String counter

**chkang0130** · December 6th, 2009, 08:07 PM

1 Parallel motif counting of strings.

Recently, there has been some interest in Computational Linguistics and Bioinformatics in counting motifs in strings. In particular, counting all of the instances of a particular sub-string in a given string. For example, if we have a string ATATATAGTATA then there are 3 instances of the motif TATA in it.

1.1 Implementation

You will write a JCSP implementation that will count the number of occurences of a particular motif in a string, where the counting is distributed across the processes.
Specification: The main program (runMotif) will accept as input the string to be examined, the motif to be searched and the number of processes npes that will perform the search. npes instances of the class motif will be constructed, each with a local copy of the string.
Each instance of the class will examine a substring of the string and count the number of motifs that occur in each substring. Hint: think about the above example for say two processes. The number of motifs computed for each process must now be added together to compute a final sum.
Note : you will not get any marks for performing a purely serial version of this code.

can anyone help me to solve this?

**Json** · December 7th, 2009, 03:37 AM

Of coruse we can, what have you done so far and what are you stuck on?

If you need something to get you started, here you go.

    public class MyClass {
 
        public static void main(String... arguments) {
            // Put your stuff here
        }
    }

// Json

**chkang0130** · December 7th, 2009, 12:13 PM

i'm dealing with making method for count motif..

i made one main class and one other class to count the strings...

method of counting motif... that's problem... help!!!

**chkang0130** · December 7th, 2009, 01:55 PM

public class stringP {
 
 
	public static void main(String[] args) {
 
		String name = "ATATATAGTATA";
		char pattern = "TATA";
	    int occurs = 0;
 
 
	    for(int i=0; i < name.length(); i++) {
	      char next = name.charAt(i);
	      if(next == pattern) {
	        occurs++;
	      }
	    }
 
	    System.out.println("Occurs " + occurs + " time(s)");
 
 
	}
 
}

i want to find string 'TATA' and count how many of them (include overlap)!
help!!!

**copeg** · December 7th, 2009, 02:07 PM

Look into using the indexOf method of String.
Defined as:

public int indexOf(String str,
                   int fromIndex)

You can use this method to iterate through the string quite easily
You can do it a similar way that you outlined, just make sure you are comparing things that can be compared (eg char pattern = "TATA"; won't compile - do you want to compare Strings or character arrays?)

**chkang0130** · December 7th, 2009, 02:12 PM

i want to compare strings actually (TATA).
im not sure how should i use indexOf method...

can you apply that in my code?

**chkang0130** · December 7th, 2009, 03:12 PM

public class stringP {
 
 
public static void main(String[] args) {
 
String name = "ATATATAGTATA";
char pattern = "TATA";
int occurs = 0;
 
 
for(int i=0; i < name.length(); i++) {
char next = name.charAt(i);
if(next == pattern) {
occurs++;
}
}
 
System.out.println("Occurs " + occurs + " time(s)");
 
 
}
 
}

can anyone help me to make this as parallel (jcsp) ?

**copeg** · December 7th, 2009, 03:35 PM

indexOf gives you the position of the first occurrence of a string within a string, or -1 if it isn't found. Read the API for a full description. I'm not going to do it for you, but this should hopefully get you started...

int loc = 0;
loc = name.indexOf(pattern, loc);

Now do this for the entire string (hint: use a loop)

**helloworld922** · December 7th, 2009, 10:59 PM

Here's a simple algorithm for doing that. fyi, this is how a Regex Engine works for constant strings.

1. Start looking from the left (position 0)
2. See if the letter matches the first letter in the string you want to match with (in your case, "T")
3. If it does, keep matching subsequent letters until you form the string you want ("TATA")
4. If it doesn't start step 2 again, but start with the second letter, third letter, etc. until you either have found the first instance of the string you want to match, or have reached the end of the string.

To count up all the instances of the matching string, simply run the above algorithm but start each subsequent run on the start position of the previous matched string + 1

**Rouba** · December 8th, 2009, 11:56 AM

See lecture notes, page 5.

See you in class!

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