UC知道一段数学英语的翻译
来源：百度知道 编辑：UC知道 时间：2024/05/04 12:37:27
Suppose now that the points of Sn are partitioned into m paths and the total number of
successive pairs of black points in all paths is k. By breaking the paths at each pair of
successive black points, we obtain k+m paths in each of which the number of black points
exceeds the number of white points by at most one. Therefore, the total number of black
points in Sn cannot exceed the number of white points by more than k +m. On the other
hand, the total number of black points in Sn exceeds the total number of white points by
exactly 2n (there is exactly one more black point in each row of Sn).
Sn为一点集
successive pairs of black points in all paths is k. By breaking the paths at each pair of
successive black points, we obtain k+m paths in each of which the number of black points
exceeds the number of white points by at most one. Therefore, the total number of black
points in Sn cannot exceed the number of white points by more than k +m. On the other
hand, the total number of black points in Sn exceeds the total number of white points by
exactly 2n (there is exactly one more black point in each row of Sn).
Sn为一点集
假设现在的要点锡是分割成米路径和总人数
连续对黑点，在所有的路径是k.突破路径在每双
历届黑点，我们获得的K +米的路径，其中每人数黑点
人数超过白点，在最一。因此，总人数黑色
点在锡不能超过的数目，白点以上的K +米。对其他
另一方面，总数黑点在锡超过总数的白点
正是为2 n （有这正是一个更黑点，每列锡） 。
既然Sn的点被把分开成为m通道和总计,假定连续的对在所有的通道中黑点的数目是k.通过在每一对连续黑点方面打破通道,我们得到千米通道,黑点的数目在每一个上超过经过最多一白色点的数目的.因此,总的在Sn中黑点的数目不能经过比k+m更多超过白色点的数目.另一方面,总的在Sn中黑点的数目超过总的经过准确2n白色点的数目((正好更多还有一的黑点在每一排Sn)中那里存在.