Sunday Times Teaser 2891 – Nine Cut Diamonds

by Stephen Hogg

Published February 18 2018 (link)

An app “shuffles” and “deals” the “ace” (=1) to “nine” of diamonds in a line, face down. Three numbers under 30 are chosen at random and each will win if it is a factor of the hidden nine-digit value. Keying # reveals the rightmost face-down card. At such a “reveal” the app displays “won”, “lost” or “in-play” for each number. The rightmost face-down cards are revealed singly until all are known. For one deal my three numbers didn’t include a prime number and after two “reveals” were all “in-play”. After the third “reveal”, two were “won” and one “lost”.

At the third “reveal” what three-digit number was displayed?

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Brian Gladman permalink

My first attempt at this failed miserably but John Crabtree suggested an approach that worked much better than the one I had tried (thanks John!).


from itertools import combinations, permutations
from collections import defaultdict
from number_theory import is_prime

# find combinations of three random numbers in play
# after reveal <n>
def find_in_play(sf3_to_h, n, snp):
  # index hidden numbers on their last <n> digit(s)
  sf_to_h, n = defaultdict(set), 10 ** n
  for sf3, hs in sf3_to_h.items():
    sf_to_h[sf3 % n].update(hs)
  
  # for each last <n> digits find the random numbers that
  # stay in play (i.e. don't win or lose) for these digits 
  sf_to_r = defaultdict(set)
  for sf, hs in sf_to_h.items():
    for r in snp:
      # wins
      if not any(h % r for h in hs):
        continue
      # losses
      if all(h % r for h in hs):
        continue
      # still in play
      sf_to_r[sf].add(r)

  return sf_to_r

# non-primes less than 30, also eliminating those that
# always win (1, 9) and never win (10, 20)
dg = set(range(2, 30)).difference([9, 10, 20])
snp = set(x for x in dg if not is_prime(x))

# index the hidden numbers on their three digit suffix
sf3_to_nbrs = defaultdict(set)
for p in permutations('123456789'):
  h = int(''.join(p))
  sf3_to_nbrs[h % 1000].add(h)

# find sets of random numbers in play after the second reveal
# for each two digit suffix
sf_to_r = find_in_play(sf3_to_nbrs, 2, snp)

# after the third reveal we are only interested in three random 
# numbers, two of which win and one of which loses; consider
# sets of random numbers in play for each two digit suffix
for sf, rs in sf_to_r.items():
  # ... and each combination of three of them
  for r3 in combinations(rs, 3):
    # consider each value of the thousands digit of the hidden numbers 
    for d3 in range(100 + sf, 1000 + sf, 100):
      # look for a combination of three random numbers and a three
      # digit suffix that results in two wins and one loss
      wins = [all(h % r == 0 for h in sf3_to_nbrs[d3]) for r in r3]
      if wins.count(True) == 2:
        losses = [all(h % r != 0 for h in sf3_to_nbrs[d3]) for r in r3]      
        if losses.count(True) == 1:
          fs = 'The three digit suffix is {} with random numbers {}.'
          print(fs.format(d3, r3))

from itertools import combinations, permutations

from collections import defaultdict

from number_theory import is_prime

# find combinations of three random numbers in play

# after reveal <n>

def find_in_play(sf3_to_h, n, snp):

# index hidden numbers on their last <n> digit(s)

sf_to_h, n = defaultdict(set), 10 ** n

for sf3, hs in sf3_to_h.items():

sf_to_h[sf3 % n].update(hs)

# for each last <n> digits find the random numbers that

# stay in play (i.e. don't win or lose) for these digits

sf_to_r = defaultdict(set)

for sf, hs in sf_to_h.items():

for r in snp:

# wins

if not any(h % r for h in hs):

continue

# losses

if all(h % r for h in hs):

continue

# still in play

sf_to_r[sf].add(r)

return sf_to_r

# non-primes less than 30, also eliminating those that

# always win (1, 9) and never win (10, 20)

dg = set(range(2, 30)).difference([9, 10, 20])

snp = set(x for x in dg if not is_prime(x))

# index the hidden numbers on their three digit suffix

sf3_to_nbrs = defaultdict(set)

for p in permutations('123456789'):

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

sf3_to_nbrs[h % 1000].add(h)

# find sets of random numbers in play after the second reveal

# for each two digit suffix

sf_to_r = find_in_play(sf3_to_nbrs, 2, snp)

# after the third reveal we are only interested in three random

# numbers, two of which win and one of which loses; consider

# sets of random numbers in play for each two digit suffix

for sf, rs in sf_to_r.items():

# ... and each combination of three of them

for r3 in combinations(rs, 3):

# consider each value of the thousands digit of the hidden numbers

for d3 in range(100 + sf, 1000 + sf, 100):

# look for a combination of three random numbers and a three

# digit suffix that results in two wins and one loss

wins = [all(h % r == 0 for h in sf3_to_nbrs[d3]) for r in r3]

if wins.count(True) == 2:

losses = [all(h % r != 0 for h in sf3_to_nbrs[d3]) for r in r3]

if losses.count(True) == 1:

fs = 'The three digit suffix is {} with random numbers {}.'

print(fs.format(d3, r3))

Sunday Times Teaser 2891 – Nine Cut Diamonds

by Stephen Hogg

Published February 18 2018 (link)

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