Sunday Times Teaser 3220 – Graffiti

by BRG on 7 June 2024

by Angela Newing

Published Friday June 07 2024 (link)

Six pupils were in detention having been caught painting graffiti on the school wall. The teacher decided to show them some examples of abstract art and produced seven different pictures in order ABCDEFG for them to study for a few minutes. Teacher took them back, showed them again upside down and in random order, and asked them to write down which of ABCDEFG each one was.

The answers were:

Helen	A B C D E F G
Ian	A B C F E D G
Jean	A D F G C E B
Kevin	A D B F C E G
Lee	A F C G E D B
Mike	C A F B E D G

It transpired that each picture had been correctly identified by at least one pupil, and everyone had a different number of correct answers.

What was the correct order for the upside down ones?

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4 Comments Leave one →

BRG permalink


from itertools import product

answers = 'ABCDEFG ABCFEDG ADFGCEB ADBFCEG AFCGEDB CAFBEDG'.split()

# consider the all possible painting orders by combining 
# possible paintings for each displayed painting position
for p in product(*(set(x) for x in zip(*answers))):
  if len(set(p)) == 7:    
    # collect the correct number of paintings for each answer
    cnts = [sum(a == b for a, b in zip(p, ans)) for ans in answers]
    # which must all be different
    if len(cnts) == len(set(cnts)):
      # all paintings must be identifed by at least one pupil
      if all(p in ans for p, *ans in zip(p, *answers)):
        print(f"The upside down order is {''.join(p)}.")

from itertools import product

answers = 'ABCDEFG ABCFEDG ADFGCEB ADBFCEG AFCGEDB CAFBEDG'.split()

# consider the all possible painting orders by combining

# possible paintings for each displayed painting position

for p in product(*(set(x) for x in zip(*answers))):

if len(set(p)) == 7:

# collect the correct number of paintings for each answer

cnts = [sum(a == b for a, b in zip(p, ans)) for ans in answers]

# which must all be different

if len(cnts) == len(set(cnts)):

# all paintings must be identifed by at least one pupil

if all(p in ans for p, *ans in zip(p, *answers)):

print(f"The upside down order is {''.join(p)}.")

John Z permalink


from itertools import permutations

Helen =	'A B C D E F G'
Ian =	'A B C F E D G'
Jean =	'A D F G C E B'
Kevin =	'A D B F C E G'
Lee =	'A F C G E D B'
Mike =	'C A F B E D G'

ans = tuple(x.replace(' ','') for x in (Helen, Ian, Jean, Kevin, Lee, Mike))

# consider all different painting orders  
for ag in permutations('ABCDEFG'):
  caps = [0 for i in range(6)] # correct answers per student
 
  # consider each painting in the permutation
  for i, p in enumerate(ag):

    # consider each student's answers
    for j, an in enumerate(ans):

      if p == an[i]: # correct answer
        caps[j] += 1 # correct answers for this student

  if len(set(caps)) == 6: # each sudent has diff number of correct answers
    # no need to test that each painting identified correctly at least once
    print('Order of paintings:',' '.join(ag))
    print('Student correct answers:', *caps)

from itertools import permutations

Helen = 'A B C D E F G'

Ian = 'A B C F E D G'

Jean = 'A D F G C E B'

Kevin = 'A D B F C E G'

Lee = 'A F C G E D B'

Mike = 'C A F B E D G'

ans = tuple(x.replace(' ','') for x in (Helen, Ian, Jean, Kevin, Lee, Mike))

# consider all different painting orders

for ag in permutations('ABCDEFG'):

caps = [0 for i in range(6)] # correct answers per student

# consider each painting in the permutation

for i, p in enumerate(ag):

# consider each student's answers

for j, an in enumerate(ans):

if p == an[i]: # correct answer

caps[j] += 1 # correct answers for this student

if len(set(caps)) == 6: # each sudent has diff number of correct answers

# no need to test that each painting identified correctly at least once

print('Order of paintings:',' '.join(ag))

print('Student correct answers:', *caps)

John Z. permalink

Slightly different approach, should be faster


from itertools import product

Helen =	'A B C D E F G'
Ian =	'A B C F E D G'
Jean =	'A D F G C E B'
Kevin =	'A D B F C E G'
Lee =	'A F C G E D B'
Mike =	'C A F B E D G'

ans = tuple(x.replace(' ','') for x in (Helen, Ian, Jean, Kevin, Lee, Mike))

ansz = tuple(zip(*ans)) # transpose matrix

anzs = tuple(set(az) for az in ansz) # paintings identified in each position

# consider all painting orders where a student has identified it as present 
for ag in product(*anzs):

  if len(set(ag)) == 7: # all paintings must be present

    # get correct answers per student
    caps = [sum(a == an[i] for i, a in enumerate(ag)) for an in(ans)]

    if len(set(caps)) == 6: # each student has diff numb of correct answers
      print('Order of paintings:',' '.join(ag))
      print('Student correct answers:', *caps)

from itertools import product

Helen = 'A B C D E F G'

Ian = 'A B C F E D G'

Jean = 'A D F G C E B'

Kevin = 'A D B F C E G'

Lee = 'A F C G E D B'

Mike = 'C A F B E D G'

ans = tuple(x.replace(' ','') for x in (Helen, Ian, Jean, Kevin, Lee, Mike))

ansz = tuple(zip(*ans)) # transpose matrix

anzs = tuple(set(az) for az in ansz) # paintings identified in each position

# consider all painting orders where a student has identified it as present

for ag in product(*anzs):

if len(set(ag)) == 7: # all paintings must be present

# get correct answers per student

caps = [sum(a == an[i] for i, a in enumerate(ag)) for an in(ans)]

if len(set(caps)) == 6: # each student has diff numb of correct answers

print('Order of paintings:',' '.join(ag))

print('Student correct answers:', *caps)

John Z permalink

And another approach more akin to the manual solution. It should be even faster.


Helen =	'A B C D E F G'
Ian =	'A B C F E D G'
Jean =	'A D F G C E B'
Kevin =	'A D B F C E G'
Lee =	'A F C G E D B'
Mike =	'C A F B E D G'

ans = tuple(x.replace(' ','') for x in (Helen, Ian, Jean, Kevin, Lee, Mike))

ansz = tuple(zip(*ans)) # transpose matrix

anzs = tuple(set(az) for az in ansz) # paintings identified in each position

def paintings(n = 0, ord = ()):
  if n == 7: 
    yield ord
  else:
    # consider paintings in column less those already found 
    for p in anzs[n] - set(ord):
      yield from paintings(n + 1, ord + (p,))
              
# consider all painting orders where a student has identified it as present
for ag in paintings():

  # get correct answers per student
  caps = [sum(a == an[i] for i, a in enumerate(ag)) for an in(ans)]

  if len(set(caps)) == 6: # each student has diff numb of correct answers
    print('Order of paintings:', ' '.join(ag))
    print('Student correct answers:', *caps)

Helen = 'A B C D E F G'

Ian = 'A B C F E D G'

Jean = 'A D F G C E B'

Kevin = 'A D B F C E G'

Lee = 'A F C G E D B'

Mike = 'C A F B E D G'

ans = tuple(x.replace(' ','') for x in (Helen, Ian, Jean, Kevin, Lee, Mike))

ansz = tuple(zip(*ans)) # transpose matrix

anzs = tuple(set(az) for az in ansz) # paintings identified in each position

def paintings(n = 0, ord = ()):

if n == 7:

yield ord

else:

# consider paintings in column less those already found

for p in anzs[n] - set(ord):

yield from paintings(n + 1, ord + (p,))

# consider all painting orders where a student has identified it as present

for ag in paintings():

# get correct answers per student

caps = [sum(a == an[i] for i, a in enumerate(ag)) for an in(ans)]

if len(set(caps)) == 6: # each student has diff numb of correct answers

print('Order of paintings:', ' '.join(ag))

print('Student correct answers:', *caps)

Sunday Times Teaser 3220 – Graffiti

Published Friday June 07 2024 (link)

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