Dada una string de texto y una string de patrones, compruebe si existe un patrón en el texto o no.
Ya se han discutido algunos algoritmos de búsqueda de patrones ( KMP , Rabin-Karp , Naive Algorithm , Finite Automata ) que se pueden usar para esta verificación.
Aquí discutiremos el algoritmo basado en el árbol de sufijos.
Como requisito previo, debemos saber cómo construir un árbol de sufijos de una u otra manera.
Una vez que tenemos un árbol de sufijos construido para un texto dado, necesitamos atravesar el árbol desde la raíz hasta la hoja contra los caracteres en el patrón. Si no nos caemos del árbol (es decir, hay un camino desde la raíz hasta la hoja o en algún lugar en el medio) durante el recorrido, entonces el patrón existe en el texto como una substring.
Aquí construiremos un árbol de sufijos utilizando el algoritmo de Ukkonen, que ya se analiza a continuación:
Construcción del árbol de sufijos de Ukkonen: parte 1
Construcción del árbol de sufijos de Ukkonen: parte 2
Construcción del árbol de sufijos de Ukkonen: parte 3
Construcción del árbol de sufijos de Ukkonen: parte 4
Construcción del árbol de sufijos de Ukkonen: parte 5
Construcción del árbol de sufijos de Ukkonen: parte 6
La implementación central transversal para la comprobación de substrings se puede modificar en consecuencia para los árboles de sufijos creados por otros algoritmos.
C
// A C program for substring check using Ukkonen's Suffix Tree Construction #include <stdio.h> #include <string.h> #include <stdlib.h> #define MAX_CHAR 256 struct SuffixTreeNode { struct SuffixTreeNode *children[MAX_CHAR]; //pointer to other node via suffix link struct SuffixTreeNode *suffixLink; /*(start, end) interval specifies the edge, by which the node is connected to its parent node. Each edge will connect two nodes, one parent and one child, and (start, end) interval of a given edge will be stored in the child node. Let's say there are two nods A and B connected by an edge with indices (5, 8) then this indices (5, 8) will be stored in node B. */ int start; int *end; /*for leaf nodes, it stores the index of suffix for the path from root to leaf*/ int suffixIndex; }; typedef struct SuffixTreeNode Node; char text[100]; //Input string Node *root = NULL; //Pointer to root node /*lastNewNode will point to newly created internal node, waiting for it's suffix link to be set, which might get a new suffix link (other than root) in next extension of same phase. lastNewNode will be set to NULL when last newly created internal node (if there is any) got it's suffix link reset to new internal node created in next extension of same phase. */ Node *lastNewNode = NULL; Node *activeNode = NULL; /*activeEdge is represented as an input string character index (not the character itself)*/ int activeEdge = -1; int activeLength = 0; // remainingSuffixCount tells how many suffixes yet to // be added in tree int remainingSuffixCount = 0; int leafEnd = -1; int *rootEnd = NULL; int *splitEnd = NULL; int size = -1; //Length of input string Node *newNode(int start, int *end) { Node *node =(Node*) malloc(sizeof(Node)); int i; for (i = 0; i < MAX_CHAR; i++) node->children[i] = NULL; /*For root node, suffixLink will be set to NULL For internal nodes, suffixLink will be set to root by default in current extension and may change in next extension*/ node->suffixLink = root; node->start = start; node->end = end; /*suffixIndex will be set to -1 by default and actual suffix index will be set later for leaves at the end of all phases*/ node->suffixIndex = -1; return node; } int edgeLength(Node *n) { if(n == root) return 0; return *(n->end) - (n->start) + 1; } int walkDown(Node *currNode) { /*activePoint change for walk down (APCFWD) using Skip/Count Trick (Trick 1). If activeLength is greater than current edge length, set next internal node as activeNode and adjust activeEdge and activeLength accordingly to represent same activePoint*/ if (activeLength >= edgeLength(currNode)) { activeEdge += edgeLength(currNode); activeLength -= edgeLength(currNode); activeNode = currNode; return 1; } return 0; } void extendSuffixTree(int pos) { /*Extension Rule 1, this takes care of extending all leaves created so far in tree*/ leafEnd = pos; /*Increment remainingSuffixCount indicating that a new suffix added to the list of suffixes yet to be added in tree*/ remainingSuffixCount++; /*set lastNewNode to NULL while starting a new phase, indicating there is no internal node waiting for it's suffix link reset in current phase*/ lastNewNode = NULL; //Add all suffixes (yet to be added) one by one in tree while(remainingSuffixCount > 0) { if (activeLength == 0) activeEdge = pos; //APCFALZ // There is no outgoing edge starting with // activeEdge from activeNode if (activeNode->children] == NULL) { //Extension Rule 2 (A new leaf edge gets created) activeNode->children] = newNode(pos, &leafEnd); /*A new leaf edge is created in above line starting from an existing node (the current activeNode), and if there is any internal node waiting for it's suffix link get reset, point the suffix link from that last internal node to current activeNode. Then set lastNewNode to NULL indicating no more node waiting for suffix link reset.*/ if (lastNewNode != NULL) { lastNewNode->suffixLink = activeNode; lastNewNode = NULL; } } // There is an outgoing edge starting with activeEdge // from activeNode else { // Get the next node at the end of edge starting // with activeEdge Node *next = activeNode->children]; if (walkDown(next))//Do walkdown { //Start from next node (the new activeNode) continue; } /*Extension Rule 3 (current character being processed is already on the edge)*/ if (text[next->start + activeLength] == text[pos]) { //If a newly created node waiting for it's //suffix link to be set, then set suffix link //of that waiting node to current active node if(lastNewNode != NULL && activeNode != root) { lastNewNode->suffixLink = activeNode; lastNewNode = NULL; } //APCFER3 activeLength++; /*STOP all further processing in this phase and move on to next phase*/ break; } /*We will be here when activePoint is in the middle of the edge being traversed and current character being processed is not on the edge (we fall off the tree). In this case, we add a new internal node and a new leaf edge going out of that new node. This is Extension Rule 2, where a new leaf edge and a new internal node get created*/ splitEnd = (int*) malloc(sizeof(int)); *splitEnd = next->start + activeLength - 1; //New internal node Node *split = newNode(next->start, splitEnd); activeNode->children] = split; //New leaf coming out of new internal node split->children] = newNode(pos, &leafEnd); next->start += activeLength; split->children] = next; /*We got a new internal node here. If there is any internal node created in last extensions of same phase which is still waiting for it's suffix link reset, do it now.*/ if (lastNewNode != NULL) { /*suffixLink of lastNewNode points to current newly created internal node*/ lastNewNode->suffixLink = split; } /*Make the current newly created internal node waiting for it's suffix link reset (which is pointing to root at present). If we come across any other internal node (existing or newly created) in next extension of same phase, when a new leaf edge gets added (i.e. when Extension Rule 2 applies is any of the next extension of same phase) at that point, suffixLink of this node will point to that internal node.*/ lastNewNode = split; } /* One suffix got added in tree, decrement the count of suffixes yet to be added.*/ remainingSuffixCount--; if (activeNode == root && activeLength > 0) //APCFER2C1 { activeLength--; activeEdge = pos - remainingSuffixCount + 1; } else if (activeNode != root) //APCFER2C2 { activeNode = activeNode->suffixLink; } } } void print(int i, int j) { int k; for (k=i; k<=j; k++) printf("%c", text[k]); } //Print the suffix tree as well along with setting suffix index //So tree will be printed in DFS manner //Each edge along with it's suffix index will be printed void setSuffixIndexByDFS(Node *n, int labelHeight) { if (n == NULL) return; if (n->start != -1) //A non-root node { //Print the label on edge from parent to current node //Uncomment below line to print suffix tree // print(n->start, *(n->end)); } int leaf = 1; int i; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { //Uncomment below two lines to print suffix index // if (leaf == 1 && n->start != -1) // printf(" [%d]\n", n->suffixIndex); //Current node is not a leaf as it has outgoing //edges from it. leaf = 0; setSuffixIndexByDFS(n->children[i], labelHeight + edgeLength(n->children[i])); } } if (leaf == 1) { n->suffixIndex = size - labelHeight; //Uncomment below line to print suffix index //printf(" [%d]\n", n->suffixIndex); } } void freeSuffixTreeByPostOrder(Node *n) { if (n == NULL) return; int i; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { freeSuffixTreeByPostOrder(n->children[i]); } } if (n->suffixIndex == -1) free(n->end); free(n); } /*Build the suffix tree and print the edge labels along with suffixIndex. suffixIndex for leaf edges will be >= 0 and for non-leaf edges will be -1*/ void buildSuffixTree() { size = strlen(text); int i; rootEnd = (int*) malloc(sizeof(int)); *rootEnd = - 1; /*Root is a special node with start and end indices as -1, as it has no parent from where an edge comes to root*/ root = newNode(-1, rootEnd); activeNode = root; //First activeNode will be root for (i=0; i<size; i++) extendSuffixTree(i); int labelHeight = 0; setSuffixIndexByDFS(root, labelHeight); } int traverseEdge(char *str, int idx, int start, int end) { int k = 0; //Traverse the edge with character by character matching for(k=start; k<=end && str[idx] != '\0'; k++, idx++) { if(text[k] != str[idx]) return -1; // mo match } if(str[idx] == '\0') return 1; // match return 0; // more characters yet to match } int doTraversal(Node *n, char* str, int idx) { if(n == NULL) { return -1; // no match } int res = -1; //If node n is not root node, then traverse edge //from node n's parent to node n. if(n->start != -1) { res = traverseEdge(str, idx, n->start, *(n->end)); if(res != 0) return res; // match (res = 1) or no match (res = -1) } //Get the character index to search idx = idx + edgeLength(n); //If there is an edge from node n going out //with current character str[idx], traverse that edge if(n->children[str[idx]] != NULL) return doTraversal(n->children[str[idx]], str, idx); else return -1; // no match } void checkForSubString(char* str) { int res = doTraversal(root, str, 0); if(res == 1) printf("Pattern <%s> is a Substring\n", str); else printf("Pattern <%s> is NOT a Substring\n", str); } // driver program to test above functions int main(int argc, char *argv[]) { strcpy(text, "THIS IS A TEST TEXT$"); buildSuffixTree(); checkForSubString("TEST"); checkForSubString("A"); checkForSubString(" "); checkForSubString("IS A"); checkForSubString(" IS A "); checkForSubString("TEST1"); checkForSubString("THIS IS GOOD"); checkForSubString("TES"); checkForSubString("TESA"); checkForSubString("ISB"); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); return 0; }
Producción:
Pattern <TEST> is a Substring Pattern <A> is a Substring Pattern < > is a Substring Pattern <IS A> is a Substring Pattern < IS A > is a Substring Pattern <TEST1> is NOT a Substring Pattern <THIS IS GOOD> is NOT a Substring Pattern <TES> is a Substring Pattern <TESA> is NOT a Substring Pattern <ISB> is NOT a Substring
La construcción del árbol de sufijos de Ukkonen toma tiempo y espacio O(N) para construir el árbol de sufijos para una string de longitud N y, después de eso, el recorrido para la verificación de substrings toma O(M) para un patrón de longitud M.
Con una ligera modificación en el algoritmo transversal discutido aquí, podemos responder lo siguiente:
- Encuentre todas las ocurrencias de un patrón dado P presente en el texto T.
- ¿Cómo comprobar si un patrón es prefijo de un texto?
- ¿Cómo comprobar si un patrón es sufijo de un texto?
Hemos publicado los siguientes artículos sobre aplicaciones de árboles de sufijos:
- Aplicación de árbol de sufijos 2: búsqueda de todos los patrones
- Aplicación de árbol de sufijos 3: substring repetida más larga
- Aplicación de árbol de sufijos 4: construir una array de sufijos de tiempo lineal
- Árbol de sufijos generalizados 1
- Aplicación de árbol de sufijos 5: substring común más larga
- Aplicación 6 del árbol de sufijos: la substring palindrómica más larga
Este artículo es una contribución de Anurag Singh . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA