isOnBoard: 35  points total

	how to grade:  First go in BogglePlayer and change the customized  
board to contain a dice  "Qu" and other dice of > 1 letter.
	For example, you can have a board that looks like this:
	
	a  b  c  d
	e  f   g  h
	qu 1 2 3
	ab  p  q  xyz

       Then go  in BoggleGUI.java,  in checkAndAddWordHuman   and  
comment out the line that checks if its in lexicon:
	
	if(!computerPlayer.isInLexicon(wordToCheck))          {
      		chideUser(wordToCheck, "Not In Lexicon");
      		return;
    	}
	Now you should be able to check random combinations of letters  at  
leisure  without  seeing the "Not in lexicon message".

	Try a systematic sample of words that should be on board: 	abcd,  
abcdh, abcdh3, af23, and so on.
	
	Then try words including Qu:   qu123, qufgh, quea, etc

	Then try words including a different dice of 2 letters (in this case  
"ab"):  ab123

	Then try words including dice of 3 letters "xyz":  xyzqp

	Then try some words that should not be on the board.


	How to divide the points:  handling 1-letter dice:   that's 23 points
			if this does not work correctly, subtract based on gravity of error.
	If it handles Qu:  7 points
	If it handles dice of arbitrary length:     5 points





getAllValidWords: 35 points total

	Same as above,  set a custom board, pick one that has some good words  
that you can see.
	
	here is what I would do: I would write a test method that gets  all  
the valid words in a different way, so that you know what the answer  
is.   then check that they get all words with their method.

      You can test yourself and write getAllValidWords(), or you can
      cheat and do exactly what we told the students NOt to do: you
      can iterate through the dictionary, and collect all words that
      are on board.

      In our own copy of the BogglePlayer, fill in getAllValidWords as
follows:

	Vector<String> getAllValidWords() {

		Iterator i =  lexicon.iterator();
		Vector<String> result = new Vector(); 	

		while (i.hasNext()) {

			String word = (String) i.next();
			//if the word is on board add it
			 if ((word.length() >= minWordLength) && (isOnBoard(word))
					result.add(word);
			}
		}
		return result;
	}	

	After you run this version,  you'll know  what  the right answer is.   
Chose a board with at least  30 words.
	Give points depending of what percentage of the total number of words  
they got.



General style:   30 points
	
	how complicated the code is
	how readable the code is