Sunday Times Teaser 3298 – Bletchley Park

by BRG on December 7, 2025

by Peter Good

Published Sunday December 07 2025 (link)

Secret agent Robert Holmes was searching the hotel room of a foreign agent who was downstairs having breakfast. Holmes discovered a piece of paper containing the text DKCCVTCSZQRZYTAZXTTX and he thought this might be a coded message about the foreign agent’s mission so he sent it to his code-breaking experts.

They discovered that it was a message that had been scrambled by consistently replacing each letter of the alphabet with a different letter (no letter being used to replace more than one different letter). They decoded the message as a sentence containing four words, which they sent back to Holmes with spaces inserted between words. Holmes realised that his life was in imminent danger as soon as he read it.

What was the decoded message?

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BRG permalink

To make a programmed solution feasible in a reasonable time we have to assume that the encoded plaintext includes the word HOLMES and this, in turn, makes it possible to recognise that C must stand for L with the first four letters spelling the word KILL.


from collections import defaultdict
from itertools import permutations

letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

# The encoded text - if we guess that the first encoded word is KILL
# then C codes as L and HOLMES, if present, must be the second word
code = "DKCCVTCSZQRZYTAZXTTX"

# read words into a dictionary indexed on word length
# I ued the file ""google-10000-english.txt from here:
# https://github.com/first20hours/google-10000-english
path = "words_in_frequency_order.txt"
words = defaultdict(set)
with open(path, "r") as inf:
  for line in inf:
    if len(ln := line.strip()):
      words[len(ln)] |= {ln.upper()}

# the initial decoding dictionary from the known words (KILL and HOLMES)
d = dict(zip('DKCVTSZQ', 'KILHOMES'))
# the encoded letters which still need to be mapped for decoding
rct_ltrs = ''.join(sorted(set(c for c in code if c not in 'DKCVTSZQ')))
# the plaintext letters that are still uncoded
rpt_ltrs = set(letters).difference('KILLHOLMES')

# consider all mappings between ciphertext and plaintext letters
for ltrs in permutations(rpt_ltrs, len(rct_ltrs)):
  d_ = d.copy()
  d_.update(zip(rct_ltrs, ltrs))
  # the final ten letters of the ciphertext (encoding two words)
  pt = ''.join(d_[c] for c in code[-10:])
  # split the result into two words and see if both
  # the resulting words are in the word list
  for i in range(1, len(pt) - 1):
    fw, sw = pt[:i], pt[i:]
    try:
      if fw in words[i] and sw in words[10 - i]:
        print('KILL', 'HOLMES', fw, sw)
    except KeyError:
      break

from collections import defaultdict

from itertools import permutations

letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

# The encoded text - if we guess that the first encoded word is KILL

# then C codes as L and HOLMES, if present, must be the second word

code = "DKCCVTCSZQRZYTAZXTTX"

# read words into a dictionary indexed on word length

# I ued the file ""google-10000-english.txt from here:

# https://github.com/first20hours/google-10000-english

path = "words_in_frequency_order.txt"

words = defaultdict(set)

with open(path, "r") as inf:

for line in inf:

if len(ln := line.strip()):

words[len(ln)] |= {ln.upper()}

# the initial decoding dictionary from the known words (KILL and HOLMES)

d = dict(zip('DKCVTSZQ', 'KILHOMES'))

# the encoded letters which still need to be mapped for decoding

rct_ltrs = ''.join(sorted(set(c for c in code if c not in 'DKCVTSZQ')))

# the plaintext letters that are still uncoded

rpt_ltrs = set(letters).difference('KILLHOLMES')

# consider all mappings between ciphertext and plaintext letters

for ltrs in permutations(rpt_ltrs, len(rct_ltrs)):

d_ = d.copy()

d_.update(zip(rct_ltrs, ltrs))

# the final ten letters of the ciphertext (encoding two words)

pt = ''.join(d_[c] for c in code[-10:])

# split the result into two words and see if both

# the resulting words are in the word list

for i in range(1, len(pt) - 1):

fw, sw = pt[:i], pt[i:]

try:

if fw in words[i] and sw in words[10 - i]:

print('KILL', 'HOLMES', fw, sw)

except KeyError:

break

John Z. permalink


AZ = 'ABCDEFGHIJKLMNOPQRDTUVWXYZ'

# coded message
cm = 'DKCCVTCSZQRZYTAZXTTX'
# consider displacements from 1 - 25 (Ceasar code)
for d in range(1, 26):
  tt = str.maketrans(AZ, AZ[d:] + AZ[:d])
  dc = cm.translate(tt)
  # we assume the word 'HOLMES' appears in the message
  x = dc.find('HOLMES')
  if x > 0:
    print(dc)
    exit()
    
# Ceasar code has not worked; try another approach
for i in range(len(cm) - 6):
  # slide 'HOLMES' along the coded message; there must be 6 different letters
  if len(set(cm[i: i + 6])) != 6:
    continue
  # make translation table, lower case to better see decoded part
  tt = str.maketrans( cm[i: i + 6], 'holmes')
  dc = cm.translate(tt)
  
  # the coded message containing 'holmes' and other decoded letters
  print(dc, '\n')

# mental pattern matching on the first line of printout
# --ll holmes -e-o-e -oo-
# makes the message apparent (spaces added manually)
# access to an online English word list is needed to formalise

AZ = 'ABCDEFGHIJKLMNOPQRDTUVWXYZ'

# coded message

cm = 'DKCCVTCSZQRZYTAZXTTX'

# consider displacements from 1 - 25 (Ceasar code)

for d in range(1, 26):

tt = str.maketrans(AZ, AZ[d:] + AZ[:d])

dc = cm.translate(tt)

# we assume the word 'HOLMES' appears in the message

x = dc.find('HOLMES')

if x > 0:

print(dc)

exit()

# Ceasar code has not worked; try another approach

for i in range(len(cm) - 6):

# slide 'HOLMES' along the coded message; there must be 6 different letters

if len(set(cm[i: i + 6])) != 6:

continue

# make translation table, lower case to better see decoded part

tt = str.maketrans( cm[i: i + 6], 'holmes')

dc = cm.translate(tt)

# the coded message containing 'holmes' and other decoded letters

print(dc, '\n')

# mental pattern matching on the first line of printout

# --ll holmes -e-o-e -oo-

# makes the message apparent (spaces added manually)

# access to an online English word list is needed to formalise

Sunday Times Teaser 3298 – Bletchley Park

by Peter Good

Published Sunday December 07 2025 (link)

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