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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.