### Spheniscine's blog

By Spheniscine, history, 3 years ago, Spoiler

Spoiler

Spoiler

Spoiler

Spoiler

### F – Random Max

Spoiler Comments (7)
| Write comment?
 » Great! Thank you for the editorial. For solution D, I am sorry for a stupid question, but can you please explain why squaring v gives the number of overlapping configurations?I understand that v gives the number of ways an A-length segment can overlap with a B-length segment within a N-length interval. But how does squaring v gives the overlapping configurations of squares?
•  » » 3 years ago, # ^ | ← Rev. 3 →   If you have two overlapping squares, and projected them onto the $x$ axis, their segments will overlap, likewise the $y$ axis.Perhaps this diagram could help illustrate: Diagram Thus there is a bijection between two choices of overlapping segments ($v^2$) to choices of overlapping squares.
•  » » » Thank you for the effort. Really useful, I understand now.
 » 3 years ago, # | ← Rev. 2 →   Just a small note, about the cumulative distribution formula. I will just consider positive numbers for now. $\displaystyle\int_{0}^{\infty} xP(x)dx = \displaystyle\int_{0}^{\infty} P(\ge x)dx = \displaystyle\int_{0}^{\infty} 1 - P(\le x)dx$This is because the probability $P(x)$ is added in the range $[0,x]$. So they are equal. You can solve similarly for -ve numbers.
 » 3 years ago, # | ← Rev. 2 →   In E why did not you just consider 2^(number of configurations in which it can be lighted) -1. (when not lighted at all). -- yup got it -- say in x number of ways a square can be ignited then considering all the squares which do not have influence on current square, our answer also includes those cases say(y) when one of those squares changes its state hence total ways would be x*y. Alternatively we can consider the above way you have mentioned.
•  » » You mean something like $(2^{p} - 1) \cdot 2^{K - p}$? That's the same formula, just expressed differently
•  » » » I got it now. but previously i was reffering to the case when there are two sets {A,B} for given square , the one igniting any one from A lights current square and from other set does not. So I was initially thinking answer like 2^(size(A)) -1 for each square. Nice editorial btw. hopefully you will keep on writing in future.