Sunday, January 27, 2008
Ubuntu Satanic Edition and Compiz
Sunday, January 20, 2008
SEcure PUblic Key TRAnsfer
 Each of the two parts generate a pair of public  private keys.
 The two parts exchange their public keys over a possibly insecure communication channel.
 Now if one part wants to send a message to the other it simply encrypts it with the public key of the other part and then sends it. An eavesdropper cannot decrypt this message because he has not got the private key needed.
 Bob generates a keypair. (a public and a private key) and sends his public key to Alice.
 Mario, who is an employer in Bob's ISP, has already generated a keypair and while the packet with Bob's key is transient he replaces Bob's key with his own (he also notes down Bob's public key).
 Alice receives Mario's public key but she believes it's Bob's.
 Alice generates a keypair and sends her public key to Bob but Mario replaces it with his own, just like he did with Bob's.
 Now Bob wants to send some data to Alice. He encrypts it with Mario's key believing that it's Alice's key and sends the encrypted data.
 The encrypted data is peaked by Mario who is able to decrypt them with his own private key. He then reencodes them with Alice's correct public key and passes them on Alice.
 The same thing happens when Alice replies to Bob. Mario is able to eavesdrop on the connection even though both part's think that is secure.
I tried to understand the algorithm but I couldn't! And I think that it's too complex for no reason... They could use this protocol to achieve the same thing:While data messages are being exchanged, either Alice or Bob may run SMP to detect impersonation or maninthemiddle attacks. As above, all exponentiations are done modulo a particular 1536bit prime, and g_{1} is a generator of that group. All sent values include zeroknowledge proofs that they were generated according to this protocol, as indicated in the detailed description below.
Suppose Alice and Bob have secret information x and y respectively, and they wish to know whether x = y. The Socialist Millionaires' Protocol allows them to compare x and y without revealing any other information than the value of (x == y). For OTR, the secrets contain information about both parties' longterm authentication public keys, as well as information entered by the users themselves. If x = y, this means that Alice and Bob entered the same secret information, and so must be the same entities who established that secret to begin with.
 Bob knows X and Alice knows Y and they want to compare them.
 Bob sends some random data to Alice. Alice does the same with some other random data.
 Bob appends X to the end of the data he received and then computes the hash of the new data. Alice appends Y to the end of the data she sent to Bob and computes the hash of the new data. Bob sends the hash to Alice. Alice checks whether the two hashes are the same.
 Alice appends Y to the end of the data she received and then computes the hash of the new data. Bob appends X to the end of the data he sent to Alice and computes the hash of the new data. Alice sends the hash to Bob. Bob checks whether the hashes are the same.
 H(x) is the hash of x. Any hash function can be used. Take a look at MD5.
 E(x,k) is the symmetrically encrypted data of x with k used as the encryption key.
 DE(x,k) is the symmetrically decrypted data of x with k used as the decryption key.
 PK is the public key of the sender.
 RDP means Random Data Padding.
 & is the concatenation symbol (VBstyle :P).
 len(x) is the length in bytes of x.
 D: The Data packet. It is composed of 4 parts: len(RDP) & RDP & H(K) & E(PK & data,k)
 A: The Acknowledge packet. It is composed of 3 parts: RDP & H(D)
 K: The Key packet. It is composed of 4 parts: RDP & H(A) & k
 Exchange public keys with the other part and start an encrypted connection.
 If you are the part that initiated the connection then set time1=0 time2=inf and jump to step 11.
 Generate a random key (k) for a symmetric encryption.
 Read data (data) from user.
 Calculate E(data,k) and H(K) (where K is the last K packet received).
 Note down the current time (time1).
 Send a D packet (with the structure described above).
 Wait for an A packet.
 Note down the current time (time2).
 Send a K packet containing the H(A) and the key (k) you used in step 5.
 Wait for a D packet and store the encrypted data (data) it contains.
 Note down the current time (time3).
 If (time2time1)*2<=(time3time2) then either there is a MITM or you connection is really unstable.
 Send an A packet containing H(K).
 Wait for a K packet and store the key (k) it contains.
 Show DE(data,k) to the user.
 Jump to step 3.
Popular Posts

The other day, while browsing Wikipedia, I wondered how hard it must be to have weird fetishes without being able to share them with someone...

It seems that the most popular article on this blog is " How to make portable applications " written back in 2007. Reading it aga...

This article probably also applies to: VGNFZ38M Yeap I bought it. I though that since it had an NVidia card and it was Centrino based etc i...