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From: crypt-comments@math.ncsu.edu
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Subject: Cryptography FAQ 04/10 - Mathematical Cryptology
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Date: 17 Apr 1993 00:00:20 -0400
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Summary: Part 4 of 10 of the sci.crypt FAQ, Mathematical Cryptology.
 Private-key systems. Types of attacks (brute-force, cyphertext, 
 plaintext, etc.). Mathematical formulation of cryptography. Security 
 of one-time pads.
X-Last-Updated: 1993/04/16

Archive-name: cryptography-faq/part04
Last-modified: 1993/4/15


FAQ for sci.crypt, part 4: Mathematical Cryptology

This is the fourth of ten parts of the sci.crypt FAQ. The parts are
mostly independent, but you should read the first part before the rest.
We don't have the time to send out missing parts by mail, so don't ask.
Notes such as ``[KAH67]'' refer to the reference list in the last part.

The sections of this FAQ are available via anonymous FTP to rtfm.mit.edu 
as /pub/usenet/news.answers/cryptography-faq/part[xx].  The Cryptography 
FAQ is posted to the newsgroups sci.crypt, sci.answers, and news.answers
every 21 days.


Contents:

* In mathematical terms, what is a private-key cryptosystem?
* What is an attack?
* What's the advantage of formulating all this mathematically?
* Why is the one-time pad secure?
* What's a ciphertext-only attack?
* What's a known-plaintext attack?
* What's a chosen-plaintext attack?
* In mathematical terms, what can you say about brute-force attacks?
* What's a key-guessing attack? What's entropy?


* In mathematical terms, what is a private-key cryptosystem?

  A private-key cryptosystem consists of an encryption system E and a
  decryption system D. The encryption system E is a collection of
  functions E_K, indexed by ``keys'' K, mapping some set of
  ``plaintexts'' P to some set of ``ciphertexts'' C. Similarly the
  decryption system D is a collection of functions D_K such that
  D_K(E_K(P)) = P for every plaintext P.  That is, succesful decryption
  of ciphertext into plaintext is accomplished using the same key
  (index) as was used for the corresponding encryption of plaintext
  into ciphertext.  Such systems, wherein the same key value is used to
  encrypt and decrypt, are also known as ``symmetric'' cryptoystems.

* What is an attack?

  In intuitive terms a (passive) attack on a cryptosystem is any method
  of starting with some information about plaintexts and their
  corresponding ciphertexts under some (unknown) key, and figuring out
  more information about the plaintexts. It's possible to state
  mathematically what this means. Here we go.

  Fix functions F, G, and H of n variables. Fix an encryption system E,
  and fix a distribution of plaintexts and keys.

  An attack on E using G assuming F giving H with probability p is an
  algorithm A with a pair f, g of inputs and one output h, such that
  there is probability p of computing h = H(P_1,...,P_n), if we have
  f = F(P_1,...,P_n) and g = G(E_K(P_1),...,E_K(P_n)). Note that this
  probability depends on the distribution of the vector (K,P_1,...,P_n).

  The attack is trivial (or ``pointless'') if there is probability at
  least p of computing h = H(P_1,...,P_n) if f = F(P_1,...,P_n) and
  g = G(C_1,...,C_n). Here C_1,...,C_n range uniformly over the possible
  ciphertexts, and have no particular relation to P_1,...,P_n. In other
  words, an attack is trivial if it doesn't actually use the encryptions
  E_K(P_1),...,E_K(P_n).

  An attack is called ``one-ciphertext'' if n = 1, ``two-ciphertext'' if
  n = 2, and so on.

* What's the advantage of formulating all this mathematically?

  In basic cryptology you can never prove that a cryptosystem is secure.
  Read part 3: we keep saying ``a strong cryptosystem must have this
  property, but having this property is no guarantee that a cryptosystem
  is strong!''

  In contrast, the purpose of mathematical cryptology is to precisely
  formulate and, if possible, prove the statement that a cryptosystem is
  strong. We say, for example, that a cryptosystem is secure against
  all (passive) attacks if any nontrivial attack against the system (as
  defined above) is too slow to be practical. If we can prove this
  statement then we have confidence that our cryptosystem will resist
  any (passive) cryptanalytic technique. If we can reduce this statement
  to some well-known unsolved problem then we still have confidence that
  the cryptosystem isn't easy to break.

  Other parts of cryptology are also amenable to mathematical
  definition. Again the point is to explicitly identify what assumptions
  we're making and prove that they produce the desired results. We can
  figure out what it means for a particular cryptosystem to be used
  properly: it just means that the assumptions are valid.

  The same methodology is useful for cryptanalysis too. The cryptanalyst
  can take advantage of incorrect assumptions. Often he can try to
  construct a proof of security for a system, see where the proof fails,
  and use these failures as the starting points for his analysis.
  
* Why is the one-time pad secure?

  By definition, the one-time pad is a cryptosystem where the
  plaintexts, ciphertexts, and keys are all strings (say byte strings)
  of some length m, and E_K(P) is just the sum (let's say the exclusive
  or) of K and P.

  It is easy to prove mathematically that there are _no_ nontrivial
  single-ciphertext attacks on the one-time pad, assuming a uniform
  distribution of keys. Note that we don't have to assume a uniform
  distribution of plaintexts. (Here's the proof: Let A be an attack,
  i.e., an algorithm taking two inputs f, g and producing one output h,
  with some probability p that h = H(P) whenever f = F(P) and
  g = G(E_K(P)) (i.e., g = G(K + P)). Then, because the distribution of
  K is uniform and independent of P, the distribution of K + P must also
  be uniform and independent of P. But also the distribution of C is
  uniform and independent of P. Hence there is probability exactly p
  that h = H(P) whenever f = F(P) and g = G(C), over all P and C. Thus
  a fortiori A is trivial.)

  On the other hand the one-time pad is _not_ secure if a key K is used
  for more than one plaintext: i.e., there are nontrivial
  multiple-ciphertext attacks. So to be properly used a key K must be
  thrown away after one encryption. The key is also called a ``pad'';
  this explains the name ``one-time pad.''

* What's a ciphertext-only attack?

  In the notation above, a ciphertext-only attack is one where F is
  constant. Given only some information G(E_K(P_1),...,E_K(P_n)) about
  n ciphertexts, the attack has to have some chance of producing some
  information H(P_1,...,P_n) about the plaintexts. The attack is trivial
  if it has just as good a chance of producing H(P_1,...,P_n) when given
  G(C_1,...,C_n) for random C_1,...,C_n.

  For example, say G(C) = C, and say H(P) is the first bit of P. We can
  easily write down an attack---the ``guessing attack,'' which simply
  guesses that H(P) is 1. This attack is trivial because it doesn't use
  the ciphertext: it has a fifty-fifty chance of guessing correctly no
  matter what. On the other hand there is an attack on RSA which
  produces one bit of information about P, with 100% success, using C.
  If it is fed a random C then the success rate drops to 50%. So this is
  a nontrivial attack.

* What's a known-plaintext attack?

  The classic known-plaintext attack has F(P_1,P_2) = P_1,
  G(C_1,C_2) = (C_1,C_2), and H(P_1,P_2) depending only on P_2.
  In other words, given two ciphertexts C_1 and C_2 and one decryption
  P_1, the known-plaintext attack should produce information about the
  other decryption P_2.

  Note that known-plaintext attacks are often defined in the literature
  as producing information about the key, but this is pointless: the
  cryptanalyst generally cares about the key only insofar as it lets him
  decrypt further messages.

* What's a chosen-plaintext attack?

  A chosen-plaintext attack is the first of an increasingly impractical
  series of _active_ attacks on a cryptosystem: attacks where the
  cryptanalyst feeds data to the encryptor. These attacks don't fit into
  our model of passive attacks explained above. Anyway, a
  chosen-plaintext attack lets the cryptanalyst choose a plaintext and
  look at the corresponding ciphertext, then repeat until he has figured
  out how to decrypt any message. More absurd examples of this sort of
  attack are the ``chosen-key attack'' and ``chosen-system attack.''

  A much more important form of active attack is a message corruption
  attack, where the attacker tries to change the ciphertext in such a
  way as to make a useful change in the plaintext.

  There are many easy ways to throw kinks into all of these attacks:
  for instance, automatically encrypting any plaintext P as
  T,E_K(h(T+R+P),R,P), where T is a time-key (sequence number) chosen anew
  for each message, R is a random number, and h is a one-way hash
  function. Here comma means concatenation and plus means exclusive-or.

* In mathematical terms, what can you say about brute-force attacks?

  Consider the following known-plaintext attack. We are given some
  plaintexts P_1,...,P_{n-1} and ciphertexts C_1,...,C_{n-1}. We're
  also given a ciphertext C_n. We run through every key K. When we find
  K such that E_K(P_i) = C_i for every i < n, we print D_K(C_n).

  If n is big enough that only one key works, this attack will succeed
  on valid inputs all the time, while it will produce correct results
  only once in a blue moon for random inputs. Thus this is a nontrivial
  attack. Its only problem is that it is very slow if there are many
  possible keys.

* What's a key-guessing attack? What's entropy?

  Say somebody is using the one-time pad---but isn't choosing keys
  randomly and uniformly from all m-bit messages, as he was supposed to
  for our security proof. In fact say he's known to prefer keys which
  are English words. Then a cryptanalyst can run through all English
  words as possible keys. This attack will often succeed, and it's much
  faster than a brute-force search of the entire keyspace.

  We can measure how bad a key distribution is by calculating its
  entropy. This number E is the number of ``real bits of information''
  of the key: a cryptanalyst will typically happen across the key within
  2^E guesses. E is defined as the sum of -p_K log_2 p_K, where p_K is
  the probability of key K.
