Ukkonen's Suffix Tree Construction - Part 1 - GeeksforGeeks
A suffix tree T for a m-character string S is a rooted directed tree with exactly m leaves numbered 1 to m. (Given that last string character is unique in string)
- Root can have zero, one or more children.
- Each internal node, other than the root, has at least two children.
- Each edge is labelled with a nonempty substring of S.
- No two edges coming out of same node can have edge-labels beginning with the same character.
Concatenation of the edge-labels on the path from the root to leaf i gives the suffix of S that starts at position i, i.e. S[i…m].
For String S = xabxa, with m = 5, following is the suffix tree:
Here we will have 5 suffixes: xabxa, abxa, bxa, xa and a.
Path for suffixes ‘xa’ and ‘a’ do not end at a leaf. A tree like above (Figure 2) is called implicit suffix tree as some suffixes (‘xa’ and ‘a’) are not seen explicitly in tree.
Path for suffixes ‘xa’ and ‘a’ do not end at a leaf. A tree like above (Figure 2) is called implicit suffix tree as some suffixes (‘xa’ and ‘a’) are not seen explicitly in tree.
To avoid this problem, we add a character which is not present in string already. We normally use $,
Following is the suffix tree for string S = xabxa$ with m = 6 and now all 6 suffixes end at leaf.
A naive algorithm to build a suffix tree Given a string S of length m, enter a single edge for suffix S[l ..m]$ (the entire string) into the tree, then successively enter suffix S[i..m]$ into the growing tree, for i increasing from 2 to m. Let Nidenote the intermediate tree that encodes all the suffixes from 1 to i. So Ni+1 is constructed from Ni as follows:
- Start at the root of Ni
- Find the longest path from the root which matches a prefix of S[i+1..m]$
- Match ends either at the node (say w) or in the middle of an edge [say (u, v)].
- If it is in the middle of an edge (u, v), break the edge (u, v) into two edges by inserting a new node w just after the last character on the edge that matched a character in S[i+l..m] and just before the first character on the edge that mismatched. The new edge (u, w) is labelled with the part of the (u, v) label that matched with S[i+1..m], and the new edge (w, v) is labelled with the remaining part of the (u, v) label.
- Create a new edge (w, i+1) from w to a new leaf labelled i+1 and it labels the new edge with the unmatched part of suffix S[i+1..m]
This takes O(m2) to build the suffix tree for the string S of length m.
High Level Description of Ukkonen’s algorithm
Ukkonen’s algorithm constructs an implicit suffix tree Ti for each prefix S[l ..i] of S (of length m).
It first builds T1 using 1st character, then T2 using 2nd character, then T3 using 3rd character, …, Tmusing mth character.
Implicit suffix tree Ti+1 is built on top of implicit suffix tree Ti.
The true suffix tree for S is built from Tm by adding $.
At any time, Ukkonen’s algorithm builds the suffix tree for the characters seen so far and so it hason-line property that may be useful in some situations.
Time taken is O(m).
Ukkonen’s algorithm constructs an implicit suffix tree Ti for each prefix S[l ..i] of S (of length m).
It first builds T1 using 1st character, then T2 using 2nd character, then T3 using 3rd character, …, Tmusing mth character.
Implicit suffix tree Ti+1 is built on top of implicit suffix tree Ti.
The true suffix tree for S is built from Tm by adding $.
At any time, Ukkonen’s algorithm builds the suffix tree for the characters seen so far and so it hason-line property that may be useful in some situations.
Time taken is O(m).
Ukkonen’s algorithm is divided into m phases (one phase for each character in the string with length m)
In phase i+1, tree Ti+1 is built from tree Ti.
In phase i+1, tree Ti+1 is built from tree Ti.
High Level Description of Ukkonen’s algorithm
Ukkonen’s algorithm constructs an implicit suffix tree Ti for each prefix S[l ..i] of S (of length m).
It first builds T1 using 1st character, then T2 using 2nd character, then T3 using 3rd character, …, Tmusing mth character.
Implicit suffix tree Ti+1 is built on top of implicit suffix tree Ti.
The true suffix tree for S is built from Tm by adding $.
At any time, Ukkonen’s algorithm builds the suffix tree for the characters seen so far and so it hason-line property that may be useful in some situations.
Time taken is O(m).
Ukkonen’s algorithm constructs an implicit suffix tree Ti for each prefix S[l ..i] of S (of length m).
It first builds T1 using 1st character, then T2 using 2nd character, then T3 using 3rd character, …, Tmusing mth character.
Implicit suffix tree Ti+1 is built on top of implicit suffix tree Ti.
The true suffix tree for S is built from Tm by adding $.
At any time, Ukkonen’s algorithm builds the suffix tree for the characters seen so far and so it hason-line property that may be useful in some situations.
Time taken is O(m).
Ukkonen’s algorithm is divided into m phases (one phase for each character in the string with length m)
In phase i+1, tree Ti+1 is built from tree Ti.
In phase i+1, tree Ti+1 is built from tree Ti.
Each phase i+1 is further divided into i+1 extensions, one for each of the i+1 suffixes of S[1..i+1]
In extension j of phase i+1, the algorithm first finds the end of the path from the root labelled with substring S[j..i].
It then extends the substring by adding the character S(i+1) to its end (if it is not there already).
In extension 1 of phase i+1, we put string S[1..i+1] in the tree. Here S[1..i] will already be present in tree due to previous phase i. We just need to add S[i+1]th character in tree (if not there already).
In extension 2 of phase i+1, we put string S[2..i+1] in the tree. Here S[2..i] will already be present in tree due to previous phase i. We just need to add S[i+1]th character in tree (if not there already)
In extension 3 of phase i+1, we put string S[3..i+1] in the tree. Here S[3..i] will already be present in tree due to previous phase i. We just need to add S[i+1]th character in tree (if not there already)
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In extension i+1 of phase i+1, we put string S[i+1..i+1] in the tree. This is just one character which may not be in tree (if character is seen first time so far). If so, we just add a new leaf edge with label S[i+1].
In extension j of phase i+1, the algorithm first finds the end of the path from the root labelled with substring S[j..i].
It then extends the substring by adding the character S(i+1) to its end (if it is not there already).
In extension 1 of phase i+1, we put string S[1..i+1] in the tree. Here S[1..i] will already be present in tree due to previous phase i. We just need to add S[i+1]th character in tree (if not there already).
In extension 2 of phase i+1, we put string S[2..i+1] in the tree. Here S[2..i] will already be present in tree due to previous phase i. We just need to add S[i+1]th character in tree (if not there already)
In extension 3 of phase i+1, we put string S[3..i+1] in the tree. Here S[3..i] will already be present in tree due to previous phase i. We just need to add S[i+1]th character in tree (if not there already)
.
.
In extension i+1 of phase i+1, we put string S[i+1..i+1] in the tree. This is just one character which may not be in tree (if character is seen first time so far). If so, we just add a new leaf edge with label S[i+1].
High Level Ukkonen’s algorithm
Construct tree T1
For i from 1 to m-1 do
begin {phase i+1}
For j from 1 to i+1
begin {extension j}
Find the end of the path from the root labelled S[j..i] in the current tree.
Extend that path by adding character S[i+l] if it is not there already
end;
end;
Construct tree T1
For i from 1 to m-1 do
begin {phase i+1}
For j from 1 to i+1
begin {extension j}
Find the end of the path from the root labelled S[j..i] in the current tree.
Extend that path by adding character S[i+l] if it is not there already
end;
end;
Suffix extension is all about adding the next character into the suffix tree built so far.
In extension j of phase i+1, algorithm finds the end of S[j..i] (which is already in the tree due to previous phase i) and then it extends S[j..i] to be sure the suffix S[j..i+1] is in the tree.
In extension j of phase i+1, algorithm finds the end of S[j..i] (which is already in the tree due to previous phase i) and then it extends S[j..i] to be sure the suffix S[j..i+1] is in the tree.
There are 3 extension rules:
Rule 1: If the path from the root labelled S[j..i] ends at leaf edge (i.e. S[i] is last character on leaf edge) then character S[i+1] is just added to the end of the label on that leaf edge.
Rule 1: If the path from the root labelled S[j..i] ends at leaf edge (i.e. S[i] is last character on leaf edge) then character S[i+1] is just added to the end of the label on that leaf edge.
Rule 2: If the path from the root labelled S[j..i] ends at non-leaf edge (i.e. there are more characters after S[i] on path) and next character is not s[i+1], then a new leaf edge with label s{i+1] and number j is created starting from character S[i+1].
A new internal node will also be created if s[1..i] ends inside (in-between) a non-leaf edge.
A new internal node will also be created if s[1..i] ends inside (in-between) a non-leaf edge.
Rule 3: If the path from the root labelled S[j..i] ends at non-leaf edge (i.e. there are more characters after S[i] on path) and next character is s[i+1] (already in tree), do nothing.
One important point to note here is that from a given node (root or internal), there will be one and only one edge starting from one character. There will not be more than one edges going out of any node, starting with same character.
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