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					        General Concepts

Players: Alice, Bob and Trudy.

How to communicate securely over an insecure medium?

Alice should be able to send a message to Bob That
Trudy can't understand or modify and
Bob is assured that Alice is the sender.
  Types of Attaches

Passive Attacks:
 The attacker eavesdrops and
  read/record messages in transit.

Active Attacks: The attacker may:
   Transmit new messages,
   Replay old essages,
   Modify/Delete messages on transit.
         Fundamental Tenet of Cryptography


 Iflots of smart people failed to solve a problem,
  then it probably won't be solved (soon).

 The time required to break a code should
  be longer than the time the encrypted data
  must remain secret.

 The   value of most data decreases overtime.
         Cryptographic System:
         Algorithm + Key

   It is perfectly OK to let everyone know the algorithm.
    Knowledge of the algorithm without the key does not
    help unmangle the information.

   Publishing the algorithm provides an enormous
    amount of free consulting to uncover weaknesses.
    Layers and Cryptography


   Application (e.g., PEM),
   Transport (e.g., SSL),
   Network     (e.g., IPsec).
      Trojan horse/virus/worm:
Malicious code written by bad guys.
Modern mail systems & Internet connectivity
  (Cable Modems/DSL) contribute to its spread.
Virus Checkers: looks for instruction sequences
  for known viruses
  and uses message digests for files.
Covert Channels :
Very low bandwidth
(e.g., 1 bit every 10 seconds),
but can be used to steal cryptographic keys.

Steganography:
Hide secret messages in other messages.
        Traditional use of
        cryptography:

plaintext >>>>>>>> ciphertext >>>>>>> plaintext

        (encryption)         (decryption)


cryptographer: invent clever secret codes.
cryptanalyst:   attempt to break these codes.
   Fundamental Tenet of Cryptography:

   If lots of smart people failed to solve a
    problem,
    then it probably won't be solved (soon).
    The time required to break a code should be longer
    than

    the time the encrypted data must remain secret.
    The value of most data decreases overtime.

 Cryptographic
  System: Algorithm + Key
 It is perfectly OK to let
  everyone know the algorithm
  because knowledge of the
  algorithm without the key does
  not help unmangle the
  information.
 Publishing the algorithm provides an
 enormous amount of free consulting to
 uncover weaknesses.
   Computational Difficulty: Example:
    combination lock
   Typically require 3 numbers between 1 and 40. If it takes

    10 seconds for a good guy,
    it would take 10*(40**3) seconds or about 1 week for the
    bad guy.By requiring 4 numbers, If it takes 13 seconds for

    the good guy,
    it would take 13*(40**4) seconds or about 1 year for the bad
    guy!In general, increasing the key length by 1 bit makes the
    good guy's job just a little bit harder,
    but makes the bad guy's job twice as hard!
   Example of Secret Codes:
   Caesar cipher: substitute each letter with another
    letter which is 3 letters away in the alphabet (with wrap
    around). E.g., dozen >>> grcho. Extension: Instead of 3
    use any number n between 1 and 25. E.g., for n=1, HAL
    >>> IBM.
   Monoalphabetic cipher: arbitrary map one
    letter to another.

    There are 26!=4*(10**26) possibilities.
    If each possibility takes 1 microsecond it would take 10
    trillion years to try all possibilities. However statistical
    analysis of language makes it much easier to break.
   Secret Key Cryptography (symmetric
    cryptography)


                (encryption)

    plaintext >>>>>>>>>ciphertext

                     |
                     key

                      |
    ciphertext   >>>>>>>> plaintext
                 (decryption)
   Can be used for:
     •     Transmission Over an Insecure Channel:                   An eavesdropper will only
           see unintelligible data.



     •     Secure Storage on Insecure Media:                 Forgetting the key makes the data
           irrevocably lost.


     •     Authentication:       Alice authenticating Bob:
           Alice                 Bob

   challenge:        r >>>>>>>  r
    response:         K{r}  <<<<<<<            K{r}

        - r     is a random number,
         - K{r} is the secret key encryption of r using shared key K.
   Public Key Cryptography (asymmetric cryptography)
   Each individual has two keys:
   private key (not revealed to anyone)
    public key (make it known to everyone)              (encryption)

    plaintext    >>>>>>>>>> ciphertext
                       |
                   public key

                private key

                   |
    ciphertext    >>>>>>>>> plaintext
                 (decryption)
   The reverse process is called digital signature:

                (signing)

    plaintext   >>>>>>>>> ciphertext
                    |
                 private key           public key

                   |
    ciphertext >>>>>>>> plaintext
              (verification)


   Public key cryptographic algorithms are orders of magnitude
    slower than the best known secret key cryptographic
    algorithms. Thus they normally used to established temporary
    shared secret key for use during a session.
   Uses of Public Key Cryptography:

   Transmission Over an Insecure
    Channel:            <>
                Alice                      Bob

                {K}eB    >>>>>>>>>     [K]dB

        K{mB}      >>>>>>>>>   K{mB}
        K{mA}      <<<<<<<<<   K{mA}
    Secure Storage on Insecure Media:
    Alice generates a random key K and save:




1.   F= K{File}
     KF= {K}
             eA
 To restore the file:




1. K= [KF]dA
     File = K{F}
 Authentication: Alice authenticating
    Bob:
                  Alice              Bo
    b
         challenge:       c={r

    }eB   >>>>>            c
          response:        r      <<<<<
           r = [c]dB
 Hash     Algorithms
    (also known as message
    digest/fingerprint, one-way functions)
    The hash of a message m, h=H(m) has
    the following properties:
   Given m, it is easy to compute h.
   Given h, it is hard to compute m.
   Given m, it is hard to find another m'
    such that H(m) = H(m').
   It is hard to find m1 and m2 such that
    H(m1) = H(m2).
   Uses of Hash Algorithms:
     •   MAC/MIC (Message Authentication/Integrity Code)
   Using Secret Key:


             Alice sends                     Bob receives
         m,h where h = H(m|K) >>          m,h , OK if h = H (m|K)
        -K is the shared secret between Alice and Bob
         Bob is sure that Alice sent the message, since she knows
    K.
       Bob can NOT prove to any one that Alice sent him
    message m, since he also knows K.
    •   Password Hashing:
   OS like UNIX stores the hash of
    passwords instead of storing the actual
    passwords.
    For each user U, there is a tuple <U, h>
    where h = H(P) is the hash
    of password P of user U.
    When a user U types a password, P, the
    OS compute H(P) and
    if it is equal to the saved value h in the
    tuple <U,h> the user is OK.
   The magic of XOR:
    A Simple XOR symmetric algorithm:                             (from   Bruce Shneier   textbook)


   0®0=0

    0®1=1

    1®0=1
    1®1=0

    Note that:


   a ®a=0
    a ® b ® b = a (since b ® b = 0)

   The following program is a very simple symmetric algorithm.

    (see /home/cs772/public_html/demos/xor )

    To encrypt, the plaintext P is XORed with a key K to produce a ciphertext C.

    To decrypt, the ciphertext C is XORed with a key K to produce a plaintext P.

    P®K=C
    C ® K = P (since (P ® K) ® K = P)
           Key Cryptography
       Secret
 General Block Encryption:
   Secret key cryptographic systems take a
    reasonable length key (e.g., 64 bits) and
    generate a one-one mapping that looks, to
    someone who does not know the key,
    completely random.
    I.e., any single bit change in the input result
    in a totally independent random number
    output.
   Types of transformation for k-bit blocks:


   Substitution:
    For small values of k, specify for each of the 2k possible
    values of the input, the k-bit output.


   Permutation:
    Specify for each of the I input bits, the output position to
    which it goes.
                           Hashes and Message Digests
   A hash or message digest, is a one-way function since it is not practical to
    reverse. A function is cryptographicaly secure if it is computationally infeasible to
    find:
     •   a message that has a given message digest.
     •   a different message with the same message digest.
     •   two messages that have the same message digest.
   Major Algorithms:

   Ron Rivest Message Digest MD-family      (MD2, MD4 and MD5):
    128-bit.
   NIST Secure Hash Algorithm SHA-1: 160-bit.
   They take an arbitrary-length string and map it to a fixed-length quantity that appears to
    be randomly chosen.
    For example, two inputs that differ by only one bit should have outputs that look
    like completely independently chosen random numbers. Ideally, the message digest
    function should be easy to compute.
    Like secret key algorithms. digest algorithms tends to be computed in rounds.
    The designers finds the smallest number of rounds necessary before the output passes
    various randomness tests and then add few more to be safe.
   Things to do with a Hash


   Authentication: Alice authenticating Bob:
      Alice             Bob
    challenge:     r >>>>>>>    r
     response:       d   <<<<<<< d=MD{K|r}
      - r     is a random number,
       - MD{K|r} is the message digest of K concatenated with r.
         Alice computes MD{K|r} and if equal d, then Bob must know K.

   Computing a MAC:         Using Secret Key K between Alice and Bob

              Alice sends               Bob receives
       m,d where d = MD(K|m) >> m,d , OK if d = MD (K|m)
   Encryption:
        Generating one-time pad:          Both Alice
    and Bob knows the shared secret K and generates:

         b1= MD(K)
          bi = MD(K|bi-1), i=2,3, ....         Alice
    sends                                Bob receives
                  ci = mi ® bi   >>        ci and computes
    mi= ci ® bi
       Public    Key Cryptography

      All secret key algorithms & hash
    algorithms do the same thing but public key
    algorithms look very different from each
    other.

   The thing that is common among all of them
    is that each participant has two keys, public
    and private, and most of them are based on
    modular arithmetic.
        Modular     Arithmetic
   x mod n is the remainder of x when
    divided by n. e.g., 8 mod 10 =
    8,     18 mod 10 = 8,       24 mod 10
    =4
   Multiplication:

   Example: multiplication mod 10
     8 x 8 = 4, 1 x 9 = 9 , 7 x 6 = 2

   Multiplication by 1, 3, 7 and 9 works as a cipher since
    it performs 1-1 mapping.
   Example: if k = 7, then 1987 is encrypted to 7369
   decryption is done by multiplying each digit by k-1 , the
    multiplicative inverse of k. A multiplicative inverse of k
    is the number to multiply by k to get 1.
   Example: if k = 7, then k-1 is 3 since 7x3 = 1
   In the above table (Fig. 6-2), each "1" is the
    intersection of k and k-1.
    Only the numbers {1,3,7,9} have multiplicative inverse
    mod 10.
   What is so special about the set {1,3,7,9}?
   These numbers are relatively prime to 10, i.e., they do not
    share with 10 any common factors other than 1.
    Note that 9 is not a prime number but it is relatively prime to
    10.
   How many numbers less than n are relatively prime to n?
    This quantity is referred to as Ø(n) and is called the totient
    function.
     o If n is prime:
   then {1,2, ..., n-1} are all relatively prime and
    thus Ø(n) = n-1.
     o If n = p.q where p and q are two distinct primes,
   then Ø(n) = (p-1)(q-1).

    Example: for n = 10 = 2.5, Ø(10) =(2-1).(5-1)=1.4=4,
    which is the set {1,3,7,9}.
 Exponentation:
 Example: exponentiation mod 10

     4 2 = 6, 8 8 = 6, 19 = 9 , 76 = 9
 An exponentiative inverse of e is the
  number d such that:
 e.d = 1 mod Ø(n)


    Example: For n= 10, Ø(10)=4:
   e=3 and d=7 are exponentiative inverses since 3.7=21= 1
    mod 4
   Encrypt/Decrypt:
   To encrypt m: compute c = me mod n
   To decrypt c: compute m = cd mod n
   Example:
   encrypt m = 8: c = 83 = 2
   decrypt c=2: m = 27 = 8
   Sign/Verify:
   To sign m: compute s = md mod n
   To verify s: compute m = se mod n
   Example:
   sign m = 8: s = 87 = 2
       verify s=2: m = 23 = 8
   In public cryptography:

   <e,   n> is public key & <d,n> is private key
   RSA Algorithm:

    generate public & private keys pair:
   1. choose two large primes p and q.
       (typically 256 bits each & keep them secret).
    2. compute n = p.q & Ø(n) = (p-1)(q-1).
       (it is very hard to factor n into p & q).
    3. choose a number e that is relatively prime to Ø(n).
    4. find a number d that is the multiplicative inverse of
       e mod Ø(n), i.e., e.d = 1 mod Ø(n).
    5. your public key: <e,n> & private key: <d,n>.
   encrypt/decrypt:
   To encrypt a message m (<n):
   c = me mod n
   & To decrypt c: m = cd mod n
   This works since:
    cd mod n = (me)d mod n
             = me.d mod n
              = m mod n // since e.d = 1 mod Ø(n)
              =m            // since m < n


    sign/verify:
   To sign a message m (<n):
   s = md mod n
   & To verify s: m = se mod n
    This also works since: se mod n = me.d mod n = m mod n = m

   Why is RSA Secure:

   Every one knows the public key: <e, n>.
    To find the private key <d,n> you need to know Ø(n)
    since
    e.d = 1 mod Ø(n).
    To know Ø(n) you need to p and q since Ø(n) = (p-
    1).(q-1).
    Thus to break RSA you should know how to factor n to
    find p and q.
    Factoring a big number like n is hard.
    (the best technique to factor 512 bit number will take
    30,000 MIPS-years!)
   Efficiency of RSA Operations:
   Exponentiation
   How to compute 12354 mod 678?
   1232 = 123.123 = 15129 = 213 mod 678
    1233 = 123.213 = 26199 = 435 mod 678
    1234 = 123.435 = 53505 = 621 mod 678
    ......
    12354 = ......          = 87 mod 678
   This requires 54 small number multiplications and 54 small number divisions.

   How to compute 12332 mod 678?
   1232 = 123.123 = 15129 = 213 mod 678
    1234 = 213.213 = 45369 = 621 mod 678
    1238 = 621.621 = 385641 = 537 mod 678
    12316 = 537.537 = 288369 = 219 mod 678
    12332 = 219.219 = 47961 = 501 mod 678
   This requires 5 multiplications and 5 divisions instead of 32.

    To efficiently compute 12354 :                54 is represented in binary as:
   1            1                      0             1          1          0
                    |             |              |        |      |
     (((( (1232)123                )2             )2123            )2123       )2


    This requires 8 multiplications and 8 divisions instead of 32.
    Each 1 requires two multipliactions and two divisions
    and each 0 requires one multipliaction and one division.
    Thus in the above we have three 1s and two 0s that yeilds 3.2+2.1=8
    (we ignore the leading 1).
   Another example: y14 , 14 is represented in binary as:

            1                 1                    1                      0
                           |            |                 |
        ((               ( y2) y            )2y               )2


   This requires 5 multiplication's and 5 divisions instead of 32.
The RSA keys:
 public key: <3|65537, n>   private
 key: <d , n>.


                           Diffie-Hellman
          Alice and Bob agree on:                                p (large
    prime) & g < p.

             Alice                                       Bob
   Pick SA (512-bit random number)             Pick SB (512-bit random
    number)
    Compute TA = ( gSA) mod p                   Compute TB = (gSB) mod p
                TA          >>>   <<<            TB
   Compute X = TB SA mod p                    Compute Y = TA SB mod p
   X is the same as Y! why?
          X=    TBSA = gSBSA
           Y=    TASB = gSASB
   No one can compute   g (SASB ) by knowing g (SA ) & g (SB )
                       Email Security Protocols:

                               PEM & S/MIME
   PEM (Privacy Enhanced Mail): Add encryption, authentication and integrity
    to ordinary text messages.


   MIME (Multipurpose Internet Mail Extensions): Is a standard for encoding
    arbitrary data in email (images, video, etc.).


   S/MIME:
    Incorporated many principles of PEM into MIME.

   1. MIC-CLEAR From: Alice

    To: Bob

    Subject: Colloquium

    Date: Tue Oct 26, 2005 -----BEGIN PRIVACY ENHANCED MESSAGE-----

    Proc-Type: 4, MIC-CLEAR

    Content-Type: RFC822

    Originator-ID-Asymmetric: <certificate ID>
    MIC-Info: RSA-MD5, RSA, <encoded MIC>

   Dear Bob:

    I would like to invite you to give a colloquium next Fall,

    if you accept, let us talk about the details.

    Alice
    -----END PRIVACY ENHANCED MESSAGE-----



   3. ENCRYPTED From: Alice

    To: Bob

    Subject: Colloquium

    Date: Tue Oct 26, 2005 -----BEGIN PRIVACY ENHANCED MESSAGE-----

    Proc-Type: 4, ENCRYPTED

    Content-Type: RFC822

    DEK-Info: DES-CBC, IV

    Originator-ID-Asymmetric: <Originator certificate ID>

    Key-Info: RSA, <encoded message key encrypted with originator public key>

    MIC-Info: RSA-MD5, RSA, <encoded encrypted MIC>

    Recipient-ID-Asymmetric: <Recipient certificate ID>
    Key-Info: RSA, <encoded message key encrypted with recipient public key>
   <encoded encrypted message using DES-CBC>

   -----END PRIVACY ENHANCED MESSAGE-----
 SSL/TLS                         Protocols
 SSL (Secure Socket Layer, developed by Netscape ) &
    TLS (Transport Layer Security, is an IETF standard) are almost the
    same.
    They run as a user-level processes on top of TCP/IP.
   The Basic Protocol:
   {========================================
    Alice                                                      Bob
   I want to talk, ciphers I support, Ra                       >
    <                         certificate, cipher I choose, Rb
    choose secret S, compute K= f (S,Ra,Rb):
    {S}Bob , {keyed hash of handshake msgs}                  >
                                            compute K= f(S,Ra,Rb):
    <                               {keyed hash of handshake msgs}
   <                             >
           data protected with keys derived from K
   =======================================}

 Keys:
 Alice chooses a random number               S, known as the pre-master
    secret.
   It is shuffled with Ra and Rb to produce a master secret K.
   Ra and Rb are 32 octets long, the first 4 are the UNIX time (seconds since
    Jan 1, 1970).
   This ensures that Rs are always different.
   The master secret is shuffled with the two Rs to produce six (6) keys:
   Three for each side for encryption, integrity, and IV.
    The three keys used for transmission are known as the write keys
    while the three used for receipt are known as the read keys
    Thus Alice's write keys are Bob's read keys and vice versa.
   To ensure that the keyed hash Alice sends is different from the keyed hash
    Bob sends,
   Alice include the string "CLNT" and the Bob include "SRVR" in the hash.
   Note that Alice has authenticated Bob, but Bob has no idea to whom he's
    talking
    In SSL it is optional for the server to authenticate the client, if he has a
    certificate.
    Normally the server authenticates the user using:
   <name, password>
   sent securely over the ssl connection.
                    Authentication Systems




   Password-based Authentication
   It's not who you know. It's what you know
   On-line Password attack:
    Easy to defend, limit and slow down the number of guesses.


   Off-line Password attack:
    Capture a quantity X derived from the password and take your time
    to guess (e.g., use a dictionary) the passwd that produces X.
   Address-based Authentication

    It's not what you know. It's where you
    are


    In Unix implementations:

   /etc/hosts.equiv: Contains a list of computers that

    have identical user accounts.
    allow users on these hosts to login (rsh) without
    providing passwords.
   Trusted Intermediaries

   If we have N nodes:
   If each nodes keeps N-1 secrets,
    then adding a new node involves adding N new secrets,
    one at each node.
    Clearly not practical for large N.

    KDC (Key Distribution Center):
      KDC knows N keys, one for each node.
    Adding a new node involves only adding one key at KDC.
    If Alice like to talk to Bob:   Alice          KDC       Bob

    Need to talk to Bob --->
                        random R

    R= KA[X] <---    X= KA{R}
                      Y= KB{R} ---> R= KB[Y]
   C1 = R{M1} ---------------------> M1 = R[C1]
    M2 = R[C2] <--------------------- C2 = R{M2}

    Disadvantages of KDC:
   If compromised, all Keys are compromised.
   Single point of failure
   Performance bottleneck.
   CA (Certificate Authority):

   Each node keeps its private key.
   The CA certifies (sign) that the public key belong to the
    node and everyone trust the CA that he checked this fact for
    each node.
   All public key certificates may be kept in one place or
    each node keeps its own certificate and presents it to
    whoever asks for it.
   Certifies expire after a reasonable period (e.g., 1 year) but
    can be revoked at any time and the CA periodically publish
    a CRL (certificate revocation list) that contains all the
    revoked certificates.
   Clients should check the latest CRL before trusting a
    certificate.

 Session         Key Establishment
   It is a good idea to generate a separate key
    for each session to use for
    encryption/decryiption of session data
    following the session authentication phase.
    Why?
    •   Keys a kind of "wear out" if used a lot! The
        availability of more cipher text, the more
        likely an intruder may find the key.
    •   Prevent replay and decryption of previously
        recorded message.
   Delegation
      It's not who you are. It's who you're working for
    Sometime it is necessary to have some entity act on your
    behave.
    One possible means of allowing this is to give your
    password to this entity. This is not usually a good idea
    (please never do that! oducsc).
   The best mechanism to achieve that is delegation
    (or authentication forwarding).
    Generate a special message, signed by you (using public
    key cryptography, or through the use of KDC), specifying:
   To whom you are delegating the rights,
   Which rights are being delegated &
   For how long.
 Passwords
        Problems:
    •   Eavesdropping.
    •   Read stored file.
    •   Easy to guess on-line.
    •   Easy to crack off-line.
    •   Users may write it down.
   On-Line Password Guessing
   Helpful Tips:
    •   Set limit on the number of trials.
    •   Process incorrect passwords
        s l o w ly
    •   Report to users of unsuccessful
        attempts.
    •   Assign users an easy to pronounce
        strings as passwords.
    •   Do not let users choose easy-to-
        guess passwords.
    •   Force users to change passwords
   Off-Line Password Guessing
   Obtaining a hash of a password h,
    an attacker can guess the password w and
    checks to see if h = MD (w).


    If some one obtains a file F containing the hashes of many passwords,
    e.g., /etc/passwd he can perform a dictionary attack :

   for each word w in dictionary D do
    compute h = MD (w)
   for each e in F do
     if e = h then w as a password
   done
   done

    The number of performed hashes is: |D|
   Storing a random number s (salt) with e = MD (w|s)
    makes it harder for a dictionary attack:

   for each entry <s, e> in F do
         for each word w in the dictionary D do

      compute h = MD (w|s)
      if e = h then w as a password
    done

   done The number of performed hashes is: |D|.|F|
   How long should a password be?

   To protect against on-line attack:
   short password is fine.
    E.g., ATM systems have 4 digits (10,000 different PIDs), it is
    OK
    since you only have 3 guesses before rejecting your card.
   To protect against off-line attack:
   64 bits of randomness makes the number of trials 264
    which is considered computationally hard:

   In decimal this is about 20 digits to remember.
   If we select random characters (from a 64 chars of upper case, lower case, digits,
    punctuations) we need 11 characters.
   If generate pronounceable passwords (case-insensitive and every third char is one of
    the 6 vowels) we need 16 characters.

   If we allow humans generated passwords, we need 32 characters.
 General Tips:
 Do not exchange passwords using
  email.
 Use different passwords on different
  systems or accounts.
 Change your password frequently.

 Abort Login Trojan Horses (e.g., type
  Alt-Ctrl-Del).
                          Mutual Authentication




   Shared Secret


   Protocol 7:
    {===============================
    Alice                              Bob I'm
    Alice                          >
    <                                  Rb
    f(K, Rb)                            >
    Ra                                   >
    <                          f(K, Ra)
    ===============================}
   Protocol 8:
    Reduce number of messages in Protocol by putting
    more than one item of information into each message:


    {================================
    Alice                               Bob I'm Alice,
    Ra                           >
    <                            Rb, f(K, Ra)
    f(K, Rb)                          >
    ================================}
   Pitfall 1: Reflection Attack

   Trudy can impersonate Alice to Bob by oppening a second connection to Bob (or to another sever that
    share the same secret with Alice):
   Session1: {=================================
    Trudy                            Bob
   I'm Alice, Ra                               >
    <                                Rb, f(K, Ra)
    suspend session 1......


    Session 2:

    {=================================

    Trudy                                 Bob I'm Alice, Rb                   >

    <                          Rb', f(K, Rb)

    abort session 2.......
    =================================}



   continue session 1......

    f(K, Rb)                                      > =================================}

   Pitfall 2: Passwod guessing
    Trudy mount an off-line password guessing
    attack:     {=====================================
    ===
   Trudy                                       Bob I'm
    Alice, Ra                                   >
    <                                   Rb, f(K, Ra)
    .........
    suspend session and use: Ra, and f(K,Ra) to guess K.
    =======================================}
   Protocol 10:    We can use time
    stamps to reduce the number of
    messages to two:


    {=================================
    Alice                            B
    ob I'm Alice, f(K,
    timestamp)              >
    <                    f(K,
    timestamp++)
    =================================}
Mediated Authentication


   The Basic Needham-Shroeder Protocol


   {======================================
   Alice             KDC                Bob

   N1, Alice wants Bob        >
    <              Ka {N1,"Bob", Kab, ticket to Bob},
                                              where ticket to Bob   = Kb {Kab, "Alice"}
   ticket to Bob, Kab{N2}                                          >
   <                                    Kab{N2--, N3}
   Kab {N3--}                                      >
   ======================================}
   N is a "nonce", a number that is used only once (e.g., a sequence numer, random
    number, timestamp).
   N1: to prevent Trudy from impersonating KDC and replaying old replies to Alice.
   N2 and N3 are challenges for mutual authentication.

   The Kerberos Authentication Protocol:
   It is based on Needham-Shroered protocol, but is much simpler since it
    is based on timestamp and the ticket includes expiration date.

    {=====================================
   Alice             KDC              Bob

   N1, Alice wants Bob         >
    <                     Ka{N1,"Bob", Kab, ticket to Bob},
                                           = Kb {Kab, "Alice", expiration time}
                                    where ticket to Bob
   ticket to Bob, Kab{timestamp}                 >
   <                             Kab{timestamp++}
   =====================================}


				
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