Ukkonen's suffix tree algorithm in plain English
I feel a bit thick at this point. I've spent days trying to fully wrap my head around suffix tree construction, but because I don't have a mathematical background, many of the explanations elude me as they start to make excessive use of mathematical symbology. The closest to a good explanation that I've found is Fast String Searching With Suffix Trees, but he glosses over various points and some aspects of the algorithm remain unclear.
A step-by-step explanation of this algorithm here on Stack Overflow would be invaluable for many others besides me, I'm sure.
For reference, here's Ukkonen's paper on the algorithm: http://www.cs.helsinki.fi/u/ukkonen/SuffixT1withFigs.pdf
My basic understanding, so far:
The basic algorithm appears to be O(n2), as is pointed out in most explanations, as we need to step through all of the prefixes, then we need to step through each of the suffixes for each prefix. Ukkonen's algorithm is apparently unique because of the suffix pointer technique he uses, though I think that is what I'm having trouble understanding.
I'm also having trouble understanding:
Here is the completed C# source code. It not only works correctly, but supports automatic canonization and renders a nicer looking text graph of the output. Source code and sample output is at:
https://gist.github.com/2373868
Update 2017-11-04
After many years I've found a new use for suffix trees, and have implemented the algorithm in JavaScript . Gist is below. It should be bug-free. Dump it into a js file, npm install chalk from the same location, and then run with node.js to see some colourful output. There's a stripped down version in the same Gist, without any of the debugging code.
https://gist.github.com/axefrog/c347bf0f5e0723cbd09b1aaed6ec6fc6
The following is an attempt to describe the Ukkonen algorithm by first showing what it does when the string is simple (ie does not contain any repeated characters), and then extending it to the full algorithm.
First, a few preliminary statements.
What we are building, is basically like a search trie. So there is a root node, edges going out of it leading to new nodes, and further edges going out of those, and so forth
But : Unlike in a search trie, the edge labels are not single characters. Instead, each edge is labeled using a pair of integers [from,to] . These are pointers into the text. In this sense, each edge carries a string label of arbitrary length, but takes only O(1) space (two pointers).
Basic principle
I would like to first demonstrate how to create the suffix tree of a particularly simple string, a string with no repeated characters:
abc
The algorithm works in steps, from left to right . There is one step for every character of the string . Each step might involve more than one individual operation, but we will see (see the final observations at the end) that the total number of operations is O(n).
So, we start from the left, and first insert only the single character a by creating an edge from the root node (on the left) to a leaf, and labeling it as [0,#] , which means the edge represents the substring starting at position 0 and ending at the current end. I use the symbol # to mean the current end, which is at position 1 (right after a ).
So we have an initial tree, which looks like this:
And what it means is this:
Now we progress to position 2 (right after b ). Our goal at each step is to insert all suffixes up to the current position . We do this by
a -edge to ab b In our representation this looks like
And what it means is:
We observe two things:
ab is the same as it used to be in the initial tree: [0,#] . Its meaning has automatically changed because we updated the current position # from 1 to 2. Next we increment the position again and update the tree by appending a c to every existing edge and inserting one new edge for the new suffix c .
In our representation this looks like
And what it means is:
We observe:
# , and inserting the one new edge for the final character can be done in O(1) time. Hence for a string of length n, only O(n) time is required. First extension: Simple repetitions
Of course this works so nicely only because our string does not contain any repetitions. We now look at a more realistic string:
abcabxabcd
It starts with abc as in the previous example, then ab is repeated and followed by x , and then abc is repeated followed by d .
Steps 1 through 3: After the first 3 steps we have the tree from the previous example:
Step 4: We move # to position 4. This implicitly updates all existing edges to this:
and we need to insert the final suffix of the current step, a , at the root.
Before we do this, we introduce two more variables (in addition to # ), which of course have been there all the time but we haven't used them so far:
(active_node,active_edge,active_length) remainder , which is an integer indicating how many new suffixes we need to insert The exact meaning of these two will become clear soon, but for now let's just say:
abc example, the active point was always (root,' x',0) , ie active_node was the root node, active_edge was specified as the null character ' x' , and active_length was zero. The effect of this was that the one new edge that we inserted in every step was inserted at the root node as a freshly created edge. We will see soon why a triple is necessary to represent this information. remainder was always set to 1 at the beginning of each step. The meaning of this was that the number of suffixes we had to actively insert at the end of each step was 1 (always just the final character). Now this is going to change. When we insert the current final character a at the root, we notice that there is already an outgoing edge starting with a , specifically: abca . Here is what we do in such a case:
[4,#] at the root node. Instead we simply notice that the suffix a is already in our tree. It ends in the middle of a longer edge, but we are not bothered by that. We just leave things the way they are. (root,'a',1) . That means the active point is now somewhere in the middle of outgoing edge of the root node that starts with a , specifically, after position 1 on that edge. We notice that the edge is specified simply by its first character a . That suffices because there can be only one edge starting with any particular character (confirm that this is true after reading through the entire description). remainder , so at the beginning of the next step it will be 2. Observation: When the final suffix we need to insert is found to exist in the tree already , the tree itself is not changed at all (we only update the active point and remainder ). The tree is then not an accurate representation of the suffix tree up to the current position any more, but it contains all suffixes (because the final suffix a is contained implicitly). Hence, apart from updating the variables (which are all of fixed length, so this is O(1)), there was no work done in this step.
Step 5: We update the current position # to 5. This automatically updates the tree to this:
And because remainder is 2 , we need to insert two final suffixes of the current position: ab and b . This is basically because:
a suffix from the previous step has never been properly inserted. So it has remained, and since we have progressed one step, it has now grown from a to ab . b . In practice this means that we go to the active point (which points to behind the a on what is now the abcab edge), and insert the current final character b . But: Again, it turns out that b is also already present on that same edge.
So, again, we do not change the tree. We simply:
(root,'a',2) (same node and edge as before, but now we point to behind the b ) remainder to 3 because we still have not properly inserted the final edge from the previous step, and we don't insert the current final edge either. To be clear: We had to insert ab and b in the current step, but because ab was already found, we updated the active point and did not even attempt to insert b . Why? Because if ab is in the tree, every suffix of it (including b ) must be in the tree, too. Perhaps only implicitly, but it must be there, because of the way we have built the tree so far.
We proceed to step 6 by incrementing # . The tree is automatically updated to:
Because remainder is 3 , we have to insert abx , bx and x . The active point tells us where ab ends, so we only need to jump there and insert the x . Indeed, x is not there yet, so we split the abcabx edge and insert an internal node:
The edge representations are still pointers into the text, so splitting and inserting an internal node can be done in O(1) time.
So we have dealt with abx and decrement remainder to 2. Now we need to insert the next remaining suffix, bx . But before we do that we need to update the active point. The rule for this, after splitting and inserting an edge, will be called Rule 1 below, and it applies whenever the active_node is root (we will learn rule 3 for other cases further below). Here is rule 1:
After an insertion from root,
active_node remains root active_edge is set to the first character of the new suffix we need to insert, ie b active_length is reduced by 1 Hence, the new active-point triple (root,'b',1) indicates that the next insert has to be made at the bcabx edge, behind 1 character, ie behind b . We can identify the insertion point in O(1) time and check whether x is already present or not. If it was present, we would end the current step and leave everything the way it is. But x is not present, so we insert it by splitting the edge:
Again, this took O(1) time and we update remainder to 1 and the active point to (root,'x',0) as rule 1 states.
But there is one more thing we need to do. We'll call this Rule 2:
If we split an edge and insert a new node, and if that is not the first node created during the current step, we connect the previously inserted node and the new node through a special pointer, a suffix link . We will later see why that is useful. Here is what we get, the suffix link is represented as a dotted edge:
We still need to insert the final suffix of the current step, x . Since the active_length component of the active node has fallen to 0, the final insert is made at the root directly. Since there is no outgoing edge at the root node starting with x , we insert a new edge:
As we can see, in the current step all remaining inserts were made.
We proceed to step 7 by setting # =7, which automatically appends the next character, a , to all leaf edges, as always. Then we attempt to insert the new final character to the active point (the root), and find that it is there already. So we end the current step without inserting anything and update the active point to (root,'a',1) .
In step 8 , # =8, we append b , and as seen before, this only means we update the active point to (root,'a',2) and increment remainder without doing anything else, because b is already present. However, we notice (in O(1) time) that the active point is now at the end of an edge. We reflect this by re-setting it to (node1,' x',0) . Here, I use node1 to refer to the internal node the ab edge ends at.
Then, in step # =9 , we need to insert 'c' and this will help us to understand the final trick:
Second extension: Using suffix links
As always, the # update appends c automatically to the leaf edges and we go to the active point to see if we can insert 'c'. It turns out 'c' exists already at that edge, so we set the active point to (node1,'c',1) , increment remainder and do nothing else.
Now in step # =10 , remainder is 4 , and so we first need to insert abcd (which remains from 3 steps ago) by inserting d at the active point.
Attempting to insert d at the active point causes an edge split in O(1) time:
The active_node , from which the split was initiated, is marked in red above. Here is the final rule, Rule 3:
After splitting an edge from an active_node that is not the root node, we follow the suffix link going out of that node, if there is any, and reset the active_node to the node it points to. If there is no suffix link, we set the active_node to the root. active_edge and active_length remain unchanged.
So the active point is now (node2,'c',1) , and node2 is marked in red below:
Since the insertion of abcd is complete, we decrement remainder to 3 and consider the next remaining suffix of the current step, bcd . Rule 3 has set the active point to just the right node and edge so inserting bcd can be done by simply inserting its final character d at the active point.
Doing this causes another edge split, and because of rule 2 , we must create a suffix link from the previously inserted node to the new one:
We observe: Suffix links enable us to reset the active point so we can make the next remaining insert at O(1) effort. Look at the graph above to confirm that indeed node at label ab is linked to the node at b (its suffix), and the node at abc is linked to bc .
The current step is not finished yet. remainder is now 2, and we need to follow rule 3 to reset the active point again. Since the current active_node (red above) has no suffix link, we reset to root. The active point is now (root,'c',1) .
Hence the next insert occurs at the one outgoing edge of the root node whose label starts with c : cabxabcd , behind the first character, ie behind c . This causes another split:
And since this involves the creation of a new internal node,we follow rule 2 and set a new suffix link from the previously created internal node:
(I am using Graphviz Dot for these little graphs. The new suffix link caused dot to re-arrange the existing edges, so check carefully to confirm that the only thing that was inserted above is a new suffix link.)
With this, remainder can be set to 1 and since the active_node is root, we use rule 1 to update the active point to (root,'d',0) . This means the final insert of the current step is to insert a single d at root:
That was the final step and we are done. There are number of final observations , though:
In each step we move # forward by 1 position. This automatically updates all leaf nodes in O(1) time.
But it does not deal with a) any suffixes remaining from previous steps, and b) with the one final character of the current step.
remainder tells us how many additional inserts we need to make. These inserts correspond one-to-one to the final suffixes of the string that ends at the current position # . We consider one after the other and make the insert. Important: Each insert is done in O(1) time since the active point tells us exactly where to go, and we need to add only one single character at the active point. Why? Because the other characters are contained implicitly (otherwise the active point would not be where it is).
After each such insert, we decrement remainder and follow the suffix link if there is any. If not we go to root (rule 3). If we are at root already, we modify the active point using rule 1. In any case, it takes only O(1) time.
If, during one of these inserts, we find that the character we want to insert is already there, we don't do anything and end the current step, even if remainder >0. The reason is that any inserts that remain will be suffixes of the one we just tried to make. Hence they are all implicit in the current tree. The fact that remainder >0 makes sure we deal with the remaining suffixes later.
What if at the end of the algorithm remainder >0? This will be the case whenever the end of the text is a substring that occurred somewhere before. In that case we must append one extra character at the end of the string that has not occurred before. In the literature, usually the dollar sign $ is used as a symbol for that. Why does that matter? --> If later we use the completed suffix tree to search for suffixes, we must accept matches only if they end at a leaf. Otherwise we would get a lot of spurious matches, because there are many strings implicitly contained in the tree that are not actual suffixes of the main string. Forcing remainder to be 0 at the end is essentially a way to ensure that all suffixes end at a leaf node. However, if we want to use the tree to search for general substrings, not only suffixes of the main string, this final step is indeed not required, as suggested by the OP's comment below.
So what is the complexity of the entire algorithm? If the text is n characters in length, there are obviously n steps (or n+1 if we add the dollar sign). In each step we either do nothing (other than updating the variables), or we make remainder inserts, each taking O(1) time. Since remainder indicates how many times we have done nothing in previous steps, and is decremented for every insert that we make now, the total number of times we do something is exactly n (or n+1). Hence, the total complexity is O(n).
However, there is one small thing that I did not properly explain: It can happen that we follow a suffix link, update the active point, and then find that its active_length component does not work well with the new active_node . For example, consider a situation like this:
(The dashed lines indicate the rest of the tree. The dotted line is a suffix link.)
Now let the active point be (red,'d',3) , so it points to the place behind the f on the defg edge. Now assume we made the necessary updates and now follow the suffix link to update the active point according to rule 3. The new active point is (green,'d',3) . However, the d -edge going out of the green node is de , so it has only 2 characters. In order to find the correct active point, we obviously need to follow that edge to the blue node and reset to (blue,'f',1) .
In a particularly bad case, the active_length could be as large as remainder , which can be as large as n. And it might very well happen that to find the correct active point, we need not only jump over one internal node, but perhaps many, up to n in the worst case. Does that mean the algorithm has a hidden O(n2) complexity, because in each step remainder is generally O(n), and the post-adjustments to the active node after following a suffix link could be O(n), too?
No. The reason is that if indeed we have to adjust the active point (eg from green to blue as above), that brings us to a new node that has its own suffix link, and active_length will be reduced. As we follow down the chain of suffix links we make the remaining inserts, active_length can only decrease, and the number of active-point adjustments we can make on the way can't be larger than active_length at any given time. Since active_length can never be larger than remainder , and remainder is O(n) not only in every single step, but the total sum of increments ever made to remainder over the course of the entire process is O(n) too, the number of active point adjustments is also bounded by O(n).
I tried to implement the Suffix Tree with the approach given in jogojapan's answer, but it didn't work for some cases due to wording used for the rules. Moreover, I've mentioned that nobody managed to implement an absolutely correct suffix tree using this approach. Below I will write an "overview" of jogojapan's answer with some modifications to the rules. I will also describe the case when we forget to create important suffix links.
Additional variables used
Let's use a concept of an internal node - all the nodes, except the root and the leafs are internal nodes .
Observation 1
When the final suffix we need to insert is found to exist in the tree already, the tree itself is not changed at all (we only update the active point and remainder ).
Observation 2
If at some point active_length is greater or equal to the length of current edge ( edge_length ), we move our active point down until edge_length is strictly greater than active_length .
Now, let's redefine the rules:
Rule 1
If after an insertion from the active node = root, the active length is greater than 0, then:
Rule 2
If we create a new internal node OR make an inserter from an internal node, and this is not the first SUCH internal node at current step, then we link the previous SUCH node with THIS one through a suffix link.
This definition of the Rule 2 is different from jogojapan', as here we take into account not only the newly created internal nodes, but also the internal nodes, from which we make an insertion.
Rule 3
After an insert from the active node which is not the root node, we must follow the suffix link and set the active node to the node it points to. If there is no a suffix link, set the active node to the root node. Either way, active edge and active length stay unchanged.
In this definition of Rule 3 we also consider the inserts of leaf nodes (not only split-nodes).
And finally, Observation 3:
When the symbol we want to add to the tree is already on the edge, we, according to Observation 1 , update only active point and remainder , leaving the tree unchanged. BUT if there is an internal node marked as needing suffix link, we must connect that node with our current active node through a suffix link.
Let's look at the example of a suffix tree for cdddcdc if we add a suffix link in such case and if we don't:
If we DON'T connect the nodes through a suffix link:
If we DO connect the nodes through a suffix link:
Seems like there is no significant difference: in the second case there are two more suffix links. But these suffix links are correct, and one of them - from the blue node to the red one - is very important for our approach with active point . The problem is that if we don't put a suffix link here, later, when we add some new letters to the tree, we might omit adding some nodes to the tree due to the Rule 3 , because, according to it, if there's no a suffix link, then we must put the active_node to the root.
When we were adding the last letter to the tree, the red node had already existed before we made an insert from the blue node (the edge labled 'c' ). As there was an insert from the blue node, we mark it as needing a suffix link. Then, relying on the active point approach, the active node was set to the red node. But we don't make an insert from the red node, as the letter 'c' is already on the edge. Does it mean that the blue node must be left without a suffix link? No, we must connect the blue node with the red one through a suffix link. Why is it correct? Because the active point approach guarantees that we get to a right place, ie, to the next place where we must process an insert of a shorter suffix.
Finally, here are my implementations of the Suffix Tree:
Hope that this "overview" combined with jogojapan's detailed answer will help somebody to implement his own Suffix Tree.
My intuition is as follows:
After k iterations of the main loop you have constructed a suffix tree which contains all suffixes of the complete string that start in the first k characters.
At the start, this means the suffix tree contains a single root node that represents the entire string (this is the only suffix that starts at 0).
After len(string) iterations you have a suffix tree that contains all suffixes.
During the loop the key is the active point. My guess is that this represents the deepest point in the suffix tree that corresponds to a proper suffix of the first k characters of the string. (I think proper means that the suffix cannot be the entire string.)
For example, suppose you have seen characters 'abcabc'. The active point would represent the point in the tree corresponding to the suffix 'abc'.
The active point is represented by (origin,first,last). This means that you are currently at the point in the tree that you get to by starting at node origin and then feeding in the characters in string[first:last]
When you add a new character you look to see whether the active point is still in the existing tree. If it is then you are done. Otherwise you need to add a new node to the suffix tree at the active point, fallback to the next shortest match, and check again.
Note 1: The suffix pointers give a link to the next shortest match for each node.
Note 2: When you add a new node and fallback you add a new suffix pointer for the new node. The destination for this suffix pointer will be the node at the shortened active point. This node will either already exist, or be created on the next iteration of this fallback loop.
Note 3: The canonization part simply saves time in checking the active point. For example, suppose you always used origin=0, and just changed first and last. To check the active point you would have to follow the suffix tree each time along all the intermediate nodes. It makes sense to cache the result of following this path by recording just the distance from the last node.
Can you give a code example of what you mean by "fix" bounding variables?
Health warning: I also found this algorithm particularly hard to understand so please realise that this intuition is likely to be incorrect in all important details...
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