Java Dynamic Programming-Example code for distance editing and java sample code

Java Dynamic Programming-Example code for distance editing and java sample code

The dynamic planning process is: each decision depends on the current state, and then causes the state transfer. A decision sequence is generated in a changed state. Therefore, this multi-stage optimization decision-making process is called dynamic planning.

Dynamic Planning is actually a general term for a category of questions, not a fixed algorithm. The significance of dynamic planning is to solve the overall practice by adopting a recursive (or divide and conquer) strategy to solve the subproblems of big problems. The core idea of dynamic planning is to skillfully split the problem into multiple sub-problems and solve the overall problem by calculating sub-problems. Sub-problems can be split into more sub-problems, so as to solve the required problems using a method similar to recursive iteration. Problem description:

For the sequence S and T, the distance between them is defined as: perform the following operations on either of them: 1, delete a character; 2, insert a character; 3, change a character. each operation increases the count by 1. the minimum count for converting S and T into equal sequences is the editdistance or similarity between the two. provide the corresponding algorithm and its implementation.

Analysis:

Assume that the length of the sequence S and T is m and n respectively, and the editing distance between the two is represented as edit [m] [n]. There are several situations when operating the sequence:

A. When the end characters of S and T are equal, no one of the above definition operations (I .e. "edit") is required for the end characters, that is, no additional count is required. the condition is met: edit [m] [n] = edit [M-1] [n-1].

B. When the end characters of S and T are not equal, edit the end of one of the two. The corresponding count will increase by 1.

B1: Modify the end of S or T to make it equal to T or S. In this case, edit [m] [n] = edit M-1] [n-1] + 1;

B2: Delete the element at the end of S to make S equal to T. In this case, edit [m] [n] = edit M-1] [n] + 1;

B3: Delete the element at the end of T to make T equal to S. In this case, edit [m] [n] = edit [m] [n-1] + 1;

B4: add the tail element T at the end of S to make S and T equal. Then, the length of S is changed to m + 1, but at this time, the end elements of S and T are equal, only the first m elements of S need to be compared with the first n-1 elements of T. Therefore, edit [m] [n] = edit [m] [n-1] + 1 is required;

B5: add the end element of S at the end of T to make T and S equal. In this case, it is the same as b4, meeting the requirements of edit [m] [n] = edit [M-1] [n] + 1;

C. In special cases, when S is null, edit [0] [n] = n; when T is null, edit [m] [0] = m; this is easy to understand. For example, for sequences "" and "abc", the minimum operation of the two is 3, that is, the sequence "" Performs 3 insert operations, or the sequence "abc" Performs 3 Delete operations.

Therefore, it is not difficult to introduce the dynamic planning equation for editing distance as follows:

Therefore, the Recursive Implementation of the Dynamic Programming Algorithm for string editing distance can be expressed using the following Java code:

public static int editDistance(String a, String b) {    if (a == null || b == null) {      return -1;    }    return editDistance(a, a.length() - 1, b, b.length() - 1);  }  public static int editDistance(String a, int m, String b, int n) {    if (m < 0 || n < 0) {      return 1;    } else if (a.charAt(m) == b.charAt(n)) {      return editDistance(a, m - 1, b, n - 1);    } else {      return Math.min(Math.min(editDistance(a, m - 1, b, n) + 1, editDistance(a, m, b, n - 1) + 1), editDistance(a, m - 1, b, n - 1) + 1);    }  }

UPDATE:

At the same time, we can see from the dynamic planning equation of the editing distance that edit [m] [n] can be changed from edit [m-1] [n-1], edit [m-1] [n], edit [m] [n-1] is obtained. If edit is a two-dimensional array, edit [m] [n] can be determined by conditions from the elements at the upper, left, and left positions. that is, we can traverse the two-dimensional array and calculate the current value through backtracking.

For example, for strings S = "sailn" and T = "failing", initialize the two-dimensional array as follows:

M \ n		F	A	I	L	I	N	G
	0	1	2	3	4	5	6	7
S	1	1
A	2
I	3
L	4
N	5

Because S [0] = s, T [0] = f, S [0]! = T [0], corresponds to the preceding two-dimensional matrix. edit [1] [1] = min (edit [0] [0], edit [0] [1], edit [1] [0]) + 1: edit [1] [1] = min (0, 1, 1) + 1: edit [1] [1] = 0 + 1 = 1.

M \ n		F	A	I	L	I	N	G
	0	1	2	3	4	5	6	7
S	1	1	2	3	4	5	6	7
A	2	2	1
I	3
L	4
N	5

For S [1] = a, T [1] = a, S [1] = T [1], it corresponds to a two-dimensional matrix, edit [2] [2] = edit [1] [1], so edit [2] [2] = 1. according to this rule, the above two-dimensional matrix is filled as follows:

M \ n		F	A	I	L	I	N	G
	0	1	2	3	4	5	6	7
S	1	1	2	3	4	5	6	7
A	2	2	1	2	3	4	5	6
I	3	3	2	1	2	3	4	5
L	4	4	3	2	1	2	3	4
N	5	5	4	3	2	2	2	3

Therefore, the editing distance between the two is edit [m] [n] = edit [5] [7] = 3.

Therefore, the Java version of the Backtracking solution dynamically planned based on the above idea can be as follows:

public static int editDistance(String a, String b) {    if (a == null || b == null) {      return -1;    }    int[][] matrix = new int[a.length() + 1][b.length() + 1];    for (int i = 0; i < a.length() + 1; i++) {      for (int j = 0; j < b.length() + 1; j++) {        if (i == 0) {          matrix[i][j] = j;        } else if (j == 0) {          matrix[i][j] = i;        } else {          if (a.charAt(i - 1) == b.charAt(j - 1)) {            matrix[i][j] = matrix[i - 1][j - 1];          } else {            matrix[i][j] = 1 + Math.min(Math.min(matrix[i - 1][j], matrix[i][j - 1]), matrix[i - 1][j - 1]);          }        }      }    }    return matrix[a.length()][b.length()];  }

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