Unoptimized solution for Problem-74.

p74.py

import time

##
# @Fle: p74.py
# @Author: PJensen
# @Version: Oct 31, 2010
# @Description: Unoptimized solution for Problem-74.
# http://projecteuler.net/index.php?section=problems&id=74

# n-factorial
def n_fact(n):
   n_result = 1
   for index in range(1, n + 1):
      n_result *= index
   return n_result

# n-factorial digit sum
def fact_digit_chain_sum(gen):
   def fact_digit_chain(n):
      for c in str(n):
         yield n_fact(int(c))
   return sum(fact_digit_chain(gen))

# generate the full non-repeating sequence
# but do not return the full chain
def fact_digit_chain_seq_len(n):
   c = [n]; current = fact_digit_chain_sum(n)
   while current not in c:
      c.append(current);
      current = fact_digit_chain_sum(current);
   return len(c)

# assertion as to the chain length of 69
assert((fact_digit_chain_seq_len(69) == 5))

# declare constant max and start time
N_MAX = 1000000; tStart = time.time()

# lc that generates all chains w/ length 60 for generators up to N_MAX
chains_with_length_60 = [x for x in range(1, N_MAX) if fact_digit_chain_seq_len(x) == 60]

# show that exact number that meet the above criterion.
print ('chains of length 60 from 0 to %s: %s, calculated in %s') %\
   (N_MAX, str(len(chains_with_length_60)) ,time.time() - tStart)

chains of length 60 from 0 to 1000000: xxxxx, calculated in 1053
(took ~17 minutes to complete, revise to get solution under 1 the minute mark)

Optimizations:

1) Solution is a *very* good candidate for memoization.
2) Since **n_fact(n):** works on the unit digits [0, 9] skip computation return hard-coded factorial.
3) In-lining of functions.
4) Remove iterator in favor of straight function.

Note: (2), (3) and (4) yield a 30% improvement, leaving running time around 738.