最接近点对问题是求二维坐标中的点对问题，该算法是为了将平面上点集S线性分割为大小大致相等的2个子集S1和S2，我们选取一垂直线l:x=m来作为分割直线。其中m为S中各点x坐标的中位数。由此将S分割为S1={p∈S|px≤m}和S2={p∈S|px>m}。从而使S1和S2分别位于直线l的左侧和右侧，且S=S1∪S2 。由于m是S中各点x坐标值的中位数，因此S1和S2中的点数大致相等。

递归地在S1和S2上解最接近点对问题，我们分别得到S1和S2中的最小距离δ1和δ2。现设δ=min(δ1,δ1)。若S的最接近点对(p,q)之间的距离d(p,q)<δ则p和q必分属于S1和S2。不妨设p∈S1，q∈S2。那么p和q距直线l的距离均小于δ。因此，我们若用P1和P2分别表示直线l的左边和右边的宽为δ的2个垂直长条，则p∈S1，q∈S2。

-closest point to the problem is for two-dimensional coordinates of the point of the problem, the algorithm is to be planar point set on the S linear divided roughly equal to the size of two sub-sets of S1 and S2, we selected a vertical line l : x = m as a separate line. Where m for S x coordinates of the points of the median. This will be divided into S S1 = (p S | px m) and (p = S2 S | pxgt; M). So that S1 and S2 are located in the linear l left and right, and S = S1 S2. M is due to S x coordinates of the points of the median value, S1 and S2 2186 roughly equal. Recursive in S1 and S2 the nearest point on the solution of the problem, we won the S1 and S2 the minimum distance between 1 and 2. The existing = min (1, 1). If S is the closest point of (p, q) the distance between d (p, q)