Sunday Times Teaser 2685 – Not the Gold Cup

by Angela Newing

Five horses took part in a race, one falling at each of the first five fences. Dave rode Egg Nog. Wally rode Nig Nag. Bill’s horse fell at fence two. Dave lasted longer. Big Gun fell at fence three. The jockey in mauve lasted the longest. The jockey riding Dragon wore red. The horse ridden by the jockey in yellow fell before Long Gone. The horse ridden by the jockey in green fell one fence before the horse ridden by Chris. The horses ridden by Fred and his friend in blue did not fall at adjacent fences.

Which jockey wore yellow and which horse did he ride?

3 Comments Leave one →

brian gladman permalink

This one is pretty tedious to program. It is possible that this could be faster if the order of the permuations is changed. But I doubt that its worth the effort.


from itertools import permutations

jockeys = ( 'Dave', 'Wally', 'Bill', 'Chris', 'Fred' )
horses =  ( 'Egg Nog', 'Nig Nag', 'Big Gun', 'Dragon', 'Long Gone' )
colours = ( 'Blue', 'Green', 'Mauve', 'Red', 'Yellow' )
fences =  ( 1, 2, 3, 4, 5 )

# Dave rode Egg Nog and Wally rode Nig Nag (only permute others)
for hh in permutations(horses[2:]):
  h = horses[:2] + hh
  j2h = dict(zip(jockeys, h))
  
  # permute the colours
  for c in permutations(colours):
    h2c = dict(zip(h, c))

    # permute the fences at which horses fell
    for f in permutations(fences):
      
      # map jockeys to the fences at which they fell
      j2f = dict(zip(jockeys, f))
      
      # Bill fell at the second fence
      if j2f['Bill'] != 2:
        continue      

      # Dave lasted longer than Bill
      if j2f['Dave'] < 3:
        continue
      
      # map colours to fences 
      c2f = dict(zip(c, f))
      
      # Big Gun fell at the third fence
      if c2f[h2c['Big Gun']] != 3:
        continue

      # Mauve fell at the fifth fence
      if c2f['Mauve'] != 5:
        continue
      
      # Long Gone lasted longer than yellow
      if c2f[h2c['Long Gone']] <= c2f['Yellow']:
        continue
      
      # Chris fell one fence later than Green
      if j2f['Chris'] != c2f['Green'] + 1:
        continue
      
      # Fred (not Blue) and Blue did not fall at adjacent fences
      if h2c[j2h['Fred']] == 'Blue' or abs(j2f['Fred'] - c2f['Blue']) == 1:
        continue
      
      # Dragon was Red
      if h2c['Dragon'] != 'Red':
        continue
      
      f2j, t = dict(zip(f, jockeys)), colours[4]
      print('{} wore {} riding {}.'.format(f2j[c2f[t]], t, j2h[f2j[c2f[t]]]))

from itertools import permutations

jockeys = ( 'Dave', 'Wally', 'Bill', 'Chris', 'Fred' )

horses = ( 'Egg Nog', 'Nig Nag', 'Big Gun', 'Dragon', 'Long Gone' )

colours = ( 'Blue', 'Green', 'Mauve', 'Red', 'Yellow' )

fences = ( 1, 2, 3, 4, 5 )

# Dave rode Egg Nog and Wally rode Nig Nag (only permute others)

for hh in permutations(horses[2:]):

h = horses[:2] + hh

j2h = dict(zip(jockeys, h))

# permute the colours

for c in permutations(colours):

h2c = dict(zip(h, c))

# permute the fences at which horses fell

for f in permutations(fences):

# map jockeys to the fences at which they fell

j2f = dict(zip(jockeys, f))

# Bill fell at the second fence

if j2f['Bill'] != 2:

continue

# Dave lasted longer than Bill

if j2f['Dave'] < 3:

continue

# map colours to fences

c2f = dict(zip(c, f))

# Big Gun fell at the third fence

if c2f[h2c['Big Gun']] != 3:

continue

# Mauve fell at the fifth fence

if c2f['Mauve'] != 5:

continue

# Long Gone lasted longer than yellow

if c2f[h2c['Long Gone']] <= c2f['Yellow']:

continue

# Chris fell one fence later than Green

if j2f['Chris'] != c2f['Green'] + 1:

continue

# Fred (not Blue) and Blue did not fall at adjacent fences

if h2c[j2h['Fred']] == 'Blue' or abs(j2f['Fred'] - c2f['Blue']) == 1:

continue

# Dragon was Red

if h2c['Dragon'] != 'Red':

continue

f2j, t = dict(zip(f, jockeys)), colours[4]

print('{} wore {} riding {}.'.format(f2j[c2f[t]], t, j2h[f2j[c2f[t]]]))

Arthur Vause permalink

Slightly more elegant by adapting code from Who Owns the Zebra, but essentially the same idea as Brian’s solution:


from itertools import permutations as perm

def dict_perms( category_list ):
  for p in perm(range(1,6)): yield {a:b for (a,b) in zip(category_list,p)}

def key(d, v): # find a key in dictionary d with value v
  return [name for name, val in d.items() if val == v][0]

horses = [ horse for horse in dict_perms(['Egg Nog','Nig Nag','Big Gun','Dragon','Long Gone'])
  if horse['Big Gun'] == 3 ]

horses_jockeys = [ (horse,jockey) for horse in horses
  for jockey in dict_perms(['Dave','Wally','Bill','Chris','Fred'])
  if  jockey['Dave']==horse['Egg Nog']
  and jockey['Wally']==horse['Nig Nag']
  and jockey['Bill'] ==2
  and jockey['Dave'] > 2                 ]

solutions = [ (horse,jockey,colour) for horse, jockey in horses_jockeys
  for colour in dict_perms(['mauve','red','yellow','green','blue'])
  if  colour['mauve']==5
  and colour['red'] == horse['Dragon'] 
  and colour['yellow'] < horse['Long Gone'] 
  and colour['green'] == jockey['Chris'] -1
  and abs( colour['blue'] - jockey['Fred']) > 1      ]


if len(solutions)>1:
  print "multiple solutions"
elif len(solutions)==0:
  print "no solutions"
else:
  (horse, jockey, colour) = solutions[0]
  for pos in range(1,6):
    print pos, ":", key(horse,pos), ",",key(jockey,pos), ",",key(colour,pos)

from itertools import permutations as perm

def dict_perms( category_list ):

for p in perm(range(1,6)): yield {a:b for (a,b) in zip(category_list,p)}

def key(d, v): # find a key in dictionary d with value v

return [name for name, val in d.items() if val == v][0]

horses = [ horse for horse in dict_perms(['Egg Nog','Nig Nag','Big Gun','Dragon','Long Gone'])

if horse['Big Gun'] == 3 ]

horses_jockeys = [ (horse,jockey) for horse in horses

for jockey in dict_perms(['Dave','Wally','Bill','Chris','Fred'])

if jockey['Dave']==horse['Egg Nog']

and jockey['Wally']==horse['Nig Nag']

and jockey['Bill'] ==2

and jockey['Dave'] > 2 ]

solutions = [ (horse,jockey,colour) for horse, jockey in horses_jockeys

for colour in dict_perms(['mauve','red','yellow','green','blue'])

if colour['mauve']==5

and colour['red'] == horse['Dragon']

and colour['yellow'] < horse['Long Gone']

and colour['green'] == jockey['Chris'] -1

and abs( colour['blue'] - jockey['Fred']) > 1 ]

if len(solutions)>1:

print "multiple solutions"

elif len(solutions)==0:

print "no solutions"

else:

(horse, jockey, colour) = solutions[0]

for pos in range(1,6):

print pos, ":", key(horse,pos), ",",key(jockey,pos), ",",key(colour,pos)

brian gladman permalink

That’s a neat approach Arthur – I like it!

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Sunday Times Teaser 2685 – Not the Gold Cup

by Angela Newing

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