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(){}
40 //********************************************************************************************************************
42 SuffixTree::~SuffixTree(){
43 for(int i=0;i<nodeVector.size();i++){ delete nodeVector[i]; }
46 //********************************************************************************************************************
48 void SuffixTree::loadSequence(Sequence* seq){
49 nodeCounter = 0; // initially there are 0 nodes in the tree
50 activeStartPosition = 0;
51 activeEndPosition = -1;
52 seqName = seq->getName();
53 sequence = seq->convert2ints();
54 sequence += '5'; // this essentially concatenates a '$' to the end of the sequence to
55 int seqLength = sequence.length(); // make it a cononical suffix tree
57 nodeVector.push_back(new SuffixBranch(-1, 0, -1)); // enter the root of the suffix tree
59 activeNode = root = 0;
61 for(int i=0;i<seqLength;i++){
62 addPrefix(i); // step through the sequence adding each prefix
66 //********************************************************************************************************************
68 string SuffixTree::getSeqName() {
72 //********************************************************************************************************************
74 void SuffixTree::print(){
75 vector<SuffixNode*> hold = nodeVector;
76 sort(hold.begin(), hold.end(), compareParents);
77 cout << "Address\t\tParent\tNode\tSuffix\tStartC\tEndC\tSuffix" << endl;
78 for(int i=1;i<=nodeCounter;i++){
79 hold[i]->print(sequence, i);
83 //********************************************************************************************************************
85 int SuffixTree::countSuffixes(string compareSequence, int& minValue){ // here we count the number of suffix parts
86 // we need to rewrite a user supplied sequence. if the
87 int numSuffixes = 0; // count exceeds the supplied minValue, bail out. The
88 int seqLength = compareSequence.length(); // time complexity should be O(L)
93 while(position < seqLength){ // while the position in the query sequence isn't at the end...
95 if(numSuffixes > minValue) { return 1000000; } // bail if the count gets too high
97 int newNode = nodeVector[presentNode]->getChild(compareSequence[position]); // see if the current node has a
98 // child that matches the next character in the query
100 if(presentNode == 0){ position++; } // if not, go back to the root and increase the count
101 numSuffixes++; // by one.
104 else{ // if there is, move to that node and see how far down
105 presentNode = newNode; // it we can get
107 for(int i=nodeVector[newNode]->getStartCharPos(); i<=nodeVector[newNode]->getEndCharPos(); i++){
108 if(compareSequence[position] == sequence[i]){
109 position++; // as long as the query and branch agree, keep going
112 numSuffixes++; // if there is a mismatch, increase the number of
113 presentNode = 0; // suffixes and go back to the root
118 // if we get all the way through the node we'll go to the top of the while loop and find the child node
119 // that corresponds to what we are interested in
121 numSuffixes--; // the method puts an extra count on numSuffixes
123 if(numSuffixes < minValue) { minValue = numSuffixes; } // if the count is less than the previous minValue,
124 return numSuffixes; // change the value and return the number of suffixes
128 //********************************************************************************************************************
130 void SuffixTree::canonize(){ // if you have to ask how this works, you don't really want to know and this really
131 // isn't the place to ask.
132 if ( isExplicit() == 0 ) { // if the node has no children...
134 int tempNodeIndex = nodeVector[activeNode]->getChild(sequence[activeStartPosition]);
135 SuffixNode* tempNode = nodeVector[tempNodeIndex];
137 int span = tempNode->getEndCharPos() - tempNode->getStartCharPos();
139 while ( span <= ( activeEndPosition - activeStartPosition ) ) {
141 activeStartPosition = activeStartPosition + span + 1;
143 activeNode = tempNodeIndex;
145 if ( activeStartPosition <= activeEndPosition ) {
146 tempNodeIndex = nodeVector[tempNodeIndex]->getChild(sequence[activeStartPosition]);
147 tempNode = nodeVector[tempNodeIndex];
148 span = tempNode->getEndCharPos() - tempNode->getStartCharPos();
155 //********************************************************************************************************************
157 int SuffixTree::split(int nodeIndex, int position){ // leaves stay leaves, etc, to split a leaf we make a new interior
158 // node and reconnect everything
159 SuffixNode* node = nodeVector[nodeIndex]; // get the node that needs to be split
160 SuffixNode* parentNode = nodeVector[node->getParentNode()]; // get it's parent node
162 parentNode->eraseChild(sequence[node->getStartCharPos()]); // erase the present node from the registry of its parent
165 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
166 parentNode->setChildren(sequence[newNode->getStartCharPos()], nodeCounter);// give the parent the new child
167 nodeVector.push_back(newNode);
169 node->setParentNode(nodeCounter); // give the original node the new node as its parent
170 newNode->setChildren(sequence[node->getStartCharPos() + activeEndPosition - activeStartPosition + 1], nodeIndex);
171 // put the original node in the registry of the new node's children
172 newNode->setSuffixNode(activeNode);//link the new node with the old active node
174 // recalculate the startCharPosition of the outermost node
175 node->setStartCharPos(node->getStartCharPos() + activeEndPosition - activeStartPosition + 1 );
177 return node->getParentNode();
180 //********************************************************************************************************************
182 void SuffixTree::makeSuffixLink(int& previous, int present){
184 // here we link the nodes that are suffixes of one another to rapidly speed through the tree
185 if ( previous > 0 ) { nodeVector[previous]->setSuffixNode(present); }
186 else { /* do nothing */ }
191 //********************************************************************************************************************
193 void SuffixTree::addPrefix(int prefixPosition){
195 int lastParentNode = -1; // we need to place a new prefix in the suffix tree
200 parentNode = activeNode;
202 if(isExplicit() == 1){ // if the node is explicit (has kids), try to follow it down the branch if its there...
203 if(nodeVector[activeNode]->getChild(sequence[prefixPosition]) != -1){ // break out and get next prefix...
206 else{ // ...otherwise continue, we'll need to make a new node later on...
209 else{ // if it's not explicit (no kids), read through and see if all of the chars agree...
210 int tempNode = nodeVector[activeNode]->getChild(sequence[activeStartPosition]);
211 int span = activeEndPosition - activeStartPosition;
213 if(sequence[nodeVector[tempNode]->getStartCharPos() + span + 1] == sequence[prefixPosition] ){
214 break; // if the existing suffix agrees with the new one, grab a new prefix...
217 parentNode = split(tempNode, prefixPosition); // ... otherwise we need to split the node
222 nodeCounter++; // we need to generate a new node here if the kid didn't exist, or we split a node
223 SuffixNode* newSuffixLeaf = new SuffixLeaf(parentNode, prefixPosition, sequence.length()-1);
224 nodeVector[parentNode]->setChildren(sequence[prefixPosition], nodeCounter);
225 nodeVector.push_back(newSuffixLeaf);
227 makeSuffixLink( lastParentNode, parentNode ); // make a suffix link for the parent node
229 if(nodeVector[activeNode]->getParentNode() == -1){ // move along the start position for the tree
230 activeStartPosition++;
233 activeNode = nodeVector[activeNode]->getSuffixNode();
235 canonize(); // frankly, i'm not entirely clear on what canonize does.
238 makeSuffixLink( lastParentNode, parentNode );
239 activeEndPosition++; // move along the end position for the tree
241 canonize(); // frankly, i'm not entirely clear on what canonize does.
245 //********************************************************************************************************************