WordPress and woocommerce impossible to set variable in admin ajax in plugi devlopment

i got a problem in my plugin development. My class got some variable and i want to stock product from woocommerce on it.

I use ajax to search product on my database and it work fine, but when i call my class function and try to set it with $ this->myarray[$ key] = $ anotherarray;

my variable myarray stay empty.

I try to know why but impossible to fixe it. Thinking because i add my action before calling the ajax and it dont get my instance of my class but i dont found the way to fixe it.

Best regards

How is consensus impossible under FLP even when no fault exists at that moment?

Could some give an illustration of the following sentence?

In fact, for every purported consensus protocol that guarantees agreement and validity, there is some execution in which there are no failures and yet the algorithm never terminates.

Source: http://ieeexplore.ieee.org/xpl/login.jsp?arnumber=6122006&tp=&url=http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber%3D6122006

Exponential amount of information in polynomial size? Impossible!

I’m reading A note on succinct representations of graphs by Papadimitriou and Yannakakis. Let me quote the following paragraph on page 183:

Formula $ F$ has a highly regular structure. It has $ |x|$ clauses stating that the input to the computation of $ U$ is $ x$ , and a finite number of other types of clauses reflecting the moves of $ U$ , repeated in a regular way an exponential number of times (for each possible time-square combination of the computation of $ U$ on $ x$ ). In particular, it is very easy to see that, given two $ c|x|$ -bit integers, the indices of a literal and a clause, it can be determined in polynomial time whether the literal appears in the clause. Let us call $ B$ the polynomial-time algorithm computing this literal-clause relation.

Though $ B$ is polynomial-time, it’s not polynomial-size because it contains an exponential amount of information, i.e., literal-clause pairs.

Then it continues:

Combining algorithms $ A$ and $ B$ , we obtain an algorithm $ C$ which, given two integers with $ ck|x|$ bits, determines in polynomial time whether the two nodes are adjacent in the graph $ G(F)$ , based only on the bits of $ x$ . For fixed $ x$ this algorithm can be rendered as a polynomial-size circuit $ C_{G(F)}$ with $ 2ck|x|$ inputs, which is therefore a succinct representation of $ G(F)$ . Now, it is easy to see that, given $ x$ , we can construct $ C_{G(F)}$ in polynomial time.

I don’t understand how you can pack an exponential amount of information in “a polynomial-size circuit $ C_{G(F)}$ .” Can you explain? This is the hardest part of the paper, or it is wrong.

Is it impossible for an ISP to block access to a site that has a dynamic IP?

I was reading on how ISPs in many countries block access to many sites. This got me wondering — How do ISPs block sites in the first place? I found answers here, and reading through, it became evident that ISPs usually drop connections going to these blocked domains on their DNS servers, which can be easily circumvented by using a different DNS (optionally also using DoH to a non-compromised server).

I then read on to find that ISPs also block domains by their known IPs, so using another DNS wouldn’t fix this issue. This led me to think — what if the server was on a dynamic IP?

The devil’s advocate of my mind quickly told me that all the ISP had to do was to run a dns lookup on the domain itself to find the current IP and block that. But then again, what do you guys say?

Can you wipe a USB flash drive so securely that it’s impossible to recover deleted files through forensic analysis?

Internal SSD can be wiped with TRIM, but USB sticks are external SSD. They’re apparently difficult to wipe securely enough to make forensic analysis of the device impossible. This question has been asked before, about 4 or 5 years ago and the information is probably outdated. Looking for up-to-date advice on how to securely wipe files on a USB from a bootable media.

Is it possible to have a setup that can securely wipe USB with TRIM? Anyone know of any programs that will securely shred files on USB, making them impossible to recover via forensic recovery software?

How can I make an app impossible to uninstall or least very hard too?

I want “DisableChromeIncognitoMode” to be uninstallable or at least very hard to uninstall. I would also like to be unable to change its settings to “off” (meaning i can use incognito mode). Or i would just like to be unable to use incognito mode on chrome. (It wastes my time when I am using it for…( ͡° ͜ʖ ͡°)…purposes)

I will only root my phone as a last resort but breaking the warranty seems a bit much just for removing incognito

Automatic resolution of checks that are impossible to fail or succeed at?

If an ability check DC is lower than your bonus, or if a check DC is higher than (20 + your bonus), and the check is a pure binary check that’s pass/fail with no side effects depending on the end result of the check, is there anything in the rules that permits, discourages or denies just resolving the check without rolling?


  • DC = 5, bonus on the check = +4 → lowest possible result is 1+4, which would make the check.

  • DC = 30, bonus on the check = less than 10 → success is impossible.

Critical fails and wins (natural 1 or 20) are not included here.

Prove it is impossible for every male to be paired with their last choice in the stable marriage algorithm (male proposing)

I’m trying to prove that it is impossible for every man to be paired with their last choice in the stable marriage algorithm. I have a general idea of how to approach it, which is:

Suppose there exists a stable pairing T in which every man is paired with his least favorite woman, also suppose it took K days for the algorithm to halt. This means that on every day before the Kth day every man had been rejected by every woman. However, it is impossible for every woman to reject every n-1 men. If the algorithm halts on the Kth day then there has to be at least one woman who has not yet received a proposal on the K-1th day. Therefore, on the last day there is a woman who receives only one proposal and cannot reject it. Consequently, such stable pairing T does not exist.

Is this logically correct? I know I probably need to refine it a bit.