两条线之间的最近距离

Question

我想创建一个函数来获得任意两条线之间的最近/最短距离（假设它们不相交）。我知道有一些关于点和线之间最短距离的帖子，但我找不到任何两条线。

我咨询过几个数学和矢量学习网站，并设法破解下面的算法。它给出了答案，但不是正确答案。例如，当找到以下之间的最短距离时：

l1 = [(5,6),(5,10)]
l2 = [(0,0),(10,10)]
print _line2line(l1,l2)

......它给出了答案......

5.75223741636 

从两条线的图中可以看出（它显示了一个10x10帧，每1个单位有一个刻度标记），从理论上讲，点（6,5）之间的距离应该小得多，大约为1。 （5,5）。

所以我想知道是否有人能在我的代码中发现我做错了什么？ （如果可能的话，我也想知道如何获得两条线最接近的实际点......）

以下代码的注释：L1和L2代表line1和line2，x / y1 vs x / y2分别代表每一行的起点和终点。后缀dx和dy代表delta或x和y，即矢量，如果它的原点是0,0。

def _line2line(line1, line2):

    """
    - line1 is a list of two xy tuples
    - line2 is a list of two xy tuples
    References consulted:
    http://mathforum.org/library/drmath/view/51980.html
    and http://mathforum.org/library/drmath/view/51926.html
    and https://answers.yahoo.com/question/index?qid=20110507163534AAgvfQF
    """

    import math
    #step1: cross prod the two lines to find common perp vector
    (L1x1,L1y1),(L1x2,L1y2) = line1
    (L2x1,L2y1),(L2x2,L2y2) = line2
    L1dx,L1dy = L1x2-L1x1,L1y2-L1y1
    L2dx,L2dy = L2x2-L2x1,L2y2-L2y1
    commonperp_dx,commonperp_dy = (L1dy - L2dy, L2dx-L1dx)

    #step2: normalized_perp = perp vector / distance of common perp
    commonperp_length = math.hypot(commonperp_dx,commonperp_dy)
    commonperp_normalized_dx = commonperp_dx/float(commonperp_length)
    commonperp_normalized_dy = commonperp_dy/float(commonperp_length)

    #step3: length of (pointonline1-pointonline2 dotprod normalized_perp).
    # Note: According to the first link above, it's sufficient to
    #    "Take any point m on line 1 and any point n on line 2."
    #    Here I chose the startpoint of both lines
    shortestvector_dx = (L1x1-L2x1)*commonperp_normalized_dx
    shortestvector_dy = (L1y1-L2y1)*commonperp_normalized_dy
    mindist = math.hypot(shortestvector_dx,shortestvector_dy)

    #return results
    result = mindist
    return result