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copied!<p>As an alternative to the above approach, here is code that will perform the exhaustive search in Java. It is not quite as efficient - but it has the advantage of being OS-independent.</p> <p>There are three conceptual blocks here: First, I create a 10x10 grid (using code from your application above... I needed to get a table[][] somehow). I added a little "trick" here - instead of picking letters randomly from the alphabet, I pick them randomly from a string with "approximate" English letter frequencies. We should get more "English sounding" word combinations in the resulting grid. </p> <p>Next, I read the grid left to right, right to left, top to bottom, and bottom to top. This creates a total of 40 strings to search. </p> <p>Third, I open the input file and look at one string at a time; I attempt to match this against the 40 strings in turn, and as soon as I find a match I stop looking (in other words, it doesn't matter if there is more than one match).</p> <p>I ran this with a very small input file, and it seemed to work. When you run it against a full dictionary it may be quite slow - but note that I do attempt to break out of the search loops as soon as I -know I have a match, or -know I don't have a match.</p> <p>There are no doubt ways this could be further optimized.</p> <pre><code>import java.util.Random; import java.io.*; public class myGrep { public static void main(String[] args){ char t40[][] = new char[40][10]; char table[][] = new char[10][10]; //Creates the array for the 10x10 grid int ii=0, jj=0; char c; int ctr = 0; char m; Random r = new Random();//for the random characters // First, I create a random table of 10x10 characters to work with // To make it more interesting, I use a string with approximate frequencies // for characters in the English language - this should yield more, and longer // "findable" words in the grid // source: http://www.math.cornell.edu/~mec/2003-2004/cryptography/subs/frequencies.html String cfreq = "eeeeeeeeeeeetttttttttaaaaaaaaoooooooo" + "iiiiiiinnnnnnnssssssrrrrrrhhhhhhddddl" + "llluuucccmmmffyywwggppbbvkxqjz"; for (int x = 0; x < 10; x++) { for (int y = 0; y < 10; y++) { table[x][y] = cfreq.charAt(r.nextInt(cfreq.length())); } } // Now, make a second table of 40 lines; this speeds up the search // since I won't have to think about "direction" any more ... just search L to R for (int x = 0; x < 10; x++) { for (int y = 0; y < 10; y++) { m = table[x][y]; t40[x][y]=m; // L to R t40[x+10][9-y]=m; // R to L t40[y+20][x]=m; // T to B t40[y+30][9-x]=m; // B to T } } // confirm that we now have 40 rows of strings: System.out.println("The following 40 strings represent searching the 10x10 grid in all directions:"); for( ii = 0; ii<40; ii++) { System.out.println(t40[ii]); } // Now starts the fun... for each word in the input file, look for a match try { FileInputStream fstream = new FileInputStream("textfile.txt"); DataInputStream in = new DataInputStream(fstream); BufferedReader br = new BufferedReader(new InputStreamReader(in)); String strLine; //Read File Line By Line while ((strLine = br.readLine()) != null) { if(canFind(strLine, t40)) { System.out.println (strLine); } } //Close the input stream in.close(); } catch (Exception e) { //Catch exception if any System.err.println("Error: " + e.getMessage()); } } // end of main() private static boolean canFind(String findme, char[][] searchme) { boolean isfound = false; int li = 0; // loop over all lines in the searchme table: we know there are 40 while (li < 40 && !isfound) { // look for a match of the first character: for(int ii = 0; ii<11-findme.length(); ii++) { if(searchme[li][ii] == findme.charAt(0)) { isfound = true; for(int jj = 1; (jj < findme.length()) && isfound; jj++) { // **** this line could probably be optimized more: **** isfound = (isfound && (findme.charAt(jj) == searchme[li][ii+jj])); } ii = 10; // exit for(int ii = 0;..) loop... } } li++; } return isfound; } } </code></pre>

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