Features of the Armory wallet and its telephone number for technical support

Armory is not a wallet, it is a wallet management application for Bitcoins. It is an open source wallet application based on Python. The arsenal was built on January 3, 2012. It is available for several operating systems such as Mac OSX and Linux. There is a new feature in Armory that is an encrypted wallet. Protects portfolios offline for 100% security of online hackers. Help in secure printing for all types of backup. If you are using an armory wallet, you have the option to resolve your problem in Telephone number of the Armory + 1-888-712-3146. This is a free number for customers of the Armory portfolio.

Security: What is the best way to show a partial mobile number?

Check the following links:

Fuzzing for obfuscated phone numbers

https://www.alldigitaltricks.com/wp-content/uploads/2017/12/kotak-bank-change-mobile-netbanking.png

https://www.trickolla.com/2016/12/Register-Mobile-Number-With-Bank-of-Baroda.html

The use case in question is to show the partial mobile number to the user:
I like to continue with the mobile phone number that ends with * * * * * * 1 2 3 4 & # 39;

The problem is that we can have colide of the last 4 digits because users have the option to select the last 4 digits of the mobile phone and we are showing multiple options of this type to uniquely identify the option we need from some other mechanism .

The options that I am thinking are:

Option 1: sample 1.2 and the last two digits & # 39; 1 2 * * * * * * 3 4 & # 39;

Option 2: shows the fifth, the sixth, the last two digits & # 39; * * * * 1 2 * * 3 4 & # 39;

Option 3: show the alternative digits & # 39; 1 * 2 * 3 * 4 * 5 * & # 39;

Option 4: shows the last 4 digits & # 39; * * * * * * * 1 2 3 4 & # 39;

Problem in option 4 for our country / target market, there is a high collision for a given user, since you can choose the last 4 digits.

———EDIT————-
In my use case, in the user interface I plan to show multiple options to the client to use one of the inscriptions made previously.

Something like this:
1. Use your account with mobile number 5647 234 4567.
2. Use your account with mobile number 7868 456 7891.
3. Use your account with the mobile number 4321 321 1989.

But instead of showing the full mobile number on the screen, we want to show the queue or partial number. The question here is in which part the best option to show it to the user, so if we select option 1, that is, we show the first and the last two and we mask others. It will look like this.

  1. Use your account with mobile number 56 ** *** ** 67.
  2. Use your account with mobile number 78 ** *** ** 91.
  3. Use your account with the mobile phone number 43 ** *** ** 89.

Binance phone number Problem due to inability to sign in to Binance

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Can I use Boolean algebra to reduce the number of lines in my code?

If you can. But should you?

Boolean algebra is used to reduce the logical expressions to their minimum form, but if this is good or bad it is left to the programmer. Let's take this validation code:

...
if (person.Money == 0) return;
if (person.Money! = 0 && person.ge) < 15) return; 
if (person.Age > 90) return;
if (person.Children> 3) return;
if (person.HasWhiteShirt () && person.HasBlueSocks ()) return;
...

This is a list of checks on the object. person, and will leave the function as soon as one of them is true. The meaning of that is pretty clear, and anyone who reads it, even a programmer, can understand what controls we are doing.

However, there are different ways of writing that. Let's go to the lines of code and simplify them, as you did in your question, and let's see what happens:

if (... || person.Money == 0 || person.Age < 15 || person.Age > 90 || person.Children> 3 || (person.HasWhiteShirt () && person.HasBlueSocks ()) || ...) he came back;

We now have a unique line of illegible conditionals, and by simplifying we also lost the information that that verification of age <15 years is related in some way to money. Editing this code will be much more difficult in the future, so reducing it was a mistake.

In conclusion: always try to readability, regardless of the minimum form of a logical expression.

Publications – Multiple site number in Subdomain article pages

I created an article under the subdomain that is in my multisite.
Let's call the subdomain ztak.mydomain.com
When I go to the main page of ztak.mydomain.com, the page is displayed correctly.
When I click on any of the items that are on my home page, they redirect me to a blank blank page (500 Error).

I checked to make sure that my virtual host is configured correctly
.htaccess is configured correctly
the database is configured correctly

I'm not exactly sure what is causing this. My development works well. I copied the code and they are exact. My only assumption is that there is something else on the server, where ztak.domain.com lives, which is causing the 500 error, but has not been able to reduce it yet.

Has anyone found this? If so, how did you solve the problem?

Lost phone number and PayPal need verification by phone number

My account was blocked. I can not log in to my account and my recovery option was just my email, but I opened it when I still had my previous number, I have no way to access that number and I can not send to another account. phone number.

What are my options?

np complete – Demonstrating the NP hardness on the problem of creating graphics with the triangle number

I have a problem creating graphics.

Given a set of graph nodes and node constraints, such as the number of neighbors (degree) of each node. They also provide me with the total number of triangles in the graph. Given the previous restrictions ($ n + 1 $), I have to generate a graph that satisfies all the restrictions.

I know that given the degrees of the nodes, it is possible to construct using a Havel-Hakimi algorithm a graph that satisfies the degrees of the nodes. But I'm not sure how to comply with the total triangulation count restriction and if it makes the problem NP-difficult.

c ++ – Find the minimum of a function dependent on a constant number

Question-
enter the description of the image here

My approach-
I noticed that the minimum of a row is when t = k-one of the numbers in that row.
Suppose in the case of the previous example,
t = 4, which is 6-2.
Since we have to find the minimum number of tests, everything will not give the solution.
Any advice to improve this technique in cases where each number is unique and the number of rows is high?

Code-

#include
#include
using namespace std;
int main ()
{
setplus;
int n, k;
cin >> n >> k;
int a[n],second[n],do[n];
for (int i = 0; i> a[i]>> b[i]>> c[i];
plus.insert (k-a[i]);
plus.insert (k-b[i]);
plus.insert (k-c[i]);
}
int ans = 1000000000, curmax = -1, tempsum;
for (auto x: plus)
{
curmax = -1;
for (int i = 0; icurmax)
curmax = tempsum;
}
yes (curmax <ans)
ans = curmax;
}
cout << ans;
}

more unique stores k-no

curmax stores the current maximum for the & # 39; t & # 39; current.

ans is the minimum of all those maximums.

Tools – Number of nodes and channels in Lightning Network?

(one) Is there a tool with which we can know the current number of nodes (participants as payers / beneficiaries) and the number of channels in Lightning Network with an accurate estimate?

(two) As a more advanced query, is it also possible to know the topology of this chart? (vertices as participants and edges as channels)

nt.number theory – A new formula for the class number of the quadratic field $ mathbb Q ( sqrt {(- 1) ^ {(p-1) / 2} p}) $?

I have the following conjecture that implies a possible new formula for the class number of the quadratic field $ mathbb Q ( sqrt {(- 1) ^ {(p-1) / 2} p}) $ with $ p $ an odd cousin

Guess. Leave $ p $ be an odd cousin and leave $ p ^ * = (- 1) ^ {(p-1) / 2} p $. Then the class number $ h (p ^ *) $ of the quadratic field $ mathbb Q ( sqrt {p ^ *}) $ matches the number
$$ frac {( frac {-2} p)} {2 ^ {(p-3) / 2} p ^ {(p-5) / 4}} det left[cotpifrac{jk}pright]_ {1 le j, k le (p-1) / 2}, $$
where $ ( frac { cdot} p) $ It is the symbol of Legendre.

This is Conjecture 5.1 in my prepress arXiv: 1901.04837. I've checked it for all odd prime numbers $ p <29 $. Note that $ h (p ^ *) = 1 $ for each odd cousin $ p <23 $Y $ h (-23) = $ 3.

Here I invite some of you to review this conjecture more thoroughly. My computer can not verify it $ p = 29 $.