[ab initio mathematics] 193th section reasoning and Proof

plot summary:
[Machine Xiao Wei] in the [engineer Ah Wei] accompanied by the [nine turn elixir] of the five-turn of the cultivation.
This is a study of [reasoning and proof].

Drama Start:

Star Calendar April 24, 2016 12:16:07, the Milky Way Galaxy Earles the Chinese Empire Jiangnan Line province.
[Engineer Ah Wei] is working with [machine Xiao Wei] to study [reasoning and proof].

<span style= "FONT-SIZE:18PX;" >>>> 23.5 13.865424623862042# Helen-Qin Jiushao formula  def hqformula (A, B, c):      p = (a+b+c)/2;      S = Math.sqrt (p* (p-a) * (p-b) * (p-c));        return S; #例3def tmp ():    #a, B, C = 3, 4, 5    A, B, C = 3, 4, 5    s1= a*b/2    S2 = a*c/2    S3 = b*c/2    S4 = H Qformula ((a*a+b*b) **0.5, (A*a+c*c) **0.5, (b*b+c*c) **0.5);    Print (S1+S2+S3, S4);</span>

<span style= "FONT-SIZE:18PX;" >if (1) {var array = new Array (); array[0] = [1];array[1] = [1,1];for (var i = 2; i <; i++) {Array[i] = [1];for (v Ar j = 1; J < I; J + +) {Array[i].push (array[i-1][j-1]+array[i-1][j]);} Array[i].push (1);} var width = n, height = 400;var x = width/2, y = 30;var measure = 0, s = ' '; var len = Array.length;for (var i = 0; I < ; Len i++) {s = array[i].join ('    ); measure = Plot.measuretext (s);p Lot.filltext (S, x-measure/2, y, measure); y + = 30;}} </span>

About seeking dihedral angle, Xiao Wei found this formula:

<span style= "FONT-SIZE:18PX;" >>>> cos = 0.9218142082600806 angle is: 22.80723538641745 degrees #[] dihedral angle; #二面角余弦def dihedral (): xyz = [-1.           2999760,0.0173840,-0.7162670, 1.0107690,1.5229620,-0.0945670, 0.0932470,-1.0086090,1.6265070,    -1.2280340,-1.4934410,1.8123150, 0.0932470,-1.0086090,1.6265070, 1.0734260,-1.5230970,2.5155820] #六个点 ax1 = xyz[0] Ay1 = xyz[1] az1 = xyz[2] bx2 = xyz[3] by2 = xyz[4] bz2 = xyz[5] cx3 = xyz[6] Cy 3 = xyz[7] cz3 = xyz[8] dx1 = xyz[9] dy1 = xyz[10] dz1 = xyz[11] ex2 = xyz[12] Ey2 = xyz[13] EZ2 = XY Z[14] fx3 = xyz[15] Fy3 = xyz[16] fz3 = xyz[17] #面一法线 NX = ((CZ3-AZ1)/(Cy3-ay1)-(BZ2-AZ1)/(By2-ay1))/((bx2- AX1)/(By2-ay1)-(CX3-AX1)/(cy3-ay1)) NY = ((CZ3-AZ1)/(CX3-AX1)-(BZ2-AZ1)/(BX2-AX1))/((By2-ay1)/(BX2-AX1)-(Cy3-ay1) /(CX3-AX1)) NZ = 1 #面二法线 mx = ((FZ3-DZ1)/(fy3-dy1)-(EZ2-DZ1)/(EY2-DY1))/((EX2-DX1)/(ey2-dy1)-(FX3-DX1)/(fy3-dy1)) my = ((FZ3-DZ1)/(FX3-DX1)-(EZ2-DZ1)/(EX2-DX1))/((ey2-dy1)/(EX2-DX1)-(fy3-dy1)/(fx3-dx1)) mz = 1 Cosangle = (NX    *MX+NY*MY+NZ*MZ)/((Math.sqrt (nx**2+ny**2+nz**2)) * (Math.sqrt (mx**2+my**2+mz**2))); Print (' cos = ', cosangle, ' angle is: ', 180/math.pi*math.acos (cosangle), ' degrees ');</span>

But in the process of verification, Wei found that the algorithm seems to have problems:

For example, the following:

Ask for VAB and ABC's dihedral angle.

<span style= "FONT-SIZE:18PX;" >>>> [2, 5, 2, 5, 0, 1, 0, 0, 0, 1, 1, 5, 5, 0, 1, 0, 0, 0]cos = 0.4911436350228293 angle: 60.5842228640166 degrees #[] Dihedral angle; #二面角余弦 # temporarily can only be counted not perpendicular or parallel to XY, XZ, yz either plane of the two planes of the dihedral angle # formula looking in ... def dihedral (points): #points格式是 []*18, six points, each point X, y, z-coordinate "' XYZ = [ -1.2999760,0.0173840,-0.7162670, 1.0107690,1.5229620,-0.0945670, 0.093247 0,-1.0086090,1.6265070, -1.2280340,-1.4934410,1.8123150, 0.0932470,-1.0086090,1.6265070, 1.0    734260,-1.5230970,2.5155820] "if (len (points)! =): Return ' INF ';    xyz = points; #六个点 ax1 = xyz[0] Ay1 = xyz[1] az1 = xyz[2] bx2 = xyz[3] by2 = xyz[4] bz2 = xyz[5] cx3 = xyz[6] Cy 3 = xyz[7] cz3 = xyz[8] dx1 = xyz[9] dy1 = xyz[10] dz1 = xyz[11] ex2 = xyz[12] Ey2 = xyz[13] EZ2 = XY Z[14] fx3 = xyz[15] Fy3 = xyz[16] fz3 = xyz[17] #面一法线 NX = ((CZ3-AZ1)/(Cy3-ay1)-(BZ2-AZ1)/(By2-ay1))/((bx2- AX1)/(by2-aY1)-(CX3-AX1)/(cy3-ay1)) NY = ((CZ3-AZ1)/(CX3-AX1)-(BZ2-AZ1)/(BX2-AX1))/((By2-ay1)/(BX2-AX1)-(cy3-ay1)/(CX3-AX1)) NZ = 1 #面二法线 mx = ((FZ3-DZ1)/(fy3-dy1)-(EZ2-DZ1)/(EY2-DY1))/((EX2-DX1)/(ey2-dy1)-(FX3-DX1)/(fy3-dy1)) my = ((fz3- DZ1)/(FX3-DX1)-(EZ2-DZ1)/(EX2-DX1))/((ey2-dy1)/(EX2-DX1)-(fy3-dy1)/(fx3-dx1)) mz = 1 Cosangle = (NX*MX+NY*MY+NZ*MZ)/    ((Math.sqrt (nx**2+ny**2+nz**2)) * (Math.sqrt (mx**2+my**2+mz**2)));    Print (' cos = ', cosangle, ' angle is: ', 180/math.pi*math.acos (cosangle), ' degrees '), #求二面角def tmp (): V = [2, 5, 2] A = [5, 0, 1]    B = [0, 0, 0] C = [1, 1, 5] points = v+ A + b + c+a+b;    #print (len (points));        print (points); Dihedral (points);</span>

Ask VBC and ABC, or VAB and vbc.

<span style= "FONT-SIZE:18PX;" >>>> [2, 5, 2, 0, 0, 0, 1, 1, 5, 1, 1, 5, 5, 0, 1, 0, 0, 0]cos =  -0.2558139534883721 angle: 104.821821062050 12 degrees >>> ================================ RESTART ================================>>> [2, 5, 2, 5, 0, 1 , 1, 1, 5, 1, 1, 5, 5, 0, 1, 0, 0, 0]cos =  0.19218663979154185 angle: 101.08042157684446 degrees </span>

How can be more than 100 degrees, look not like Ah, but little Wei also can not say is right is wrong.

Let's put it on first.

<span style= "FONT-SIZE:18PX;"                    >if (1) {var r = 20;                      Config.setsector (1,1,1,1);                    Config.graphpaper2d (0, 0, R);                                      Config.axis2d (0, 0,190);                  Axis set var ScaleX = 4*r, ScaleY = 4*r;                   var SpaceX = 2, SpaceY = 2;                  var XS = -10, XE = 10;                  var YS = -10, YE = 10;                    Config.axisspacing (XS, XE, SpaceX, ScaleX, ' X ');                                                     Config.axisspacing (YS, YE, SpaceY, ScaleY, ' Y ');            var array = [[2, 5, 2], [5, 0, 1], [0, 0, 0], [1, 1, 5];          Array = shape.xyzsort (array);                      var size = Array.Length;          var array2d = [];          for (var i = 0; i < size; i++) {Array2d.push (Shape.point3d (array[i][0], array[i][1], array[i][2])); }//Remove duplicate point var Pointarray = RemoveduplicaTedpoint (Array2D);                        Number of non-repeating points var points = pointarray.length;            Get the distance array//format as [[Point 1 ordinal, point 2 ordinal, distance value], ...]            var Distancearray = Distancesort (Pointarray);                        The number of sides var edges = distancearray.length;            Store the side var linkedarray = [] that need to be connected;                        Number of connected edges var links = 0;                        The set of edges associated with each vertex var edgeofvertex = []; for (var i = 0; i < points; i++) {//Gets a collection of vertex-related edges edgeofve                Rtex = [];                        for (var j = 0; J < edges; J + +) {if (distancearray[j][0] = = I | |                    DISTANCEARRAY[J][1] = = i) {Edgeofvertex.push (distancearray[j]);                    }}//Based on the starting point, look for the shortest length of two edges edgeofvertex.sort (function (A, b) { return A[2]- B[2];                                });                var choice = 4;                if (Edgeofvertex.length > Choice) {edgeofvertex = Edgeofvertex.slice (0, choice);            } Linkedarray = Linkedarray.concat (Edgeofvertex);            }//document.write (Linkedarray.join (', ') + ' <br/> ');            Linkedarray = Removeduplicatedpoint (Linkedarray);                        Links = linkedarray.length;                            document.write (Linkedarray.join (', ') + ' <br/> ');            var startPoint, EndPoint, x1, y1, x2, y2;                        Scaled var scale = 40;                for (var i = 0; i < links; i++) {startPoint = linkedarray[i][0];                EndPoint = linkedarray[i][1];                x1 = pointarray[startpoint][0];                y1 = pointarray[startpoint][1];                x2 = pointarray[endpoint][0]; y2 = poINTARRAY[ENDPOINT][1];            Shape.vectordraw ([[X1,y1], [X2, Y2]], ' red ', scale);                        } shape.pointdraw (Pointarray, ' blue ', scale, 1, ' vabc ');            Plot.setfillstyle (' Blue ');                  Plot.filltext (' Vector graph ',-270,-170, 300); }</span>

The end of this section, to know how to funeral, please see tell.

[ab initio mathematics] 193th section reasoning and Proof