在剩余的10个点中，离（10,5）或（2,2）的最短距离最大的那个点是

在剩余的10个点中，离（10,5）或（2,2）的最短距离最大的那个点是（4,10）该点到（10,5）的距离是√61，到（2,2）的距离是√68，可以检查到其他点到（10,5）或（2,2）中至少一个点的最短距离都小于√61。因此，最终选择的三个初始点为（10,5），（2,2）和（4,10）。这三个点属于不同的簇。

結果 (英文) 1:

In the remaining 10 points short of (10,5) or (2,2) is the shortest distance of that point (4,10) that point to (10,5) is the distance √ 61, (2,2) √ 68 is the distance, you can check to another point to (10,5) or (2,2) of at least one point in the shortest distance is less than √ 61. Therefore, the final selection of three initial points for (10,5), (2,2) and (4,10). The three belonging to different clusters.

結果 (英文) 3:

In the remaining 10 points, (10,5) or (2,2) away from the shortest distance the point at which the maximum is 4,10) the point to (10,5) distance is tick 61, to (2,2) distance tick 68, can check to the other points to (10,5) or (2,2) in at least one point of the shortest distance is smaller than that of the tick,. Therefore, the final selection of the three initial points is (10,5), (2,2) and (4,10). These three points belong to different clusters.