Re: Question on puzzle from lecture #5
Ok, I got it now. I hadn't watched the Q&A after the lectures (until now) so for anybody who's still having a problem, here's what I've understood.
* You're given 3 points and the break point k (=2 in this case).
* Start enumerating the possibilities and for convenience, lets do this in a binary sequence.
* Note that you don't know which 4th pattern of any two points is not allowed. You just know that one of the four is not allowed (because of the breakpoint).
* (In the lecture), black dots and white dots can be +1 or 1 ( or 1 and +1  it doesn't matter). All that matters is that not all four combinations of 2 points are allowed. (In my explanation I use o and x)
* So let's start enumerating the possibilities
Possibility 1 (all white)
x1 x2 x3
0 0 0 (allowed because irrespective of which 2 points you take, you've seen just one pattern).
0 0 x (allowed because if you take x1 and x2, you've seen one pattern so far (00) and for x1 and x2 (or x2 and x3) you've seen two patterns (00 and 0x)
0 x 0 (allowed because for x1 and x2, you've seen two patterns so far (00 and 0x) and for x2 and x3 you've seen three patterns so far(00, 0x and x0) (still not 4!))
0 x x (BAM! If you take points x2 and x3, you've seen 4 patterns thus far i.e (00, 0x, x0 and xx) and this contradicts the behavior that is indicated by the breakpoint =2).
Similarly , all other combinations (in increasing binary sequence  if continued from the sequence above) except x 0 0 are 'illegal'.
