对三次贝塞尔曲线过点平滑中尖角和交叉现象的优化

目前在做等值线等值面相关的功能，用户可拖拽控制点修改等值线，再用等值线生成等值面。因为初始的等值线点数据太多，不利于用户操作，所以先使用道格拉斯-普克算法(Douglas–Peucker)进行等值线抽稀，再将抽稀后的控制点使用贝塞尔曲线算法进行平滑。

对于贝塞尔曲线算法的平滑过程，有人做了很详细的示意图，推荐大家看下贝塞尔曲线算法之JS获取点

可以了解到贝赛尔曲线算法平滑得到的曲线是经过起始点的，同时二阶算法需要三个点，三阶算法需要四个点，四阶算法需要五个点，以此类推。

一般的来说，三阶贝塞尔曲线就已经够用了，而且效果还不错，所以我选择了三次贝塞尔曲线平滑算法来进行控制点的平滑处理。

贝塞尔曲线平滑后的等值线是基本不经过控制点的，考虑到用户操作逻辑，以及点线关系（我的控制点是等值线抽稀得到的，所以等值线是经过控制点的），所以采用三次贝塞尔曲线过点平滑算法来进行控制点的平滑处理。

过点平滑的原理就是以相邻两个控制点为起始点，然后往起始点中间插入其他过程点（不是在起始点直线上选择点），这样平滑得到的曲线是经过起始点的，而曲线如何平滑是由插入的点来控制的，三次贝塞尔曲线需要四个点，那就需要在起始点中间插入两个点。

大致思路就是，先算出相邻原始点的中点，在把相邻中点连成的线段平移到对应的原始点，以平移后的中点作为控制点，相邻原始点为起始点画贝塞尔曲线，这样就保证了连接处的光滑。而贝塞尔曲线本身是光滑的，所以就把这些原始点用光滑曲线连起来了。具体代码及示意图如下：

代码：

function createCurve(originPoint, option){ 

 //控制点收缩系数 ，经调试0.6较好

 let scale = option.tension || 0.6; 

 //平滑插值插入的最大点数

 let maxpoints = option.pointsPerSeg

 let originCount = originPoint.length

 let curvePoint = []

 let midpoints = []

 //生成中点 

 for(let i = 0 ;i < originCount - 1 ; i++){ 

     midpoints.push([

         (originPoint[i][0] + originPoint[i + 1][0])/2.0,
         (originPoint[i][1] + originPoint[i + 1][1])/2.0

     ])

 } 

 //平移中点 

 let extrapoints = []

 for(let i = 1 ;i < originCount - 1 ; i++){ 

     let backi = i - 1; 

     let midinmid = [

         (midpoints[i][0] + midpoints[backi][0])/2.0,

         (midpoints[i][1] + midpoints[backi][1])/2.0

     ]

     let offsetx = originPoint[i][0] - midinmid[0]; 

     let offsety = originPoint[i][1] - midinmid[1]; 

     let extraindex = 2 * i; 

     extrapoints[extraindex] = [

         midpoints[backi][0] + offsetx,

         midpoints[backi][1] + offsety

     ]

     //朝 originPoint[i]方向收缩 

     let addx = (extrapoints[extraindex][0] - originPoint[i][0]) * scale; 

     let addy = (extrapoints[extraindex][1] - originPoint[i][1]) * scale; 

     extrapoints[extraindex] = [

         originPoint[i][0] + addx,

         originPoint[i][1] + addy

     ]

     let extranexti = extraindex + 1; 

     extrapoints[extranexti] = [

         midpoints[i][0] + offsetx,

         midpoints[i][1] + offsety

     ]

     //朝 originPoint[i]方向收缩 

     addx = (extrapoints[extranexti][0] - originPoint[i][0]) * scale; 

     addy = (extrapoints[extranexti][1] - originPoint[i][1]) * scale; 

     extrapoints[extranexti] = [

         originPoint[i][0] + addx,

         originPoint[i][1] + addy

     ]

 } 

 let controlPoint = []

 //生成4控制点，产生贝塞尔曲线 

 for(let i = 1 ;i < originCount - 2 ; i++){ 

    controlPoint[0] = originPoint[i]; 

    let extraindex = 2 * i; 

    controlPoint[1] = extrapoints[extraindex + 1]; 

    let extranexti = extraindex + 2; 

    controlPoint[2] = extrapoints[extranexti]; 

    let nexti = i + 1; 

    controlPoint[3] = originPoint[nexti]; 

    for(let n = maxpoints; n >= 0; n--){

         //存入曲线点 

        curvePoint.push( bezier3func(n / maxpoints, controlPoint) );

     }

 } 

 return curvePoint

} 

//三次贝塞尔曲线 

function bezier3func(uu, controlP){ 

 let partX0 = controlP[0][0] * uu * uu * uu; 

 let partX1 = 3 * controlP[1][0] * uu * uu * (1 - uu); 

 let partX2 = 3 * controlP[2][0] * uu * (1 - uu) * (1 - uu); 

 let partX3 = controlP[3][0] * (1 - uu) * (1 - uu) * (1 - uu); 

 let partX = partX0 + partX1 + partX2 + partX3; 

 let partY0 = controlP[0][1] * uu * uu * uu; 

 let partY1 = 3 * controlP[1][1] * uu * uu * (1 - uu); 

 let partY2 = 3 * controlP[2][1] * uu * (1 - uu) * (1 - uu); 

 let partY3 = controlP[3][1] * (1 - uu) * (1 - uu) * (1 - uu); 

 let partY = partY0 + partY1 + partY2 + partY3; 

 return [partX, partY]

}

c++版的可以看穿过已知点画平滑曲线（3次贝塞尔曲线）

然而事情到这里并没有结束，还有坑需要填，直接使用该算法进行平滑，在实际应用中发现控制点距离近的话，平滑的曲线会有尖角或交叉现象。

比如这样的

还有这样的

这是因为插入的两个新控制点和起始控制点位置特殊，平滑后产生尖角或交叉，如下所示：

优化思路是判断四个点的关系，获取直线控制点1-新控制点1与直线控制点2-新控制点2之间的交点（如果有的话），如果交点在线段控制点1-点1中时，用交点替换点1，线段控制点2-点2同理。或者以控制点1、交点、控制点2三点，然后使用二次贝赛尔曲线算法进行平滑。新的效果示意图如下所示：

优化前与优化后的效果对比如下：左边为优化前的平滑效果，右边是优化后的平滑效果：

优化后的代码：

function createCurve(originPoint, option){ 

 //控制点收缩系数 ，经调试0.6较好

 let scale = option.tension || 0.6; 

 //平滑插值插入的最大点数

 let maxpoints = option.pointsPerSeg

 let originCount = originPoint.length

 let curvePoint = []

 let midpoints = []

 //生成中点 

 for(let i = 0 ;i < originCount - 1 ; i++){ 

     midpoints.push([

         (originPoint[i][0] + originPoint[i + 1][0])/2.0,

         (originPoint[i][1] + originPoint[i + 1][1])/2.0

     ])

 } 

 //平移中点 

 let extrapoints = []

 for(let i = 1 ;i < originCount - 1 ; i++){ 

     let backi = i - 1; 

     let midinmid = [

         (midpoints[i][0] + midpoints[backi][0])/2.0,

         (midpoints[i][1] + midpoints[backi][1])/2.0

     ]

     let offsetx = originPoint[i][0] - midinmid[0]; 

     let offsety = originPoint[i][1] - midinmid[1]; 

     let extraindex = 2 * i; 

     extrapoints[extraindex] = [

         midpoints[backi][0] + offsetx,

         midpoints[backi][1] + offsety

     ]

     //朝 originPoint[i]方向收缩 

     let addx = (extrapoints[extraindex][0] - originPoint[i][0]) * scale; 

     let addy = (extrapoints[extraindex][1] - originPoint[i][1]) * scale; 

     extrapoints[extraindex] = [

         originPoint[i][0] + addx,

         originPoint[i][1] + addy

     ]

     let extranexti = extraindex + 1; 

     extrapoints[extranexti] = [

         midpoints[i][0] + offsetx,

         midpoints[i][1] + offsety

     ]

     //朝 originPoint[i]方向收缩 

     addx = (extrapoints[extranexti][0] - originPoint[i][0]) * scale; 

     addy = (extrapoints[extranexti][1] - originPoint[i][1]) * scale; 

     extrapoints[extranexti] = [

         originPoint[i][0] + addx,

         originPoint[i][1] + addy

     ]

 } 

 let controlPoint = []

 //生成4控制点，产生贝塞尔曲线 

 for(let i = 1 ;i < originCount - 2 ; i++){ 

     controlPoint[0] = originPoint[i]; 

     let extraindex = 2 * i; 

     controlPoint[1] = extrapoints[extraindex + 1]; 

     let extranexti = extraindex + 2; 

     controlPoint[2] = extrapoints[extranexti]; 

     let nexti = i + 1; 

     controlPoint[3] = originPoint[nexti]; 

     let fn = bezier3func; 

     let cp = intersects(controlPoint.slice(0, 2), controlPoint.slice(-2))

     if(cp && isContains(controlPoint[0], controlPoint[1], cp)){

        controlPoint[1] = cp

     }

     if(cp && isContains(controlPoint[2], controlPoint[3], cp)){

        controlPoint[2] = cp

     }

     if(controlPoint[1][0] == controlPoint[2][0] && controlPoint[1][1] == controlPoint[2][1]){

         fn = bezier2func

         controlPoint.splice(1, 1)

     }

     for(var n = maxpoints; n >= 0; n--){

        //存入曲线点 

        curvePoint.push( fn(n / maxpoints, controlPoint) );

     }

 } 

 return curvePoint

} 

//三次贝塞尔曲线 

function bezier3func(uu, controlP){ 

 let partX0 = controlP[0][0] * uu * uu * uu; 

 let partX1 = 3 * controlP[1][0] * uu * uu * (1 - uu); 

 let partX2 = 3 * controlP[2][0] * uu * (1 - uu) * (1 - uu); 

 let partX3 = controlP[3][0] * (1 - uu) * (1 - uu) * (1 - uu); 

 let partX = partX0 + partX1 + partX2 + partX3; 

 let partY0 = controlP[0][1] * uu * uu * uu; 

 let partY1 = 3 * controlP[1][1] * uu * uu * (1 - uu); 

 let partY2 = 3 * controlP[2][1] * uu * (1 - uu) * (1 - uu); 

 let partY3 = controlP[3][1] * (1 - uu) * (1 - uu) * (1 - uu); 

 let partY = partY0 + partY1 + partY2 + partY3; 

 return [partX, partY]

} 

//二次贝塞尔曲线 

function bezier2func(uu, controlP){ 

 let partX0 = controlP[0][0] * uu * uu; 

 let partX1 = 2 * controlP[1][0] * uu * (1 - uu); 

 let partX2 = controlP[2][0] * (1 - uu) * (1 - uu); 

 let partX = partX0 + partX1 + partX2;

 let partY0 = controlP[0][1] * uu * uu; 

 let partY1 = 2 * controlP[1][1] * uu * (1 - uu); 

 let partY2 = controlP[2][1] * (1 - uu) * (1 - uu); 

 let partY = partY0 + partY1 + partY2; 

 return [partX, partY]

} 

/**

 * Find a point that intersects LineStrings with two coordinates each

 * 找到一个点，该点与每个线串有两个坐标相交

 */

function intersects(coords1, coords2) {

 if (coords1.length !== 2) {

    throw new Error("<intersects> line1 must only contain 2 coordinates");

 }

 if (coords2.length !== 2) {

    throw new Error("<intersects> line2 must only contain 2 coordinates");

 }

 const x1 = coords1[0][0];

 const y1 = coords1[0][1];

 const x2 = coords1[1][0];

 const y2 = coords1[1][1];

 const x3 = coords2[0][0];

 const y3 = coords2[0][1];

 const x4 = coords2[1][0];

 const y4 = coords2[1][1];

 //斜率交叉相乘 k1 = (y4 - y3) / (x4 - x3) ... k2 = (y2 - y1) / (x2 - x1)

 //k1 k2 同乘 (x4 - x3) * (x2 - x1) 得到denom

 const denom = ((y4 - y3) * (x2 - x1)) - ((x4 - x3) * (y2 - y1)); 

 const numeA = ((x4 - x3) * (y1 - y3)) - ((y4 - y3) * (x1 - x3));

 const numeB = ((x2 - x1) * (y1 - y3)) - ((y2 - y1) * (x1 - x3));

 if (denom === 0) { //斜率一样，平行线

    return null;

 }

 const uA = numeA / denom;

 const uB = numeB / denom;

 const x = x1 + (uA * (x2 - x1));

 const y = y1 + (uA * (y2 - y1));

 return [x, y];

}

function isContains(sp, ep, p) {

 return (

 (p[0] > ep[0] && p[0] < sp[0]) || (p[0] > sp[0] && p[0] < ep[0])

 ) && (

 (p[1] > ep[1] && p[1] < sp[1]) || (p[1] > sp[1] && p[1] < ep[1])

 )

}