New Scientist Enigma No 526 – Ewe to Move

by Keith Austin

From Issue #1678, 19th August 1989

Each of the four fields at Sunny Meadows Farm contains some sheep and some cows. On the gate of each field is hung a sign saying what fraction of the animals in that field are sheep. The signs are 1/4, 1/3, 1/2, 2/3.

Farmer Gillian explained that if she exchanged the signs on two of the fields then, by simply moving some sheep from one of the two fields to the other, she could return to a situation where each sign again correctly indicated the fraction of the animals that were sheep in that field.

As I walked round I noticed that the total number of animals on the farm was between 300 and 350.

How many sheep, and how many cows, were on the farm?

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Brian Gladman permalink

The ratios (sheep / cows) in each field are as follows:\[1/3, 1/2, 1/1, 2/1\] The fractions of sheep to move between any two fields in order to exchange their two ratios is given by \[\begin{array}{|r|rrrr|} \hline \text{from} & & \text{to:} & 1/3 & 1/2 & 1/1 & 2/1\\ \hline 1/3 & & & & 1/6 & 2/3 & 5/3 \\ \hline 1/2 & & & & & 1/2 & 3/2 \\ \hline 1/1 & & & & & & 1/1 \\ \hline \end{array}\]Since we can only move whole sheep we need to scale this up to give:

\[\begin{array}{|l|rrrr|r|} \hline \text{field (sheep / cows)} & 1/3 & 1/2 & 1/1 & 2/1 & \text{totals} \\ \hline \text{sheep} & 2 & 3 & 6 & 12 & 23\\ \hline \text{cows} & 6 & 6 & 6 & 6 & 24 \\ \hline \text{animals} & 8 & 9 & 12 & 18 & 47 \\ \hline \end{array}\]

To obtain a total of between 300 and 350 animals we have to scale these numbers up by a factor of seven to obtain the initial numbers of sheep and cows in the four fileds and the totals:

\[\begin{array}{|l|rrrr|r|} \hline \text{field (sheep / animals)} & 1/4 & 1/3 & 1/2 & 2/3 & \text{totals} \\ \hline \text{sheep} & 14 & 21 & 42 & 84 & 161\\ \hline \text{cows} & 42 & 42 & 42 & 42 & 168 \\ \hline \text{animals} & 56 & 63 & 84 & 126 & 329 \\ \hline \end{array}\]

Or we can solve it with a Python program as follows:


from itertools import combinations

# check that we can swap sheep between all pairs of
# fields while keeping the sheep/cow ratios unchanged

def check_swap(f):
  # check all pairs of fields
  for f1, f2 in combinations(f, 2):
    # fraction (t / b), sheep, cows for two fields 
    t1, b1, s1, c1 = f1
    t2, b2, s2, c2 = f2
    # find the number of sheep needed to swap ratios
    x, r = divmod(s1 * b2 - (s1 + c1) * t2, b2 - t2)
    y, s = divmod(s2 * b1 - (s2 + c2) * t1, b1 - t1)
    # which must be integer, equal in magnitude and
    # of opposite sign
    if r or s or x + y:
      return False
  return True    
    
maxv = 51
f = [0] * 4
for s1 in range(1, maxv):
  # 1/ 4 are sheep
  f[0] = (1, 4, s1, 3 * s1)
  
  for s2 in range(1, maxv):
    # 1 / 3 are sheep 
    f[1] = (1, 3, s2, 2 * s2)
  
    # check that we can swap sheep while keeping
    # the ratios unchanged in the first two fields
    if not check_swap(f[:2]):
      continue
    
    for s3 in range(1, maxv):
      # 1 / 2 are sheep
      f[2] = (1, 2, s3, s3)
  
      # check that we can swap sheep while keeping
      # the ratios unchanged in the first three fields
      if not check_swap(f[:3]):
        continue
      
      for c4 in range(1, maxv):
        # 2 / 3 are sheep 
        f[3] = (2, 3, 2 * c4, c4)
        # check that we have an admissable number of animals
        ss = s1 + s2 + s3 + 2 * c4
        cc = 3 * s1 + 2 * s2 + s3 + c4
        if 300 <= 4 * s1 + 3 * s2 + 2 * s3 + 3 * c4 <= 350:
          # and we can swap sheep while keeping ratios unchanged
          if check_swap(f):
            ss = s1 + s2 + s3 + 2 * c4
            cc = 3 * s1 + 2 * s2 + s3 + c4
            print(ss, cc, [(s1, 3 * s1), (s2, 2 * s2), (s3, s3), (2 * c4, c4)])

from itertools import combinations

# check that we can swap sheep between all pairs of

# fields while keeping the sheep/cow ratios unchanged

def check_swap(f):

# check all pairs of fields

for f1, f2 in combinations(f, 2):

# fraction (t / b), sheep, cows for two fields

t1, b1, s1, c1 = f1

t2, b2, s2, c2 = f2

# find the number of sheep needed to swap ratios

x, r = divmod(s1 * b2 - (s1 + c1) * t2, b2 - t2)

y, s = divmod(s2 * b1 - (s2 + c2) * t1, b1 - t1)

# which must be integer, equal in magnitude and

# of opposite sign

if r or s or x + y:

return False

return True

maxv = 51

f = [0] * 4

for s1 in range(1, maxv):

# 1/ 4 are sheep

f[0] = (1, 4, s1, 3 * s1)

for s2 in range(1, maxv):

# 1 / 3 are sheep

f[1] = (1, 3, s2, 2 * s2)

# check that we can swap sheep while keeping

# the ratios unchanged in the first two fields

if not check_swap(f[:2]):

continue

for s3 in range(1, maxv):

# 1 / 2 are sheep

f[2] = (1, 2, s3, s3)

# check that we can swap sheep while keeping

# the ratios unchanged in the first three fields

if not check_swap(f[:3]):

continue

for c4 in range(1, maxv):

# 2 / 3 are sheep

f[3] = (2, 3, 2 * c4, c4)

# check that we have an admissable number of animals

ss = s1 + s2 + s3 + 2 * c4

cc = 3 * s1 + 2 * s2 + s3 + c4

if 300 <= 4 * s1 + 3 * s2 + 2 * s3 + 3 * c4 <= 350:

# and we can swap sheep while keeping ratios unchanged

if check_swap(f):

ss = s1 + s2 + s3 + 2 * c4

cc = 3 * s1 + 2 * s2 + s3 + c4

print(ss, cc, [(s1, 3 * s1), (s2, 2 * s2), (s3, s3), (2 * c4, c4)])

New Scientist Enigma No 526 – Ewe to Move

by Keith Austin

From Issue #1678, 19th August 1989

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