Sunday Times Teaser 2182 – Team Bonding

by Ahmet Cetinbudaklar

The 21 members of a football squad, who wear the numbers 1 to 21, were asked to attend a training session but some were missing as a result of injuries. They were asked to stand in a circle in such a way that, for any adjacent players, one of their numbers was a multiple of the other.

It was then noticed that this was the largest number of players from the squad for which this was possible. One player also noticed that the sum of the numbers of his immediate neighbours was equal to the lowest of the numbers among the missing players. A little further clockwise around the circle, another player found that the sum of the numbers of his immediate neighbours was equal to the highest number among the missing players. In no case did any player’s two immediate neighbours numbers add to 20.

Starting at player 1, list the numbers of players around the circle clockwise.

3 Comments Leave one →

brian gladman permalink


from collections import defaultdict

# list for each player number the possible adjacent player numbers
dadj = defaultdict(set)
for nbr in range(1, 22):
  # form all multiples less than 22
  for m in range(2 * nbr, 22, nbr):
    # adjacent numbers can be multiples or sub-multiples
    dadj[nbr].add(m)
    dadj[m].add(nbr)

max_seen = 0
def place_next(used, unused):
  global max_seen
  # try to add the next value still unused
  for nxt in unused:
    # check that it can be adjacent to the last one added
    if nxt in dadj[used[-1]]:
      place_next(used + [nxt], unused - set((nxt,)))
  else:
    # if we can't add any more see if we have a longer
    # circular list than we have seen previously
    ln = len(used)
    if ln > max_seen and used[0] in dadj[used[-1]]:
      # find the sums of immediate neighbours
      sums = [used[i - 1] + used[(i + 1) % ln] for i in range(1, ln)]
      # check that none sum to 20
      if any(x == 20 for x in sums):
        return
      # find any sums that match the min and max unused numbers
      try:
        i_min = sums.index(min(unused))
        i_max = sums.index(max(unused))
      except ValueError:
        # exit if there is no match
        return
      # print the circular list and increase the maximum length seen
      print(used)
      max_seen = ln

# start the list at player number 1
place_next([1], set(range(2, 22)))

from collections import defaultdict

# list for each player number the possible adjacent player numbers

dadj = defaultdict(set)

for nbr in range(1, 22):

# form all multiples less than 22

for m in range(2 * nbr, 22, nbr):

# adjacent numbers can be multiples or sub-multiples

dadj[nbr].add(m)

dadj[m].add(nbr)

max_seen = 0

def place_next(used, unused):

global max_seen

# try to add the next value still unused

for nxt in unused:

# check that it can be adjacent to the last one added

if nxt in dadj[used[-1]]:

place_next(used + [nxt], unused - set((nxt,)))

else:

# if we can't add any more see if we have a longer

# circular list than we have seen previously

ln = len(used)

if ln > max_seen and used[0] in dadj[used[-1]]:

# find the sums of immediate neighbours

sums = [used[i - 1] + used[(i + 1) % ln] for i in range(1, ln)]

# check that none sum to 20

if any(x == 20 for x in sums):

return

# find any sums that match the min and max unused numbers

try:

i_min = sums.index(min(unused))

i_max = sums.index(max(unused))

except ValueError:

# exit if there is no match

return

# print the circular list and increase the maximum length seen

print(used)

max_seen = ln

# start the list at player number 1

place_next([1], set(range(2, 22)))

ahmet cetinbudaklar permalink

Dear Brian

You are so kind to? remind me of my teaser with a programming solution, one which I shall keep as a souvenir.

Reply
Ogee permalink

Players with prime numbers > 10 must be injured. I quickly found a loop for the remaining players, with 2 neighbours summing to the largest injury, but my first attempt didn’t have neighbours summing to the smallest. But there’s only one way to achieve that, so the change was easy, just swapping 2 adjacent players. A similar change creates or removes a 20 summation. I’m fairly sure that my final loop is unique.

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Sunday Times Teaser 2182 – Team Bonding

by Ahmet Cetinbudaklar

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