[从头学数学] 第193节 推理与证明

剧情提要：
[机器小伟]在[工程师阿伟]的陪同下进入了[九转金丹]之第五转的修炼。
这次要研究的是[推理与证明]。

正剧开始：

星历2016年04月24日 12:16:07, 银河系厄尔斯星球中华帝国江南行省。
[工程师阿伟]正在和[机器小伟]一起研究[推理与证明]。

<span style="font-size:18px;">>>> 
23.5 13.865424623862042

#海伦-秦九韶公式  
def HQFormula(a, b, c):  
    p = (a+b+c)/2;  
    S = math.sqrt(p*(p-a)*(p-b)*(p-c));  
  
    return S;

#例3
def 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 = HQFormula((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 < 10; i++) {
			array[i] = [1];
			for (var j = 1; j < i; j++) {
				array[i].push(array[i-1][j-1]+array[i-1][j]);
			}
			array[i].push(1);
		}
		
		var width = 600, 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);
			
			plot.fillText(s, x-measure/2, y, measure);
			y += 30;
		
		
		}
	}</span>

关于求二面角，小伟找到了这个公式：

<span style="font-size:18px;">>>> 
cos =  0.9218142082600806 角度是： 22.80723538641745 度

#[数] 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]
    cy3 = xyz[7]
    cz3 = xyz[8]
    dx1 = xyz[9]
    dy1 = xyz[10]
    dz1 = xyz[11]
    ex2 = xyz[12]
    ey2 = xyz[13]
    ez2 = xyz[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, '角度是：', 180/math.pi*math.acos(cosAngle), '度');
	</span>

但是在验证的过程中小伟发现这个算法好像有问题：

比如对于下面这个例子：

求一下VAB和ABC的二面角

<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 角度是： 60.584222864016645 度


#[数] dihedral angle;
#二面角余弦
#暂时只能算不垂直或平行于xy, xz, yz任一平面的两平面的二面角
#公式寻找中...
def dihedral(points):
    #points格式是 []*18, 分六个点，每个点x, y, z坐标排列
    '''
    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]
    '''

    if (len(points) != 18):
        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]
    cy3 = xyz[7]
    cz3 = xyz[8]
    dx1 = xyz[9]
    dy1 = xyz[10]
    dz1 = xyz[11]
    ex2 = xyz[12]
    ey2 = xyz[13]
    ez2 = xyz[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, '角度是：', 180/math.pi*math.acos(cosAngle), '度');


#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>

再求一下VBC与ABC的，或是VAB与VBC的

<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 角度是： 104.82182106205012 度
>>> ================================ RESTART ================================
>>> 
[2, 5, 2, 5, 0, 1, 1, 1, 5, 1, 1, 5, 5, 0, 1, 0, 0, 0]
cos =  -0.19218663979154185 角度是： 101.08042157684446 度</span>

怎么都会是100多度呢，看着不像啊，但小伟也说不准是对是错。

先放着吧。

<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);              
                
        //坐标轴设定        
        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]));  
        }  
          
          
        //去除重复点    
        var pointArray = removeDuplicatedPoint(array2D);    
        //无重复的点的数量    
        var points = pointArray.length;    
            
        //得到距离阵列    
        //格式为[[点1序号，点2序号, 距离值], ...]    
        var distanceArray = distanceSort(pointArray);    
        //边的数量    
        var edges = distanceArray.length;    
            
        //存放需要连通的边    
        var linkedArray = [];    
        //连通的边的数量    
        var links = 0;    
            
        //每个顶点相关的边的集合    
        var edgeOfVertex = [];    
            
        for (var i = 0; i < points; i++) {    
                
                
            //获得顶点相关的边的集合    
            edgeOfVertex = [];    
            for (var j = 0; j < edges; j++) {    
                if (distanceArray[j][0] == i ||    
                    distanceArray[j][1] == i) {    
                    edgeOfVertex.push(distanceArray[j]);    
                }    
            }    
                
            //根据起始点寻找最短长度的两条边    
            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;    
        //比例缩放    
        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('向量图', -270, -170, 300);    
          
    }</span>

本节到此结束，欲知后事如何，请看下回分解。