Quotient space of harmonic functions on punctured plane

Let $ a_1,\ldots,a_n$ be $ n$ distinct points in $ \mathbb{C}$ and let $ \Omega := \mathbb{C} \setminus \{a_1,\ldots,a_n\}$ . Define $ H(\Omega)$ to be the space of harmonic functions on $ \Omega$ and $ R(\Omega)$ the subspace of $ H(\Omega)$ consisting of real parts of holomorphic functions on $ \Omega$ . I want to show that the equivalence classes of $ $ \log\vert z – a_1 \vert,\ldots, \log \vert z – a_n \vert$ $ form a basis for the quotient space $ H(\Omega)/R(\Omega)$ . I know how to show linear independence. I have seen rather lengthy arguments that the above functions span the space, but I am wondering if there is a way to show that the dimension of $ H(\Omega)/R(\Omega)$ must be $ n$ without actually computing a basis. That way I can conclude the above functions form a basis based on their linear independence.

Is is possible to get the TLS implemetation information of the remote server?

I am trying to devise a way to find side-channels that reveal exact TLS library (and its version) running on the remote server by adopting similar approach to JavaScript Template Attacks

However, if there exist an explicit channel, then I might not dive into this problem.

Are TLS libraries reveal their implementation information (I don’t mean the protocol version) through the messages or during the key exchange?

Systematic way to find a matrix that satisfies an equation


It is relatively easier to find a solution to a system of linear equations in the form of $ A\textbf{v}=\textbf{b}$ given the matrix $ A$ . But what systematic ways are there that allows us to obtain a matrix given a equation?

For example, consider the following equations:

$ $ \begin{bmatrix} a&b&c\ d&e&f\ g&h&i\ \end{bmatrix}\begin{bmatrix}2\3\4\end{bmatrix}=\begin{bmatrix}1\1\1\end{bmatrix} $ $

Although it is easy to see that $ a=\frac{1}{2}, e=\frac{1}{3}, i=\frac{1}{4}$ with all other terms being $ 0$ is a viable solution, I am curious if there is a more systematic way of finding a matrix that satisfies a equation. Even more importantly, how should these methods be adapted when there are added constraints on the properties of the matrix? For example, if we require that the matrix of interest should be invertible, or of rank = $ k$ ?

Minimization problem

We have a positive integer N and we can use the following operations to minize our positive number N to 1. We want to minize N to 1 with fewest steps possible.

  1. N / 2
  2. N / 3
  3. N – 1

I came up with some idea given below and not sure about correctness of the way I go ahead.

First, I decided variables. Here the only variable is number, obviously.

Second, I try to give a recursion.

minimize(n) = min {minimize(n/2), minimize(n/3) ,minimize(n-1)} 

Third, I try to see base cases.

minimize(1) = 0 minimize(2) = 1 minimize(3) = 1 

Here is my algorithm

A[1] = 0; A[2] = 1; A[3] = 1;  minimizeTo1(n):   for(i = 4 to n)     if(i is dividable by 2 and i is dividable by 3)       a[i] = min{minimizeTo1(i/2) + 1, minimizeTo1(i/3) + 1, minimizeTo1(i-1) + 1};     if(i is dividable by 2)       a[i] = min{minimizeTo1(i/2) + 1, minimizeTo1(i-1) + 1};     if(i is dividable by 3)       a[i] = min{minimizeTo1(i/3) + 1, minimizeTo1(i-1) + 1};     if(i is not dividable by 2 and i is not dividable by 3)       a[i] = minimizeTo1(i-1) + 1; 

bcrypt.checkpw() returning true for different salt+hash

new user here. I’ve tried searching for an answer but I don’t think I know the correct terms to search for. I’ve tried things like “bcrypt same password different hash” but can’t find an answer to my question below.

Preamble: From what I have searched online, bcrypt returns the hashpw function with a byte-type string concatenation of the salt appended with the actual hash(salt+password).

My Question: Why does bcrypt recognize it as a password match when I provide it with another generated hash from the website https://bcrypt-generator.com? What is bcrypt.checkpw() actually comparing?

If I were to check if the user’s input matches the salt+hash on the server, how should I do it? bcrypt provides a different salt+hash every time the function runs, and apparently salt1+hash1 matches salt2+hash2 when generated from the same string. I’m really confused with how this works.

import bcrypt # import public crypto hash function bcrypt import time from datetime import datetime  password = b"test password" # convert password from string type to bytes type  i = 13 # work factor to determine computation time start = datetime.fromtimestamp(time.time()) # mark computation start time salt = bcrypt.gensalt(i) end = datetime.fromtimestamp(time.time()) # mark computation end time hashSPW = bcrypt.hashpw(password, salt) print("Start: ", start) print("End: ", end) print("Duration: ", end-start) # print duration print("Salt: ", salt) print("HashSPW: ", hashSPW)  user_input = b"test password" # change input here to test #if bcrypt.checkpw(user_input, hashSPW): if bcrypt.checkpw(user_input, b"$  2y$  13$  iynZoPYY5DL3TjnAJkOcbevq0QJBfDPMShN27aSG5wzL7MaRza.Sa"): # value got from https://bcrypt-generator.com/     print("Password Matched!") else:     print("Password Does Not Match!") 

Signed curvature of catenary involving turning/tangential angle

Suppose we want to find the signed curvature of the catenary $ $ \gamma(t)=(t,\cosh t)$ $ where $ \mathcal{k}_n=\frac{d\phi}{ds}$ and $ \phi(s)$ is the turning angle of $ \gamma$ such that$ $ \dot\gamma(s)=(\cos\phi(s), \sin\phi(s))$ $

We proceed: $ $ s=\int_{s_0}^{s}|\dot\gamma(t)|dt=\int_{s_0}^s\sqrt{1+\sinh^2t}dt=\sinh t$ $ so if $ \phi$ is the angle between $ \dot\gamma$ and the $ x$ -axis, then $ $ \tan\phi=\sinh t=s\implies\sec^2\phi\frac{d\phi}{ds}=1\implies k_s=\frac{1}{1+s^2}$ $

Can someone explain how we proceed in that last line of calculation? Why does $ \tan\phi=\sinh t=s? $