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Exact String Matching: the brute force method

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• The Exact String Matching problem
  • Exact String Matching:

    Given: A text string T = t0 t1 t2 ... tn-1                   length(T) = n A pattern string P = p0 p1 p2 ... pm-1          length(P) = m Problem: Determine if P is a substring of T: Example: T = acgttagatactaggatgcca P = gata Solution: acgttagatactaggatgcca gata

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• The brute force approach
  • Brute Force:

    Generate every possible outcome           Test if a solution is correct

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  • Psuedo code for any brute force algorithm:

    for ( every possible outcome i ) do { if ( i is a correct solution ) { print solution i; (exit if you only need one solution) } }

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• Brute force exact string matching
  • Try every spot in the text string T to find P:

    T = acgttagatactaggatgcca P = gata Try: acgttagatactaggatgcca gata acgttagatactaggatgcca gata acgttagatactaggatgcca gata acgttagatactaggatgcca gata and so on...

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  • Psuedo code:

    for ( pos = 0; pos < n-m; pos++ ) { if ( T[pos..pos+(m-1)] == P ) { return(pos); // Found match.... } }

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  • Java code:

    public static int BruteForce(String T, String P) { int i, j, m, n; n = T.length(); m = P.length(); T_ruler = ruler(n); P_ruler = ruler(m); i = 0; j = 0; for ( i = 0; i < n-m; i++ ) { if ( P.equals( T.substring(i, i+m) ) ) { return( i ); // Found P at position i in T } } return(-1); }

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  • Example Program: (Demo above code)                                                
    • Prog file: click here

    How to run the program:

    Right click on link(s) and save in a scratch directory To compile:   javac BruteForce.java To run:          java BruteForce

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• Performance of the Brute Force Exact String matching
  • Simple loop counting will give:

    Running time = (n-m+1)*m (You need m compare operations to test: T[pos..pos+(m-1)] == P) Running time = O(n*m)

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• Can we do better ???
  • Usual answer:

    Yes, but it will cost you....

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  • Another way to say this:

    "Do it" is easy, "do it well" is hard

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  • My college professor (Herman Bavinck) at TH Delft has a funny way to express this.

    He has an interesting law of conservation, like the law of conservation of matter and energy in Physics.

    His law is:

    The law of conservation of misery No matter which route you take to solve a problem (or prove a theorem - he was a Mathematician), it will be equally miserable Herman Bavinck

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  • In terms of algorithms:

    If you write a simple but inefficient algorithm, then: Your level of misery in writing the code is lower, but your level of misery in waiting for the code to complete is higher, If you write a harder but more efficient algorithm, then: Your level of misery in writing the code is higher, but your level of misery in waiting for the code to complete is lower,