blocks|key|60178|text|在几个阶段中，假设最终结果需要4对点(和/或它们之间的直线方程)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|60179|取任何三个点，然后形成一个三角形。|ordered-list-item|60180|如果第四个点在套路内，则将其与三个中的ny交换。|60181|只剩下最后一个点，计算出你想把它附加到哪两个点。这是通过播放线方程和找到相交点的位置来完成的。下文对此作出澄清。|60182|返回三角形的两边%2B第四点和第二阶段中选择的两条线。|60183|第2步澄清：|60184|让我们说，A点不是在三角形(即BCD)中。每条线都把平原分成两面。我们想找出点(B，C或D)+s.t。它和A之间的线是在另外两条线之间运行的(它们在两条线的对立面)。这就是我们不想附加在A上的一点。|60185|例子:给定A(0,0)，B(10,0)，C(10,10)和D(0,10)。我们有三角BCD。BC线在平原的同一侧离开A&D。DC也是。线路AC离开B&D在平原的对面，所以我们想连接A和B&+C。|60186|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|U|8|@]|9|@]|A|$]]|$1|G|3|H|5|D|7|V|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|W|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|X|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Y|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Z|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|R|$]]

A few stages, assuming the end result needs to be 4 couples of points (and/or the equation of the line between them):

<ol>
<li>Take any three points, and make a triangle. </li>
<li>If the forth point is inside the traingle, swap it with ny of the three. </li>
<li>Having only the last point left, calculate which two point you want to attach it to. This is done by playing with the line eqations and finding where the intersecting points are. Clarification below. </li>
<li>Return two sides of the triangle + the 2 lines between the forth point and the ones you chose from stage 2. </li>
</ol>

Step 2 clarification: 
Let us say that point A is not in the triangle (which is BCD). Every line devides the plain into two sides. we want to find the point (B, C or D) s.t. the line between it and A runs between the other two (they are on opposite sides of the line). This is the point we DO NOT want to attach to A.

Example:
Given A(0,0), B(10,0), C(10,10) and D(0,10). We have the triangle BCD. The line BC leaves A &amp; D on the same side of the plain. So does DC. The line AC leaves B &amp; D on opposite sides of the plain - so we want to connect A to B &amp; C.

blocks|key|60192|text|编辑:正如评论中指出的，它只在特定情况下起作用，因此是一个糟糕的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|60193|四边形的绘制方法是找出这四个点中哪一对点被最大距离隔开。找到这对后，通过将剩余的两个点与对的每个点连接起来，绘制出四回线。|60194|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

EDIT: As pointed out in the comments, it only works in specific situations and thus is a bad answer.

The quadrilateral can be drawn by finding which pair of points amongst the 4 points is separated by the largest distance. Once this pair is found, the quadrileteral is drawn by linking the two remaining points with each point of the pair.

blocks|key|906952|text|编辑：这个答案已经被证明是错误的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|906953|计算这四个点的中心。|ordered-list-item|906954|枚举逆时针方向上的点(或顺时针方向)。|906955|把他们联系在一起。|906956|entityMap^0|0|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$]]|$1|I|3|J|5|H|7|S|8|@]|D|@]|E|$]]|$1|K|3|L|5|H|7|T|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|N|$]]

Edit: This answer has already proven to be wrong. 

<ol>
<li>Calculate the center of the four points. </li>
<li>Enumerate the points in the counter-clock wise direction (or clockwise).</li>
<li>Link them together.</li>
</ol>

blocks|key|907047|text|考虑通过绘制前三个点定义的三条线来划分平面。它界定了7个区域。通过三个符号区域测试(三角形的代数面积ABD、BCD、CAD)，您可以很容易地找到第四个点属于哪个区域。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|907048|在每一种情况下绘制一个四边形是很简单的(每个案例可以有一个、两个或三个解决方案)。|907049|在下面的例子中，使用区域-%2B%2B中的D，ADBC就可以了。|offset|length|style|CODE|907050|​|907051|📷|atomic|907052|907053|实际上，两个区域评估就足够了:如果第一个测试返回-+(regions+-%2B-，-%2B%2B或--%2B)，则ADBC是一个解决方案，如果第二个测试返回-+(Regions%2B-%2B或%2B--)，则ABDC是一个解决方案，否则(Regions%2B%2B-或%2B%2B%2B)是一个解决方案。|907054|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/68d68ee5e5b6531244cde48239e25112.png|imageAlt^0|0|0|C|3|0|0|0|1|0|0|0|O|1|Z|3|13|3|17|3|1Y|1|28|3|2C|3|34|3|38|3|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@$F|13|G|14|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|16|8|@]|9|@$F|17|G|18|1|19]]|A|$]]|$1|O|3|K|5|6|7|1A|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1B|8|@$F|1C|G|1D|H|I]|$F|1E|G|1F|H|I]|$F|1G|G|1H|H|I]|$F|1I|G|1J|H|I]|$F|1K|G|1L|H|I]|$F|1M|G|1N|H|I]|$F|1O|G|1P|H|I]|$F|1Q|G|1R|H|I]|$F|1S|G|1T|H|I]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y|Z|-4]]]]

Consider the partition of the plane obtained by drawing the three lines defined by the first three points. It defines 7 regions. You can easily find to which region the fourth point belongs by means of three signed area tests (algebraic area of the triangles ABD, BCD, CAD).

Drawing a quadrilateral in every case is straightforward (there can be one, two or three solutions per case).

In the example below, with D in the region <code>-++</code>, ADBC will do.

<img src="https://i.stack.imgur.com/b5tvF.png" alt="enter image description here">

Actually two area evaluations are enough: if the first test returns <code>-</code> (regions <code>-+-</code>, <code>-++</code> or <code>--+</code>), ADBC is a solution, else if the second test returns <code>-</code> (regions <code>+-+</code> or <code>+--</code>), ABDC is a solution, else (regions <code>++-</code> or <code>+++</code>) ABCD is a solution .

blocks|key|907110|text|您可以使用任何凸壳的著名算法的简化。贾维斯会很容易。如果凸包是三角形，则四边形是凸的。只需在边缘列表中的任何位置插入缺失点即可。如果凸包是一条线(两个端点)，只需对x-或y-坐标上的所有点进行排序，得到退化的四边形。(如果更接近水平(abs增量x>+abs增量y)，则使用x进行排序，否则使用y。)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|907111|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Gift_wrapping_algorithm^0|7|7|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You can use a simplification of any of the <a href="http://en.wikipedia.org/wiki/Gift_wrapping_algorithm" rel="nofollow">well-known algorithms for convex hull</a>. Jarvis would be easy. If the convex hull is a triangle, the quadrilateral is convex. Just insert the missing point anywhere in the edge list. If the convex hull is a line (2 endpoints), just sort all the points on either x- or y-coordinate to get a degenerate quadrilateral. (If closer to horizontal (abs delta x > abs delta y), use x for sorting, else use y.)

blocks|key|907153|text|我认为，你每次只需要连接两个点，我们就会得到一个线段，并检查剩下的两个点是否在该线段的同一边，如果不是，那不是一个所需的四边形线段，如果是，那么处理不同的点集(即点上可以是这段的坐标之一，另一个点会形成其余的两个点)，并检查相同的东西，直到你得到4个线段为止。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|907154|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

i think, you just need to join two points each time, we will get a line segment and check if the remaining other two points are on the same side of that line segments, if not thats not a line segment of required quadrilateral, if yes then proceed with different set of points (i.e on point can be one of the coordinate of this segment and other one would be form remaining two points) and check the same thing until u get 4 line segments

blocks|key|907215|text|考虑仿射变换|type|unstyled|depth|inlineStyleRanges|entityRanges|data|907216|Px+=+Ax+%2B+u+(Bx+-+Ax)+%2B+v+(Cx+-+Ax)
Py+=+Ay+%2B+u+(By+-+Ay)+%2B+v+(Cy+-+Ay)|code-block|syntax|javascript|907217|它将(0,+0)映射到A，(1,+0)映射到B，(0,+1)映射到C。(这将三角形ABC置于标准位置。)|offset|length|style|CODE|907218|求解2x2线性系统|907219|Dx+=+Ax+%2B+u+(Bx+-+Ax)+%2B+v+(Cx+-+Ax)
Dy+=+Ay+%2B+u+(By+-+Ay)+%2B+v+(Cy+-+Ay)|907220|给出与(u,+v)对应的D值。|907221|然后,|907222|if+u+<+0+=>+ABCD
else+if+v+<+0+=>+BCAD
else+=>+CABD|907223|得到的四边形与三角形ABC具有相同的方向。|907224|​|907225|📷|atomic|907226|907227|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/98fa6320c604b9a75ac771a86ff9a46c.png|imageAlt^0|0|0|2|6|B|1|D|6|M|1|O|6|X|1|0|0|0|3|6|C|1|0|0|0|A|3|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1E|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1F|8|@$I|1G|J|1H|K|L]|$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]|$I|1M|J|1N|K|L]|$I|1O|J|1P|K|L]|$I|1Q|J|1R|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1S|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1T|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1U|8|@$I|1V|J|1W|K|L]|$I|1X|J|1Y|K|L]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|20|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|21|8|@$I|22|J|23|K|L]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|24|8|@]|9|@]|A|$]]|$1|10|3|11|5|12|7|25|8|@]|9|@$I|26|J|27|1|28]]|A|$]]|$1|13|3|Z|5|6|7|29|8|@]|9|@]|A|$]]|$1|14|3|-4|5|6|7|2A|8|@]|9|@]|A|$]]]|15|$16|$5|17|18|19|A|$1A|1B|1C|-4]]]]

Consider the affine transform 

<pre><code>Px = Ax + u (Bx - Ax) + v (Cx - Ax)
Py = Ay + u (By - Ay) + v (Cy - Ay)
</code></pre>

It maps <code>(0, 0)</code> to <code>A</code>, <code>(1, 0)</code> to <code>B</code> and <code>(0, 1)</code> to <code>C</code>. (This puts the triangle ABC in a canonical position.)

Solving the 2x2 linear system

<pre><code>Dx = Ax + u (Bx - Ax) + v (Cx - Ax)
Dy = Ay + u (By - Ay) + v (Cy - Ay)
</code></pre>

gives you the values of <code>(u, v)</code> corresponding to <code>D</code>.

Then,

<pre><code>if u &lt; 0 =&gt; ABCD
else if v &lt; 0 =&gt; BCAD
else =&gt; CABD
</code></pre>

The resulting quadrilateral has the same orientation as the triangle <code>ABC</code>.

<img src="https://i.stack.imgur.com/Ucn8k.png" alt="enter image description here">

blocks|key|60445|text|为了清晰起见，我将坐标为p_n的点考虑为(x_n,+y_n)点。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|60446|为了连接4点，您可以遵循以下步骤：|60447|用最小的p_1得到点x。|ordered-list-item|60448|计算从坡度到其余每个点的3条线的p_1。|60449|将p_1与构成最大斜率线的点p_2连接起来。|60450|将p_1与构成最小斜率的直线的点p_3连接。|60451|将剩余的点p_4与p_2和p_3连接起来。|60452|如果有什么不清楚的话请告诉我。|60453|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Slope^0|C|3|K|A|0|0|4|3|A|1|0|G|3|3|2|0|0|1|3|E|3|0|1|3|G|3|0|5|3|9|3|D|3|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|K|3|L|5|J|7|1D|8|@$9|1E|A|1F|B|C]]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|M|3|N|5|J|7|1J|8|@$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1O|8|@$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1T|8|@$9|1U|A|1V|B|C]|$9|1W|A|1X|B|C]|$9|1Y|A|1Z|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|20|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|21|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

For the sake of clarity, I'm considering as a point <code>p_n</code> a point with coordinates <code>(x_n, y_n)</code>.

In order to connect 4 points you could follow these steps:

<ol>
<li>Get the point <code>p_1</code> with the smallest <code>x</code>.</li>
<li>Calculate the <a href="http://en.wikipedia.org/wiki/Slope" rel="nofollow">slope</a> of the 3 lines that go from <code>p_1</code> to each of the remaining points. </li>
<li>Connect <code>p_1</code> with the point <code>p_2</code> that composes the line with the greatest slope.</li>
<li>Connect <code>p_1</code> with the point <code>p_3</code> that composes the line with the smallest slope.</li>
<li>Connect the remaining point <code>p_4</code> with <code>p_2</code> and <code>p_3</code>.</li>
</ol>

Let me know if something is unclear.

blocks|key|60455|text|四处看看凸包算法。其中一个步骤包括两个步骤:在给定的顶点集(可能是凹的)上建立一个普通多边形，然后删除“凹点”，使剩余的多边形变为凸的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|60456|第一步是解决你的问题。|60457|当然这有点过分了。对于4个顶点，只需按顺序(任意顺序)设置它们，然后验证线段是否连接点1-2和3-4相交；如果是，交换点2和3；或可能边2-3和1-4相交-然后交换点3和4。完成。|60458|检验AB段与CD段是否相交，A点和B点是否位于线CD的相对边，C点和D点是否位于AB线的对面。|60459|若要识别点K所在的直线PQ的边，请计算PQ×PK向量积的Z-部分：(xq-xp)(yk-yp)-(yq-yp)(xk-xp)。这个表达式在线的一边是正的，在另一边是负的(在线上是零的)。|offset|length|style|CODE|60460|entityMap^0|0|0|0|0|X|T|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|S|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@$J|U|K|V|L|M]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|O|$]]

Look around for convex hull algorithms. One of them consists of two steps: building an ordinary polygon on a given set of vertices (which may be concave), then removing 'concave vertices' so that the remaining polygon becomes convex.

The first step is a solution for your problem.

Of course it is kind of overkill. For 4 vertices just set them in a sequence (in any order) then verify if line segments joining points 1-2 and 3-4 intersect; if so, swap points 2 and 3; or possibly edges 2-3 and 1-4 intersect – then swap points 3 and 4. Done.

To verify if segment AB and CD intersect test if points A and B are on opposite sides of the line CD and points C and D are on opposite sides of line AB.

To identify the side of a line PQ where a point K lies, calculate the Z-part of the PQ×PK vector product: <code>(xq-xp)(yk-yp)-(yq-yp)(xk-xp)</code>. The expression is positive on one side of the line and negative on the other one (and zero on the line).

blocks|key|60469|text|通过一个函数可以很容易地解决这个问题，这个函数可以说明两个线段是否相交。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|60470|给定点A，B，C，D，只有三种不同的顺序连接顶点:+ABCD，ABDC和ACBD+(顶点A要么连接到顶点B，要么不连接。如果连接到顶点B，则有两种方式来排序C和D。如果没有，则A连接到C和D，它们都必须连接到B)。|60471|这四个点的顺序产生一个四边形，如果没有任何边缘相交(除了角)。这就为查找工作命令提供了以下程序：|60472|If+AB+intersects+with+CD+then+return+ACBD.
If+AD+intersects+with+BC+then+return+ABDC.
Otherwise+return+ABCD.|code-block|syntax|javascript|60473|证明这一点是容易的：|60474|ABCD和ABDC都包含边AB和CD，所以如果这对边相交，那么良好的顺序必须是ACBD。|ordered-list-item|60475|ABCD和ACBD都包含边AD和BC，所以如果这对边相交，那么良好的顺序必须是ABDC。|60476|如果AB/CD和AD/BC都不相交，那么ABCD顺序就会产生一个四边形。|60477|如果您自己找不出，那么确定两个线段是否相交的代码可以在线找到。|60478|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Z|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|10|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|12|8|@]|9|@]|A|$]]|$1|P|3|Q|5|O|7|13|8|@]|9|@]|A|$]]|$1|R|3|S|5|O|7|14|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|15|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|W|$]]

One can solve this problem very easily with the help of a function that says whether two line segments intersect.

Given points A, B, C, D, there's only three different orders to join up the vertices: ABCD, ABDC and ACBD (vertex A either connects to vertex B or it doesn't. If it does, there's two ways to order C and D. If it doesn't then A connects to both C and D and each of those must connect to B).

An ordering of the four points produces a quadrilateral if none of the edges intersect (except at the corners). That gives this procedure for finding a working order:

<pre><code>If AB intersects with CD then return ACBD.
If AD intersects with BC then return ABDC.
Otherwise return ABCD.
</code></pre>

The proof that this works is easy:

<ol>
<li>both ABCD and ABDC include the edges AB and CD, so if those pair of edges intersect, the good order must be ACBD.</li>
<li>both ABCD and ACBD include the edges AD and BC, so if those pair of edges intersect, the good order must be ABDC.</li>
<li>if neither AB/CD and AD/BC intersect, then the order ABCD produces a quadrilateral.</li>
</ol>

The code for determining if two line segments intersect can be found online if you can't figure it out yourself.

blocks|key|1949492|text|这是一种琐碎的，非数学的方法。我试过几个例子，但都无法证明它是错误的。如果你这么做或知道如何做得更好，请告诉我：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1949493|选择其他点右边的两个点，比如第1点和第2点。|ordered-list-item|1949494|连接点1-2在一起，以及3-4点.|1949495|每个点必须只与另外两个点相连，所以只有两个可能性:连接点1-3和2-4，或者点1-4和2-3。|1949496|检查每一行是否与另一行相交。|1949497|这样就行了。|1949498|注意:在选择前两点时，以下是一些特殊情况：|1949499|点1位于所有其他坐标的右边，但点2和3位于相同的x坐标上。选择最接近第1点的点。|1949500|如果三个点1,2和3具有相同的x坐标，则将最接近的两个点连接起来。如果点是均匀分开的，选择两种可能性中的一种。|1949501|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|W|8|@]|9|@]|A|$]]|$1|G|3|H|5|D|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|10|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|11|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|12|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|T|$]]

Here is a trivial, non-mathematical way of doing it. I've tried it on several examples and could not prove it wrong. Please let me know if you do or know how to make it better:

<ol>
<li>Choose the two points that are on the right of the others, say point 1 and 2. </li>
<li>Link points 1-2 together, as well as points 3-4.</li>
<li>Each point must be linked to only two other points so there are only 2 possibilities: linking points 1-3 and 2-4, or points 1-4 and 2-3. </li>
<li>Check if each line intersects another.</li>
</ol>

That should do it.

Note: when choosing the first two points, here are some special cases:

<ol>
<li>Point 1 is on the right of all others, but points 2 and 3 are on the same x coordinate. Pick the point that's closest to point 1.</li>
<li>If three points 1,2 and 3 have the same x-coordinate, link the two that are closest to each other. If the points are evenly separated, pick one of the two possibilities.</li>
</ol>

We're given the coordinates of 4 points on the 2D plane. How can we find an order to join them with lines to form a <a href="http://en.wikipedia.org/wiki/Quadrilateral" rel="nofollow">quadilateral</a> (whenever it's possible)?

Algorithm to join 4 co-ordinates such that it always results in a quadilateral

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我们得到了二维平面上4个点的坐标。我们如何找到一个订单，将它们与行连接起来，形成一个 (只要有可能)？

问将4个协调连接在一起的算法，使其总是产生四边形。
EN

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问将4个协调连接在一起的算法，使其总是产生四边形。EN