Mini dvi-vga problem - Need help

When I connect my mac mini to a 32 inch Samsung HDTV via a dvi-vga adapter, the source is not recognized. I jsut get a black screen. However, when I connect it to the HDMI port, it works great. But, I wantto use the HDMI port for the Television source, not the computer source.
Any ideas?
Thanks

Did you get the mini-DVI to VGA adapter from Apple? Are your other cables okay?
Perhaps a mini-DVI to video adapter would be a better choice for you.
~Lyssa

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    9 5 2 2 8
    6 5 4 5 3
    C0 C1 C2 C3 C4
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    NP
    What I have tried till now.
    I have one more question could you help me with this.
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    import java.util.*;
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    private  int P;
    private  int C;
    private int weight;
    public void set_p(int P1)
    P=P1;
    public void set_c(int C1)
    C=C1;
    public void set_weight(int W1)
    weight=W1;
    public int get_p()
    return P;
    public int get_c()
    return C;
    public int get_x()
    return weight;
    public static void main(String args[])
    ArrayList a=new ArrayList();
    ArrayList min_weights_int=new ArrayList();
    ArrayList rows=new ArrayList();
    ArrayList temp=new ArrayList();
    Hashtable h=new Hashtable();
    String v;
    int o=0;
    DD[] d=new DD[5];
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    d[i].set_weight(0);
    System.out.println("Producers");
    for(int i=0;i<4;i++)
    System.out.println(d[i].get_p());
    System.out.println("Consumers");
    for(int i=0;i<4;i++)
    System.out.println(d[i].get_c());
    System.out.println("Weights");
    for(int i=0;i<4;i++)
    System.out.println(d[i].get_x());
    for(int i=0;i<4;i++ )
    int bi =d[i].get_c();
    ArrayList row=new ArrayList();
    for(int j=0;j<4;j++)
    if( d[j].get_p() >=bi)
    d[j].set_weight((int)(StrictMath.random()*10 + 1));
    row.add("(" + bi + "," + d[j].get_p() + "," +d[j].get_x() + ")");
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    d[j].set_weight(0);
    row.add("null");
    rows.add(row);
    System.out.println(rows);
    int f=0;
    for(Iterator p=rows.iterator();p.hasNext();)
    temp=(ArrayList)p.next();
    String S="S" +f;
    h.put(S,temp);
    String tt=new String();
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    //System.out.println(temp.get(j).toString());
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    System.out.println(l);
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    for(Enumeration e=h.keys();e.hasMoreElements();)
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    int smallest=0;
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    if(l!="null")
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    [\code]

    In your example you selected the pair with the greatest
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    function was used.
    Also it's not clear to me that there is always a solution,
    and, if there is, whether consistently choosing the
    furthest or the closest pairs will always work.
    The most obvious approach is to systematically try
    all possibilities until the answer is reached, or there
    are no possibilities left. This means backtracking whenever
    a point is reached where you cannot continue. In this case
    backtrack one step and try another possibility at this
    step. After all possible choices of the previous step,
    backtrack one more step and so on.
    This seems rather involved, and it probably is.
    Interestingly, if you know Prolog, it is ridiculously
    easy because Prolog does all the backtracking for you.
    In Java, you can implement the algorithm in much the same
    way as Prolog implementations do it--keep a list of all the
    choice points and work through them until success or there
    are none left.
    If you do know Prolog, you could generate lots of random
    problems and see if there is always a solution.

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