Question on puzzle from lecture #5

[jcatanz]

In considering the possible dichotomies of 3 points (subject to the restriction of a breakpoint at k=2) you said that

o o o
o o x
o x o

are allowed. Here, o and x can represent -1 and +1.

Then you said

o x x is not allowed,

x o o is allowed,

x o x,

x x o, and

x x x are not allowed.

So that for 3 points, half of the possible dichotomies are eliminated.

If there is a breakpoint at k = 2, all we can say is that not all of the 2^2 = 4 possible dichotomies {+1,-1}, {+1,+1} {-1,+1} {-1,-1} for two points are allowed.

How did you determine which dichotomies are not allowed?