def squares():
for x in euler.integers(1):
yield x**2
"""
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x,
contain the same digits.
"""
# very easy once we have a neat way to check the digits of two numbers
from lib import euler
def have_same_digits(a, b):
a_, b_ = str(a), str(b)
if len(a_) != len(b_): return False
a_map, b_map = {}, {}
for n1, n2 in zip(a_, b_):
a_map[n1] = 1 + a_map.get(n1, 0)
b_map[n2] = 1 + b_map.get(n2, 0)
return a_map == b_map
for x in euler.integers(2):
same = True
for m in xrange(2, 6):
if not have_same_digits(x, x*m):
same = False
break
if not same: continue
print x
break
integer, and seeing if the difference can be made up in the list of squares
HURRAH FOR GENERATORS
"""
def squares():
for x in euler.integers(1):
yield x**2
prime_generator = euler.primes()
square_generator = squares()
"""Using dicts for speedy lookup"""
primes = {}
squares = {}
for i in euler.integers(9, 2):
while 1:
p = prime_generator.next()
primes[p] = True
if p > i: break
while 1:
s = square_generator.next()
squares[s] = True
if s > i: break
if i in primes: continue
found = False
for p in primes.keys():
diff = i - p
if diff > 0:
diff/=2
"""
How many n-digit positive integers exist which are also an nth power?
"""
# easy to calculate, but I'm not sure what the upper bound is so this will
# just run until it's interrupted.
# clearly though, the exponential will grow a lot faster than the number of
# digits so I guess you could come up with some way of figuring out the
# upper bound
from lib import euler
integers = []
for power in euler.integers(1):
for base in euler.integers(1):
x = base ** power
if len(str(x)) == power:
integers += [x]
print len(list(set(integers)))
elif len(str(x)) > power:
break
