Question

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(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:

• 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

Answer 1

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.

1. 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

2. 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

• expanding the existing a-edge to ab
• inserting one new edge for b

In our representation this looks like

And what it means is:

We observe two things:

• The edge representation for 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.
• Each edge consumes O(1) space, because it consists of only two
  pointers into the text, regardless of how many characters it
  represents.

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:

• The tree is the correct suffix tree up to the current position
  after each step
• There are as many steps as there are characters in the text
• The amount of work in each step is O(1), because all existing edges
  are updated automatically by incrementing #, 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:

• The active point, which is a triple
  (active_node,active_edge,active_length)
• The 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:

• In the simple abc example, the active point was always
  (root,'\0x',0), i.e. active_node was the root node, active_edge was specified as the null character '\0x', 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.
• The 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:

• We do not insert 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.
• We set the active point to (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).
• We also increment 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:

• The 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.
• And we need to insert the new final edge 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:

• Update the active point to (root,'a',2) (same node and edge
  as before, but now we point to behind the b)
• Increment the 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, i.e. 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,
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:

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,'\0x',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,
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 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
(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).