Sunday Times Teaser 3197 – Three or Four?

by BRG on December 29, 2023

by Victor Bryant

Published Friday December 29 2023 (link)

Here are some clues about my PIN:

• It has 3 digits (or it might be 4).

• The first digit is 3 (or it might be 4).

• The last digit is 3 (or it might be 4).

• It is divisible by 3 (or it might be 4).

• It differs from a square by 3 (or it might be 4).

In each of those statements above just one of the guesses is true, and the lower guess is true in 3 cases (or it might be 4).

That should enable you to get down to 3 (or it might be 4) possibilities for my PIN. But, even if you could choose any one of the statements, knowing which guess was correct in that statement would not enable you to determine the PIN.

What is my PIN?

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BRG permalink


is_sqr = lambda n: round(n ** (1 / 2)) ** 2 == n

pins = dict()
# the possible number of digits in the PIN
for nd in (3, 4):
  # the possible middle digit(s) in the PIN
  for mdgts in range(10 ** (nd - 2)):
    # the possible first and last digits in the PIN
    for fd in (3, 4):
      for ld in (3, 4):
        pin = int(str(fd) + f"{mdgts:0{nd - 2}}" + str(ld))
        
        # the PIN is divisible by 3 or 4
        if (bd3 := bool(int(pin) % 3) == 0) == bool(int(pin) % 4 == 0):
          continue
        
        # the PIN is 3 or 4 away from a square
        sqs = set(abs(s) for s in (-4, -3, 3, 4) if is_sqr(pin + s))
        if (sq3 := (3 in sqs)) == (4 in sqs):
          continue
        
        # collect possible PINs and the five statement results
        cnt3 = (nd == 3, fd == 3, ld == 3, bd3, sq3)
        if cnt3.count(True) in (3, 4):
          pins[pin] = tuple([2, 1][b] for b in cnt3)
          print(f"{pin:>4} {pins[pin]}")

# the number of possible PINs is 3 or 4
if len(pins) in (3, 4):
  # find the results for the remaining pins from each of the five statements
    s5 = list(zip(*pins.values()))
    for p, r5 in pins.items():
      # we now know the correct answer for each statement (in r5) but this does
      # not allow us to find the PIN using any individual statement;  so in all
      # statements the correct PIN's answer must be shared with another PIN 
      if all(s.count(r) > 1 for s, r in zip(s5, r5)):
        print(f"Victor's PIN is {p}")

is_sqr = lambda n: round(n ** (1 / 2)) ** 2 == n

pins = dict()

# the possible number of digits in the PIN

for nd in (3, 4):

# the possible middle digit(s) in the PIN

for mdgts in range(10 ** (nd - 2)):

# the possible first and last digits in the PIN

for fd in (3, 4):

for ld in (3, 4):

pin = int(str(fd) + f"{mdgts:0{nd - 2}}" + str(ld))

# the PIN is divisible by 3 or 4

if (bd3 := bool(int(pin) % 3) == 0) == bool(int(pin) % 4 == 0):

continue

# the PIN is 3 or 4 away from a square

sqs = set(abs(s) for s in (-4, -3, 3, 4) if is_sqr(pin + s))

if (sq3 := (3 in sqs)) == (4 in sqs):

continue

# collect possible PINs and the five statement results

cnt3 = (nd == 3, fd == 3, ld == 3, bd3, sq3)

if cnt3.count(True) in (3, 4):

pins[pin] = tuple([2, 1][b] for b in cnt3)

print(f"{pin:>4} {pins[pin]}")

# the number of possible PINs is 3 or 4

if len(pins) in (3, 4):

# find the results for the remaining pins from each of the five statements

s5 = list(zip(*pins.values()))

for p, r5 in pins.items():

# we now know the correct answer for each statement (in r5) but this does

# not allow us to find the PIN using any individual statement; so in all

# statements the correct PIN's answer must be shared with another PIN

if all(s.count(r) > 1 for s, r in zip(s5, r5)):

print(f"Victor's PIN is {p}")

John Z permalink


sq = set(i * i for i in range(100))

resn = [] # results: possible PIN values
rest = [] # results: number of statements True for value = 3
for n in range(100, 10000): # it has 3 or 4 digits
  ns = str(n)
  if not (f31 := ('3' == ns[0])) ^ ('4' == ns[0]): # first digit 3 or 4
    continue
  if not (f3L := ('3' == ns[-1])) ^ ('4' == ns[-1]): # last  digit 3 or 4
    continue
  if not (f3d := ((n % 3) == 0)) ^ ((n % 4) == 0): # divisible by 3 or 4
    continue
  # It differs from a square by 3 (or it might be 4)
  if not ((f3sq := bool({n + 3, n - 3} & sq)) ^ bool({n + 4, n - 4} & sq)):
    continue


  f3dg = bool(n < 1000) # 3 digits

  # In each of those statements above just one of the guesses is true, and the
  # lower guess is true in 3 cases (or it might be 4).
  f3t = f3dg + f31 + f3L + f3d + f3sq # total number of statements true for 3

  
  if f3t in (3, 4):
    print(f'{n}, {f3t}, {f3dg=}, {f31=}, {f3L=}, {f3d=}, {f3sq=}')
    resn += [n]
    rest += [f3t]

  # The only possible differentiation is on the total number of statements that
  # are true
for nn, tt in zip(resn, rest):
  if rest.count(tt) == 1: # PIN is the one with a unique number of statements
    #                       correct
    print(f"Victor's PIN: {nn}")

sq = set(i * i for i in range(100))

resn = [] # results: possible PIN values

rest = [] # results: number of statements True for value = 3

for n in range(100, 10000): # it has 3 or 4 digits

ns = str(n)

if not (f31 := ('3' == ns[0])) ^ ('4' == ns[0]): # first digit 3 or 4

continue

if not (f3L := ('3' == ns[-1])) ^ ('4' == ns[-1]): # last digit 3 or 4

continue

if not (f3d := ((n % 3) == 0)) ^ ((n % 4) == 0): # divisible by 3 or 4

continue

# It differs from a square by 3 (or it might be 4)

if not ((f3sq := bool({n + 3, n - 3} & sq)) ^ bool({n + 4, n - 4} & sq)):

continue

f3dg = bool(n < 1000) # 3 digits

# In each of those statements above just one of the guesses is true, and the

# lower guess is true in 3 cases (or it might be 4).

f3t = f3dg + f31 + f3L + f3d + f3sq # total number of statements true for 3

if f3t in (3, 4):

print(f'{n}, {f3t}, {f3dg=}, {f31=}, {f3L=}, {f3d=}, {f3sq=}')

resn += [n]

rest += [f3t]

# The only possible differentiation is on the total number of statements that

# are true

for nn, tt in zip(resn, rest):

if rest.count(tt) == 1: # PIN is the one with a unique number of statements

# correct

print(f"Victor's PIN: {nn}")

John Z permalink

The logic in my code lines 31 — 36 used to determine Victor’s PIN from the 3 (or 4) candidates is flawed and while it may produce the ‘right answer’ this is a coincidence.

Reply

John Z permalink

Borrowing heavily from Brian’s code lines 32-38


sq = set(i * i for i in range(100))

res = []# results: (possible PIN values, statements T/F for value = 3)
for n in range(100, 10000): # it has 3 or 4 digits
  ns = str(n)
  if ((f3f := ('3' == ns[0])) ^ ('4' == ns[0]) and  # first digit 3 or 4
      (f3L := ('3' == ns[-1])) ^ ('4' == ns[-1]) and # last  digit 3 or 4
      (f3d := ((n % 3) == 0)) ^ ((n % 4) == 0) and # divisible by 3 or 4
      # It differs from a square by 3 (or it might be 4)
      ((f3sq := bool({n + 3, n - 3} & sq)) ^ bool({n + 4, n - 4} & sq))):

    f3dg = 1000 > n  # 3 digits

    # In each of those statements above just one of the guesses is true, and
    # the lower guess is true in 3 cases (or it might be 4).
    f3t = f3dg + f3f + f3L + f3d + f3sq # total number of statements true for 3
    
    if f3t in (3, 4):
      print(f'{n}, {f3t=}, {f3dg=}, {f3f=}, {f3L=}, {f3d=}, {f3sq=}')

      # store PIN candidate and its 5 statement results
      res += [(n, f3dg, f3f, f3L, f3d, f3sq)]

# This bit of code is 'cribbed' from Brian's lines 32-38
for i in range(len(res)):
  if all(x.count(x[i]) > 1 for x in tuple(zip(*res))[1:]): # consider T/F only
    print(f"Victor's pin: {res[i][0]}")

sq = set(i * i for i in range(100))

res = []# results: (possible PIN values, statements T/F for value = 3)

for n in range(100, 10000): # it has 3 or 4 digits

ns = str(n)

if ((f3f := ('3' == ns[0])) ^ ('4' == ns[0]) and # first digit 3 or 4

(f3L := ('3' == ns[-1])) ^ ('4' == ns[-1]) and # last digit 3 or 4

(f3d := ((n % 3) == 0)) ^ ((n % 4) == 0) and # divisible by 3 or 4

# It differs from a square by 3 (or it might be 4)

((f3sq := bool({n + 3, n - 3} & sq)) ^ bool({n + 4, n - 4} & sq))):

f3dg = 1000 > n # 3 digits

# In each of those statements above just one of the guesses is true, and

# the lower guess is true in 3 cases (or it might be 4).

f3t = f3dg + f3f + f3L + f3d + f3sq # total number of statements true for 3

if f3t in (3, 4):

print(f'{n}, {f3t=}, {f3dg=}, {f3f=}, {f3L=}, {f3d=}, {f3sq=}')

# store PIN candidate and its 5 statement results

res += [(n, f3dg, f3f, f3L, f3d, f3sq)]

# This bit of code is 'cribbed' from Brian's lines 32-38

for i in range(len(res)):

if all(x.count(x[i]) > 1 for x in tuple(zip(*res))[1:]): # consider T/F only

print(f"Victor's pin: {res[i][0]}")

Sunday Times Teaser 3197 – Three or Four?

by Victor Bryant

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