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Height of epsilon-balanced binary search tree

In, Hannes says:

For example, one can say, a BST is balanced, if each subtree has at most epsilon * n nodes, where epsilon < 1 (for example epsilon = 3/4 or even epsilon = 0.999 — which are practically not balanced at all). The reason for that is that the height of such a BST is roughly log_{1/epsilon}

I am a bit puzzled on the last statement — how do we know that the height is roughly 1/epsilon?

Why does going from 2’s complement (in binary) to the positive value by completing to 1 then adding 1 work?

I’m studying Computer science and this has confused me for a long time since our professor didn’t give any proof.

When changing from 2’s complement to the positive value, we can go in reverse (by subtracting 1, then using 1’s complement), and that’s clear why it works.

But our professor told us another method which is taking the number, using 1’s complement, THEN adding 1.

I don’t understand why the second method works.

in order of binary search tree

This is what I got for the in-order of the bst but it’s wrong because I’m answering some questions about some successors of some of the letters and I got them wrong. so I’m wondering where in this in-order i’ve gone wrong? enter image description here

d, b, m, h, i, e, a, j, k, f, g, c

(Sorry if these questions aren’t allowed here, please let me know where I can ask it if not!)

partition binary strings [on hold]

You are given a binary string of 0s and 1s.

Your task is to calculate the number of ways so that a string can be partitioned by satisfying the following constraints:

  1. The length of each partition must be in non-decreasing format. Therefore, the length of the previous partition must be less than or equal to the length of the current partition.

  2. The number of set bits in each partition must be in non-decreasing format. Therefore, the number of 1’s that are available in the previous partition must be less than or equal to the number of 1’s that are available in the current partition.

  3. There should be no leading zeroes available in each partition.

example: string= 101110 The three valid partitions are 101110, 10|1110, 101|110

Note: number of 0s need not be in non-decreasing order

// I want to know how to tackle this type of problems.

I have tried bruteforce methods by running loop for 2 groups 3 groups etc. But

they are taking long time.//

I want to compute it in feasible time.//

please: try to explain methods and suggestions in simpler terms.

Unique property of a full binary tree

I read a statement that was unclear to me, and was hoping to get some clarification.

It said that given a full binary tree with $ n > 2$ leaves, there exists some internal node such that one third to two thirds of all $ n$ leaves in the tree are its descendants.

From my understanding, I know each internal node in a full binary tree has 2 children. That said, the whole tree has $ 2n – 1$ nodes, which means $ n – 1$ of them are internal nodes (not leaves).

I can come up with drawn examples where this is always the case, but I am not sure how to formally reason it. Any help would be greatly appreciated.

Having a code-signed binary, how can I tell if it’s signed with an Extended Validation (EV) certificate?

I can’t seem to find an answer to this seemingly simple question. Say, on Windows, if I have a binary file:

enter image description here

How can I tell if it was signed with an extended validation (EV) code-signing certificate?

Say, the file above, being a Windows driver on a 64-bit Windows 10 has to have an EV signature to be able to load. So I can’t seem to find anything in its properties that can indicate that it’s an EV:

enter image description here

And since the OS can clearly tell the difference between EV and OV cert, how does it know?