Sunday Times Teaser 3257 – Powers of Deduction

by BRG on February 23, 2025

by John Fletcher

Published Sunday February 23 2025 (link)

One evening, Sherlock Holmes challenges Watson and Mycroft to a game of Cluedo.

He places one each of the six “person” cards, six “weapon” cards and nine “room” cards into the murder bag, shuffles the rest and deals them out equally.

Mycroft sees his cards and says, “If I guessed what was in the murder bag now, my chance of being correct is one in an integer that is a power greater than two.”

“I’m saying nothing.” Watson replies. “I’d only give something away.”

“Commendably prudent”, Sherlock says, “but the same is true for you.”

Watson studies his cards, then looks agitated. “How could you know that?”

“Sherlock isn’t cheating.” Mycroft reassures him. “He didn’t even know how many person cards you’re holding.”

How many room cards is Mycroft holding?

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4 Comments Leave one →

BRG permalink


# find integer roots <x> (greater than 2) of integer <n>
# such that n = r^x when n <= 324
def find_power(n):
  for r in range(2, 7):
    for x in range(3, 10):
      if n == r ** x:
        yield r, x

# split the integer <n> into <m> parts in any order
def split(n, m, ms=()):
  if m == 1:
    yield ms + (n,)
  else:
    for i in range(n + 1):
      yield from split(n - i, m - 1, ms + (i,)) 

pwrs = set()
# find numbers of people, weapon and room cards held
# by a person such that their chances of predicting
# the contents of the bag is (1 / N) where N is a 
# power greater than 2 of an integer 
for p, w, r in split(6, 3):
  if (N := (6 - p) * (6 - w) * (9 - r)):
    for n, x in find_power(N):
      print(f"{p = } {w = } {r = } ==> {N:>3} = {n}^{x}")
      pwrs.add((p, w, r))

assert len(pwrs) == 2
# Analysis: At this point there are only two solutions for
# points (p), weapons (w) and rooms (r)
# 
#     p, w, r = (1, 1, 4)   and   (3, 3, 0)
#
# and each player has one of these. Neither 2 people with 
# (3, 3, 0) nor 3 people with (1, 1, 4) are possible since
# there are not enough cards in play. So one has (3, 3, 0)
# and two have (1, 1, 4). Anyone who has (3, 3, 0) knows
# that both other players have (1, 1, 4). For this reason
# Sherlock cannot have (3, 3, 0) because he doesn't know 
# Watson's number of person cards. Also, since Mycoft knows
# Sherlock's holding, he must hold (3, 3, 0). Hence the 3
# hands are  M = (3, 3, 0), W = (1, 1, 4), S = (1, 1, 4)]
knows_hands = (3, 3, 0)

s0 = []
# Mycroft's possible holdings
for Mp, Mw, Mr in pwrs:
  # Watson's possible holdings
  for Wp, Ww, Wr in pwrs:
    # Sherlock's possible holdings
    Spwr = 5 - (Mp + Wp), 5 - (Mw + Ww), 8 - (Mr + Wr)
    # values cannot be negative and Sherlock does not
    # know the hands (see above)
    if all(s >= 0 for s in Spwr) and Spwr != knows_hands:
      s0 += [((Mp, Mw, Mr), (Wp, Ww, Wr), Spwr)]

for M, W, S in s0:
  # Mycroft knows the hands (see above)
  if M == knows_hands:
    print(f"\nMycroft holds {M[2]} room cards [{M = }, {W = } {S = }].")

# find integer roots <x> (greater than 2) of integer <n>

# such that n = r^x when n <= 324

def find_power(n):

for r in range(2, 7):

for x in range(3, 10):

if n == r ** x:

yield r, x

# split the integer <n> into <m> parts in any order

def split(n, m, ms=()):

if m == 1:

yield ms + (n,)

else:

for i in range(n + 1):

yield from split(n - i, m - 1, ms + (i,))

pwrs = set()

# find numbers of people, weapon and room cards held

# by a person such that their chances of predicting

# the contents of the bag is (1 / N) where N is a

# power greater than 2 of an integer

for p, w, r in split(6, 3):

if (N := (6 - p) * (6 - w) * (9 - r)):

for n, x in find_power(N):

print(f"{p = } {w = } {r = } ==> {N:>3} = {n}^{x}")

pwrs.add((p, w, r))

assert len(pwrs) == 2

# Analysis: At this point there are only two solutions for

# points (p), weapons (w) and rooms (r)

# p, w, r = (1, 1, 4) and (3, 3, 0)

# and each player has one of these. Neither 2 people with

# (3, 3, 0) nor 3 people with (1, 1, 4) are possible since

# there are not enough cards in play. So one has (3, 3, 0)

# and two have (1, 1, 4). Anyone who has (3, 3, 0) knows

# that both other players have (1, 1, 4). For this reason

# Sherlock cannot have (3, 3, 0) because he doesn't know

# Watson's number of person cards. Also, since Mycoft knows

# Sherlock's holding, he must hold (3, 3, 0). Hence the 3

# hands are M = (3, 3, 0), W = (1, 1, 4), S = (1, 1, 4)]

knows_hands = (3, 3, 0)

s0 = []

# Mycroft's possible holdings

for Mp, Mw, Mr in pwrs:

# Watson's possible holdings

for Wp, Ww, Wr in pwrs:

# Sherlock's possible holdings

Spwr = 5 - (Mp + Wp), 5 - (Mw + Ww), 8 - (Mr + Wr)

# values cannot be negative and Sherlock does not

# know the hands (see above)

if all(s >= 0 for s in Spwr) and Spwr != knows_hands:

s0 += [((Mp, Mw, Mr), (Wp, Ww, Wr), Spwr)]

for M, W, S in s0:

# Mycroft knows the hands (see above)

if M == knows_hands:

print(f"\nMycroft holds {M[2]} room cards [{M = }, {W = } {S = }].")

Frits permalink


from math import log

def is_power(n, m=2):  
  if n < 2: return 0
  max_base = round(n** (1 / m))
  max_base1 = round(n** (1 / (m + 1)))
  logn = log(n)
  b = 2 + (n % 2)
  while b <= max_base: 
    # if log(n) / log(b) = e then b^e = n  
    if b ** (e := round(f := logn / log(b))) == n:
      return e
    # skip ranges  
    if b < max_base and f < m + 1:
      b = max_base
    elif b < max_base1 and f < m + 2:
      b = max_base1 + 1
    else:  
      b += 2
      
  return 0    
  
sols = set()
# determine combinations of Mycroft's hand
for p in range(7):
  for w in range(7 - p):
    r = 6 - p - w
    n = ((6 - p) if p < 6 else 1) * ((6 - w) if w < 6 else 1) * (9 - r)
    # is <n> a power greater than 2 of an integer
    if is_power(n, 3):
      sols.add((p, w, r))
      
Sherlock_knowperson_cards = set()
for m_p, m_w, m_r in sols:   # Mycroft
  # assume Dr Watson has a similar hand as Mycroft
  for d_p, d_w, d_r in sols: # Dr Watson
    # see if Sherlock might know how many person cards Watson is holding
    if len({m_p, d_p}) == 1: 
      # calculate what remains for Sherlock
      p, w, r = (5 - m_p - d_p, 5 - m_w - d_w, 8 - m_r - d_r) 
      if all(x >= 0 for x in (p, w, r)): 
        Sherlock_knowperson_cards.add((p, w, r))

if len(Sherlock_knowperson_cards) == 1:
  # Mycroft can only make the remark if he himself has the combination of
  # which Sherlock might know how many person cards Watson is holding
  # (and two of such combinations isn't possible)   
  p, w, r = Sherlock_knowperson_cards.pop()
  if 2 * p > 5 or 2 * r > 5 or 2 * r > 8:
    print("answer:", r)

from math import log

def is_power(n, m=2):

if n < 2: return 0

max_base = round(n** (1 / m))

max_base1 = round(n** (1 / (m + 1)))

logn = log(n)

b = 2 + (n % 2)

while b <= max_base:

# if log(n) / log(b) = e then b^e = n

if b ** (e := round(f := logn / log(b))) == n:

return e

# skip ranges

if b < max_base and f < m + 1:

b = max_base

elif b < max_base1 and f < m + 2:

b = max_base1 + 1

else:

b += 2

return 0

sols = set()

# determine combinations of Mycroft's hand

for p in range(7):

for w in range(7 - p):

r = 6 - p - w

n = ((6 - p) if p < 6 else 1) * ((6 - w) if w < 6 else 1) * (9 - r)

# is <n> a power greater than 2 of an integer

if is_power(n, 3):

sols.add((p, w, r))

Sherlock_knowperson_cards = set()

for m_p, m_w, m_r in sols: # Mycroft

# assume Dr Watson has a similar hand as Mycroft

for d_p, d_w, d_r in sols: # Dr Watson

# see if Sherlock might know how many person cards Watson is holding

if len({m_p, d_p}) == 1:

# calculate what remains for Sherlock

p, w, r = (5 - m_p - d_p, 5 - m_w - d_w, 8 - m_r - d_r)

if all(x >= 0 for x in (p, w, r)):

Sherlock_knowperson_cards.add((p, w, r))

if len(Sherlock_knowperson_cards) == 1:

# Mycroft can only make the remark if he himself has the combination of

# which Sherlock might know how many person cards Watson is holding

# (and two of such combinations isn't possible)

p, w, r = Sherlock_knowperson_cards.pop()

if 2 * p > 5 or 2 * r > 5 or 2 * r > 8:

print("answer:", r)

Sonam Tamang permalink

This is such an insightful and thought-provoking post! I really enjoyed reading your perspective, and it’s clear that you’ve put a lot of effort into researching and presenting your ideas. The way you’ve structured the article makes it easy to follow, and I appreciate how you’ve supported your points with relevant examples and evidence. It’s refreshing to see such a well-balanced and nuanced discussion, especially on a subject that can often be polarizing. I also found the visuals and links you included to be very helpful they added depth to the content and made it even more engaging. Your writing style is both informative and approachable, which makes complex topics easier to understand. Thanks for sharing your knowledge and insights it’s clear that you’re passionate about the subject, and that passion shines through in your work. I’ll definitely be sharing with others and checking out more of your posts.

Reply
- BRG permalink
  
  Hi Sonam,
  
  Thank you for your kind words, which I much appreciate. This site has a small
  but dedicated group of contributors who I believe are as passionate as I am
  about encouraging others to look at Python as a way of ‘getting into’ and
  developing their programming skills.
  
  Welcome to our world!
  
  Reply

Sunday Times Teaser 3257 – Powers of Deduction

by John Fletcher

Published Sunday February 23 2025 (link)

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