Graal Forums - Programming Exercise #5: Solving Boggle?!

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Programming Exercise #5: Solving Boggle?!

For those of you who know not what Boggle is:
http://en.wikipedia.org/wiki/Boggle

So today boys and... boys, our task is to make a word finder in Boggle.
Given a word list, find all the words from that list you can make on an x by x Boggle board (the game itsself is a 4x4 board, but we can do better :D)

Expert Level: Do the above task with recursion :)

Addeum: We can use the TWL06 as our word list.
http://www.cs.bu.edu/fac/byers/cours...ndouts/TWL.txt

So, the algorithm to find all words in a Boggle board in the most efficient way?

And, just pick random letters and find words by them? Are there any rules (Hard letters like K (2 or more), U, X, Z etc.) of letters that can be used to produce the board?

Quote:

Originally Posted by zokemon (Post 1398009)

Arguments please.

If someone wants to go grab a boggle board and put up a random example, feel free to do so.

As for the rules, that is the reason I linked the wiki...

I'm asking what should be passed to the function x_x.

Quote:

Originally Posted by zokemon (Post 1398217)

I'm asking what should be passed to the function x_x.

I left that up to you guys since the answer was obvious i thought.

You pass the board its self. That is the only thing you would ever need. If you want to give it pointers to different word lists too that is your choice.

Do 2 letters have to link to form a word or can you pick any letter from any position to form a word?

Ahh, didn't see that you added a wordlist above.

Sudoku is better than Boggle!

Quote:

Originally Posted by Rufus (Post 1398279)

Sudoku

Crosswords for dyslexic people.

Quote:

Originally Posted by Dan (Post 1398273)

Do 2 letters have to link to form a word or can you pick any letter from any position to form a word?

have to link, and you can't reuse letters.

Quote:

Originally Posted by Kristi (Post 1397978)

Given a word list, find all the words from that list you can make on an x by x Boggle board (the game itsself is a 4x4 board, but we can do better :D)

Addeum: We can use the TWL06 as our word list.
http://www.cs.bu.edu/fac/byers/cours...ndouts/TWL.txt

Expert Level: Do the above task with recursion

you can find all of the words on an x by x boggle board, as long as you can find all of the words on an x by x boggle board.

edit- answer:

Let M be an n x n matrix. (meaning, let M be an n-1 x n-1 array)
Let W be the set of accepted words.

if a set of entries of M, called X, where each entry of M in X is adjacent to the one both preceding and following, is an element of W, then there is a word in the matrix.

Problem: define the set of all possible values for X
(note: the maximum length of X is n*n)

how to do this:

loop through all entries of M sub 'i,j', (we don't need to check if it is apart of the alphabet because this is a boggle board) then check for each entry that:

M sub 'i+1,j' is an element of W.substring(0,1)
M sub 'i-1,j' is an element of W.substring(0,1)
M sub 'i,j+1' is an element of W.substring(0,1)
M sub 'i,j-1' is an element of W.substring(0,1)
M sub 'i+1,j+1' is an element of W.substring(0,1)
M sub 'i-1,j-1' is an element of W.substring(0,1)
M sub 'i+1,j-1' is an element of W.substring(0,1)
M sub 'i-1,j+1' is an element of W.substring(0,1)

if any of those are true (and if not then move onto the next entry in the matrix), repeat the process with the new entry and see if it is an element of W.substring(1,2). if that is true, do it again with W.substring(2,3), then (3,4), and so on.

how all of that looks in code talk

PHP Code:


		
			
M = 2-dimensional string array of length n-1 by n-1 containing only letters of the alphabet because arrays start at index 0 and not 1;

W = string array containing all of the acceptable words, one word per entry;

for (i=0;i<n-1;i++) { //row # of the matrix
  for (j=0;j<n-1;j++) { //column # of the matrix
    check(i,j,0);
  }
}
check(r, s, p) {
  t = p;
  for (k=0;k<8 && t==p;k++) { //8 because there are 8 directions to check
    if (k==0 && r-1>=0) { //look above the entry, if its not at the top
      for (l=0;l<W.length();l++) { //for each word in W
        if (M[r][s] in W[l].substring(p,p+1)) { 
          //p is used to check the pth letter of the lth string in W.
          p++; //<--move onto the next letter and...
          check(r-1, s, p); //start over
        }
      }
    } else if (k==1 && r+1<=n-1) {//look below, if its not at the bottom
      ...
    } else if (k==2 && s-1>=0) {//look left, if its not at the left edge
      ...
    }

    //dot dot dot

  }
}

since i'm being asked to show what words can be found:

add the string W[l] to a separate array every time
M[r][s] in W[l].substring(p,p+1) && p == W[l].size(), and then break; so that the search can continue.

Quote:

Originally Posted by _Z3phyr_ (Post 1410788)

stuffs

I just quickly glanced this over because I am at work but two things...

I don't see a return value.

I don't see handling of already used letters (you can't use the same letter/piece twice in boggle)