找到没有后缀数组或后缀树的最长重复子字符串

Question

在最长的重复子问题如下:

给定字符串w,找到至少出现在两个位置的w的最长子字符串.

这个问题可以使用后缀树在线性时间内解决,在使用增强后缀数组的线性时间内解决.

我的问题是 - 这个问题是否有线性时间算法,不涉及后缀树或后缀数组？我很好奇,因为后缀树和后缀数组很难编码和操作,如果这个问题的算法不需要这些其他结构的编码或内存开销,那就太好了.

谢谢!

Answer 1

After digging on this for a while, I did find an alternative to suffix trees and suffix arrays. The implementation itself is simple, like you wanted, but even though the code is (roughly) high on brevity, it is extremely low on intuition.

There is a paper, The Suffix Binary Search Tree and Suffix AVL Tree, by Robert W. Irving and Lorna Love, from University of Glasgow, that presents an alternative solution to suffix trees and suffix arrays. The authors claim that managing and building a suffix binary search tree is substantially simpler when compared to suffix trees, and show that construction can be guaranteed to take no longer than O(n log(n)) in the case of a suffix AVL tree (this is also true for the standard, non-balanced original BST in the average case, but it can degrade to O(n^2) in the worst case).

This is, of course, worse than traditional suffix trees, since the time to find the longest repeated substring in an SBST is bounded by the time of building it. So, strictly speaking, this doesn't answer your question, but I decided to post this for future reference and for interested readers.

Furthermore, the paper states that the resulting tree is a strong candidate alternative to traditional suffix trees and suffix arrays. The paper is written around the problem of finding a pattern in logarithmic time after building the tree. You can download it from CiteSeerX.

The idea is that each node is associated with an integer i, where i is the offset of the first character in the target text of the suffix that that node represents. For a given string [a1,a2,a3,...,an] of length n, there will be n nodes, and node i represents the suffix [ai,ai+1,...,an].

The paper doesn't address the problem of determining the longest repeated substring, but the proposed way to build the binary search tree makes it easy to solve the longest repeated substring problem. I will go through that extension once I talk about the building process. It is crucial to understand how to build the tree in order to understand the extension to solve the longest repeated string problem (and otherwise the code will look like magic).

Hence, I will explain first how the tree is built, and will do so based on the ideas I got from the paper. Formal proofs, along with other details (including how to turn this into an AVL tree) are discussed in the paper. You are free to jump ahead to the section where I talk about how I extended the algorithm, but keep in mind that it may be hard to understand if you do so.

The paper describes an optimized algorithm for searching a pattern in a suffix binary search tree that avoids comparing the pattern with the suffix from a node from scratch on each node we visit. Doing so would, in the worst case, require l*m character comparisons, where m is the size of the pattern, and l is the search path. We don't want to compare characters from scratch on each node to decide on which direction to branch; instead, it is desirable to compare each character in the pattern at most once. This is possible if we store two additional attributes per node. The process of building the tree is tightly bound with this optimization (and with my proposed extension), so it is very important to understand it.

First, some terminology: for a given node i, a node j is said to be a right ancestor of i if i is in the left subtree of j. The left ancestor is defined similarly. The closest right ancestor of i is the node j such that no descendant of j is a right ancestor of i; again, a similar definition applies for the closest left ancestor. From now on, I will refer to the closest right ancestor of i using the the abbreviated form cra(i), and the closest left ancestor of i is abbreviated to cla(i). We also define the longest common prefix between two nodes j and i, and we represent it as lcp(i, j).

For a given node i, we will store two attributes, m(i) and d(i). m(i) represents the value for which lcp(i, j) is highest, where j is the set of ancestors of i. Note that, due to the binary tree's properties, node j is either cla(i) or cra(i). d(i) is an attribute to keep track of where m(i) comes from; if j == cla(i), then d(i) is RIGHT (meaning that i is in the right subtree of the node j for which lcp(i, j) is maximized), otherwise, d(i) is LEFT.

What follows is a set of theorems that together build the basic algorithm to perform a search of a given pattern in an SBST. The theorems describe what to do when the search for a pattern arrives at a node i. Please refer to the paper for the formal proofs, I will try to provide an intuitive proof of why the rules are like this. Together, these theorems form a set of rules that allows the algorithm to search for a pattern by comparing each character in the pattern at most once, which is pretty neat!

When a search arrives at node i, we make use of 2 values, llcp and rlcp. llcp is the value for which lcp(pattern, j) is highest, where j is the set of all right ancestors of i. rlcp is the same, but the maximum is taken over all left ancestors of i. Again, due do the binary tree's properties, llcp is just lcp(pattern, cra(i)), and rlcp is lcp(pattern, cla(i)).

Before diving into the theorems, I think it is a good idea to draw a sample SBST on a piece of paper and visualize the semantical meaning of each theorem on the tree.

Theorem 1 is the simplest and covers the case m(i) > max(llcp, rlcp). If that happens, llcp and rlcp don't change, because we will be able to match as much with i as we did with its ancestors, and the search branches in the same direction as d(i). To see why, consider the case that d(i) == LEFT. If d(i) == LEFT, then it means that m(i) comes from a match with cra(i). If we're visiting i, it is because we already know that the pattern is lower than cra(i), and lcp(cra(i), pattern) < lcp(i, cra(i)), which makes the pattern less than i, so we move left. The same process can be applied for the case d(i) == RIGHT, so, indeed, we just need to follow the direction of d(i).

定理2涉及案例mi < max(llcp, rlcp).这个很难理解.让我们看看时间max(llcp, rlcp) == llcp和时间会发生什么max(llcp, rlcp) == rlcp.

案例1:max(llcp,rlcp)== llcp

如果max(llcp, rlcp) == llcp,那么这意味着该模式与节点i最接近的右祖先与其最近的左祖先有更多共同之处.此外,因为llcp > m(i),模式与cra(i)节点i有更多共同之处cra(i).这与i低于cra(i)(根据BST的定义)的事实一起暗示该模式大于i.因此,我们分支正确.

那些更新llcp和rlcp？因为我们分支正确,cra(i)在下一次迭代中将是相同的,所以llcp保持不变.更新rlcp有点棘手.当我们分支正确时,新的cla将是节点i.接下来会发生什么取决于m(i)是来自cra(i)还是来自cla(i).我们可以d(i)用来知道:如果m(i)来自cra(i),那么d(i) == LEFT,否则,d(i) == RIGHT.

案例1.1:max(llcp,rlcp)== llcp && d(i)==对

In this case, we know that m(i) comes from cla(i), which means that node i has more in common with cla(i) than with cra(i) (and remember, the pattern has more in common with cra(i)). As we saw before, we'll be branching right, which makes the pattern greater than i. This implies that cla < i < pattern, and, therefore, the longest common prefix between pattern and i is the same as between pattern and cla; or in other words, lcp(i, cla) > lcp(i, pattern) (otherwise, pattern would have to be less than i), so rlcp, which is lcp(i, pattern), remains the same.

Case 1.2: max(llcp, rlcp) == llcp && d(i) == LEFT

Now, we know that m(i) comes from cra(i), but the pattern has more in common with cra(i) than i has with cra(i). This means that node i acts as a "bottleneck", making rlcp smaller - rlcp can only be as big as the common prefix between i and cra(i), which is equal to m(i). So, in this case, rlcp is the same as m(i).

A similar analysis can be performed for the reverse situation, where max(llcp, rlcp) == rlcp (and then consider the sub-cases d(i) == RIGHT and d(i) == LEFT). The actions to perform are the reverse version of the previous cases: we branch left, rlcp remains unchanged, and llcp becomes m(i) if d(i) == RIGHT, otherwise, it remains unchanged.

In short:

Theorem 2 results

                                   d(i) == RIGHT                d(i) == LEFT                      

max(llcp, rlcp) == llcp |            branch right        |   branch right; rlcp = m(i)
max(llcp, rlcp) == rlcp |       branch left; llcp = m(i) |         branch left

Theorem 3 explores two cases when m(i) is equal to the longest common prefix of the pattern with another ancestor. In particular, if m(i) == llcp && llcp > rlcp && d(i) == RIGHT, we know that i matches as much with cla as the pattern with cra. Since d(i) == RIGHT, and m(i) == llcp, it follows that lcp(i, cra) < llcp, and i < cra, implying that the pattern is greater than i - we branch right. A similar argument holds for the opposite case; if m(i) == rlcp && rlcp > llcp && d(i) == LEFT, we branch left. In either case, both llcp and rlcp will stay the same: in the former, llcp would never change because cra is still the same, and rlcp won't change because d(i) == RIGHT && m(i) == llcp, i.e., matching the pattern with cla or with i is the same; the same happens in the latter case.

Theorem 4 is where we have to perform actual character comparisons. This happens every time we cannot infer anything about the relative order between the pattern and the current node, that is, when m(i) == rlcp == llcp, or m(i) == llcp && llcp > rlcp && d(i) == LEFT, or the reverse m(i) == rlcp && rlcp > llcp && d(i) == RIGHT. Intuitively, we know that nothing can be infered about the order, and character comparisons are performed beginning on the first character that is not part of the longest common prefix, and stopping on the first that is different (at which point we can reason about the order). Again, if we branch right, then cra remains the same, so llcp doesn't change, and rlcp will now hold the value of the newly computed longest common prefix between the pattern and node i. A similar process happens for the reverse case: we branch left, rlcp stays untouched, and llcp becomes the previously computed value of the longest common prefix. This theorem is intuitively correct, so I won't go any further.

Theorem 3 & 4 results

                                     d(i) == RIGHT                d(i) == LEFT                      
m(i) == llcp && llcp > rlcp |         branch right       |               *
m(i) == rlcp && rlcp > llcp |             *              |         branch left

* = compare(); if branch == left then rlcp = computed_lcp else llcp = computed_lcp

Here's the pseudo-code that brings all of this together, taken from page 9:

My modifications to the algorithm

It may take a little while to wrap your head around the awesomeness of this algorithm, but once you grasp all the details, it can be seen that the problem of finding the longest repeated substring is equivalent to finding the node i for which m(i) is greatest. After all, the longest repeated substring is the longest common prefix that can ever be found between any two nodes. This can be found without any significant overhead if we keep track of it while building the tree: the maximum m(i) seen so far must be kept, and whenever a new node j is inserted, we compare m(j) with the maximum seen so far, and update it if m(j) is greater. In fact, this approach is nothing more than a fancy way to implement an algorithm that is equivalent to sorting all the suffixes and finding the longest common prefix between any two consecutive suffixes, with the advantage of not performing unnecessary character comparisons. That's a considerably good improvement.

The pseudo-code shown above is almost enough to build the standard SBST. We start by adding a root node, with i == 1, representing the whole text. Suffixes are then added left to right. To insert a new suffix i, we search for a pattern that is suffix i. Doing so will make the algorithm stop at the exact spot of insertion. However, the paper doesn't go into much detail on the insertion process. We must be careful with the return i; in the final else for Theorem 4. We can only return if we're searching. If we're doing an insertion and reach a node that brings us to that case of theorem 4, then it means that all of the characters in the new suffix match with some previously inserted suffix. Because suffixes are being inserted from left to right, we also know that the new suffix has less characters than the other one, which means that the new suffix is lower than the other: the correct move is to branch left. Since we branched left, the closest left ancestor stays the same, so we only need to update llcp. llcp becomes the size of the suffix itself, because as we saw, all of it is matched with the node that is now the closest right ancestor.

Obviously, the new node will have a value of m(i) equal to max(llcp, rlcp), and d(i) will, by definition, be RIGHT if max(llcp, rlcp) == rlcp, and LEFT otherwise.

My implementation in C boils down to the pseudo-code, together with the insertion logic. There are two data structures: struct sbst represents a suffix binary search tree along with the maximum m(i) seen so far; and struct node, which is the descriptor for a tree's node.

Here's the full program listing:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define LEFT 0
#define RIGHT 1
#define MODE_INSERT 1
#define MODE_FIND 2
#define MAX_TEXT_SIZE 1024
#define max(a,b) ((a)>(b)?(a):(b))

struct node {
  int m;
  int d;
  int i;
  struct node *left;
  struct node *right;
};

struct sbst {
  struct node *root;
  int max_mi; /* The maximum lcp in the tree */
  int max_i; /* The node i with m(i) == max_mi */
};

struct node *allocate_node(int m, int d, int i) {
  struct node *new_node = malloc(sizeof(*new_node));
  new_node->m = m;
  new_node->d = d;
  new_node->i = i;
  new_node->left = new_node->right = NULL;
  return new_node;
}

int lcp(char *str1, char *str2);

/* The core where all the work takes place.
   It is assumed that when this function is called, tree->root always points to a valid, allocated root. That is, it is assumed
   that the tree contains at least one node.
   This function provides both a find and an insert algorithm. To find a pattern, <mode> must be MODE_FIND. To insert,
   <mode> must be MODE_INSERT. In the latter case, the parameter <text_i> corresponds to the index in the original string
   of the suffix being inserted (index starts counting at 1, as described in the paper). Also, in MODE_INSERT, the pattern is
   the suffix being inserted.
   If in MODE_FIND, this function returns the index (starting at 1) in the text where <pattern> can be found, or 0 if no such
   pattern could be found.
*/
int find_insert_aux(struct sbst *tree, char *pattern, size_t pattern_len, char *text, size_t text_len, char mode, int text_i) {
  struct node *current, *prev;
  int llcp, rlcp;
  int next_dir;

  current = tree->root;
  llcp = rlcp = 0;

  while (current != NULL) {
    int max_pattern = max(llcp, rlcp);
    if (current->m > max_pattern) {
      next_dir = current->d;
    } else if (current->m < max_pattern) {
      if (llcp > rlcp) {
    next_dir = RIGHT;
    if (current->d == LEFT) {
      rlcp = current->m;
    }
      } else if (rlcp > llcp) {
    next_dir = LEFT;
    if (current->d == RIGHT) {
      llcp = current->m;
    }
      }
    } else if (current->m == llcp && llcp > rlcp && current->d == RIGHT) {
      next_dir = RIGHT;
    } else if (current->m == rlcp && rlcp > llcp && current->d == LEFT) {
      next_dir = LEFT;
    } else {
      int sub_lcp = lcp(pattern+current->m, text+current->m+current->i-1);
      int t = current->m + sub_lcp;
      if (t == pattern_len) {
    if (mode == MODE_FIND) {
      return current->i;
    } else {
      next_dir = LEFT;
      llcp = t;
    }
      } else if (current->i+t-1 == text_len || pattern[t] > text[t+current->i-1]) {
    next_dir = RIGHT;
    rlcp = t;
      } else {
    next_dir = LEFT;
    llcp = t;
      }
    }
    prev = current;
    current = (next_dir == RIGHT ? current->right : current->left);
  }
  if (mode == MODE_INSERT) {
    struct node *new_node = allocate_node(max(llcp, rlcp), (llcp > rlcp ? LEFT : RIGHT), text_i);
    if (next_dir == LEFT)
      prev->left = new_node;
    else
      prev->right = new_node;
    if (new_node->m > tree->max_mi) {
      tree->max_mi = new_node->m;
      tree->max_i = new_node->i;
    }
  }
  return 0;
}

void sbst_insert(struct sbst *tree, char *text, size_t text_size, int i) {
  (void) find_insert_aux(tree, text+i-1, text_size-i+1, text, text_size, MODE_INSERT, i);
}

int sbst_find(struct sbst *tree, char *text, size_t text_size, char *pattern, size_t pattern_size) {
  return find_insert_aux(tree, pattern, pattern_size, text, text_size, MODE_FIND, 0);
}

/* Builds a Suffix Binary Search Tree that keeps track of the highest m(i) as it is built. */
struct sbst *build_sbst(char *text, size_t text_size) {
  if (*text == '\0')
    return NULL;

  struct sbst *tree = malloc(sizeof(*tree));
  tree->root = allocate_node(0, 0, 1);
  tree->max_mi = 0;
  tree->max_i = 1;

 for (int i = 1; text[i] != '\0'; i++)
   sbst_insert(tree, text, text_size, i+1);

  return tree;
}

/* Given an SBST for the input, finds the longest repeated substring in O(1)
   Stores the offset in *offset, and the size of the lrs in *size
*/
void find_lrs(struct sbst *tree, int *offset, int *size) {
  *offset = tree->max_i-1;
  *size = tree->max_mi;
}

/* Debug section */
void dump(struct node *n, char *text, int depth) {
  if (!n)
    return;
  for (int i = 0; i < depth; i++)
    putchar(' '), putchar(' ');
  printf("%d|%d|%d|%s\n", n->m, n->d, n->i, text+n->i-1);
  dump(n->left, text, depth+1);
  dump(n->right, text, depth+1);
}

void dump_sorted(struct node *n, char *text) {
  if (!n)
    return;
  dump_sorted(n->left, text);
  printf("%s\n", text+n->i-1);
  dump_sorted(n->right, text);
}
/* End debug section */

int lcp(char *str1, char *str2) {
  int i;
  for (i = 0; str1[i] != '\0' && str1[i] == str2[i]; i++);
  return i;
}

int main(void) {
  char text[MAX_TEXT_SIZE];
  printf("Enter text, hit RETURN to terminate (max. %d chars): ", MAX_TEXT_SIZE-1);

  fgets(text, sizeof text, stdin);
  size_t text_size = strlen(text);

  /* Trim newline */
  text[--text_size] = '\0';

  struct sbst *tree = build_sbst(text, text_size);

  /* Debug */
  #ifdef DEBUG_MODE
  dump(tree->root, text, 0);
  printf("\n");
  dump_sorted(tree->root, text);
  #endif

  int lrs_offset, lrs_size;
  find_lrs(tree, &lrs_offset, &lrs_size);
  if (lrs_size == 0)
    printf("No longest repeated substring.\n");
  else {
    printf("Longest repeated substring found at offset %d with size %d: %.*s\n", lrs_offset, lrs_size, lrs_size, text+lrs_offset);
  }
  return 0;
}

Had I not read the paper, I would have looked upon this code as black magic.

Note that this is not exactly a good demonstration of the best software engineering practices: someone who is not familiar with the algorithm will have a very hard time reading and understanding it, it leaks memory, it doesn't check the return value of malloc, and has some other flaws too, but I think it's enough to make my point.

While this may not be as optimal as a suffix tree, it is clearly simple to build and provides a good starting point. For example, as a by-product, it can perform pattern matching in logarithmic time - that's pretty good!

NOTE: I haven't had much time to test the implementation. I did some basic testing and it seems to be working, but I can't assure there are no errors.