Here is my practise with boggle letter square non-overlapping word finder after some drastic debugging and cleaning.
I think it is reasonable speed also. Comments wellcome.
Yet another boggle word game
from time import clock
DICT = 'unixdict.txt'
def word_path(word,used=[],pos=None):
if word:
correct_neighbour = [neigh
for p,neigh in neighbour_set
if ((pos is None or pos==p) and
(neigh not in used) and
boggle[neigh]==word[0]
)
]
for i in correct_neighbour:
used_copy=used[:]+[i]
if len(word)==1:
if boggle[i]==word:
return (used_copy,)
else:
for solution in word_path(word[1:],used_copy,pos=i):
if solution:
return (solution,)
return (False,)
neighbours=(
(-1,-1), (-1,0), (-1,1), ## above line
(0,-1), (0,1), ## same line
(1,-1), (1,0), (1,1) ## next line
)
wantquit = False
while wantquit != 'q':
boggle=''
while not boggle:
boggle=raw_input('Give your boggle:').lower() ## CATXANTXTREEEBXY
dimension=int(len(boggle)**0.5)
if dimension**2!=len(boggle):
print 'Not square boggle, give again!'
boggle=''
for i in range(dimension,len(boggle)+1,dimension):
print "%2i: %s" % (i-dimension,boggle[i-dimension:i])
neighbour_indexpairs = set(((i,j), (i+di,j+dj))
for i in range(dimension)
for j in range(dimension)
for di,dj in neighbours if 0<=dj+j<dimension and 0<=di+i<dimension
)
# change to one dimension
neighbour_set=set((a*dimension+b,c*dimension+d) for (a,b),(c,d) in neighbour_indexpairs)
t0=clock()
letterpairs = set((boggle[a]+boggle[b]) for a,b in neighbour_set)
words = set(word for word in open(DICT).read().split()
if (word.lower()==word and
len(word)>dimension-2 and
all((word[i]+word[i+1] in letterpairs) for i in range(len(word)-1))
)
)
print len(words),'candidates'
t1=clock()
result = [word for word in words
for solution in word_path(word)
if solution]
result.sort()
result.sort(key=len, reverse=True)
print "\nTotal", len(result),"solutions."
print "\nFound in %.3f ms" % (1000*(clock()-t0)),
print "of which final check took %.3f ms" % (1000*(clock()-t1))
print '\t'.join(result)
wantquit=raw_input('Ready! Push q to finish something to continue.\n').lower()
TrustyTony 888 pyMod Team Colleague Featured Poster
I did little optimizing after it is working and got "slightly" better performance:
(with example box: CATXANTXTREEEBXY)
Previous version:
Found in 229.725 ms of which final check took 102.581 ms
First improvements (dictionary only split once etc):
Found in 168.441 ms of which final check took 46.105 ms
This version:
Found in 127.655 ms of which final check took 5.148 ms
from time import clock
DICT = 'unixdict.txt'
def word_path(word,used=[],pos=None):
if word:
correct_neighbour = [neigh
for neigh in (neighbourlist[pos] if pos else range(dimension*dimension))
if ((neigh not in used) and
boggle[neigh]==word[0])
]
for i in correct_neighbour:
used_copy=used[:]+[i]
if len(word)==1:
if boggle[i]==word:
return (used_copy,)
else:
for solution in word_path(word[1:],used_copy,pos=i):
if solution:
return (solution,)
return (False,)
neighbours=(
(-1,-1), (0,-1), (1,-1), ## left column
(-1,0), (1,0), ## same column
(-1,1), (0,1), (1,1) ## next column
)
wordlist=open(DICT).read().split()
wantquit = False
while wantquit != 'q':
boggle=''
while not boggle:
boggle=raw_input('Give your boggle:').lower() ## CATXANTXTREEEBXY
dimension=int(len(boggle)**0.5)
if dimension**2!=len(boggle):
print 'Not square boggle, give again!'
boggle=''
for i in range(0,len(boggle)-dimension+1,dimension):
print "%2i: %s" % (i,boggle[i:i+dimension])
neighbour_indexpairs = set(((i,j), (i+di,j+dj))
for i in range(dimension)
for j in range(dimension)
for di,dj in neighbours if 0<=dj+j<dimension and 0<=di+i<dimension
)
# change to one dimension
neighbour_set=set((a*dimension+b,c*dimension+d) for (a,b),(c,d) in neighbour_indexpairs)
neighbourlist = [[] for _ in boggle]
[neighbourlist[char_index].append(b) for char_index,b in neighbour_set]
t0=clock()
letterpairs = set((boggle[a]+boggle[b]) for a,b in neighbour_set)
words = set(word for word in wordlist
if (word.lower()==word and
len(word)>dimension-2 and
all((word[i]+word[i+1] in letterpairs) for i in range(len(word)-1))
)
)
print len(words),'candidates'
t1=clock()
result = [word for word in words
for solution in word_path(word)
if solution]
result.sort()
result.sort(key=len, reverse=True)
print "\nTotal", len(result),"solutions."
print "\nFound in %.3f ms" % (1000*(clock()-t0)),
print "of which final check took %.3f ms" % (1000*(clock()-t1))
print '\t'.join(result)
wantquit=raw_input('Ready! Push q to finish something to continue.\n').lower() TrustyTony 888 pyMod Team Colleague Featured Poster
Line 7 should be
for neigh in (neighbourlist[pos] if pos is not None else range(dimension*dimension))0 is valid index of neighbour list.
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