如何使用javascript HTML5 canvas通过N点绘制平滑曲线?


133

对于绘图应用程序,我将鼠标移动坐标保存到数组中,然后使用lineTo进行绘制。结果线不平滑。如何在所有收集的点之间生成一条曲线?

我已经用谷歌搜索过,但是我只发现了3个绘制线条的函数:对于2个采样点,只需使用lineTo。对于3个采样点quadraticCurveTo,对于4个采样点,bezierCurveTo

(我尝试bezierCurveTo为阵列中的每4个点绘制一个,但是这导致每4个采样点出现扭结,而不是连续的平滑曲线。)

我该如何编写一个函数绘制5个采样点及更多的平滑曲线?


5
“平滑”是什么意思?无限微分?两次可区分?三次样条曲线(“贝塞尔曲线”)具有许多良好的特性,并且可以二次微分,并且易于计算。
Kerrek SB 2011年

7
@Kerrek SB,“平滑”,我的意思是视觉上无法检测到任何拐角/尖端。–
Homan

@sketchfemme,您是实时渲染线条,还是将渲染推迟到收集了很多点之后?
Crashalot 2012年

@Crashalot我正在将点收集到一个数组中。您至少需要4分才能使用此算法。之后,您可以通过清除每个mouseMove调用的屏幕来实时在画布上进行渲染
Homan

1
@sketchfemme:不要忘记接受答案。如果是您自己的,那很好
TJ Crowder 2013年

Answers:


130

将随后的采样点与不连续的“ curveTo”类型的函数连接在一起的问题是曲线相交的位置不平滑。这是因为两条曲线共享一个端点,但受到完全不相交的控制点的影响。一种解决方案是“弯曲到”下两个后续采样点之间的中点。使用这些新的内插点将曲线连接起来,可以在端点处实现平滑过渡(一次迭代的终点变为下一次迭代的控制点。)换句话说,两条不相交的曲线现在有很多共同点。

该解决方案摘自《 Foundation ActionScript 3.0动画:使事物移动》一书。第95页-渲染技术:创建多条曲线。

注意:此解决方案实际上并未绘制每个点,这是我的问题的标题(而是通过采样点近似绘制曲线,但从未穿过采样点),但出于我的目的(绘制应用程序),这对我来说已经足够好了,从视觉上您看不出区别。这里一个解决方案要经过所有的采样点,但它要复杂得多(见http://www.cartogrammar.com/blog/actionscript-curves-update/

这是近似方法的图形代码:

// move to the first point
   ctx.moveTo(points[0].x, points[0].y);


   for (i = 1; i < points.length - 2; i ++)
   {
      var xc = (points[i].x + points[i + 1].x) / 2;
      var yc = (points[i].y + points[i + 1].y) / 2;
      ctx.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
   }
 // curve through the last two points
 ctx.quadraticCurveTo(points[i].x, points[i].y, points[i+1].x,points[i+1].y);

+1对于我正在处理的JavaScript /画布项目非常
Matt Matt

1
很高兴有帮助。仅供参考,我已经启动了一个开源html5 canvas绘图板,它是一个jQuery插件。这应该是一个有用的起点。 github.com/homanchou/sketchyPad
Homan

4
很好,但是您将如何制作曲线以使其通过所有点?
理查德

使用此算法,每条连续的曲线是否都应从先前的曲线终点开始?
Lee Brindley 2013年

非常感谢Homan!有用!我花了很多天来解决它。嗨,来自Delphi Android / iOS社区!
alitrun

104

有点晚了,但是为了记录。

您可以通过使用基本样条曲线(也称为规范样条曲线)绘制穿过点的平滑曲线来获得平滑线。

我为画布制作了此功能-它分为三个功能以增加多功能性。主要包装函数如下所示:

function drawCurve(ctx, ptsa, tension, isClosed, numOfSegments, showPoints) {

    showPoints  = showPoints ? showPoints : false;

    ctx.beginPath();

    drawLines(ctx, getCurvePoints(ptsa, tension, isClosed, numOfSegments));

    if (showPoints) {
        ctx.stroke();
        ctx.beginPath();
        for(var i=0;i<ptsa.length-1;i+=2) 
                ctx.rect(ptsa[i] - 2, ptsa[i+1] - 2, 4, 4);
    }
}

要绘制曲线,请按以下顺序排列一个具有x,y点的数组:x1,y1, x2,y2, ...xn,yn

像这样使用它:

var myPoints = [10,10, 40,30, 100,10]; //minimum two points
var tension = 1;

drawCurve(ctx, myPoints); //default tension=0.5
drawCurve(ctx, myPoints, tension);

上面的函数调用两个子函数,一个子函数计算平滑点。这将返回一个包含新点的数组-这是计算平滑点的核心功能:

function getCurvePoints(pts, tension, isClosed, numOfSegments) {

    // use input value if provided, or use a default value   
    tension = (typeof tension != 'undefined') ? tension : 0.5;
    isClosed = isClosed ? isClosed : false;
    numOfSegments = numOfSegments ? numOfSegments : 16;

    var _pts = [], res = [],    // clone array
        x, y,           // our x,y coords
        t1x, t2x, t1y, t2y, // tension vectors
        c1, c2, c3, c4,     // cardinal points
        st, t, i;       // steps based on num. of segments

    // clone array so we don't change the original
    //
    _pts = pts.slice(0);

    // The algorithm require a previous and next point to the actual point array.
    // Check if we will draw closed or open curve.
    // If closed, copy end points to beginning and first points to end
    // If open, duplicate first points to befinning, end points to end
    if (isClosed) {
        _pts.unshift(pts[pts.length - 1]);
        _pts.unshift(pts[pts.length - 2]);
        _pts.unshift(pts[pts.length - 1]);
        _pts.unshift(pts[pts.length - 2]);
        _pts.push(pts[0]);
        _pts.push(pts[1]);
    }
    else {
        _pts.unshift(pts[1]);   //copy 1. point and insert at beginning
        _pts.unshift(pts[0]);
        _pts.push(pts[pts.length - 2]); //copy last point and append
        _pts.push(pts[pts.length - 1]);
    }

    // ok, lets start..

    // 1. loop goes through point array
    // 2. loop goes through each segment between the 2 pts + 1e point before and after
    for (i=2; i < (_pts.length - 4); i+=2) {
        for (t=0; t <= numOfSegments; t++) {

            // calc tension vectors
            t1x = (_pts[i+2] - _pts[i-2]) * tension;
            t2x = (_pts[i+4] - _pts[i]) * tension;

            t1y = (_pts[i+3] - _pts[i-1]) * tension;
            t2y = (_pts[i+5] - _pts[i+1]) * tension;

            // calc step
            st = t / numOfSegments;

            // calc cardinals
            c1 =   2 * Math.pow(st, 3)  - 3 * Math.pow(st, 2) + 1; 
            c2 = -(2 * Math.pow(st, 3)) + 3 * Math.pow(st, 2); 
            c3 =       Math.pow(st, 3)  - 2 * Math.pow(st, 2) + st; 
            c4 =       Math.pow(st, 3)  -     Math.pow(st, 2);

            // calc x and y cords with common control vectors
            x = c1 * _pts[i]    + c2 * _pts[i+2] + c3 * t1x + c4 * t2x;
            y = c1 * _pts[i+1]  + c2 * _pts[i+3] + c3 * t1y + c4 * t2y;

            //store points in array
            res.push(x);
            res.push(y);

        }
    }

    return res;
}

并将点实际绘制为平滑曲线(或任何其他分段线,只要您具有x,y数组):

function drawLines(ctx, pts) {
    ctx.moveTo(pts[0], pts[1]);
    for(i=2;i<pts.length-1;i+=2) ctx.lineTo(pts[i], pts[i+1]);
}

结果是:

pix实例

您可以轻松扩展画布,因此可以这样调用它:

ctx.drawCurve(myPoints);

将以下内容添加到javascript中:

if (CanvasRenderingContext2D != 'undefined') {
    CanvasRenderingContext2D.prototype.drawCurve = 
        function(pts, tension, isClosed, numOfSegments, showPoints) {
       drawCurve(this, pts, tension, isClosed, numOfSegments, showPoints)}
}

您可以在NPM(npm i cardinal-spline-js)或GitLab上找到更优化的版本。


3
首先:这太好了。:-)但是看着那张图像,难道不是给人一种(误导的)印象,即在#9和#10之间途中,这些值实际上低于值#10吗?(我是从可以看到的实际点开始计数的,因此#1将是初始向下轨迹顶部附近的那个,#2将是最底部[图形中的最低点]的那个,依此类推... )
TJ Crowder 2013年

6
只想说,搜索的天,这是唯一的util的实际工作正是我想要的。非常感谢
cnp 2014年

4
是是是谢谢!我跳起来跳起舞来。
Jeffrey Sun

1
您的代码中存在类型错误。参数ptsa应为pts,否则将引发错误。
gfaceless 2015年

2
很久以前,您发布了此解决方案,今天您帮助我解决了一个大问题。非常感谢你!
ÂlexBay

19

第一个答案将不会贯穿所有要点。该图将精确地穿过所有点,并且将是一条完美曲线,其中点[[x:,y:}] n个点。

var points = [{x:1,y:1},{x:2,y:3},{x:3,y:4},{x:4,y:2},{x:5,y:6}] //took 5 example points
ctx.moveTo((points[0].x), points[0].y);

for(var i = 0; i < points.length-1; i ++)
{

  var x_mid = (points[i].x + points[i+1].x) / 2;
  var y_mid = (points[i].y + points[i+1].y) / 2;
  var cp_x1 = (x_mid + points[i].x) / 2;
  var cp_x2 = (x_mid + points[i+1].x) / 2;
  ctx.quadraticCurveTo(cp_x1,points[i].y ,x_mid, y_mid);
  ctx.quadraticCurveTo(cp_x2,points[i+1].y ,points[i+1].x,points[i+1].y);
}

1
到目前为止,这是最简单,最正确的方法。
haymez

10

正如Daniel Howard所指出的,Rob Spencer在http://scaledinnovation.com/analytics/splines/aboutSplines.html上描述了您想要的东西。

这是一个交互式演示:http : //jsbin.com/ApitIxo/2/

这是jsbin关闭时的摘要。

<!DOCTYPE html>
    <html>
      <head>
        <meta charset=utf-8 />
        <title>Demo smooth connection</title>
      </head>
      <body>
        <div id="display">
          Click to build a smooth path. 
          (See Rob Spencer's <a href="http://scaledinnovation.com/analytics/splines/aboutSplines.html">article</a>)
          <br><label><input type="checkbox" id="showPoints" checked> Show points</label>
          <br><label><input type="checkbox" id="showControlLines" checked> Show control lines</label>
          <br>
          <label>
            <input type="range" id="tension" min="-1" max="2" step=".1" value=".5" > Tension <span id="tensionvalue">(0.5)</span>
          </label>
        <div id="mouse"></div>
        </div>
        <canvas id="canvas"></canvas>
        <style>
          html { position: relative; height: 100%; width: 100%; }
          body { position: absolute; left: 0; right: 0; top: 0; bottom: 0; } 
          canvas { outline: 1px solid red; }
          #display { position: fixed; margin: 8px; background: white; z-index: 1; }
        </style>
        <script>
          function update() {
            $("tensionvalue").innerHTML="("+$("tension").value+")";
            drawSplines();
          }
          $("showPoints").onchange = $("showControlLines").onchange = $("tension").onchange = update;
      
          // utility function
          function $(id){ return document.getElementById(id); }
          var canvas=$("canvas"), ctx=canvas.getContext("2d");

          function setCanvasSize() {
            canvas.width = parseInt(window.getComputedStyle(document.body).width);
            canvas.height = parseInt(window.getComputedStyle(document.body).height);
          }
          window.onload = window.onresize = setCanvasSize();
      
          function mousePositionOnCanvas(e) {
            var el=e.target, c=el;
            var scaleX = c.width/c.offsetWidth || 1;
            var scaleY = c.height/c.offsetHeight || 1;
          
            if (!isNaN(e.offsetX)) 
              return { x:e.offsetX*scaleX, y:e.offsetY*scaleY };
          
            var x=e.pageX, y=e.pageY;
            do {
              x -= el.offsetLeft;
              y -= el.offsetTop;
              el = el.offsetParent;
            } while (el);
            return { x: x*scaleX, y: y*scaleY };
          }
      
          canvas.onclick = function(e){
            var p = mousePositionOnCanvas(e);
            addSplinePoint(p.x, p.y);
          };
      
          function drawPoint(x,y,color){
            ctx.save();
            ctx.fillStyle=color;
            ctx.beginPath();
            ctx.arc(x,y,3,0,2*Math.PI);
            ctx.fill()
            ctx.restore();
          }
          canvas.onmousemove = function(e) {
            var p = mousePositionOnCanvas(e);
            $("mouse").innerHTML = p.x+","+p.y;
          };
      
          var pts=[]; // a list of x and ys

          // given an array of x,y's, return distance between any two,
          // note that i and j are indexes to the points, not directly into the array.
          function dista(arr, i, j) {
            return Math.sqrt(Math.pow(arr[2*i]-arr[2*j], 2) + Math.pow(arr[2*i+1]-arr[2*j+1], 2));
          }

          // return vector from i to j where i and j are indexes pointing into an array of points.
          function va(arr, i, j){
            return [arr[2*j]-arr[2*i], arr[2*j+1]-arr[2*i+1]]
          }
      
          function ctlpts(x1,y1,x2,y2,x3,y3) {
            var t = $("tension").value;
            var v = va(arguments, 0, 2);
            var d01 = dista(arguments, 0, 1);
            var d12 = dista(arguments, 1, 2);
            var d012 = d01 + d12;
            return [x2 - v[0] * t * d01 / d012, y2 - v[1] * t * d01 / d012,
                    x2 + v[0] * t * d12 / d012, y2 + v[1] * t * d12 / d012 ];
          }

          function addSplinePoint(x, y){
            pts.push(x); pts.push(y);
            drawSplines();
          }
          function drawSplines() {
            clear();
            cps = []; // There will be two control points for each "middle" point, 1 ... len-2e
            for (var i = 0; i < pts.length - 2; i += 1) {
              cps = cps.concat(ctlpts(pts[2*i], pts[2*i+1], 
                                      pts[2*i+2], pts[2*i+3], 
                                      pts[2*i+4], pts[2*i+5]));
            }
            if ($("showControlLines").checked) drawControlPoints(cps);
            if ($("showPoints").checked) drawPoints(pts);
    
            drawCurvedPath(cps, pts);
 
          }
          function drawControlPoints(cps) {
            for (var i = 0; i < cps.length; i += 4) {
              showPt(cps[i], cps[i+1], "pink");
              showPt(cps[i+2], cps[i+3], "pink");
              drawLine(cps[i], cps[i+1], cps[i+2], cps[i+3], "pink");
            } 
          }
      
          function drawPoints(pts) {
            for (var i = 0; i < pts.length; i += 2) {
              showPt(pts[i], pts[i+1], "black");
            } 
          }
      
          function drawCurvedPath(cps, pts){
            var len = pts.length / 2; // number of points
            if (len < 2) return;
            if (len == 2) {
              ctx.beginPath();
              ctx.moveTo(pts[0], pts[1]);
              ctx.lineTo(pts[2], pts[3]);
              ctx.stroke();
            }
            else {
              ctx.beginPath();
              ctx.moveTo(pts[0], pts[1]);
              // from point 0 to point 1 is a quadratic
              ctx.quadraticCurveTo(cps[0], cps[1], pts[2], pts[3]);
              // for all middle points, connect with bezier
              for (var i = 2; i < len-1; i += 1) {
                // console.log("to", pts[2*i], pts[2*i+1]);
                ctx.bezierCurveTo(
                  cps[(2*(i-1)-1)*2], cps[(2*(i-1)-1)*2+1],
                  cps[(2*(i-1))*2], cps[(2*(i-1))*2+1],
                  pts[i*2], pts[i*2+1]);
              }
              ctx.quadraticCurveTo(
                cps[(2*(i-1)-1)*2], cps[(2*(i-1)-1)*2+1],
                pts[i*2], pts[i*2+1]);
              ctx.stroke();
            }
          }
          function clear() {
            ctx.save();
            // use alpha to fade out
            ctx.fillStyle = "rgba(255,255,255,.7)"; // clear screen
            ctx.fillRect(0,0,canvas.width,canvas.height);
            ctx.restore();
          }
      
          function showPt(x,y,fillStyle) {
            ctx.save();
            ctx.beginPath();
            if (fillStyle) {
              ctx.fillStyle = fillStyle;
            }
            ctx.arc(x, y, 5, 0, 2*Math.PI);
            ctx.fill();
            ctx.restore();
          }

          function drawLine(x1, y1, x2, y2, strokeStyle){
            ctx.beginPath();
            ctx.moveTo(x1, y1);
            ctx.lineTo(x2, y2);
            if (strokeStyle) {
              ctx.save();
              ctx.strokeStyle = strokeStyle;
              ctx.stroke();
              ctx.restore();
            }
            else {
              ctx.save();
              ctx.strokeStyle = "pink";
              ctx.stroke();
              ctx.restore();
            }
          }

        </script>


      </body>
    </html>


7

我发现这很好用

function drawCurve(points, tension) {
    ctx.beginPath();
    ctx.moveTo(points[0].x, points[0].y);

    var t = (tension != null) ? tension : 1;
    for (var i = 0; i < points.length - 1; i++) {
        var p0 = (i > 0) ? points[i - 1] : points[0];
        var p1 = points[i];
        var p2 = points[i + 1];
        var p3 = (i != points.length - 2) ? points[i + 2] : p2;

        var cp1x = p1.x + (p2.x - p0.x) / 6 * t;
        var cp1y = p1.y + (p2.y - p0.y) / 6 * t;

        var cp2x = p2.x - (p3.x - p1.x) / 6 * t;
        var cp2y = p2.y - (p3.y - p1.y) / 6 * t;

        ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, p2.x, p2.y);
    }
    ctx.stroke();
}

6

我决定添加内容,而不是将解决方案发布到另一篇文章中。以下是我构建的解决方案,可能并不完美,但到目前为止输出效果很好。

重要:它将贯穿所有要点!

如果您有任何想法,请使其更好,请,请与我分享。谢谢。

以下是之前和之后的比较:

在此处输入图片说明

将此代码保存为HTML进行测试。

    <!DOCTYPE html>
    <html>
    <body>
    	<canvas id="myCanvas" width="1200" height="700" style="border:1px solid #d3d3d3;">Your browser does not support the HTML5 canvas tag.</canvas>
    	<script>
    		var cv = document.getElementById("myCanvas");
    		var ctx = cv.getContext("2d");
    
    		function gradient(a, b) {
    			return (b.y-a.y)/(b.x-a.x);
    		}
    
    		function bzCurve(points, f, t) {
    			//f = 0, will be straight line
    			//t suppose to be 1, but changing the value can control the smoothness too
    			if (typeof(f) == 'undefined') f = 0.3;
    			if (typeof(t) == 'undefined') t = 0.6;
    
    			ctx.beginPath();
    			ctx.moveTo(points[0].x, points[0].y);
    
    			var m = 0;
    			var dx1 = 0;
    			var dy1 = 0;
    
    			var preP = points[0];
    			for (var i = 1; i < points.length; i++) {
    				var curP = points[i];
    				nexP = points[i + 1];
    				if (nexP) {
    					m = gradient(preP, nexP);
    					dx2 = (nexP.x - curP.x) * -f;
    					dy2 = dx2 * m * t;
    				} else {
    					dx2 = 0;
    					dy2 = 0;
    				}
    				ctx.bezierCurveTo(preP.x - dx1, preP.y - dy1, curP.x + dx2, curP.y + dy2, curP.x, curP.y);
    				dx1 = dx2;
    				dy1 = dy2;
    				preP = curP;
    			}
    			ctx.stroke();
    		}
    
    		// Generate random data
    		var lines = [];
    		var X = 10;
    		var t = 40; //to control width of X
    		for (var i = 0; i < 100; i++ ) {
    			Y = Math.floor((Math.random() * 300) + 50);
    			p = { x: X, y: Y };
    			lines.push(p);
    			X = X + t;
    		}
    
    		//draw straight line
    		ctx.beginPath();
    		ctx.setLineDash([5]);
    		ctx.lineWidth = 1;
    		bzCurve(lines, 0, 1);
    
    		//draw smooth line
    		ctx.setLineDash([0]);
    		ctx.lineWidth = 2;
    		ctx.strokeStyle = "blue";
    		bzCurve(lines, 0.3, 1);
    	</script>
    </body>
    </html>


5

尝试一下KineticJS-您可以使用点数组定义样条线。这是一个例子:

旧网址:http//www.html5canvastutorials.com/kineticjs/html5-canvas-kineticjs-spline-tutorial/

参见存档网址:https ://web.archive.org/web/20141204030628/http: //www.html5canvastutorials.com/kineticjs/html5-canvas-kineticjs-spline-tutorial/


厉害的lib!最好的任务!
Dziad Borowy

是!!我需要blob()函数来制作一个通过所有点的闭合形状。
AwokeKnowing 2014年

7
404页面不存在。
节食者

原始链接- 404 dnot发现-见web.archive.org/web/20141204030628/http://...
satels

1

太晚了,但受到Homan出色的简单答案的启发,让我发布了一个更通用的解决方案(一般而言,Homan的解决方案会在少于3个顶点的点阵列上崩溃):

function smooth(ctx, points)
{
    if(points == undefined || points.length == 0)
    {
        return true;
    }
    if(points.length == 1)
    {
        ctx.moveTo(points[0].x, points[0].y);
        ctx.lineTo(points[0].x, points[0].y);
        return true;
    }
    if(points.length == 2)
    {
        ctx.moveTo(points[0].x, points[0].y);
        ctx.lineTo(points[1].x, points[1].y);
        return true;
    }
    ctx.moveTo(points[0].x, points[0].y);
    for (var i = 1; i < points.length - 2; i ++)
    {
        var xc = (points[i].x + points[i + 1].x) / 2;
        var yc = (points[i].y + points[i + 1].y) / 2;
        ctx.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
    }
    ctx.quadraticCurveTo(points[i].x, points[i].y, points[i+1].x, points[i+1].y);
}

0

为了增加K3N的基数样条方法,也许还解决了TJ Crowder对在误导位置“浸入”曲线的担忧,我在getCurvePoints()函数中插入了以下代码,res.push(x);

if ((y < _pts[i+1] && y < _pts[i+3]) || (y > _pts[i+1] && y > _pts[i+3])) {
    y = (_pts[i+1] + _pts[i+3]) / 2;
}
if ((x < _pts[i] && x < _pts[i+2]) || (x > _pts[i] && x > _pts[i+2])) {
    x = (_pts[i] + _pts[i+2]) / 2;
}

这有效地在每对连续点之间创建一个(不可见的)边界框,并确保曲线停留在该边界框内-即。如果曲线上的某个点在两个点的上方/下方/左侧/右侧,则其位置将更改为位于框内。这里使用了中点,但是可以通过使用线性插值来进行改进。


0

如果要通过n个点确定曲线方程,则以下代码将为您提供n-1阶多项式的系数,并将这些系数保存到coefficients[]数组中(从常数项开始)。x坐标不必按顺序排列。这是Lagrange多项式的示例。

var xPoints=[2,4,3,6,7,10]; //example coordinates
var yPoints=[2,5,-2,0,2,8];
var coefficients=[];
for (var m=0; m<xPoints.length; m++) coefficients[m]=0;
    for (var m=0; m<xPoints.length; m++) {
        var newCoefficients=[];
        for (var nc=0; nc<xPoints.length; nc++) newCoefficients[nc]=0;
        if (m>0) {
            newCoefficients[0]=-xPoints[0]/(xPoints[m]-xPoints[0]);
            newCoefficients[1]=1/(xPoints[m]-xPoints[0]);
    } else {
        newCoefficients[0]=-xPoints[1]/(xPoints[m]-xPoints[1]);
        newCoefficients[1]=1/(xPoints[m]-xPoints[1]);
    }
    var startIndex=1; 
    if (m==0) startIndex=2; 
    for (var n=startIndex; n<xPoints.length; n++) {
        if (m==n) continue;
        for (var nc=xPoints.length-1; nc>=1; nc--) {
        newCoefficients[nc]=newCoefficients[nc]*(-xPoints[n]/(xPoints[m]-xPoints[n]))+newCoefficients[nc-1]/(xPoints[m]-xPoints[n]);
        }
        newCoefficients[0]=newCoefficients[0]*(-xPoints[n]/(xPoints[m]-xPoints[n]));
    }    
    for (var nc=0; nc<xPoints.length; nc++) coefficients[nc]+=yPoints[m]*newCoefficients[nc];
}
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