Let me spoof your miserable lifes. Hidden Markov models, random pseudo noise channels ...and PREDADOR...TARGET : worldwide internet virus infection

1. GET THE OPTICAL SIGNAL 

ltePBCHDecode

Physical broadcast channel decoding

collapse all in page

Syntax

[bits,symbols,nfmod4,trblk,cellrefp] = ltePBCHDecode(enb,sym)

[bits,symbols,nfmod4,trblk,cellrefp] = ltePBCHDecode(enb,sym,hest,noiseest)

[bits,symbols,nfmod4,trblk,cellrefp] = ltePBCHDecode(enb,sym,hest,noiseest,alg)

https://www.mathworks.com/help/lte/ref/ltepbchdecode.html

2. virus ....Lets encrypt some random buffer A[] with some hashing algorithm so many

times N, so it will take us time period T.
After these calculations done, we have another random buffer B[] which
is used to encrypt/decrypt our "delayed" code.

There is no way to perform required N iterations using more than one
computer ('coz each time current buffer is encrypted), so minimal
decryption time is limited with maximal CPU speed.

If you will use computer which is fast enough, and use some time
to encrypt fucking random buffer, then you may be sure that
the same operation may not be done in a less time period.

So, each time the virus is active, it iterates N encryption cycles until
buffer A[] will be converted into B[].
After time T will be spent to decryption, virus will got buffer B[]
and use it to decrypt "delayed code".

http://z0mbie.daemonlab.org/dpgn_eng.txt

3. ADC CAPTURE 

i only have this code
// defines for setting and clearing register bits
#ifndef cbi
#define cbi(sfr, bit) (_SFR_BYTE(sfr) &= ~_BV(bit))
#endif
#ifndef sbi
#define sbi(sfr, bit) (_SFR_BYTE(sfr) |= _BV(bit))
#endif

// set prescale to 16
sbi(ADCSRA,ADPS2) ;
cbi(ADCSRA,ADPS1) ;
cbi(ADCSRA,ADPS0) ;

4. AND FINALLY BIT MANIPULATION ( ATMEGA8 BIT MANIPULATION )

http://www.avrfreaks.net/forum/tut-c-bit-manipulation-aka-programming-101?page=all

Hi friends after lot of research online I got a new and working trick for making money online without any investment and free of cost. To learn this trick follow these steps:


1. Register on this website binbox. This is a shorting URL site where you can earn money through shorting URLs. When someone click on your short link and type the captchas to unlock the actual link you get paid.


2. Now short atleast 100 or more URLs by login to your account.

3. Now download Tor browser.


4. Make a blog post on blogger or bitlanders or post on bitbin and post those 100 links.


5. Now open Tor browser and Go to your blog post you have made (bookmark it).


6. Click all links one by one and after clicking every link type the captcha so that you get paid and remember after visiting every link click on the onion icon in tor browser and select new Identity otherwise you will be get banned.


You can get your payment through PayPal and bitcoins. The minimum amount for payout is 5$. You can withdraw it any time to your bitcoin or PayPal account.

Try this method if it help you then please donate some bitcoins to this address:

http://thebot.net/threads/earn-10-per-day-new-bitcoins-and-paypal-dollars-earning-trick.322801/

Any ideas on why os.system() could fail?

Though i added all the programs I needed to Path, I never took a look at the ComSpec environment variable.

Apparently, I added C:\Program Files (x86)\Java\jre7\bin to comspec in addition to the default %SystemRoot%\system32\cmd.exe.

The solution was to either add all the paths from Path to ComSpec as well, or simply remove the path to the Java binary.

http://stackoverflow.com/questions/11033703/any-ideas-on-why-os-system-could-fail

WHAT ABOUT A CODE FOR A TROJAN ?...

I want to remove everything before the wa path.
p = path.split('/')
counter = 0
while True:
if p[counter] == 'wa':
break
counter += 1
path = '/'+'/'.join(p[counter:])
For instance, I want '/book/html/wa/foo/bar/' to become '/wa/foo/bar/'

would be to use os.path.relpath:

>>> import os
>>> full_path = '/book/html/wa/foo/bar/'
>>> print os.path.relpath(full_path, '/book/html')
'wa/foo/bar'

http://stackoverflow.com/questions/8693024/how-to-remove-a-path-prefix-in-python

wake up NEO...

1.the question is in which format you record and transmit the data images files, to be then "recreated" on audio...and the same reverse engeniering, from audio back to image. For instance we have the format exiftool read/write/create like XMP, MIE, VRD
https://linux.die.net/man/1/exiftool

2. How to transport it? & DESIGN

Fractional N Frequency Synthesizer Tutorial

http://www.radio-electronics.com/info/rf-technology-design/pll-synthesizers/fractional-n-synthesis.php

3. How to begin write the biological code to be transport?

a) 

Scanning electron microscope

https://en.wikipedia.org/wiki/Scanning_electron_microscope

b) 

How to turn living cells into computers

http://www.nature.com/news/how-to-turn-living-cells-into-computers-1.12406

c) how to write the code? 

How to take a user input (from scanner) and convert the entire input into ascii

http://stackoverflow.com/questions/36175004/how-to-take-a-user-input-from-scanner-and-convert-the-entire-input-into-ascii

4. HOW TO BE BIOLOGICAL PREPARED? 

a) THE RED PILL : Electroceuticals are the electronic counterparts of pharmaceuticals. They are tiny electronic devices that interact with the electrophysiology of the human body. 

http://www.bioelectronics.tudelft.nl/Research/theme.php?id=42

AND THEN...GET INSIDE THE MATRIX

Repeated XOR

Problem

There's a secret passcode hidden in the robot's "history of cryptography" module. But it's encrypted! Here it is, hex-encoded: encrypted.txt. Can you find the hidden passcode?

Hint

Like the title suggests, this is repeating-key XOR. You should try to find the length of the key - it's probably around 10 bytes long, maybe a little less or a little more.

Answer

Overview

Preform Kasiski elimination to find the key length. For this problem, the keylength is 8. Since the cipher text is encrypted using repeateing xor, you can do a frequency analysis on the 0th, 8th, 16th characters, and then another one on the 1st, 9th, and 17th characters, adn then on the 2nd, 10th, and 18th character.

Details

About repeated XOR

In a repeating XOR cipher, if the key is shorter than the message (it almost always is), the key is duplicated in order to cover the whole message. Then each byte of the plain text is XORed with each according byte of the key. For example, suppose we are trying to encrypt the message 'THIS IS A MESSAGE', with the key 'YOU', we first convert all of the characters to integers using ASCII encoding. (More examples may be found on the Wikipedia page)

T   H   I   S       I   S       A       M   E   S   S   A   G   E
84, 72, 73, 83, 32, 73, 83, 32, 65, 32, 77, 69, 83, 83, 65, 71, 69

Then we duplicate the key and convert it to an integer sequence.

Y   O   U   Y   O   U   Y   O   U   Y   O   U   Y   O   U   Y   O
89, 79, 85, 89, 79, 85, 89, 79, 85, 89, 79, 85, 89, 79, 85, 89, 79

Then we XOR the integers together.

84, 72, 73, 83, 32, 73, 83, 32, 65, 32, 77, 69, 83, 83, 65, 71, 69
89, 79, 85, 89, 79, 85, 89, 79, 85, 89, 79, 85, 89, 79, 85, 89, 79
xor --------------------------------------------------------------
13,  7, 28, 10,111, 28, 10,111, 20,121,  2, 16, 10, 28, 20, 30, 10

Do you notice the repeating 28, 10, 111? That is because there was a repeat in the paintext that was a multiple of the keylength apart. In reality, repeats happen quite frequently, but it is just as likely to get a repeat 3 characters apart as it is to get one 4 characters apart. If they are 4 characters apart, a repeat is not shown in the cipher text, because the length of the offset (4) is not a multiple of the key length (3). For example, encrypting 'THIS MESSAGE IS NOT' with the key 'YOU' gives:

https://ehsandev.com/pico2014/cryptography/repeated_xor.html

  intercept-xks_logos_embedded_in_docs.pdf by Elsa Cristina David on Scribd

JAP Backdoor ....

From: goncalo.costa at kpnqwest.pt (Goncalo Costa)
Subject: JAP back doored

>
> Don't be a smart ass.
>

Well, good morning to you too !

> Your arguments have nothing to do with the argument at hand which is quite
> simple: Governments should have no right to force developers to trojanize
> their applications and keep silent about it.
>

Governments have a lot of powers they should not have but German government
had nothing to do with this.

I hope you can tell the difference between a government eavesdropping on
someone and a judge/court order to eavesdrop on a suspect to gather evidence
against him.

> There have been some notes come out of this:

1> Germany has now removed this legal action, which is great

I think you should stop for a minute and try to learn the difference between
Germany (country), the German state, the German government and the German
judicial system.

"Germany" did nothing.

If you want to talk about the German government you could talk about
http://www.gnupg.org/aegypten

> 2> They intended to only watch traffic to a single German server

It seems you did not follow some posts on this list. I believe someone from
Germany explained the why and how of this JAP backdoor, and mentioned that.

3> The developers may not
> have been so forced into doing this, as much as willing -- I rather doubt
> this, especially since the order was rescinded, but their culpability does
> factor into this

So you mean these guys offering a free public anonymizing service are to
blame for complying with a court order ? I'm sure you would rather go to
jail. Where's the free public anonymizing service you're providing to the
Internet ? I'd like to use it. I'm sure I can trust you to keep my id safe.

4> I, personally, admit I would not care if they did this
> for a very serious reason such as for pedophiles or terrorists... I think a
> lot of people outraged would have to agree with this... However, I am sure
> a lot would not
>

(I believe the same person also wrote) it was a pedophilia case.

> As for the US government, this is utterly unimportant. I was playing around
> even to begin to mess with that. Yes, I am unaware of the US actively
> trojanizing applications by forcing the developers to do this.

Lotus Notes NSA backdoor ?
We're not talking about a court order here.
And Notes was not free software - its customers paid for it.
Nor it was open source software as is the case with JAP.

> So are you.
> This is illegal. You wouldn't like it if it was the US doing this. So, what
> are your real motives here?
>

Besides money that is ? :-)

http://lists.openwall.net/full-disclosure/2003/08/28/6

we need this cookie ...I'll see it better tomorrow

package org.apache.directory.server.ntp.messages;
  
  
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.List;
  
  
  /**
   * Reference Identifier: This is a 32-bit bitstring identifying the
   * particular reference source. In the case of NTP Version 3 or Version
   * 4 stratum-0 (unspecified) or stratum-1 (primary) servers, this is a
   * four-character ASCII string, left justified and zero padded to 32
   * bits. In NTP Version 3 secondary servers, this is the 32-bit IPv4
   * address of the reference source. In NTP Version 4 secondary servers,
   * this is the low order 32 bits of the latest transmit timestamp of the
   * reference source. NTP primary (stratum 1) servers should set this
   * field to a code identifying the external reference source according
   * to the following list. If the external reference is one of those
   * listed, the associated code should be used. Codes for sources not
   * listed can be contrived as appropriate.

http://directory.apache.org/apacheds/gen-docs/2.0.0-M9/xref/org/apache/directory/server/ntp/messages/ReferenceIdentifier.html

Elsa David this chit works on every fuckin' frequency Hacker Anonymous Việt Nam

Deciphering a key from XOR encrypted cypher using boolean logic

following the white rabbit direct to cables i have this hotdog ..because xor is the base of all hardware encryption 

Elsa David of course i know we're talking about optic tronics

If K$K$ is random and you only know A$A$ or B$B$ (but not both) then, no, there is no way to infer anything about the key - this is the (in)famous one-time-pad.

If you know A$A$ and B$B$, then you can recover K$K$ very easily. Exclusive-or has those properties:

• ∀n$\mathrm{\forall }n$,     n⊕n=0$n\oplus n=0$
• ∀n$\mathrm{\forall }n$,     n⊕0=n$n\oplus 0=n$ (identity element)
• ∀a,b$\mathrm{\forall }a,b$,     a⊕b=b⊕a$a\oplus b=b\oplus a$ (commutativity)
• ∀a,b,c$\mathrm{\forall }a,b,c$,     a⊕b⊕c=(a⊕b)⊕c=a⊕(b⊕c)$a\oplus b\oplus c=(a\oplus b)\oplus c=a\oplus (b\oplus c)$ (associativity)

So we can do the following:

B=A⊕K   ⇒   A⊕B=A⊕(A⊕K)=(A⊕A)⊕K=0⊕K=K$B=A\oplus K\Rightarrow A\oplus B=A\oplus (A\oplus K)=(A\oplus A)\oplus K=0\oplus K=K$

So A⊕B=K$A\oplus B=K$

Viewed differently, the exclusive-or operator is invertible:

⊕01001110$\begin{array}{|ccc|}\hline \oplus & 0& 1\\ 0& 0& 1\\ 1& 1& 0\\ \hline\end{array}$

A0011B0101⊕0110$\begin{array}{|ccc|}\hline A& B& \oplus \\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\\ \hline\end{array}$

And since the truth table is symmetric, the exclusive-or operation just happens to be its own inverse, i.e. x⊕−1y=x⊕y$x{\oplus }^{-1}y=x\oplus y$. So if we take our original equation:

A⊕K=B$A\oplus K=B$

We can represent it as follows:

K⊕A=B$K\oplus A=B$

And we can then undo (invert) the exclusive-or by A:

K⊕A⊕−1A=B⊕−1A   ⇒   K=B⊕−1A$K\oplus A{\oplus }^{-1}A=B{\oplus }^{-1}A\Rightarrow K=B{\oplus }^{-1}A$

And as we found above, this is identical to:

K=B⊕A=A⊕B$K=B\oplus A=A\oplus B$

As found at the beginning.

However, this is assuming A, B and K are all the same length. If K is smaller than A and B, then it means that K will be used multiple times (repeated over the length of the plaintext, presumably). This repetition can be exploited to successfully recover K from only B provided there is enough repetition and there is enough ciphertext to work with - see Vigenere cipher.