def recurse_generate(words: list,
trie: Trie,
square_size: int,
chosen_words_length=0) -> list:
if chosen_words_length >= square_size or square_size <= 1:
if check_solution_is_valid(words, square_size):
return words
return None
# build up the solution letter by letter
# on each iteration we check if the substring is a key inside the Trie
# if not a key then we know the current permutation is not a solution so return None
# loop through the characters
for i in range(chosen_words_length, square_size):
prefix = "".join(word[i] for word in words)
# using the soon to be deprecated function because it runs ~30% faster
if not trie.has_keys_with_prefix(prefix):
return None
prefix = "".join(word[chosen_words_length] for word in words)
# we use a prefix to dictate which key to start going over
for word in trie.iterkeys(prefix):
new_list = words + [word]
res = recurse_generate(new_list, trie, square_size,
chosen_words_length + 1)
if res:
return res
return None
def test():
# 1. build a trie
d = dict(zero=0, one=1, two=2, three=3, four=4, five=5, six=6, seven=7,
eight=8, nine=9, ten=10, eleven=11, twelve=12, thirteen=13,
fourteen=10, fifteen=15, sixteen=16, seventeen=17, eighteen=18,
nineteen=19, twenty=20, thirty=30, fourty=40, fifty=50,
sixty=60, seventy=70, eighty=80, ninety=90,
hundred=100)
t = Trie(list(d.keys()))
# 2. scan 2000 "sentences" with it
for _ in range(1000):
# scanning for the longest matches only in sentence 1
i = S1[0]
#print(TEXT[i:S1[1]])
while i < S1[1]:
pfx = list(t.prefixes(TEXT[i:S1[1]]))
if pfx:
k = pfx[-1]
#print(d[k])
i += len(k)
else:
i += 1
# scanning for all matches in sentence 2
i = S2[0]
#print(TEXT[i:S2[1]])
s = 0
while i < S2[1]:
for k in t.prefixes(TEXT[i:S2[1]]):
#print(k)
s += d[k]
i += 1
if s != 142:
raise RuntimeError(str(s))
# 3. make a real list of all keys in the trie
if 'nine' not in list(t.iterkeys()):
raise RuntimeError(str(list(t.iterkeys())))
class TrieNameDB(NameDB):
def __init__(self, pair_gen):
self._dic = self._construct_dic(pair_gen)
self._index = Trie(self._dic.keys())
def _construct_dic(self, pair_gen):
dic = collections.defaultdict(list)
for k, v in pair_gen:
dic[k.lower()].append((k, v))
return dic
def find_by_prefix(self, str, limit=50):
result = []
for key in self._index.iterkeys(str.lower()):
result.extend(self._dic[key])
if limit <= len(result):
break
return result[:limit]
def test():
# 1. build a trie
d = dict(zero=0,
one=1,
two=2,
three=3,
four=4,
five=5,
six=6,
seven=7,
eight=8,
nine=9,
ten=10,
eleven=11,
twelve=12,
thirteen=13,
fourteen=10,
fifteen=15,
sixteen=16,
seventeen=17,
eighteen=18,
nineteen=19,
twenty=20,
thirty=30,
fourty=40,
fifty=50,
sixty=60,
seventy=70,
eighty=80,
ninety=90,
hundred=100)
t = Trie(list(d.keys()))
# 2. scan 2000 "sentences" with it
for _ in range(1000):
# scanning for the longest matches only in sentence 1
i = S1[0]
#print(TEXT[i:S1[1]])
while i < S1[1]:
pfx = list(t.prefixes(TEXT[i:S1[1]]))
if pfx:
k = pfx[-1]
#print(d[k])
i += len(k)
else:
i += 1
# scanning for all matches in sentence 2
i = S2[0]
#print(TEXT[i:S2[1]])
s = 0
while i < S2[1]:
for k in t.prefixes(TEXT[i:S2[1]]):
#print(k)
s += d[k]
i += 1
if s != 142:
raise RuntimeError(str(s))
# 3. make a real list of all keys in the trie
if 'nine' not in list(t.iterkeys()):
raise RuntimeError(str(list(t.iterkeys())))
