def problem60(limit=9000):
primes = list(takewhile(lambda x: x < limit, get_primes()))
primes.reverse() # we want to search smaller primes first from pop()
# Use depth-first search.
frontier = [[p] for p in primes]
while frontier:
node = frontier.pop()
if len(node) == 5:
return sum(node)
for x in primes:
child = node + [x]
if x > max(node) and all_concat_to_prime(child):
frontier.append(child)
def problem70():
# The search space is too large for brute-force. So, note that we are
# seeking roughly the inverse of the previous problem -- to minimize
# n/phi(n). Therefore, we want to maximize phi(n), which is acheived for
# numbers with the fewest and largest unique prime factors. But the number
# cannot simply be prime because in that case phi(n) == n-1 which is not a
# permutation of n. Therefore, the best candidates should have two unique
# prime factors.
def is_permutation(x, y):
return sorted(str(x)) == sorted(str(y))
# Since we are seeking large values for both prime factors, we can search
# among numbers close to the value of sqrt(1e7) ~ 3162
ps = list(takewhile(lambda x: x < 4000, get_primes(start=2000)))
ns = [x*y for x in ps
for y in ps
if x != y and x*y < 1e7]
candidates = [n for n in ns if is_permutation(n, phi(n))]
return min(candidates, key=lambda n: n/phi(n))
def problem51():
return next(filter(is_smallest_member, get_primes(start=56995)))
def candidates():
primes = get_primes()
x = next(primes)
while True:
yield x
x *= next(primes)
