Situation number two is I take the same number, and I make a row. They are all in one row.

状态二我将采用同样的数字，并且我做了一个箭头，它们全部都在一列上。

麻省理工公开课 - 固态化学导论课程节选

Let me add, let me make it 17 people so we have an odd number, so go and sit in that row.

我再加一个人这样就是奇数17个人,请到那行中

耶鲁公开课 - 博弈论课程节选

And what I am going to do is say start with this ion, add up the energy associated with the interactions between that ion and everybody else in the row and then multiply it by Avogadro's number, because that is the number of atoms there are in a row.

接下来我要从这一离子开始，加上相互作用的能量，也就是这一离子,和其它所有在这一行的离子之间的能量,再乘以阿伏加德罗常数，因为这是在一行的原子的数量。

麻省理工公开课 - 固态化学导论课程节选

There's a little icon of a hand here, at top left, you go ahead and click that, and what happens is, we, the staff, see a little blinking icon in the top left that says one out of two hands, one out of 10 hands, or some number of hands, and it maintains in the little row up there your placement in line.

这里有一个小的手势图标，在左上角,点击它，我们教员，在左上角就看到一个小的闪烁的图标，这说明,两个人中有一个举手，10个人中有一个举手，或很多人中有人举手，并且它处于一行之中,那里你们布置成一行了。

哈佛公开课 - 计算机科学课程节选

so given that we chose an odd number of people in the row, if exactly one candidate stands and that candidate is the center candidate, then that's an equilibrium.

假若我们选中行的人数为奇数,如果确实只有一个候选人参选且,那个候选人就是在中间,则那是个均衡

耶鲁公开课 - 博弈论课程节选