2004-08-07 20:10:42 +02:00
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<HEAD>
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<TITLE>Directed Acyclic Word Graphs</TITLE>
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<META NAME="authors" CONTENT="Mark & Ceal Wutka">
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<META NAME="HTML_editor" CONTENT="Arachnophilia 3.9">
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<META NAME="keywords" CONTENT="Scrabble, Dictionary, Scrabble Dictionary, Word Graphs">
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<META NAME="date" CONTENT="12 JUN 1999">
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<META NAME="date" CONTENT="15 APR 1999">
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2004-04-08 11:43:06 +02:00
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2004-04-08 11:43:06 +02:00
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This page has been stolen from <a href="http://www.wutka.com/dawg.html">here</a>.<br>
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<br>
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<H1>Directed Acyclic Word Graphs</H1>
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<P>A Directed Acyclic Word Graph, or DAWG, is a data structure that permits
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extremely fast word searches. The entry point into the graph represents
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the starting letter in the search. Each node represents a letter, and you
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can travel from the node to two other nodes, depending on whether you the
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letter matches the one you are searching for.</p>
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<P>It's a Directed graph because you can only move in a specific direction
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between two nodes. In other words, you can move from A to B, but you can't
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move from B to A. It's Acyclic because there are no cycles. You cannot have
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a path from A to B to C and then back to A. The link back to A would create
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a cycle, and probably an endless loop in your search program.</p>
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<P>The description is a little confusing without an example, so imagine we have
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a DAWG containing the words CAT, CAN, DO, and DOG. The graph woud look like this:</p>
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<P>
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<PRE>
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C --Child--> A --Child--> N (EOW)
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| |
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| Next
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Next |
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| v
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| T (EOW)
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v
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D--Child--> O (EOW) --Child --> G (EOW)
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</PRE>
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<P>Now, imagine that we want to see if CAT is in the DAWG. We start at the entry
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point (the C) in this case. Since C is also the letter we are looking for,
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we go to the child node of C. Now we are looking for the next letter in CAT,
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which is A. Again, the node we are on (A) has the letter we are looking for,
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so we again pick the child node which is now N. Since we are looking for T
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and the current node is not T, we take the Next node instead of the child.
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The Next node of N is T. T has the letter we want. Now, since we have processed
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all the letters in the word we are searching for, we need to make sure that
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the current node has an End-of-word flag (EOW) which it does, so CAT is
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stored in the graph.</p>
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<P>One of the tricks with making a DAWG is trimming it down so that words with
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common endings all end at the same node. For example, suppose we want to store
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DOG and LOG in a DAWG. The ideal would be something like this:</p>
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<P>
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<PRE>
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D --Child--> O --Child--> G(EOW)
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| ^
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Next |
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v |
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L --Child----
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</PRE></p>
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<P>In other words, the OG in DOG and LOG is defined by the same pair of nodes.</P>
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<H2>Creating a DAWG</H2>
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<P>I'm not sure this is the optimal way to create a DAWG, but I have been able
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to create a program that is reasonably efficient (160000 words in about 2
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minutes). </p>
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<P>The idea is to first create a tree, where a leaf would represent the end of
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a word and there can be multiple leaves that are identical. For example,
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DOG and LOG would be stored like this:</p>
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<P>
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<PRE>
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D --Child--> O --Child--> G (EOW)
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Next
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v
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L --Child-> O --Child--> G (EOW)
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</PRE></p>
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<P>Now, suppose you want to add DOGMA to the tree. You'd proceed as if you were
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doing a search. Once you get to G, you find it has no children, so you add
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a child M, and then add a child A to the M, making the graph look like:</p>
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<P>
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<PRE>
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D --Child--> O --Child--> G (EOW) --Child--> M --Child--> A (EOW)
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Next
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v
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L --Child-> O --Child--> G (EOW)
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</PRE></p>
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<P>As you can see, by adding nodes to the tree this way, you share common
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beginnings, but the endings are still separated. To shrink the size of the
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DAWG, you need to find common endings and combine them. To do this, you start
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at the leaf nodes (the nodes that have no children). If two leaf nodes are
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identical, you combine them, moving all the references from one node to the
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other. For two nodes to be identical, they not only must have the same letter,
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but if they have Next nodes, the Next nodes must also be identical (if they
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have child nodes, the child nodes must also be identical).</p>
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<P>Take the following tree of CITIES, CITY, PITIES and PITY:</p>
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<P>
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<PRE>
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C --Child--> I --Child--> T --Child--> I --Child--> E --Child--> S (EOW)
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| Next
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Next |
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| Y (EOW)
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P --Child--> I --Child--> T --Child--> I --Child--> E --Child--> S (EOW)
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Next
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v
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Y (EOW)
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</PRE></p>
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<P>It's probably pretty obvious to you that there is a lot of redundant
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information here. The trick is to make it obvious to a computer. While you
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can see that it would be easiest just to make C and P point to the same
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I, my algorithm for this takes a longer approach.</p>
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<P>Once I create the tree, I run through it and tag each node with a number of
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children and a child depth (the highest number of nodes you can go through
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from the current node to reach a leaf). Leaf nodes have 0 for the child depth and
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0 for the number of children. The main reason for the tagging is that when
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looking for identical nodes, you can quickly rule out nodes with a different
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number of children or a different child depth. When I tag the nodes, I also
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put them in an array of nodes with a similar child depth (again to speed
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searching).</p>
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<P>Now that the nodes have been tagged and sorted by depth, I start with the
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children that have a depth of 0. If node X and node Y are identical, I make
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any node that points to Y as a child now point to X. Originally, I also allowed
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nodes that pointed to Y as a Next node point to X. While this works from
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a data structure standpoint, it made it impossible to store the DAWG in a file
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the way I needed to.</p>
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<P>In the CITY-PITY graph, the algorithm would see that the S's at the end are the
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same. Although the Y nodes are identical, they are not referenced by any
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Child links, only Next links. As I said before, combining common next links
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make it difficult to store the graph the way I needed to.
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The graph would look like this after processing the nodes with a 0 child depth (leaf nodes):</p>
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<P>
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<PRE>
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C --Child--> I --Child--> T --Child--> I --Child--> E --Child--> S (EOW)
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| | ^
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| Next |
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Next | |
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| v |
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| Y (EOW) |
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P --Child--> I --Child--> T --Child--> I --Child--> E --Child--
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Next
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v
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Y (EOW)
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</PRE></p>
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<P>Next, the algorithm looks at nodes with a child depth of 1, which in this
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case would be the two E nodes (although T has 1-deep path to Y, it's
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child depth is 3 because the longest path is the 3-deep path to S).
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In order to test that the two E's are identical, you first see that the letters
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are the same, which they are. Now you make sure that the children are identical.
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Since the only child of E is the same node, you KNOW they are identical. So
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the E's are now combined:</p>
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<P><PRE>
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C --Child--> I --Child--> T --Child--> I --Child--> E --Child--> S (EOW)
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| Next |
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Next | |
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| v |
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| Y (EOW) |
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P --Child--> I --Child--> T --Child--> I --Child-->
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Next
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v
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Y (EOW)
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</PRE></p>
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<P>Notice that the E and S nodes that come from the PITIES word are no longer
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needed. This technique pares the tree down pretty nicely.</p>
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<P>Next come the nodes with a child depth of 2, which would be the I (the
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I's that have ES and Y as children) nodes.
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Applying the same comparison strategy, we can see that both the I nodes are
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identical, so they become combined. This procedure repeats for the T nodes
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and then the I nodes that are the parents of T. The last set of nodes, the
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C & P nodes, are not identical, so they can't be combined. The final tree
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looks like this:</p>
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<P><PRE>
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C --Child--> I --Child--> T --Child--> I --Child--> E --Child--> S (EOW)
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| | |
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| | Next
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Next | |
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| | Y (EOW)
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P --Child----
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</PRE></p>
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<H2>File Format</H2>
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<P>One of the easiest ways to store the DAWG is by writing out each node as a
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4-byte number. The number contains a reference to the first child node,
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some flags to indicate whether the node can end a word, a flag to
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indicate whether the node is the last node in a Next chain, and the letter
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the node represents. Notice that the Next pointer itself is not contained
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in the 4-byte number. Instead, when a node is written, the node that its
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Next pointer refers to is written immediately after it in the file. That's
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why you have an indicator to tell when the node is the last in a Next chain.</p>
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<P>Let's assume, for a minute, that the format used is:
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<UL>
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<LI>2-bytes for the Child pointer,
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<LI>1 byte for the flags (2 indicates end-of-list, 1 indicates end-of-word)
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<LI>1 byte for the character
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</UL>
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<P>The CITY-PITY tree could be written this way:</p>
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<P><PRE>
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00 00 03 00 Dummy null value, allows 0 to indicate no child
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00 03 00 43 C, child at 4-byte-word # 3
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00 03 02 50 P, child at 4-byte-word #3, end-of-list
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00 04 00 49 I, child at 4-byte-word #4
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00 05 00 54 T, child at 4-byte-word #5
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00 06 00 49 I, child at 4-byte-word #7
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00 00 03 59 Y, no child, end-of-word, end-of-list
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00 08 00 45 E, child at 4-byte-word #8
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00 00 03 53 S, no child, end-of-word, end of list
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</PRE></p>
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<P>Having only two bytes for the child pointer limits you to 65536 total nodes.
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The Hasbro Scrabble CDROM uses a more expanded format. The leftmost 22-bits
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indicate the offset of the child node. The next bit is the end-of-list,
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followed by the end-of-word bit. The last 8 bits are the letter (always
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lower-case). Using a 22-bit offset, you can handle 4 million nodes. If you
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ever need more than that, you can get up to 32 million and still store the
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nodes in 4 bytes. For the maximum storage, put the offset in the left-most
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25 bits. The next bit for end-of-list, followed by a bit for end-of-word.
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Finally, the last 5 bits would be the letter, numbered from 0 to 25 for A-Z.</p>
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<P>Also, it is up to you where to start the tree. In the above example, the
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tree starts at offset 1. The Hasbro Scrabble CDROM uses the last 26 nodes in
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the file as the beginning of the tree.</p>
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