## Cost effective parts – profitable-coins.com – HYIPs

I am neither owner nor administrator. The information has been published here for discussion only.

beginning: April 9, 2019

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## Keto Buzz: Burn your fat from different parts of the body

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## Portfolio recovery – Is there a way to recover a Bitcoin private key by using certain parts of the key?

It depends on how much information you are missing.

A wallet import format key is coded base58. Thus, each character of the key can be one of 58 characters.

You mention in the comments that you are missing 21 characters. Assuming you know exactly which 21 characters are missing (ie you know the first 21 are missing), you have 21 ^ 58, or `4,88336 ... × 10 ^ 76` opportunities. This is outside the range of bruteforcing.

In addition, if you do not know what 21 characters are missing, you still have many more possibilities, because you must not only iterate the 58 possibilities for each character, but also try all their possible locations in the key.

If you can not find more information, the key can not be recovered.

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## Profitable parts – Profitable-coins.com

I am neither owner nor administrator. The information has been published here for discussion only.

beginning: April 9, 2019

Characteristics: SSL encryption | Registered company | Online discussion

QUOTE

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## complex – Effective evaluation of real / imaginary parts of long expressions

I have the following expression, rather compact, but complex (see below). I just want the real part of this. Now, when I make the habit, that is to say `ComplexExpand[Re[...]]//Simplify` since all parameters and functions are real, the number of terms produced by `ComplexExpand` is apparently too large (about 6000 terms) for `Simplify` manage. It has been running for almost an hour now, with no sign of conclusion.

From similar expressions in the same context, I know that the result of the simplification is also quite compact (just like the expression I'm starting with), so I'm wondering if it's possible to skip the "Expand" part. "from `ComplexExpand`? Why develop in thousands of terms, while everything resonates in a compact expression?

There must be a more efficient way, right?

``````- ((I E ^ (I am[Phi] + t [Omega]I -
He [Omega]r) ((1 / (
1 + (i [Nu] -
I r [Nu]) ^ 2 [Chi]^ 2 Cos[[Theta]]^ 2)) (r ^ 2 + [Chi]^ 2 Cos[
[Theta]]^ 2) (-1 + (2 r) / (
r ^ 2 + [Chi]^ 2 Cos[[Theta]]^ 2)) (-m [Chi] (r (-2 +
r (I[Nu] +
r[Nu]) ^ 2 ((-2 + r) r + [Chi]^ 2)) + (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 2 (r ^ 2 + [Chi]^ 2) Cos[[Theta]]^ 2)
(I RaI[r] + RaR[r]) (I say[[Theta]]+
SaR[[Theta]]) + (I [Omega]I +[Omega]r) (-r (r ^ 3 + (2
+ r + r ^ 3 (i [Nu] - I r[Nu]) ^ 2 +
2 r ^ 2 (I i [Nu] + r[Nu]) ^ 2)[Chi]^ 2 +
r (i [Nu] - I r[Nu]) ^ 2 [Chi]^ 4) + [Chi]^ 2 (r (2 -
r + 2 r ^ 2 (I i [Nu] + r[Nu]) ^ 2) + (-1 +
r ^ 2 (I i [Nu] + r[Nu]) ^ 2)[Chi]^ 2 + (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 4) Cos[[Theta]]^ 2) (I RaI[r] +
RaR[r]) (I say[[Theta]]+ SaR[[Theta]]+
r (I[Nu] +
r[Nu]) ((-2 +
r) r + [Chi]^ 2) (r ^ 2 + [Chi]^ 2) (-1 + (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 2 Cos[[Theta]]^ 2) (I SaI[[Theta]]+
SaR[[Theta]]) (I drift[1][RaI][r]    +
Derivative[1][RaR][r]) - (I i [Nu] + r[Nu]) (1 +
r ^ 2 (I i [Nu] + r[Nu]) ^ 2)[Chi]^ 2 ((-2 +
r) r + [Chi]^ 2) Cos[[Theta]](I RaI[r] +
RaR[r]) Peach[[Theta]](I drift[1][SaI][[Theta]]+
Derivative[1][SaR][[Theta]])) + (
1 / (- 1 + (I i [Nu] + r[Nu]) ^ 2 [Chi]^ 2 Cos[[Theta]]^ 2))
2 r[Chi] Peach[[Theta]]^ 2 (1 /
2 r (I[Nu] +
r[Nu]) [Chi] ((-2 + r) r + [Chi]^ 2) (-2 + (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 2 + (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 2 Cos[2 [Theta]]) (I say[[Theta]]+
SaR[[Theta]]) (I drift[1][RaI][r]    +
Derivative[1][RaR][r]) - (I RaI[r] +
RaR[r]) ((-r [Chi] (-2 +
r (I[Nu] +
r[Nu]) ^ 2 ((-2 +
r) r + [Chi]^ 2)) (I [Omega]I +[Omega]r) +
m[Chi]^ 2 (1 + (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 2) baby crib[[Theta]]^ 2 - (I i [Nu] +
r[Nu]) ^ 2 [Chi]^ 3 Cos[[Theta]]^ 2 ((r ^ 2 +
[Chi]^ 2) (I [Omega]I +[Omega]r) + m [Chi] Crib[[Theta]]^ 2) +
m r (-2 + r + 2 r ^ 2 (i [Nu] - I r[Nu]) ^ 2 +
r ^ 3 (I i [Nu] + r[Nu]) ^ 2 +
r (I[Nu] +
r[Nu]) ^ 2 [Chi]^ 2) Csc[[Theta]]^ 2) (I SaI[
[Theta]]+ SaR[[Theta]]) + (I i [Nu] + r[Nu]) (1 +
r ^ 2 (I i [Nu] + r[Nu]) ^ 2)[Chi] (-2 +
r) r + [Chi]^ 2) baby crib[[Theta]](I drift[1][
SaI][[Theta]]+
Derivative[1][SaR][[Theta]])))))) / ((1 +
r ^ 2 (I i [Nu] + r[Nu]) ^ 2) ((-2 +
r) r + [Chi]^ 2) (r ^ 2 + [Chi]^ 2 Cos[[Theta]]^ 2) ^ 2))
``````

## How to detect all document library Web Parts and do something with CSS or JavaScript

I would like to apply some CSS / JavaScript to ALL the document libraries of the site collection. For example, I would like to add a banner at the top of each document library for an information word.

I thought that this should perhaps be done in the .master file as it is applied globally. However, I have trouble finding a detector / recorder capable of locating all the Web Parts of the document library on the page. Are there unique identifiers such as "Id =" "Class =" or other properties that I could use in CSS / JavaScript to recognize that the Web part is a library of documents and fix it?

Please note the following two conditions that the solution should also be able to overcome:

1. Distinguish between the document library v. Generic list, although they have made similar in many ways.

2. This should work when the document library is on its own aspx page or is added to an area on another page as a Web Part.

## Take the blind parts of a PHP chain

I have this string:

``````"a: 11: i: 0; s: 3:" 191 "; i: 1; s: 3:" 256 "; i: 2; s: 3:" 247 "; i: 3; s: 3:" 244 "; i: 4; s: 3:" 257 "; i: 5; s: 3:" 250 "; i: 6; s: 3:" 253 "; i: 7; s: 3:" 258 "; i: 8; s: 3: "261"; i: 9; s: 3; "259"; i: 10; s: 3: "542";
``````

I want to take the numbers in quotation marks eg "191"

I tried to explode, but it cuts in places where it does not help me.
Neither split, until strpos, but it returns the position of the first coincidence.

I'd need something like what's between "%"

## Turing machines – Is the proof of the undecidability of \$ A_ {TM} \$ still valid if we modify certain parts?

I have a question based on a question that I've seen exists on the site, but with false information and no response, so I republish it with valid information (cited incorrectly in the book).

on page 207 of the book "Introduction to the Theory of Computation", there is evidence that $$A_ {TM}$$ it's not decidable.

the proof is constructed using contradiction, assuming that there is a machine $$H$$ it's a decision maker for $$A_ {TM}$$ and takes the entrance of $$langle M, w rangle$$ or $$M$$ is a turing machine and $$w$$ is a string, and H stops and accepts if $$M$$ accepted $$w$$and assuming that $$H$$ stops and rejects if $$M$$ fails to accept $$w$$.

then, they build a new turing machine $$D$$ with $$H$$ as a subroutine that performs the following operations:

$$D$$ = "On the entrance $$langle M rangle$$ or $$M$$ is a TM:

1. Run $$H$$ entrance $$langle M, langle M rangle rangle$$
2. Take out the opposite of what $$H$$ exit. that is, if $$H$$ accept, reject; and if $$H$$ reject, accept.

the questions: will the proof still be valid if we modify independently the following modifications, that is to say that each modification is distinct and does not affect the other distinct situation:

1) if we change step 1 for it to be: Run $$H$$ entrance $$langle M, langle M rangle ^ R rangle$$,

2) if we change step 2 so that it is: Si $$H$$ accept, loop. if $$H$$ reject, accept.

my attempt:

1) the proof will remain true, because we prove by contradiction that on machine $$D$$operating $$H$$ as a subroutine will not allow us to decide $$A_ {TM}$$. in addition, being run $$D$$ with entry of itself, namely $$D ( langle D rangle)$$, forces the machine to do the opposite, thus creating the contradiction. so it does not matter if we run $$D ( langle M rangle)$$ or $$D ( langle M rangle ^ R)$$the result will be the same when we run $$D ( langle D rangle)$$ and the $$D$$ will have to decide otherwise (accept if not accept $$langle D rangle$$ and rejects if $$D$$ accepted $$langle D rangle$$).

2) Well, here I am not sure. because in the original proof if $$H$$ accept, then he rejects, and if $$H$$ reject, so he accepts. now if we say "Si $$H$$ accept, loop. if $$H$$ rejects, accepts ", well, we have a problem because now $$H$$ Stop and then it is decided, even if it is incorrect. The loop is the main problem here and I hope that I have interpreted it correctly.

did I do it well? I'm really not sure, especially about 2. Is it enough to give an acceptable explanation?

Thanks a lot for your help. do my best to develop it and solve the problem in the site that lacks correct information from the book (also gave sources from it).

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