Let's say I have two identical jars and I want to find the height that the jars will breaks when dropped from various heights. I can drop the jars from height increments using steps on a staircase. I want to solve this problem sublinearly.

I decided to increment using square numbers (1, 4, 9, 16..etc) for a complexity of O($\sqrt{n}$), where n is the number of steps on the stairs, so I know that I have an upper and lower bound once the first jar breaks. If the jar breaks at 16, I know that the height that it breaks at is between 9 + 1 and 16. If I were to run the algorithm linearly (on the second jar) from 10 to 16, would this make my algorithm linear?

  • $\begingroup$ What is $n$? Is it equal to the height you wish to find? $\endgroup$
    – dkaeae
    Jan 17, 2019 at 14:45
  • $\begingroup$ n is the number of steps on the staircase, so indirectly height, yes. $\endgroup$
    – V S
    Jan 17, 2019 at 14:46
  • $\begingroup$ Can you compute with a variable, which is $n$ here? Can you compute how many drops will be used? For simplicity, you could assume $\sqrt n$ is an integer at first, just to see what happens. $\endgroup$
    – John L.
    Jan 17, 2019 at 14:48

1 Answer 1


No, it would not make the algorithm linear. The worst-case scenario would be $n = m^2 + 1$, where $m$ is some positive integer. In this case, you have an additional $(m+1)^2 - m^2 = 2m + 1$ many tries (since the next square after $m^2$ is $(m+1)^2$ and $n < (m+1)^2$), which is roughly equal to $2\sqrt{n} \in \Theta(\sqrt{n})$.


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