Sunday Times Teaser 2830 – Time for Christmas

by Danny Roth

Published: 18 December 2016 (link)

For Christmas George has bought Martha a novelty digital 24-hour clock. It has an eight-digit display forming four two-digit numbers. The first three of these two-digit numbers are very conventional – the hours, the minutes and the seconds. However, the final two-digit display shows the sum of the first six displayed digits!

On two occasions today all eight digits displayed were different and three of the four two-digit numbers were perfect squares. Between those two occasions there was a time when the eight digits displayed were again all different, but this time the sum of the eight digits was a perfect square!

What was the eight-digit display at that in-between time?

2 Comments Leave one →

Brian Gladman permalink


from itertools import product

out = lambda v: ' ({:02d} {:02d} {:02d} {:02d})'.format(*v)

# squares below 60
sq = set(x * x for x in range(8))

# run through the hour, minute and second combinations in order
first = False
for h, m, s in product(range(24), range(60), range(60)):
  # form the resulting six digits ...
  d6 = set(divmod(h, 10) + divmod(m, 10) + divmod(s, 10))
  # ... their sum and the eight digits of the display
  u = sum(d6)
  d8 = d6.union(divmod(u, 10))
  # record if they are all different
  if len(d8) == 8:
    t = (h, m, s, u)
    
    # look for times with three perfect squares
    if len(sq.intersection(t)) == 3:
      if not first:
        # record the first appearance
        s1 = 'first' + out(t)
        first = True
        # record the second and terminate the search
      else:
        s3 = 'second' + out(t)
        break

    # if we have seen the first time with three perfect
    # squares (but not the second) look for a time when 
    # the sum of all eight digits is a perfect square
    if first and sum(d8) in sq:
      s2 = 'in-between' + out(t)

print(', '.join((s1, s2, s3)))

from itertools import product

out = lambda v: ' ({:02d} {:02d} {:02d} {:02d})'.format(*v)

# squares below 60

sq = set(x * x for x in range(8))

# run through the hour, minute and second combinations in order

first = False

for h, m, s in product(range(24), range(60), range(60)):

# form the resulting six digits ...

d6 = set(divmod(h, 10) + divmod(m, 10) + divmod(s, 10))

# ... their sum and the eight digits of the display

u = sum(d6)

d8 = d6.union(divmod(u, 10))

# record if they are all different

if len(d8) == 8:

t = (h, m, s, u)

# look for times with three perfect squares

if len(sq.intersection(t)) == 3:

if not first:

# record the first appearance

s1 = 'first' + out(t)

first = True

# record the second and terminate the search

else:

s3 = 'second' + out(t)

break

# if we have seen the first time with three perfect

# squares (but not the second) look for a time when

# the sum of all eight digits is a perfect square

if first and sum(d8) in sq:

s2 = 'in-between' + out(t)

print(', '.join((s1, s2, s3)))

Jim Randell permalink

If we accept that there are only “two occasions” in any given day, then we don’t need to consider all the times, just look until we find the second condition (p2) occurring between the two occurrences of the first condition (p1).

This code stops when it finds the second occurrence of the first condition and runs in 96ms.


from itertools import product
from enigma import irange, flatten, is_square, printf

# output format: <hour>:<minute>:<second>:<sum> [<state>]
fmt = "{h:02d}:{m:02d}:{s:02d}:{t:02d} [{state}]"

state = 0

# consider times in order
for (h, m, s) in product(irange(0, 23), irange(0, 59), irange(0, 59)):
  # collect digits
  digits = flatten((divmod(x, 10) for x in (h, m, s)), fn=list)
  t = sum(digits)
  digits.extend(divmod(t, 10))
  # all digits are different
  if len(set(digits)) != 8: continue

  # mark special states
  # p1 = "three of the four two digits numbers are perfect squares"
  p1 = (sum(bool(is_square(x)) for x in (h, m, s, t)) == 3)
  # p2 = "the sum of all 8 digits is a perfect square"
  p2 = bool(is_square(sum(digits)))
  if not(p1 or p2): continue

  # if we're in the initial state wait until we see a p1
  if state == 0:
    if p1:
      printf(fmt, state='first')
      state = 1

  # we've seen the first p1
  elif state == 1:
    # look for a p2
    if p2:
      printf(fmt, state='in-between')
    # or for the second p1
    elif p1:
      printf(fmt, state='second')
      # and we are done
      break

from itertools import product

from enigma import irange, flatten, is_square, printf

# output format: <hour>:<minute>:<second>:<sum> [<state>]

fmt = "{h:02d}:{m:02d}:{s:02d}:{t:02d} [{state}]"

state = 0

# consider times in order

for (h, m, s) in product(irange(0, 23), irange(0, 59), irange(0, 59)):

# collect digits

digits = flatten((divmod(x, 10) for x in (h, m, s)), fn=list)

t = sum(digits)

digits.extend(divmod(t, 10))

# all digits are different

if len(set(digits)) != 8: continue

# mark special states

# p1 = "three of the four two digits numbers are perfect squares"

p1 = (sum(bool(is_square(x)) for x in (h, m, s, t)) == 3)

# p2 = "the sum of all 8 digits is a perfect square"

p2 = bool(is_square(sum(digits)))

if not(p1 or p2): continue

# if we're in the initial state wait until we see a p1

if state == 0:

if p1:

printf(fmt, state='first')

state = 1

# we've seen the first p1

elif state == 1:

# look for a p2

if p2:

printf(fmt, state='in-between')

# or for the second p1

elif p1:

printf(fmt, state='second')

# and we are done

break

We could stop when we hit the first p2 condition after seeing a p1, but this gives you the times of the p1’s and also all the intermediate p2’s (although as expected there is only one).

The enigma.py library is available from [ htto://www.magwag.plus.com/jim/enigma.html ].

Sunday Times Teaser 2830 – Time for Christmas

by Danny Roth

Published: 18 December 2016 (link)

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