Sunday Times Teaser 2659 – Two by Two

by Peter G Chamberlain

The letters A to Z have each been given a different value between 0 and 25 inclusive in some order. Hence a three letter word could, for example, be represented by a number with three, four, five or six digits.

Using this representation it turns out that every digit in TWO is a prime and that TWO x TWO = FOUR.

What is the number FOUR?

4 Comments Leave one →

brian gladman permalink


from itertools import permutations

# split a string of digits into four 1 or 2 digit
# letter values each in the range 0..25 inclusive
def split(word, ltrs=[]):
  # if word is empty, see if we now have four different
  # one or two digit letter values (excluding two digit
  # values starting with '0') 
  if not word and (len(ltrs) == len(set(ltrs)) == 4
       and all(len(x) == 1 or x[0] != '0' for x in ltrs)): 
    # change the letters from strings to integers
    t = tuple(int(x) for x in ltrs)
    # yield the sequence if all its values are in range
    if all(0 <= x <= 25 for x in t):
      yield t
  else:
    # split off a one or two digit element from the string
    for ln in (1, 2):
      if len(word) >= ln:
        # add it to 'ltrs' and then split the remainder
        yield from split(word[ln:], ltrs + [word[:ln]])

# the possible letter values in TWO (all digits are prime)
twos_letters = (2, 3, 5, 7, 22, 23, 25) 

# check all permutations of these letter values for TWO
for p in permutations(twos_letters, 3):
  # form TWO and FOUR as integers
  TWO = int(''.join(str(x) for x in p))
  FOUR = TWO ** 2
  # now check all partitions of FOUR into 4 individual letter values
  for ltr_values in split(str(FOUR)):
    # and check that TWO and FOUR share only the letter 'O' 
    if len(set(p) & set(ltr_values)) == 1 and p[2] == ltr_values[1]:
      fs = 'TWO = {:d} {:s}, FOUR = {:d} {:s}'
      print(fs.format(TWO, p, FOUR, ltr_values))

from itertools import permutations

# split a string of digits into four 1 or 2 digit

# letter values each in the range 0..25 inclusive

def split(word, ltrs=[]):

# if word is empty, see if we now have four different

# one or two digit letter values (excluding two digit

# values starting with '0')

if not word and (len(ltrs) == len(set(ltrs)) == 4

and all(len(x) == 1 or x[0] != '0' for x in ltrs)):

# change the letters from strings to integers

t = tuple(int(x) for x in ltrs)

# yield the sequence if all its values are in range

if all(0 <= x <= 25 for x in t):

yield t

else:

# split off a one or two digit element from the string

for ln in (1, 2):

if len(word) >= ln:

# add it to 'ltrs' and then split the remainder

yield from split(word[ln:], ltrs + [word[:ln]])

# the possible letter values in TWO (all digits are prime)

twos_letters = (2, 3, 5, 7, 22, 23, 25)

# check all permutations of these letter values for TWO

for p in permutations(twos_letters, 3):

# form TWO and FOUR as integers

TWO = int(''.join(str(x) for x in p))

FOUR = TWO ** 2

# now check all partitions of FOUR into 4 individual letter values

for ltr_values in split(str(FOUR)):

# and check that TWO and FOUR share only the letter 'O'

if len(set(p) & set(ltr_values)) == 1 and p[2] == ltr_values[1]:

fs = 'TWO = {:d} {:s}, FOUR = {:d} {:s}'

print(fs.format(TWO, p, FOUR, ltr_values))

tonysmith permalink

It is not clear why 22 and 25 are not in line 24 – letter values in TWO are not constrained to be prime

Reply
brian gladman permalink

Hi Tony, Nice to see you back here! Good point, I’ll change it.

Reply

Frrits permalink


from itertools import permutations
 
valid = {str(x) for x in range(26)} 
# the possible letter values in TWO (all digits are prime)
twos_letters = ("2", "3", "5", "7", "22", "23", "25") 

# check all permutations of these letter values for TWO
for p in permutations(twos_letters, 3):
  # form TWO and FOUR as integers
  TWO = int(''.join(p))
  O = p[-1]
  FOUR = TWO ** 2
  if len(sFOUR := str(FOUR)) > 8 - (len(O) == 1): continue

  # find first O after first digit
  firstO = sFOUR[1:].find(O)
  firstO = (sFOUR := str(FOUR))[1:].find(O)
  if firstO not in {0, 1}: continue
  
  firstO += 1 
  URs = []
  # check where F and UR may reside in FOUR
  pos2 = (sFOUR[2:].find(O) == 0)
  for i in (1, 2):
    if firstO == i or (i == 2 and pos2): 
      if (F := sFOUR[:i]) not in valid or F in p: continue 
      URs.append(sFOUR[i + len(O):])
    
  for UR in URs:
    if not (1 < len(UR) < 5): continue
    # check for every distribution of U and R
    for i in range((ln := len(UR)) // 2, (ln + 3) // 2):
      U, R = UR[:i], UR[i:]
      if any(x not in valid for x in (U, R)): continue
      F = sFOUR[:len(sFOUR) - len(O + UR)]
      if any(x in (p + (F, )) for x in (U, R)): continue
      print(f"answer: {sFOUR}")

from itertools import permutations

valid = {str(x) for x in range(26)}

# the possible letter values in TWO (all digits are prime)

twos_letters = ("2", "3", "5", "7", "22", "23", "25")

# check all permutations of these letter values for TWO

for p in permutations(twos_letters, 3):

# form TWO and FOUR as integers

TWO = int(''.join(p))

O = p[-1]

FOUR = TWO ** 2

if len(sFOUR := str(FOUR)) > 8 - (len(O) == 1): continue

# find first O after first digit

firstO = sFOUR[1:].find(O)

firstO = (sFOUR := str(FOUR))[1:].find(O)

if firstO not in {0, 1}: continue

firstO += 1

URs = []

# check where F and UR may reside in FOUR

pos2 = (sFOUR[2:].find(O) == 0)

for i in (1, 2):

if firstO == i or (i == 2 and pos2):

if (F := sFOUR[:i]) not in valid or F in p: continue

URs.append(sFOUR[i + len(O):])

for UR in URs:

if not (1 < len(UR) < 5): continue

# check for every distribution of U and R

for i in range((ln := len(UR)) // 2, (ln + 3) // 2):

U, R = UR[:i], UR[i:]

if any(x not in valid for x in (U, R)): continue

F = sFOUR[:len(sFOUR) - len(O + UR)]

if any(x in (p + (F, )) for x in (U, R)): continue

print(f"answer: {sFOUR}")

Sunday Times Teaser 2659 – Two by Two

by Peter G Chamberlain

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