Chapter 12: Cryptography

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Chapter 12 Cryptography

• MAT 320 Spring 2008

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Cryptography Basic Ideas

• We want to encode information so that no one
  other than the intended recipient can decode it.
• Essentially we have two functions an encoding
  function E, and a decoding function, D.

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Two Functions

• If x is a message (and its not hard to express
  messages as numbers), then E(x) should be the
  encoded message
• Once the message is received, D(E(x)) is the
  original message x
• So E and D are inverse functions

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Options

• One option is to keep both functions secret
• The advantage of this method is that if
  unintended recipients do not know your functions,
  they should be unable to decode your message
• The disadvantage of this method is that the more
  people who know your functions, the less able you
  will be to keep them secret

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Personal Keys

• In a more advanced system, each person has their
  own functions.
• Anne has her functions EAnne and DAnne
• Bob has his functions EBob and DBob
• The problem with this method is that so far, Anne
  and Bob can only send messages to themselves, not
  to each other

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Public Keys

• To fix this problem, we make everyones E
  function public.
• So anyone can encode a message using anyone
  elses key.
• However, we need to keep the D functions private,
  or else our information could be stolen.

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How It Works

• Anne wants to send a message to Bob.
• Anne knows EBob, so she sends EBob(message) to
  Bob
• Only Bob knows DBob, so only Bob can compute
  DBob(EBob(message)) message

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Signatures

• Another advantage of a public-key is the ability
  to sign messages.
• Suppose that the bank receives an encoded message
  claiming to be from Anne.
• Anyone can send EBank(message) to the Bank.
• But only Anne can send EBank(DAnne(message))
• The bank knows DBank and EAnne, so they can
  decode the message by applying these functions

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Troubles

• The main problem that comes up with public-key
  cryptography is that we need to make sure that
  its very difficult to figure out how the D
  function works from knowing how the E function
  works.
• One method that accomplishes this is RSA
  cryptography.

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Intermission

• Lemma 12.1 (Limited Cancelling)
• Lemma 12.2 (Fermats Little Theorem)

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How RSA Works

• Let p and q be distinct primes. Let n pq.
• In practice, we let p and q be quite large, with
  hundreds of digits. It is difficult to factor
  large numbers, even by computer, and if someone
  were able to factor n, they would be able to
  break our code.
• Go ahead and choose primes now. For purposes
  that will become clear soon, make sure that n is
  at least 270,000.
• Again, in practice, n is much, much larger.

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How RSA Works, continued

• Let k (p 1)(q 1), and choose d so that (d,
  k) 1.
• Using Bezouts Theorem, find e so that de?? 1
  (mod k)
• The numbers e and n are made public, and the
  number d is kept private.
• The encoding function is E(x) xe mod n
• The decoding function is D(x) xd mod n

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Lets Try It

• Following the example on your handout, enter your
  numbers p, q, n, and k into Mathematica.
• Choose a number d so that (d, k) 1. You may
  have to try a few times to get a d that works.
• Once you find a value of d, use the ExtendedGCD
  command to find e so that ed ? 1 (mod k).
• If Mathematica gives you a negative value of e,
  add k to it (since were working mod k, this will
  be congruent)

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Sending Messages

• First we need to convert our message to numbers.
  Converting the entire message to a single number
  would make our calculations difficult, so instead
  we break it up into blocks.
• Using A 01, B 02, , Z 26, break your
  message into 3-letter blocks and convert them to
  numbers.
• Now you see why we needed to have n be at least
  270,000.
• Add extra zeros to the end of your message if it
  doesnt break up evenly into three-letter blocks.

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Encoding

• Now we are ready to plug these numbers into our
  encoding function.
• When x and d are large, computing xd, dividing it
  by n, and computing the remainder is very time
  consuming.
• However, there are many computational shortcuts
  Mathematica can use, including the PowerMod
  command.
• PowerModx,d,n computes xd mod n

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One More Proof

• In order to convince ourselves that RSA always
  works, we need to prove this theorem
• Theorem 12.3 (RSA Works!)Let p and q be distinct
  primes, and let n pq and k (p 1)(q 1).
  If d and e are chosen so that (d, k) 1 and ed ?
  1 (mod k), then for all integers x, xed ? x (mod
  n).