Sunday Times Teaser 2758 – Square Dance

by Angela Newing

Published: 2 August 2015 (link)

Dancers numbered from 1 to 9 were about to perform a square dance: five were dressed in red and the rest in blue. They stood in a 3-by-3 array with all three dancers in the first row wearing red and all three in another row wearing blue. Their numbers formed a magic square (ie, the sum of the three numbers in any row, column or diagonal was the same). One of the dancers in red looked around and noted that the sum of the numbers of the four other dancers in red was the same as the sum of the numbers of the four dancers in blue. One of the dancers in blue was number 8, …

… what were the numbers of the other three dancers in blue?

2 Comments Leave one →

Brian Gladman permalink


from itertools import permutations

# for the solutions
solns = set()

# generate 3 by 3 magic squares using the approach explained on the page at
# https://sites.google.com/site/sundaytimesteasers/home/magic-squares-3-x-3
for x in range(-4, 5):
  for y in range(-4, x):
    msq = [   5 - x,    5 + x + y,    5 - y,
            5 + x - y,      5,      5 - x + y,
              5 + y,    5 - x - y,    5 + x    ]
    # only generate magic squares with the values 1..9
    if set(msq) == set(range(1, 10)):
      # choose the blue that is in a line with two reds
      for b in msq[3:]:        
        # the initial index for this mixed colour row
        p = 3 * (msq.index(b) // 3)
        # collect the numbers for red and blue dancers
        red = msq[:3] + [msq[i] for i in range(p, p + 3) if msq[i] != b]
        blue = msq[9 - p : 12 - p] + [b]        
        # 8 is in blue;  sum(red) - red_observer = sum(blue), hence:
        # red_observer = sum(red) - sum(blue) = (30 - b) - (15 + b) = 15 - 2.b
        if 8 in blue and 15 - 2 * b in red:
          solns.update([(tuple(sorted(red)), tuple(sorted(blue)), tuple(msq))])

# output the results
for red, blue, msq in solns:
  fs = 'Answer = {}, Red = {}, Blue = {}nSquare = {}'   
  print(fs.format(tuple(b for b in blue if b != 8), red, blue, msq))

from itertools import permutations

# for the solutions

solns = set()

# generate 3 by 3 magic squares using the approach explained on the page at

# https://sites.google.com/site/sundaytimesteasers/home/magic-squares-3-x-3

for x in range(-4, 5):

for y in range(-4, x):

msq = [ 5 - x, 5 + x + y, 5 - y,

5 + x - y, 5, 5 - x + y,

5 + y, 5 - x - y, 5 + x ]

# only generate magic squares with the values 1..9

if set(msq) == set(range(1, 10)):

# choose the blue that is in a line with two reds

for b in msq[3:]:

# the initial index for this mixed colour row

p = 3 * (msq.index(b) // 3)

# collect the numbers for red and blue dancers

red = msq[:3] + [msq[i] for i in range(p, p + 3) if msq[i] != b]

blue = msq[9 - p : 12 - p] + [b]

# 8 is in blue; sum(red) - red_observer = sum(blue), hence:

# red_observer = sum(red) - sum(blue) = (30 - b) - (15 + b) = 15 - 2.b

if 8 in blue and 15 - 2 * b in red:

solns.update([(tuple(sorted(red)), tuple(sorted(blue)), tuple(msq))])

# output the results

for red, blue, msq in solns:

fs = 'Answer = {}, Red = {}, Blue = {}nSquare = {}'

print(fs.format(tuple(b for b in blue if b != 8), red, blue, msq))

Jim Randell permalink

At first I thought “the dancers around them” meant only those adjacent, but you can’t get a solution that way.

So I went with: “One dancer wearing red noticed that the sum of the numbers on the other dancers wearing red was the same as the sum of the numbers on the dancers wearing blue”, and that gives a unique solution. And the code is simpler.


# consider the square:
#
#   a b c
#   d e f
#   g h i
#
# where each latter stands for a number from 1 to 9
#
# the magic sum is: s = (1 + 2 + ... + 9) / 3 = 15
#
# and e = s / 3 = 5

from itertools import permutations, combinations
from enigma import irange, printf

# the numbers
numbers = set(irange(1, 9))

# total sum, magic sum, centre square
t = sum(numbers)
s = t // 3
e = s // 3

# generate magic squares
# 8 is a blue, so can't be in the first row
for (a, b) in permutations(numbers.difference((e, 8)), 2):
  c = s - (a + b)
  if c == 8: continue

  # fill out the rest of the square
  i = s - (a + e)
  f = s - (c + i)
  d = s - (e + f)
  g = s - (a + d)
  h = s - (g + i)

  # check it uses the numbers 1 - 9
  if numbers.difference([a, b, c, d, e, f, g, h, i]): continue

  # choose the row with mixed colours
  for row in ((d, e, f), (g, h, i)):
    # and choose two from that row to be red
    for red in combinations(row, 2):
      if 8 in red: continue
      # so the reds are...
      reds = set((a, b, c) + red)
      # and if the observant red dancer is x...
      # we need 2(sum(reds) - x) = (t - x)
      x = 2 * sum(reds) - t
      if x not in reds: continue
      # so the blues are...
      blues = numbers.difference(reds)
      printf("{a} {b} {c} / {d} {e} {f} / {g} {h} {i}, reds={reds}, x={x}, blues={blues}")

# consider the square:

# a b c

# d e f

# g h i

# where each latter stands for a number from 1 to 9

# the magic sum is: s = (1 + 2 + ... + 9) / 3 = 15

# and e = s / 3 = 5

from itertools import permutations, combinations

from enigma import irange, printf

# the numbers

numbers = set(irange(1, 9))

# total sum, magic sum, centre square

t = sum(numbers)

s = t // 3

e = s // 3

# generate magic squares

# 8 is a blue, so can't be in the first row

for (a, b) in permutations(numbers.difference((e, 8)), 2):

c = s - (a + b)

if c == 8: continue

# fill out the rest of the square

i = s - (a + e)

f = s - (c + i)

d = s - (e + f)

g = s - (a + d)

h = s - (g + i)

# check it uses the numbers 1 - 9

if numbers.difference([a, b, c, d, e, f, g, h, i]): continue

# choose the row with mixed colours

for row in ((d, e, f), (g, h, i)):

# and choose two from that row to be red

for red in combinations(row, 2):

if 8 in red: continue

# so the reds are...

reds = set((a, b, c) + red)

# and if the observant red dancer is x...

# we need 2(sum(reds) - x) = (t - x)

x = 2 * sum(reds) - t

if x not in reds: continue

# so the blues are...

blues = numbers.difference(reds)

printf("{a} {b} {c} / {d} {e} {f} / {g} {h} {i}, reds={reds}, x={x}, blues={blues}")

Sunday Times Teaser 2758 – Square Dance

by Angela Newing

Published: 2 August 2015 (link)

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