Drawing a Christmas tree w/ nested for loops and from JavaScript to Java

I need help in making this code compatible with Java (I use BlueJ):
It's JavaScript...
Thanks in advance!
<HTML>
<BODY>
<CENTER>
<SCRIPT>
var width=1;
for (i=0; i <= 5 ; i++)
for (x=0; x<=4; x++)
for (y=1; y<=width; y++)
var Number=Math.random()*10;
var Ornament=Math.round(Number);
if (Ornament<=1)
document.write("O");
if (Ornament>=2)
document.write(" X");
document.write("<BR>");
width=width+1;
width=width-2;
</SCRIPT>
</CENTER>
</BODY>
</HTML>

What do you mean "compatible with Java"? Java doesn't enter into this at all.
Are you asking how to translate this code to Java?
If so, where you have "<BR>", you'll want a println rather than a print (as in System.out.print).
The second tricky part is centering. The browser handled that for you here.
You'll have to either do it yourself, or find another Java API tool to do it for you.
If I were you I'd look at the API docs for PrintStream.format(), I think it is. It's possible that the %s conversion has a "center" option. Or maybe something else there does.
If there is no textual centering tool, you can build one yourself by using StringBuffer or StringBuilder and padding with spaces, I guess.
Otherwise it should map across pretty easily.

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    I have fixed a number of things in your vi.
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             E-List Master - Kudos glutton - Press the yellow button on the left...        
    Attachments:
    your_vi.vi.zip ‏13 KB

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    ___X___
    __XXX__
    something like that. Need some explaination or hint what i am missing. (underscores are not part of my program:))
    thanks

    So since the spaces look right now, lets step through the plus loop
    for(k = i; k <= i*2-1; k++)
                    System.out.print("+"); First time starting to go through the loop i is 0.
    So 0 is not <= -1 [0*2-1 = -1] so no pluses get printed on the first time through.
    the second time starting to go through the loop i is 1.
    So first time through 1 is <= 1 [1*2-1 = 1] so print a +.
    Second time through 2 is not <= 1, so we end with just one +.
    The third time starting the loop i is 2.
    So first time through 2 is <= 3 [2*2-1 = 3] so print a +.
    Second time through 3 is <= 3 so print another +.
    Third time through 4 is not <= 3 so we end with just two +s.
    Uh oh, we wanted three s not two s, where did we go wrong? Let's try one more and see if we can see it.
    The fourth time starting i is 3.
    So first time through 3 is <= 5 [3*2-1 = 5] so print a +.
    Second time through 4 is <= 5 so print another +.
    Third time through 5 is <= 5 so print another +.
    Fourth time through 6 is not <= 5 so we end with just three +s.
    Now we have a choice, we can get mad and start throwing stuff and screaming at the stupid computer for doing it all wrong, or we can realize the problem is somewhere in the code and we can find it and fix it.
    What do you think?
    JSG
    EDIT: Note: you actually have two problems displayed here, both should be obvious but one is slightly hidden in the output until you fix the other. Both should have made themselves clear if you had done a simple step through yourself. As i demonstrated it is really easy to do and something you better get real good at if you want to be a good programmer.
    Edited by: JustSomeGuy on Nov 24, 2008 11:22 AM

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