, product
from collections import defaultdict

first = ('Alan', 'Brian', 'Colin', 'Dave', 'Ed')
last = ('Smith', 'Jones', 'Rogers', 'Mason', 'Hall')
pos = ('first', 'second', 'third', 'fourth', 'fifth')

# dictionary to map Alan's last name to (multiple) queues
an_qs = defaultdict(set)

# position the first and last names in the queue
for f_name, l_name in product(permutations(first), permutations(last)):

  # we are told that they are not in the initial order
  if set(zip(f_name, l_name)) != set(zip(first, last)):
  
    # check the two orders given
    if (f_name.index('Brian') < l_name.index('Smith') < f_name.index('Ed')
        and l_name.index('Jones') < f_name.index('Colin') 
                              < f_name.index('Dave') < l_name.index('Hall')):
  
      # check Mr Mason's neighbours
      t = l_name.index('Mason')
      if abs(t - f_name.index('Alan')) == abs(t - l_name.index('Rogers')) == 1:
  
        # now form the queue
        queue = tuple(f + ' ' + l for f, l in zip(f_name, l_name))
        # accumulate the set of queues for Alan's last name
        an_qs[l_name[f_name.index('Alan')]].add((queue,))

# find a mapping from Alan's last name to a unique queue
for name in an_qs:
  if len(an_qs[name]) == 1:
    full_names, = an_qs[name]
    for i, full_name in enumerate(full_names[0]):
      if 'Alan' in full_name:
        print(f'{full_name} is {pos[i]} in the queue: {full_names[0]}')

from itertools import permutations, product

from collections import defaultdict

first = ('Alan', 'Brian', 'Colin', 'Dave', 'Ed')

last = ('Smith', 'Jones', 'Rogers', 'Mason', 'Hall')

pos = ('first', 'second', 'third', 'fourth', 'fifth')

# dictionary to map Alan's last name to (multiple) queues

an_qs = defaultdict(set)

# position the first and last names in the queue

for f_name, l_name in product(permutations(first), permutations(last)):

# we are told that they are not in the initial order

if set(zip(f_name, l_name)) != set(zip(first, last)):

# check the two orders given

if (f_name.index('Brian') < l_name.index('Smith') < f_name.index('Ed')

and l_name.index('Jones') < f_name.index('Colin')

< f_name.index('Dave') < l_name.index('Hall')):

# check Mr Mason's neighbours

t = l_name.index('Mason')

if abs(t - f_name.index('Alan')) == abs(t - l_name.index('Rogers')) == 1:

# now form the queue

queue = tuple(f + ' ' + l for f, l in zip(f_name, l_name))

# accumulate the set of queues for Alan's last name

an_qs[l_name[f_name.index('Alan')]].add((queue,))

# find a mapping from Alan's last name to a unique queue

for name in an_qs:

if len(an_qs[name]) == 1:

full_names, = an_qs[name]

for i, full_name in enumerate(full_names[0]):

if 'Alan' in full_name:

print(f'{full_name} is {pos[i]} in the queue: {full_names[0]}')

arthurvause permalink

Here’s another one (leaving the identification of Alan’s unique surname for visual inspection from 5 candidates)


from itertools import permutations as perm

candidates = [ ({'A':pf[0],'B':pf[1],'C':pf[2],'D':pf[3],'E':pf[4]},
                {'S':ps[0],'J':ps[1],'R':ps[2],'M':ps[3],'H':ps[4]}
                )
               for pf in perm(range(1,6))
               for ps in perm(range(1,6))]
    

candidates2 = [x for x in candidates
               if x[0]['B'] < x[1]['S'] < x[0]['E']
               and x[1]['J'] < x[0]['C'] < x[0]['D'] < x[1]['H']
               and abs(x[1]['M']-x[0]['A'])==1
               and abs(x[1]['M']-x[1]['R'])==1
               and x[0]['A']<>x[1]['R']
               and not (    x[0]['A']==x[1]['S']
                        and x[0]['B']==x[1]['J']
                        and x[0]['C']==x[1]['R']
                        and x[0]['D']==x[1]['M']
                        and x[0]['E']==x[1]['H']  )
               ]

for x in candidates2:
    order=[]
    for s in x[1]:
        for f in x[0]:
          if x[1][s]==x[0][f]:
            order.append( str(x[1][s])+f+s)
    order.sort()
    print order

from itertools import permutations as perm

candidates = [ ({'A':pf[0],'B':pf[1],'C':pf[2],'D':pf[3],'E':pf[4]},

{'S':ps[0],'J':ps[1],'R':ps[2],'M':ps[3],'H':ps[4]}

)

for pf in perm(range(1,6))

for ps in perm(range(1,6))]

candidates2 = [x for x in candidates

if x[0]['B'] < x[1]['S'] < x[0]['E']

and x[1]['J'] < x[0]['C'] < x[0]['D'] < x[1]['H']

and abs(x[1]['M']-x[0]['A'])==1

and abs(x[1]['M']-x[1]['R'])==1

and x[0]['A']<>x[1]['R']

and not ( x[0]['A']==x[1]['S']

and x[0]['B']==x[1]['J']

and x[0]['C']==x[1]['R']

and x[0]['D']==x[1]['M']

and x[0]['E']==x[1]['H'] )

]

for x in candidates2:

order=[]

for s in x[1]:

for f in x[0]:

if x[1][s]==x[0][f]:

order.append( str(x[1][s])+f+s)

order.sort()

print order

Sunday Times Teaser 2606 – Cue for a Queue

by Victor Bryant

Published: 2 September 2012 (link)

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