Aplicación de árbol de sufijos 1: comprobación de substrings

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. 
 

Suffix Tree Application

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: 
 

  1. Encuentre todas las ocurrencias de un patrón dado P presente en el texto T.
  2. ¿Cómo comprobar si un patrón es prefijo de un texto?
  3. ¿Cómo comprobar si un patrón es sufijo de un texto?

Hemos publicado los siguientes artículos sobre aplicaciones de árboles de sufijos: 
 

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

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