blockchain.info – send bitcoin to mobile number or email address

First of all: You do not really send the bitcoins to an email-adress or a phone number. You can only transfer them from one bitcoin adress to another one.

My understanding: The internetsite / (software behind it) creates a keypair (private and public key) and generates a bitcoin adress. Then, the site tells the user to send x Btc to that adress and provides you a textfield to enter an email-adress of your fried for example. And after that, the site sends an email to the email adress like “Hey, someone sent you bitcoins. Visit our site http://…
And finally, you need to provide the 2nd person the opportunity to import the keypair into his wallet or you ask him for a passwort that the first person gave the second person and let him/her enter a new bitcoin adress where he/she than send the bitcoins to. And you should use https for such a project.

Applications of Tits’ Alternative in algebraic number theory

I have recently studying Tits’ alternative. The theorem statement goes like the following:

Tits’ alternative: Let $G$ be any finitely generated linear group over a field. Then one of the following is true,

$(1)$ $G$ contains a solvable normal subgroup of finite index,

$(2)$ $G$ contains a non-abelian free subgroup (of rank at least $2$).

I am in search of applications of this wonderful theorem in algebraic number theory. Any help, resources or reference will be appreciated. Thanks in advance.

ubuntu – Getting current number of connections in mysql

I want to know if there is any parameter to show the current number of mysql connections (or peak value in the past 10 minutes or something else). Right now, I only see max and max_used connections.

mysql> show global status like '%connection%';
+-----------------------------------+---------------------+
| Variable_name                     | Value               |
+-----------------------------------+---------------------+
| Connection_errors_accept          | 0                   |
| Connection_errors_internal        | 0                   |
| Connection_errors_max_connections | 434                 |
| Connection_errors_peer_address    | 0                   |
| Connection_errors_select          | 0                   |
| Connection_errors_tcpwrap         | 0                   |
| Connections                       | 2380515             |
| Max_used_connections              | 152                 |
| Max_used_connections_time         | 2020-07-09 19:28:43 |
+-----------------------------------+---------------------+
9 rows in set (0.00 sec)

mysql> show global variables like '%connection%';
+--------------------------+-------------------+
| Variable_name            | Value             |
+--------------------------+-------------------+
| character_set_connection | latin1            |
| collation_connection     | latin1_swedish_ci |
| max_connections          | 300               |
| max_user_connections     | 0                 |
+--------------------------+-------------------+

probability – number of visits between recurrent and transient states

Given is a Markov chain $X=(X_n)_{ngeq0}$ on state space $V$ and transition matrix $P$. Let $V_x$ denote the number of visits to state $xin V$ and $mathbb{P}^x$ the conditioned probability on $X_0=x$.

If $x neq yin V$ are two transient states, I want to show that $mathbb{P}^x(V_y < infty) = 1$.

First I will give you my intuition and I hope that the intuition is correct.

We visit $x$ and $y$ only a finite number of times, starting in $x$ or $y$, respectively (by definition). Starting in $X_0=x$, we have two cases. The first is that $x$ and $y$ are not connected, giving that the number of visits to $y$ starting in $x$ is $0<infty$. The second case is that $x$ and $y$ are connected. Starting in $x$ and going to $y$ we will count $1$ visit in $y$. Then starting in $y$, by definition, we visit $y$ only a finite number $k<infty$ of times. All in all, starting in $x$, the number of visits to $y$ is $k+1 <infty$.

Moreover, I want to show that if if $x neq yin V$ are two recurrent states, $mathbb{P}^x(V_y = infty) = 1$.

The argument therefore will be similar as above, considering that if we are in $y$, we will return to $y$ an infinite number of times. So all in all, the number of visits to $y$ is $1 + infty = infty$, where the $1$ comes from oing from $x$ to $y$.

Do you have any tips for me how ti put this into formulas? Thank you very much in advance.

❓ASK – How to increase number of YouTube views | NewProxyLists

You are welcome here ,
You should first post good content it have a great search, for example the news of artists, celebrities, and football news. It is best to create the video on your own to be the content owner, and I advise you to learn montage and SEO because it will help you a lot to increase number of views .

 

Where is the serial number on a Tamron SP 150-600mm F/5-6.3 Di VC USD G2 (A022)?

On the first-gen (A011) model it was stamped faintly in the black plastic near the manufacturer location. I don’t see it anywhere on the G2. I did receive a 1/8 sheet printout with the serial number on it. Thanks.

d7 search api solr : partial keyword number + letter not working

We use Drupal 7.71 and use “search_api_solr”. Solr version :drupal-4.3-solr-4.x

Problem: we have an existing part of a node title: swf6p175

  • we can search for swf6p175
  • we can search for swf
  • issue: we can NOT search for swf6: : only combination of partial search term with letter + number does not work

This is schema.xml:
https://pastebin.com/GCkaim0R

thanks

javascript – Is it possible to update the database without referring to the ID number?

I am using Javascript on Applab. I have the code:

var player = {};
player.Username=" ";
onEvent("confirmUsernameButton", "click", function() {
  updateRecord("AllUserData", player, function() {
    player.Username = getText("Username_input");
  });
});

However, it is not updating the database at all. Can I update the database with an event.click without referring to the id number, or could I use each user’s specific “User ID” to update the database instead? Any help is appreciated.’

Link to app: https://studio.code.org/projects/applab/P2Vrn_Mmpb1g0cXOoTOL4XzFdIiaQcQAvwxTA_IaGlQ
To access the code and database, press “view code” at the top right corner of the screen.

usa – What reason would a stranger record my license plate number?

While traveling through the USA, my wife stopped at a shop. Due to the virus rules, only one person in the shop per group, so I waited and I read a book in the car. After about 20 minutes, some man who I had noticed in the parking lot for a while came by and wrote down my license plate number. He left, then came back and kept watching my car. The car is not a luxury car or anything special.

Why would a stranger want to write down my license plate number? Are there groups keeping a list of foreigner’s license plates? Is that something I should be concerned about?

c++ – Factorize All Numbers Up to a Given Number

this post is sort of a continuation of my answer on the following question: Fast Algorithm to Factorize All Numbers Up to a Given Number. As this post explains – We need to factorize all the numbers up to a large N.

At first I gave a python solution which was pretty slow (since – you know, python), than I decided to write it in C++. I am not that good with C++ and I would like to have a code review about that answer:

#include <math.h>
#include <unistd.h>
#include <list>
#include <vector>
#include <ctime>
#include <thread>
#include <iostream>
#include <atomic>

#ifndef MAX
#define MAX 200000000
#define TIME 10
#endif


std::atomic<bool> exit_thread_flag{false};

void timer(int *i_ptr) {
    for (int i = 1; !exit_thread_flag; i++) {
        sleep(TIME);
        if (exit_thread_flag) {
            break;
        }
        std::cout << "i = " << *i_ptr << std::endl;
        std::cout << "Time elapsed since start: " 
                  << i * TIME 
                  << " Seconds" << std::endl;
    }
}

int main(int argc, char const *argv())
{
    int i, upper_bound, j;
    std::time_t start_time;
    std::thread timer_thread;
    std::vector< std::list< int > > factors;

    std::cout << "Initiallizating" << std::endl;
    start_time = std::time(nullptr);
    timer_thread = std::thread(timer, &i);
    factors.resize(MAX);
    std::cout << "Initiallization took " 
              << std::time(nullptr) - start_time 
              << " Seconds" << std::endl;

    std::cout << "Starting calculation" << std::endl;
    start_time = std::time(nullptr);
    upper_bound = sqrt(MAX) + 1;
    for (i = 2; i < upper_bound; ++i)
    {
        if (factors(i).empty())
        {
            for (j = i * 2; j < MAX; j += i)
            {
                factors(j).push_back(i);
            }
        }
    }
    std::cout << "Calculation took " 
              << std::time(nullptr) - start_time 
              << " Seconds" << std::endl;

    // Closing timer thread
    exit_thread_flag = true;

    std::cout << "Validating results" << std::endl;
    for (i = 2; i < 20; ++i)
    {
        std::cout << i << ": ";
        if (factors(i).empty()) {
            std::cout << "Is prime";
        } else {
            for (int v : factors(i)) {
                std::cout << v << ", ";
            }
        }
        std::cout << std::endl;
    }
    
    timer_thread.join();
    return 0;
}

I would especially like a review about my usage of threads (I am afraid it might slow down the code). The performance are reaching 6619 which is the 855th (out of 1662 primes up to 14140 ~ square root of 200000000) in 1.386111 hours, if you find any way to make it faster I will be amazing! A more semantic review is also very welcome (Like #include order?).

Just for fun and a point of reference if you are trying to run the code yourself:

Progression Graph

Where X is time and Y is the prime reached (i). The orange tradeline is y = 13 * 1.00124982852632 ^ x. The graph is exponential since indeed the inner loop time is getting shorter.

The orange tradeline says I will reach 14107 (The highest prime before the square root) at x ≈ 5595.842803197861 seconds which is 1.554 hours!