Is HTML size a problem using JavaHelp?

Hi guys,
I converted a MS Word document to HTML and I'd like to display it using JavaHelp. The HTML file size is about 800k. It does work, but it takes a few minutes to be displayed!!!
I tried with HTML files with size of about 100k and it works better (it takes a few seconds).
Where is the problem? Is the HTML size? Or is MS Word that doesn't work so well?
Thanks a lot for yoour answers!
Paul.

I agree, using Ms Word to produce HTML it's not the best way. But I have a word file (more than 100 pages) and I can't spend much time to create the HTML by myself.
However I'll try to split my file and see if it could work better.
In any case, may the html size be a problem? Or a well formatted html file with size of 800k should be displayed quickly?
Thank you very much, regards
Paul.

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    -Pop items off the stack until the stack becomes empty or the top of the stack is a choice that is not in column n. If the stack is now not empty, then increase the column number of the top choice by 1.
    else if (no conflict and the stack size is n)
    -Set success to true because we have found a solution to the n-queens problem.
    else
    -Push information onto the stack indicating tat the next choice is to place a queen at row number s.size()+1 and column number 1.
    And here is my excuse for code so far. I have no idea how to check the diagonals, or how to even really make this work
    {code}import java.util.Stack;
    public class NQueens {
    int row, column, n;
    public NQueens(int n) {
    row = 0;
    column = 0;
    n = n;
    public Stack Solve(){
    boolean success, conflict;
    Stack<NQueens> Qs = new Stack<NQueens>();
    if (Qs.size() == 0)
    Qs.push(new NQueens(1));
    success = false;
    while (!success && !Qs.isEmpty())
    if (Qs.peek().row == row)
    conflict = true;
    if (Qs.peek().column == column)
    conflict = true;
    if (conflict = true)
    Qs.pop();
    Qs.peek().column += 1;
    else
    if (!conflict && Qs.size() == n)
    success = true;
    else
    Qs.push(new NQueens(Qs.size()+1));
    return Qs;
    {code}

    First off I'll address this:
        int row, column, n;
        public NQueens(int n) {
            row = 0;
            column = 0;
            n = n; //here
        }Notice the last line of that. I get what you're trying to do, but think about what the compiler sees there. If you have two variables called 'n', one at the class level, and one at the method level, the compiler must have rules so it knows which one you're talking about. And if it follows those rules, then saying 'n' inside that method must always refer to the same n. Otherwise, how would it decide which one you're talking about? The rule here is that it uses the most local variable available to it, which in this case is your method parameter. So 'n = n' is setting the parameter equal to itself. To refer to the class variable n, use "this.n", like so:
    this.n = n;Now that that's settled, let's address some logic. You'll need to figure out whether two Queens share a diagonal. I can think of at least a couple options for that. First, every time you look at a Queen you could loop through the entire board and make sure that if a piece is in r3c4, that no piece is in r2c3,r4c5,r5c6, etc. Or you could create a new variable for your class similar to your 'row' and 'column' variables that tracks the diagonal a piece is in. But that requires some calculating. And remember, there's 2 directions for diagonals, so you'll need 2 diagonal variables.
    If these are your Row and Column values:
    Row          Column
    00000000     01234567
    11111111     01234567
    22222222     01234567
    33333333     01234567
    44444444     01234567
    55555555     01234567
    66666666     01234567
    77777777     01234567And these are your diagonal values:
    Diag 1                    Diag 2
    0  1  2  3  4  5  6  7          7  6  5  4  3  2  1  0
    1  2  3  4  5  6  7  8          8  7  6  5  4  3  2  1
    2  3  4  5  6  7  8  9          9  8  7  6  5  4  3  2
    3  4  5  6  7  8  9  10          10 9  8  7  6  5  4  3
    4  5  6  7  8  9  10 11          11 10 9  8  7  6  5  4
    5  6  7  8  9  10 11 12          12 11 10 9  8  7  6  5
    6  7  8  9  10 11 12 13          13 12 11 10 9  8  7  6
    7  8  9  10 11 12 13 14          14 13 12 11 10 9  8  7Then examine the relationship between Row(R), Column(C), and Diagonals 1 and 2 (D1/D2):
    RC   D1,D2
    00 = 0,7
    01 = 1,6
    02 = 2,5
    03 = 3,4
    04 = 4,3
    05 = 5,2
    06 = 6,1
    07 = 7,0
    10 = 1,8
    11 = 2,7
    12 = 3,6
    13 = 4,5
    14 = 5,4
    15 = 6,3
    16 = 7,2
    17 = 8,1You'll notice that D1 is always the same as R + C. And D2 is always the same as C subtracted from 7 plus R, giving:
    int d1 = row + column;
    int d2 = (7 + row) - column;But remember, that 7 in the formula above, is based on how big your grid is. So whatever your 'N' is, whether its a 10x10 square or a 574x574 square, that 7 should be changed to (N-1)
    So those could be your variables to track diagonals. What I'm noticing in your current code, is that you never change your row and column variables...so every Queen is always at r0c0. You should probably put values for those in your parameters for the constructor, in addition to n. Then the diagonals can be calculated from those.
    Edited by: newark on Apr 17, 2008 10:46 AM

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