Which Pattern Matching Algorithm fits for my case?

Question

I have a project which it needs to compare to text documents and find the similarity rate between every single sentence and the general similarities of the texts. I did some transforming on texts like lowering all words,deleting duplicate words,deleting punctuations except fullstops. After doing some operations, i had 2 arraylists which include sentences and the words all seperated. It looks like

[["hello","world"],["welcome","here"]]

Then i sorted every sentence alphabetically.After all these, i'm comparing all the words one by one,doing linear search but if the word which i'm searching is bigger than i'm looking (ASCII of first character like world > burger),i'm not looking remaining part,jumping other word. It seems like complicated but i need an answer of " Is there any faster,efficient common algorithms like Boyer Moore,Hashing or other?". I'm not asking any code peace but i need some theoretical advices.Thank you.

EDIT: I should've tell the main purpose of the project. Actually it is kinda plagiarism detector.There are two txt files which are main.txt and sub.txt.The program will compare them and it gives an output something like that:

Output:
Similarity rate of two texts is: %X
{The most similar sentence}
{The most similar 2nd sentence}
{The most similar 3d sentence}
{The most similar 4th sentence}
{The most similar 5th sentence}

So i need to find out sub.txt similarity rate to main.txt file.I thought that i need to compare all the sentences in two files with each other.

For instance, main.txt has 10 sentences and sub.txt has 5 sentences, there will be 50 comparison and 50 similarity rate will be calculated and stored.

Finally i sort the similarty rates and print the most 5 sentences.Actually i've done the project,but it's not efficient. It has 4 nested for loops and compares all words uncountable times and complexity becomes like O(n^4) ( maybe not that much) but it's really huge even in the worst case. I found Levenshtein distance algorithm and Cosine similarity algorithms but i'm not sure about them. Thanks for any suggestion!

EDIT2: For my case similarity between 2 sentence is like:

main_sentence:"Hello dude how are you doing?"
sub_sentence:"Hello i'm fine dude."
Since intersection is 2 words ["hello","dude"]
The similarity is : (length of intersected words)*100/(length of main text)
For this case it's: 2*100/6 = %33,3

Answer 1

As a suggestion, and even if this is not a "complete answer" to your issue, comparing Strings is usually a "heavy" operation (even if you first check their length, which, in fact, is one of the first things the equals() method already performs when comparing Strings)

What I suggest is doing next: create a dummy hashcode()-like method. It won't be a real hashcode() , but the number associated to the order in which that word was read by your code. Something like a cryptographic method, but much simpler.

Note that string.hashCode() won't work, as the word "Hello" from the first document wouldn't return the same hashcode than the word "Hello" from the second document.

---

Data "Warming" - PreConversion

Imagine you have a shared HashMap<String,Integer> ( myMap ), which key is an String and the value an Integer. Note that HashMap's hashing in java with String keys lower than 10 characters (which usually are, in english language) is incredibly fast . Without any check, just put each word with its counter value:

  myMap.put(yourString, ++counter);  

Let's say you have 2 documents:

 1.txt-   Welcome mate what are you doing here 
     
 2.txt-   Mate I was here before are you dumb

I assume you already lowercased all words, and removed duplicates. You start reading the first document and assigning each word to a number. The map would look like:

        KEY                     VALUE
       welcome                    1
       mate                       2
       what                       3
       are                        4
       you                        5
       doing                      6
       here                       7

Now with the second document. If a key is repeated, the put() method will update its value. So:

        KEY                     VALUE
       welcome                    1
       mate                       8
       what                       3
       are                        13
       you                        14
       doing                      6
       here                       11
         I                        9
       was                        10
       before                     12
       dumb                       15

Once complete, you create another HashMap<Integer,String> ( reverseMap ), this way in reverse:

       KEY                    VALUE
       1                       welcome
       8                       mate
       3                       what
       13                       are
       14                       you
       6                       doing
       11                       here
        9                        I
       10                        was
       12                     before
       15                       dumb

You convert both documents into a List of Integers, so they look like:

 1.txt-   Welcome mate what are you doing here
 2.txt-   Mate I was here before are you dumb

to:

 listOne - [1, 8, 3, 13, 14, 6, 11]
 listTwo - [8, 9, 10, 11, 12, 13, 14, 15]

---

Duplicate words, positions and sequences

To find the duplicated within both documents:

• First, create a deep clone of one of the lists, for example, listTwo. A deep clone of a List of Integers is relatively easy to perform. Calling it listDuplicates as that will be its objective.

   List<Integer> listDuplicates = new ArrayList<>(); for (Integer i:listTwo) listDuplicates.add(new Integer(i));

• Call retainAll :

listDuplicates.retainAll(listOne);

The result would be:

listDuplicates- [8,11,13,14]

So, from a total of listOne.size()+listTwo.size() = 15 words found on 2 documents, 4 are duplicates are 11 are unique.

In order to get the converted values, just call:

for (Integer i : listDuplicates)
 System.out.println(reverseMap.get(i)); // mate , here, are, you

Now that duplicates are identified, listOne and listTwo can also be used now in order to:

• Identify the position on each list (so we can get the difference in the positions of this words). The first word would have -1 value, as its the first one and doesn't have a diff with the previous one, but won't necessarily mean they are consequent with any other (they are just the first duplicates).

If the next element has -1 value, that means the [8] and [11] would aslo be consecutive:

          doc1   doc2   difDoc1     difDoc2
    [8]     2      1     -1  (0-1)   -1  (0-1)
    [11]    7      4     -5  (2-7)   -3  (1-4)
    [13]    4      6      3  (7-4)   -2  (4-6)
    [14]    5      7     -1  (4-5)   -1  (6-7)

In this case, the distance shown in [14] with its previous duplicate (the diff between [13] and [14] ) is the same in both documents: -1 : that means that not only are duplicates, but both are consequently placed in both documents.

Hence, we've found not only duplicate words, but also a duplicate sequence of two words between those lines:

[13][14] -- are you

The same mechanism (identifying a diff of -1 for the same variable in both documents) would also help to find a complete duplicate sequence of 2 or more words. If all the duplicates show a diff of -1 in both documents, that means we've found a complete duplicate line :

In this example this is shown clearer:

   doc1- "here i am" [4,5,6]
   doc2- "here i am" [4,5,6]

listDuplicates - [4,5,6]

              doc1   doc2   difDoc1     difDoc2
        [4]    1      1      -1  (0-1)   -1  (0-1)
        [5]    2      2      -1  (1-2)   -1  (1-2)
        [6]    3      3      -1  (2-3)   -1  (2-3)
      

All the diffs are -1 for the same variable in both documents -> all duplicates are next to each other in both documents --> The sentence is exactly the same in both documents . So, this time, we've found a complete duplicate line of 3 words .

[4][5][6] -- here i am

Apart of this duplicate sequence search, this difference table would also be helpful when calculating the variance, median,... from the duplicate words, in order to get some kind of "similarity" factor (something like a basic indicative value of equity between documents. By no mean definitive, but somehow helpful)

---

Unique values - helpful in order to get a non-equity indicative?

Similar mechanisms would be used to get those unique values. For example, by removing the duplicates from the reverseMap :

for (Integer i: listDuplicates)
    reverseMap.remove(i);

Now the reverseMap only contains unique values. reverseMap.size() = 11

   KEY                     VALUE
   1                       welcome
   3                       what
   6                       doing
   9                       I
   10                      was
   12                      before
   15                      dumb

In order to get the unique words:

reverseMap.values() = {welcome,what,doing,I,was,before,dumb}

If you need to know which unique words are from which document, you could use the reverseMap (as the Lists may be altered after you execute methods such as retainAll on them):

• Count the number of words from the 1st document. This time, 7.
• If the key of the reverseMap is <=7, that unique word comes from the 1st document. {welcome,what,doing}
• If the key is >7, that unique word comes from the 2nd document. {I,was,before,dumb}

The uniqueness factor could also be another indicative, this way, a negative one (as we are searching for similarities here). Still could be really helpful.

---

equals and hashCode - avoid

As the hashcode() method for Strings won't return the same value for two same words (only for two same String Object references), wouldn't work here. String.equals() method works by comparing the chars (also checks for the length, as you do manually) which would be total overkill if used for big documents:

public boolean equals(Object anObject) {  
      if (this == anObject) {  
          return true;  
      }  
      if (anObject instanceof String) {  
          String anotherString = (String) anObject;  
          int n = value.length;  
          if (n == anotherString.value.length) {  
              char v1[] = value;  
              char v2[] = anotherString.value;  
              int i = 0;  
              while (n-- != 0) {  
                  if (v1[i] != v2[i])  
                          return false;  
                  i++;  
              }  
              return true;  
          }  
      }  
      return false;  
  }  

My oppinion is to avoid this as much as possible, specially hashCode() should never be used, as:

String one = "hello";
String two = "hello"; 

one.hashCode() != two.hashCode()

There's an exception to this, but only when the compiler interns strings; Once you load thousands of them, that won't ever again be used by the compiler. In those rare cases where both String Objects reference the same cached memory address, this will also be true:

one.hashCode() == two.hashCode() --> true
one == two --> true

But those are really unusual exceptions, and once string internship doesn't kick, those hashCodes won't be equal and the operator == to compare Strings will return false even if the Strings hold the same value (as usual, because it works comparing their memory addresses).

Answer 2

The essential technique is to see this is as a multi-stage process. The key is that you're not trying to compare every document with every other document, but rather, you have a first pass that identifies small clusters of likely matches in essentially a one-pass process:

(1) Index or cluster the documents in a way that will allow candidate matches to be identified;

(2) Identify candidate documents that may be a match based on those indexes/clusters;

(3) For each cluster or index match, have a scoring algorithm that scores the similarity of a given pair of documents.

There are a number of ways to solve (1) and (3), depending on the nature and number of the documents. Options to consider:

• For certain datasets, (1) could be as simple as indexing on unusual words/clombinations of words
• For more complex documents and/or larger datasets, you will need to do something sometimes called 'dimension reduction': rather than clustering on shared combinations of words, you'll need to cluster on combinations of shared features, where each feature is identified by a set of words. Look at a feature extraction technique sometimes referred to as "latent semantic indexing" -- essentially, you model the documents mathematically as a matrix of "words per feature" multiplied by "feature per document", and then by factorising the matrix you arrive at an approximation of a set of features, along with a weighted list of which words make up each feature
• Then, once you have a means of identifying a set of words/features to index on, you need some kind of indexing function that will mean that candidate document matches have identical/similar index keys. Look at cosine similarity and "locality-sensitive hashing" such as SimHash.
• Then for (3), given a small set of candidate documents (or documents that cluster together in your hashing system), you need a similarity metric. Again, what method is appropriate depends on your data, but conceptually, one way you could see this as "for each sentence in document X, find the most similar document in document Y and score its similarity; obtain a 'plagiarism score' that his the sum of these values". There are various ways to define 'similarity score' between two strings: eg longest common subsequence, edit distance, number of common word pairs/sequences...

As you can probably imagine from all of this, there's no single algorithm that will hand you exactly what you need on a plate. (That's why entire companies and research departments are dedicated to this problem...) But hopefully the above will give you some pointers.