Use a DAWG as a map

dawg.py

#!/usr/bin/python3
# By Steve Hanov, 2011. Released to the public domain.
# Please see http://stevehanov.ca/blog/index.php?id=115 for the accompanying article.
#
# Based on Daciuk, Jan, et al. "Incremental construction of minimal acyclic finite-state automata." 
# Computational linguistics 26.1 (2000): 3-16.
#
# Updated 2014 to use DAWG as a mapping; see 
# Kowaltowski, T.; CL. Lucchesi (1993), "Applications of finite automata representing large vocabularies", 
# Software-Practice and Experience 1993
import sys
import time

DICTIONARY = "/usr/share/dict/words"
QUERY = sys.argv[1:]

# This class represents a node in the directed acyclic word graph (DAWG). It
# has a list of edges to other nodes. It has functions for testing whether it
# is equivalent to another node. Nodes are equivalent if they have identical
# edges, and each identical edge leads to identical states. The __hash__ and
# __eq__ functions allow it to be used as a key in a python dictionary.
class DawgNode:
    NextId = 0
    
    def __init__(self):
        self.id = DawgNode.NextId
        DawgNode.NextId += 1
        self.final = False
        self.edges = {}

        # Number of end nodes reachable from this one.
        self.count = 0

    def __str__(self):        
        arr = []
        if self.final: 
            arr.append("1")
        else:
            arr.append("0")

        for (label, node) in self.edges.items():
            arr.append( label )
            arr.append( str( node.id ) )

        return "_".join(arr)

    def __hash__(self):
        return self.__str__().__hash__()

    def __eq__(self, other):
        return self.__str__() == other.__str__()

    def numReachable(self):
        # if a count is already assigned, return it
        if self.count: return self.count

        # count the number of final nodes that are reachable from this one.
        # including self
        count = 0
        if self.final: count += 1
        for label, node in self.edges.items():
            count += node.numReachable()

        self.count = count
        return count

class Dawg:
    def __init__(self):
        self.previousWord = ""
        self.root = DawgNode()

        # Here is a list of nodes that have not been checked for duplication.
        self.uncheckedNodes = []

        # Here is a list of unique nodes that have been checked for
        # duplication.
        self.minimizedNodes = {}

        # Here is the data associated with all the nodes
        self.data = []

    def insert( self, word, data ):
        if word <= self.previousWord:
            raise Exception("Error: Words must be inserted in alphabetical " +
                "order.")

        # find common prefix between word and previous word
        commonPrefix = 0
        for i in range( min( len( word ), len( self.previousWord ) ) ):
            if word[i] != self.previousWord[i]: break
            commonPrefix += 1

        # Check the uncheckedNodes for redundant nodes, proceeding from last
        # one down to the common prefix size. Then truncate the list at that
        # point.
        self._minimize( commonPrefix )

        self.data.append(data)

        # add the suffix, starting from the correct node mid-way through the
        # graph
        if len(self.uncheckedNodes) == 0:
            node = self.root
        else:
            node = self.uncheckedNodes[-1][2]

        for letter in word[commonPrefix:]:
            nextNode = DawgNode()
            node.edges[letter] = nextNode
            self.uncheckedNodes.append( (node, letter, nextNode) )
            node = nextNode

        node.final = True
        self.previousWord = word

    def finish( self ):
        # minimize all uncheckedNodes
        self._minimize( 0 );

        # go through entire structure and assign the counts to each node.
        self.root.numReachable()

    def _minimize( self, downTo ):
        # proceed from the leaf up to a certain point
        for i in range( len(self.uncheckedNodes) - 1, downTo - 1, -1 ):
            (parent, letter, child) = self.uncheckedNodes[i];
            if child in self.minimizedNodes:
                # replace the child with the previously encountered one
                parent.edges[letter] = self.minimizedNodes[child]
            else:
                # add the state to the minimized nodes.
                self.minimizedNodes[child] = child;
            self.uncheckedNodes.pop()

    def lookup( self, word ):
        node = self.root
        skipped = 0 # keep track of number of final nodes that we skipped
        for letter in word:
            if letter not in node.edges: return None
            for label, child in sorted(node.edges.items()):
                if label == letter: 
                    if node.final: skipped += 1
                    node = child
                    break
                skipped += child.count

        if node.final:
            return self.data[skipped]

    def nodeCount( self ):
        return len(self.minimizedNodes)

    def edgeCount( self ):
        count = 0
        for node in self.minimizedNodes:
            count += len(node.edges)
        return count

    def display(self):
        stack = [self.root]
        done = set()
        while stack:
            node = stack.pop()
            if node.id in done: continue
            done.add(node.id)
            print("{}: ({})".format(node.id, node))
            for label, child in node.edges.items():
                print("    {} goto {}".format(label, child.id))
                stack.append(child)

if 0:
    dawg = Dawg()
    dawg.insert("cat", 0)
    dawg.insert("catnip", 1)
    dawg.insert("zcatnip", 2)
    dawg.finish()
    dawg.display()
    sys.exit()
            
dawg = Dawg()
WordCount = 0
words = open(DICTIONARY, "rt").read().split()
words.sort()
start = time.time()    
for word in words:
    WordCount += 1
    # insert all words, using the reversed version as the data associated with
    # it
    dawg.insert(word, ''.join(reversed(word)))
    if ( WordCount % 100 ) == 0: print("{0}\r".format(WordCount), end="")
dawg.finish()
print("Dawg creation took {0} s".format(time.time()-start))

EdgeCount = dawg.edgeCount()
print("Read {0} words into {1} nodes and {2} edges".format(
    WordCount, dawg.nodeCount(), EdgeCount))

print("This could be stored in as little as {0} bytes".format(EdgeCount * 4))

for word in QUERY:
    result = dawg.lookup(word)
    if result == None:
        print("{0} not in dictionary.".format(word))
    else:
        print("{0} is in the dictionary and has data {1}".format(word, result))

Thank you, very useful

Nice. Comments are super useful.
What is the algo complexity of this approach? esp .wr.t minimize & finalize.

Is it necessary to use sorted() in numReachable? If yes, why?

Vary cool but not truly minimal, for example you could run

dawg = Dawg(); dawg.insert('cat', '') dawg.insert('catnip', '') dawg.insert('zcatnip', '') dawg.finish()
the z edge would not connect to the first c node.

millerdev: I have removed the sorted() in numReachable since it has no useful effects. It is only necessary in lookup() to calculate the true index based on lexicographic order. In a true application, the edges of each node would be instead stored on disk on the correct order.

gotenxds: I have added a display() function to confirm what you are saying. However, the dawg is still minimal. The reason is that if the dawg were to connect the z to the original c as you suggest, then this would allow the word zcat which is not in the dictionary. The minimization process correctly recognizes that it can join the endings of zcatnip and catnip at "nip".

0: (0_z_7_c_1)
    z goto 7
    c goto 1
1: (0_a_2)
    a goto 2
2: (0_t_3)
    t goto 3
3: (1_n_4)
    n goto 4
4: (0_i_5)
    i goto 5
5: (0_p_6)
    p goto 6
6: (1)
7: (0_c_8)
    c goto 8
8: (0_a_9)
    a goto 9
9: (0_t_10)
    t goto 10
10: (0_n_4)
    n goto 4

I would be interested in how you would combine this with your succinct trie. Any tips? Thanks.

Keith-mcqueen: there is a proof somewhere that succinct structures cannot encode non-planar graphs. So any graph that cannot be drawn without edge crossings won't work. Unfortunately a dawg is non planar so it cannot be represented succinctly. However, I do have code for encoding / accessing one using 4 bytes per edge, and I will post it here soon.

nodeCount and edgeCount seem to return wrong results. E.g. for cat, catnip, zcatnip the node count is off by 1 and the edge count is off by 2 (10 and 9; should be 11 and 11).

The graph: http://i.imgur.com/z8AxsoE.png

That's because the root node is not in minimizedNodes.

I came here trying to understand the minimal hashing described by Lucchesi and Kowaltowski (http://www.ic.unicamp.br/~reltech/1992/92-01.pdf) implemented in actual code (great series of posts on DAWGs btw @smhanov), but I'm still confused - it appears to consist of merely adding all the reachable leaf counts of each letter node in the word path, which, if I'm reading their example correctly, will lead to an identical hash for both "rework" and "replay": 8 + 8 + 4 + 4 + 4 + 4.

@ahamid in the paper take a look at figure 12 for the code to the hashing algorithm, it is more complicated than what you describe. There is an inner and outer for loop that ensures uniqueness.

I think the __eq__() method is flawed: it depends on __str__() which itself depends on the arbitrary order of items in a dictionary.
Changing line 41 to

    for (label, node) in sorted(self.edges.items()):

should fix this.

For my memory efficient implementation in go, see:

https://godoc.org/github.com/smhanov/dawg

Instead of storing a map for edge node, this version uses one map for all edges.

For my memory efficient implementation in go, see:

https://godoc.org/github.com/smhanov/dawg

Instead of storing a map for edge node, this version uses one map for all edges.

@smhanov - how can I modify this if I'm dealing with sentences & I want the edges to be keyed on a token/word as opposed to a character?

@dhrushilbadani - Did you try it? Because it's python, I think it will work unmodified with sentences. Instead of using a word use a sentence split on spaces. Then fix any minor problems you find.

@smhanov - how can I modify this if I'm dealing with sentences & I want the edges to be keyed on a token/word as opposed to a character?

@smhanov makes sense, thanks for the reply! Also, what can one do when sorting the input strings is too expensive?

@dhrushilbadani The original paper "Incremental construction of minimal acyclic finite-state automata." also had an algorithm for non-sorted input, but at the time I thought it was too complicated for my blog entry.

Another option: The Unix sort command is magical and optimized for very large files. Usually when something is too expensive to sort, I will pipe it through "sort" and get the results.