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