import java.applet.Applet;
import java.awt.*;
import java.awt.event.*;
import java.applet.*;
import java.awt.geom.*;
import java.math.*;


public class DocumentStats {
    Manager M;
    public DocumentStats(Manager MM) {
	this.M=MM;
    }

    public void statsInfo() {

	String S="Configuration Stats";

	S=S+"\n\nA dyadic box is the product \n\n Q=Q0 x Q1 x Q2 x Q3 \n\n where Q0 is a dyadic segment contained in [0,4] and each of Q1, Q2, Q3 is a dyadic square contained in [-2,2] x [-2,2].   Our proof performs certain tests on dyadic boxes and eliminates a box if it passes at least one of the tests.  This window displays the numerical results of the tests.  It serves as a sanity check that the program is working correctly. ";

	S=S+"\n\nBASIC INFORMATION. The first portion of this display lists some basic information about the configuration that does not directly feed into the algorithm:\n\n1. We list the energy of the bi-pyramid.\n\n2. We list the energy of the configuration made from the centers of the segments/squares.\n\n3. We list the energy of the same configuration, computed with the bond masking that is controlled by the pink/blue segments in the configuration control panel.  You should read the documentation for the configuration control to see what this means.\n\n4. We list the volume of the dyadic box.  The volume of the whole configuration space is 2^(14).\n\n5. We say whether or not the dyadic box in question confined to within 2^(-k-1) of the triangular bi-pyramid in the L_infinity norm.  Here k is controlled by the red arrow keys labelled 'confine diameter' on the left hand side of the program. You can see this in action by turning on the 'min1' feature on the 'display' panel.  The answer to the question here is 'yes' if and only if each Qj is contained within 2^(-k) of the square of side length 2^(-k) about the jth point of the triangular bi-pyramid.  If you drag the configuration around on the picture window, and note when the question changes from 'yes' to 'no' you will see what we are taking about here.";

	S=S+"\n\nThe remaining information is precisely as we have described in the paper, in section 2.5.   The final display says whether or not the program eliminates the box.   If the answer to the final question is 'no', then the program replaces the dyadic box by some subdivision.  In this case,  a colored square also appears on the display.  This colored square tells you which way the program subdivides the box.";

	M.E.setExplain(S);
	M.E.repaint();
    }






}