def test_lychrel(num):
n = 1
candidate = num
while n <= 50:
candidate += utils.reverse(candidate)
if utils.palindrome(candidate):
return False
n += 1
return True
def main1():
#fp = open('palindromes.txt','w')
cont = 0
ns = count(2)
n = ns.next()
tot = 0
while n <= 1000:
if palindrome(n):
cont+=1
#fp.write("%d\n" % (n))
if test_sum(n):
print ">", n
tot += n
n = ns.next()
print "total", tot
Can probably build up a hash map at each index if that number will eventually turn into a palindromic number
"""
CACHE = {}
lychrel_count = 0
for n in range(1, 10000):
if n in CACHE:
continue
# an array to hold all the tested values for this number
# if the number winds up being a lychrel number, all the other numbers
# should be cached as "True" to avoid recomputing in a subsequent iteration
m = n
rev_add_count = 0 # how many times has it been reversed and added to self
attempted = [m]
while rev_add_count < 50:
m += utils.reverse(m)
if utils.palindrome(m):
# update cache for each to True
for a in attempted:
CACHE[a] = True
break
else:
# try to reverse again, and add this number to list of those attempted
rev_add_count += 1
attempted.append(m)
if n not in CACHE:
lychrel_count += 1
print lychrel_count
#!/usr/bin/python
import utils
if __name__ == "__main__":
n = range(1,1000000)
p1 = filter(lambda x: utils.palindrome(str(x)), n)
nb = map(lambda x: bin(x)[2:], n)
p2 = filter(utils.palindrome, nb)
p2f = map(lambda x: int(x, 2), p2)
nums = set(p1).intersection(set(p2f))
print len(nums)
print nums
print sum(list(nums))
