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:

- I need to iterate through each prefix P of a given string T
- I need to iterate through each suffix S in prefix P and add that to tree
- To add suffix S to the tree, I need to iterate through each character in S, with the iterations consisting of either walking down an existing branch that starts with the same set of characters C in S and potentially splitting an edge into descendent nodes when I reach a differing character in the suffix, OR if there was no matching edge to walk down. When no matching edge is found to walk down for C, a new leaf edge is created for C.

The basic algorithm appears to be O(n^{2}), 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:

- exactly when and how the "active point" is assigned, used and changed
- what is going on with the canonization aspect of the algorithm
- Why the implementations I've seen need to "fix" bounding variables that they are using

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 (i.e. does not contain any repeated characters), and then extending it to the full algorithm.

First, a few preliminary statements.What we are building, is

basicallylike 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 forthBut: 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:

The algorithm

works in steps, from left to right. There isone 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 atthe current end. I use the symbol`#`

to meanthe 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 stepis to insertall 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 observetwo things:`ab`

isthe sameas 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:up to the current positionafter each step`#`

, 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:

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 point, which is a triple`(active_node,active_edge,active_length)`

`remainder`

, which is an integer indicating how many new suffixes we need to insertThe 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)`

, i.e.`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:do notinsert a fresh edge`[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.set the active pointto`(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 beonly oneedge 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 finalsuffix we need to insert is found to exist in the tree already, the tree itself isnot changedat all (we only update the active point and`remainder`

). The tree is then not an accurate representation of the suffix treeup to the current positionany more, but itcontainsall suffixes (because the final suffix`a`

is containedimplicitly). Hence, apart from updating the variables (which are all of fixed length, so this is O(1)), there wasno workdone in this step.Step 5:We update the current position`#`

to 5. This automatically updates the tree to this:And

because, we need to insert two final suffixes of the current position:`remainder`

is 2`ab`

and`b`

. This is basically because:`a`

suffix from the previous step has never been properly inserted. So it hasremained, 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 suffixof it (including`b`

) must be in the tree, too. Perhaps onlyimplicitly, but it must be there, because of the way we have built the tree so far.We proceed to

step 6by incrementing`#`

. The tree is automatically updated to:Because

, we have to insert`remainder`

is 3`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 calledRule 1below, and it applies whenever the`active_node`

is root (we will learn rule 3 for other cases further below). Here is rule 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, i.e. 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: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 7by 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, we need to insert 'c' and this will help us to understand the final trick:`#`

=9## 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: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 nextremaining insertat 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, i.e. 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

remainingfrom 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 arecontained 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 allimplicitin 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 theyend at a leaf. Otherwise we would get a lot of spurious matches, because there aremanystringsimplicitlycontained 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 forgeneral substrings, not onlysuffixesof 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(n^{2}) 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 (e.g. 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).