def num_of_single_factor_from_begin(begin_list, primes, target_num):
begin_product = product_of_a_list(begin_list)
number_of_factors = len(begin_list) + 1
cnt = 0
# print begin_product
if begin_product > target_num:
return cnt, number_of_factors
# last in begin_list
if begin_list != []:
begin_list_last = begin_list[len(begin_list) - 1]
else:
begin_list_last = 1
# an example of begin_list can be [2, 61]
# or [3, 7, 11]
for p in primes:
# next p must be larger than begin_list_last
if p <= begin_list_last:
continue
product = begin_product * p
if product <= target_num:
cnt += 1
continue
if product > target_num:
break
return cnt, number_of_factors
def get_primes_to_try(begin_list, primes, target_num):
begin_product = product_of_a_list(begin_list)
# this function shall not be called if the following condition not met
assert(begin_product <= target_num)
last_in_begin_list = begin_list[len(begin_list) - 1]
primes_to_try = []
if(begin_product == target_num):
return primes_to_try # this is []
else:
product = begin_product
for p in primes:
# find the 1st p which is larger than last_in_begin_list
if (p <= last_in_begin_list):
continue
product = begin_product*p
if (product <= target_num):
primes_to_try.append(p)
else:
break
return primes_to_try
def is_this_end_of_tree_leaf(begin_list, primes, target_num):
begin_product = product_of_a_list(begin_list)
# from our algorithm, the following must be true
assert(begin_product <= target_num)
# find end leaf and it is a unique factorization
# which means n = p1*p2*p3....*pi
if(begin_product == target_num):
is_end_leaf = g_END_LEAF_WITH_UNIQ_FACTORIZATION
return is_end_leaf
primes_to_try = get_primes_to_try(begin_list, primes, target_num)
# if can not find next_prime_to_try,
if(len(primes_to_try) == 0):
is_end_leaf = g_END_LEAF_NOT_UNIQ_FACTORIZATION
return is_end_leaf
# To test if this is the end leaf, we only need to best with
# the smallest primes in primes_to_try, because if begin_product*next_prime_to_try > target_num
# then all other primes in primes_to_try will also fail the test
next_prime_to_try = primes_to_try[0]
product_with_extra_prime = begin_product*next_prime_to_try
if (product_with_extra_prime > target_num):
is_end_leaf = g_END_LEAF_NOT_UNIQ_FACTORIZATION
# it seems that we will not reach here, let try to catch it to confirm this
# it will be detected by if(len(primes_to_try) == 0) in previous lines.
assert(0)
else: # Note, even if product_with_extra_prime == target_num, this function still return 0
is_end_leaf = g_NOT_END_OF_LEAF
return is_end_leaf
