5 * Created by Pat Schloss on 12/15/08.
6 * Copyright 2008 Patrick D. Schloss. All rights reserved.
8 * This is my half-assed attempt to implement a suffix tree. This is a cobbled together algorithm using materials that
9 * I found at http://marknelson.us/1996/08/01/suffix-trees/ and:
11 * Ukkonen E. (1995). On-line construction of suffix trees. Algorithmica 14 (3): 249--260
12 * Gusfield, Dan (1999). Algorithms on Strings, Trees and Sequences: Computer Science and Computational Biology.
13 * USA: Cambridge University Press
15 * The Ukkonen paper is the seminal paper describing the on-line method of constructing a suffix tree.
17 * I have chosen to store the nodes of the tree as a vector of pointers to SuffixNode objects. The root is stored at
18 * nodeVector[0]. Each tree also stores the sequence name and the string that corresponds to the actual sequence.
19 * Finally, this class provides a way of counting the number of suffixes that are needed in one tree to generate a new
20 * sequence (countSuffixes). This method is used to determine similarity between sequences and was inspired by the
21 * article and Perl source code provided at http://www.ddj.com/web-development/184416093.
25 #include "sequence.hpp"
26 #include "suffixnodes.hpp"
27 #include "suffixtree.hpp"
30 //********************************************************************************************************************
32 inline bool compareParents(SuffixNode* left, SuffixNode* right){// this is necessary to print the tree and to sort the
33 return (left->getParentNode() < right->getParentNode()); // nodes in order of their parent
36 //********************************************************************************************************************
38 SuffixTree::SuffixTree(){ m = MothurOut::getInstance(); }
40 //********************************************************************************************************************
42 SuffixTree::~SuffixTree(){
43 for(int i=0;i<nodeVector.size();i++){ delete nodeVector[i]; }
47 //********************************************************************************************************************
49 void SuffixTree::loadSequence(Sequence seq){
50 nodeCounter = 0; // initially there are 0 nodes in the tree
51 activeStartPosition = 0;
52 activeEndPosition = -1;
53 seqName = seq.getName();
54 sequence = seq.convert2ints();
55 sequence += '5'; // this essentially concatenates a '$' to the end of the sequence to
56 int seqLength = sequence.length(); // make it a cononical suffix tree
58 nodeVector.push_back(new SuffixBranch(-1, 0, -1)); // enter the root of the suffix tree
60 activeNode = root = 0;
62 for(int i=0;i<seqLength;i++){
63 addPrefix(i); // step through the sequence adding each prefix
67 //********************************************************************************************************************
69 string SuffixTree::getSeqName() {
73 //********************************************************************************************************************
75 void SuffixTree::print(){
76 vector<SuffixNode*> hold = nodeVector;
77 sort(hold.begin(), hold.end(), compareParents);
78 m->mothurOut("Address\t\tParent\tNode\tSuffix\tStartC\tEndC\tSuffix"); m->mothurOutEndLine();
79 for(int i=1;i<=nodeCounter;i++){
80 hold[i]->print(sequence, i);
84 //********************************************************************************************************************
86 int SuffixTree::countSuffixes(string compareSequence, int& minValue){ // here we count the number of suffix parts
87 // we need to rewrite a user supplied sequence. if the
88 int numSuffixes = 0; // count exceeds the supplied minValue, bail out. The
89 int seqLength = compareSequence.length(); // time complexity should be O(L)
94 while(position < seqLength){ // while the position in the query sequence isn't at the end...
96 if(numSuffixes > minValue) { return 1000000; } // bail if the count gets too high
98 int newNode = nodeVector[presentNode]->getChild(compareSequence[position]); // see if the current node has a
99 // child that matches the next character in the query
101 if(presentNode == 0){ position++; } // if not, go back to the root and increase the count
102 numSuffixes++; // by one.
105 else{ // if there is, move to that node and see how far down
106 presentNode = newNode; // it we can get
108 for(int i=nodeVector[newNode]->getStartCharPos(); i<=nodeVector[newNode]->getEndCharPos(); i++){
109 if(compareSequence[position] == sequence[i]){
110 position++; // as long as the query and branch agree, keep going
113 numSuffixes++; // if there is a mismatch, increase the number of
114 presentNode = 0; // suffixes and go back to the root
119 // if we get all the way through the node we'll go to the top of the while loop and find the child node
120 // that corresponds to what we are interested in
122 numSuffixes--; // the method puts an extra count on numSuffixes
124 if(numSuffixes < minValue) { minValue = numSuffixes; } // if the count is less than the previous minValue,
125 return numSuffixes; // change the value and return the number of suffixes
128 //********************************************************************************************************************
130 int SuffixTree::countSuffixes(string compareSequence){ // here we count the number of suffix parts
131 // we need to rewrite a user supplied sequence. if the
132 int numSuffixes = 0; // count exceeds the supplied minValue, bail out. The
133 int seqLength = compareSequence.length(); // time complexity should be O(L)
138 while(position < seqLength){ // while the position in the query sequence isn't at the end...
140 int newNode = nodeVector[presentNode]->getChild(compareSequence[position]); // see if the current node has a
141 // child that matches the next character in the query
143 if(presentNode == 0){ position++; } // if not, go back to the root and increase the count
144 numSuffixes++; // by one.
147 else{ // if there is, move to that node and see how far down
148 presentNode = newNode; // it we can get
150 for(int i=nodeVector[newNode]->getStartCharPos(); i<=nodeVector[newNode]->getEndCharPos(); i++){
151 if(compareSequence[position] == sequence[i]){
152 position++; // as long as the query and branch agree, keep going
155 numSuffixes++; // if there is a mismatch, increase the number of
156 presentNode = 0; // suffixes and go back to the root
161 // if we get all the way through the node we'll go to the top of the while loop and find the child node
162 // that corresponds to what we are interested in
164 numSuffixes--; // the method puts an extra count on numSuffixes
166 return numSuffixes; // change the value and return the number of suffixes
168 //********************************************************************************************************************
170 void SuffixTree::canonize(){ // if you have to ask how this works, you don't really want to know and this really
171 // isn't the place to ask.
172 if ( isExplicit() == 0 ) { // if the node has no children...
174 int tempNodeIndex = nodeVector[activeNode]->getChild(sequence[activeStartPosition]);
175 SuffixNode* tempNode = nodeVector[tempNodeIndex];
177 int span = tempNode->getEndCharPos() - tempNode->getStartCharPos();
179 while ( span <= ( activeEndPosition - activeStartPosition ) ) {
181 activeStartPosition = activeStartPosition + span + 1;
183 activeNode = tempNodeIndex;
185 if ( activeStartPosition <= activeEndPosition ) {
186 tempNodeIndex = nodeVector[tempNodeIndex]->getChild(sequence[activeStartPosition]);
187 tempNode = nodeVector[tempNodeIndex];
188 span = tempNode->getEndCharPos() - tempNode->getStartCharPos();
195 //********************************************************************************************************************
197 int SuffixTree::split(int nodeIndex, int position){ // leaves stay leaves, etc, to split a leaf we make a new interior
198 // node and reconnect everything
199 SuffixNode* node = nodeVector[nodeIndex]; // get the node that needs to be split
200 SuffixNode* parentNode = nodeVector[node->getParentNode()]; // get it's parent node
202 parentNode->eraseChild(sequence[node->getStartCharPos()]); // erase the present node from the registry of its parent
205 SuffixNode* newNode = new SuffixBranch(node->getParentNode(), node->getStartCharPos(), node->getStartCharPos() + activeEndPosition - activeStartPosition); // create a new node that will link the parent with the old child
206 parentNode->setChildren(sequence[newNode->getStartCharPos()], nodeCounter);// give the parent the new child
207 nodeVector.push_back(newNode);
209 node->setParentNode(nodeCounter); // give the original node the new node as its parent
210 newNode->setChildren(sequence[node->getStartCharPos() + activeEndPosition - activeStartPosition + 1], nodeIndex);
211 // put the original node in the registry of the new node's children
212 newNode->setSuffixNode(activeNode);//link the new node with the old active node
214 // recalculate the startCharPosition of the outermost node
215 node->setStartCharPos(node->getStartCharPos() + activeEndPosition - activeStartPosition + 1 );
217 return node->getParentNode();
220 //********************************************************************************************************************
222 void SuffixTree::makeSuffixLink(int& previous, int present){
224 // here we link the nodes that are suffixes of one another to rapidly speed through the tree
225 if ( previous > 0 ) { nodeVector[previous]->setSuffixNode(present); }
226 else { /* do nothing */ }
231 //********************************************************************************************************************
233 void SuffixTree::addPrefix(int prefixPosition){
235 int lastParentNode = -1; // we need to place a new prefix in the suffix tree
240 parentNode = activeNode;
242 if(isExplicit() == 1){ // if the node is explicit (has kids), try to follow it down the branch if its there...
243 if(nodeVector[activeNode]->getChild(sequence[prefixPosition]) != -1){ // break out and get next prefix...
246 else{ // ...otherwise continue, we'll need to make a new node later on...
249 else{ // if it's not explicit (no kids), read through and see if all of the chars agree...
250 int tempNode = nodeVector[activeNode]->getChild(sequence[activeStartPosition]);
251 int span = activeEndPosition - activeStartPosition;
253 if(sequence[nodeVector[tempNode]->getStartCharPos() + span + 1] == sequence[prefixPosition] ){
254 break; // if the existing suffix agrees with the new one, grab a new prefix...
257 parentNode = split(tempNode, prefixPosition); // ... otherwise we need to split the node
262 nodeCounter++; // we need to generate a new node here if the kid didn't exist, or we split a node
263 SuffixNode* newSuffixLeaf = new SuffixLeaf(parentNode, prefixPosition, sequence.length()-1);
264 nodeVector[parentNode]->setChildren(sequence[prefixPosition], nodeCounter);
265 nodeVector.push_back(newSuffixLeaf);
267 makeSuffixLink( lastParentNode, parentNode ); // make a suffix link for the parent node
269 if(nodeVector[activeNode]->getParentNode() == -1){ // move along the start position for the tree
270 activeStartPosition++;
273 activeNode = nodeVector[activeNode]->getSuffixNode();
275 canonize(); // frankly, i'm not entirely clear on what canonize does.
278 makeSuffixLink( lastParentNode, parentNode );
279 activeEndPosition++; // move along the end position for the tree
281 canonize(); // frankly, i'm not entirely clear on what canonize does.
285 //********************************************************************************************************************