This course will introduce you to the foundations of modern cryptography, with an eye toward practical applications.

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From the course by University of Maryland, College Park

Cryptography

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University of Maryland, College Park

348 ratings

Course 3 of 5 in the Specialization Cybersecurity

This course will introduce you to the foundations of modern cryptography, with an eye toward practical applications.

From the lesson

Week 2

Computational Secrecy and Principles of Modern Cryptography

- Jonathan KatzProfessor, University of Maryland, and Director, Maryland Cybersecurity Center

Maryland Cybersecurity Center

In the last few lectures we've explored the one-time pad encryption scheme.

Â And shown that it achieves our definition of perfect secrecy.

Â And indeed, the one-time pad has been used in the real world.

Â A famous example is the red phone that connected Washington DC and

Â Moscow in the 1980s, where the keying material was shared by trusted couriers,

Â who would take suitcases full of printouts of bits, as it were and

Â carry them between the two parties.

Â Nevertheless the one-time pad encryption scheme is not used very often nowadays.

Â Why is that?

Â Well in fact there are several limitations that prevent more widespread use of

Â the one time pad.

Â One of these which you may have noticed as we've described the scheme.

Â Is that the key is as long as the message.

Â Remember that in the description of the one-time pad encryption scheme.

Â We had a parameter n that determined both the length of

Â the messages that could be encrypted.

Â As well as the length of the key.

Â Moreover the one-time pad encryption scheme

Â is only secure if each key is used to encrypt a single message.

Â Certainly the proof of perfect secrecy that we gave for

Â the one-time pad relied on the assumption that the key was being used to encrypt

Â only a single message, and we'll see in a few moments that in fact there are some

Â simple attacks on the scheme in case a key is used to encrypt more than one message.

Â What this means is that if you want to use the one-time pad encryption scheme

Â to secure your communication, then parties need to share keys whose total

Â length is equal to the total length of all messages they might ever want to send.

Â That's because each portion of the shared key can be used only once, and

Â must be used to encrypt a message whose length is equal to the length of

Â that portion of the key.

Â It's even worse if the parties don't know in advance how much they're going to

Â want to communicate.

Â In that case they would need to share a total key.

Â Or a key rather, whose total length is at least an upper bound on the total length

Â of all messages they want to send.

Â I want to look in more detail at the problems that might arise in

Â case the one-time pad encryption scheme is used to encrypt two

Â messages using the same key.

Â So let's say we have two parties who share a key k.

Â And use this key to encrypt two messages, m1 and m2.

Â An attacker who observes the two ciphertexts, phi 1 and phi 2, can compute

Â the XOR of those two ciphertexts, and we see by performing the substitution,

Â that that XOR is equal to the XOR of m1 and m2.

Â Right c1 is equal to kx or m1.

Â C2 is equal to KX or M2.

Â And if we if the attacker XOR is C1 and C2 together the key

Â K cancels from the resulting expression and the attacker learns m1 XOR m2.

Â This is information about m1 and m2 jointly, right?

Â It might not be information about m1 or

Â m2 individually, but it does represent information about m1 and m2 jointly.

Â And that's something that should, that would be ruled out by any definition of

Â perfect secrecy for encrypting two messages using the same key.

Â Now, m1 and m1 XOR m2 is indeed some information about m1 and m2.

Â But I want to stress that this is not

Â just a technical point that we've leaked some information but in fact,

Â it can represent some significant information to an attacker.

Â Let me give a couple of examples.

Â So first of all the excor of m1 and

Â m2 reveals exactly where the messages m1 and m2 differ, right?

Â On any positions where m1 and m2 are the same their XOR will be 0.

Â And on any positions where m1 and m2 differ, their XOR will be 1.

Â So the XOR of the ciphertext c1 and c2 which is equal to the XOR of m1 and

Â m2 reveals to the attacker precisely those positions where m1 and

Â m2 differ and that can be a significant piece of information.

Â In addition, it turns out that some former frequency analysis can be applied,

Â even to the XOR of two English language messages.

Â All right. We saw in

Â an earlier lecture that we can take standard letter frequencies of,

Â of standard English text and use that to help cryptanalyze.

Â Encryptions using the vigenere cipher, or the shift cipher, and the same

Â underlying principle can be used when all you have is the XOR of, two messages.

Â It's a little bit more difficult.

Â A lot more difficult,

Â actually, because XORing two plain, two English language messages tends to

Â smooth out the distribution on the possible values of the XOR.

Â Nevertheless, if the plain texts are long enough, it turns out this information can

Â be used to derive lots of information about the underlying plain text m1 and m2.

Â Finally.

Â It turns out that, by exploiting some particular characteristics of

Â the ASCII representation.

Â The XOR of m1 and m2 can reveal lots more information.

Â I want to go through that in more detail next.

Â So here's the ASCII table that we've seen in a previous lecture, and

Â I've highlighted that portion of the table that illustrates the correspondence

Â between the letters of the English alphabet, both upper and

Â lower case, and their askew equivalent.

Â If you look at this table, what you can notice is that all letters

Â begin with 01, or rather the ASCII representation of all letters,

Â whether they're upper or lower case begin with the two bits 01, right?

Â Every letter corresponds to a single bit, to a single byte, rather, or eight bits.

Â And if you look at the eight bits that correspond to every letter,

Â you'll see that they all share the prefix 01.

Â In contrast to that, the space character begins with the prefix 00.

Â What this means is that if we have the XOR of any two letters, we get some byte.

Â With prefix 00.

Â But, if we have the XOR of a letter, and

Â a space character, then we get the prefix 01.

Â And, this allows an attacker to identify the XOR of a letter and a space.

Â Right, if the attacker has the XOR of two messages it can look byte by byte.

Â And it can guess that any byte with a prefix of

Â 01 corresponds to the XOR of a letter and a space.

Â Now this is not perfect, right it's possible to have two

Â space characters that XOR to a byte with prefix 00.

Â It's possible to have punctuation characters as

Â well that might mess this up.

Â Nevertheless, punctuation characters and pairs of spaces are expected to be

Â much less frequent than pairs of two letters and pairs of a letter and a space.

Â Just to go through this in pictures to illustrate the point more clearly.

Â So here we have, for example, two ciphertexts.

Â That were both encrypted using the one time pad and the same key.

Â And I just highlighted here the initial two bit prefix of every byte.

Â So the rectangles they represent bites, and

Â the bits that are displayed there are the two bit prefix of each bite.

Â If the attacker XOR's these two cypher text together.

Â Then it obtains what, what I've displayed on the bottom here.

Â And what you can see, right,

Â is that the two bit prefix of the first three bytes is 00.

Â And the two bit prefix of the fourth byte is 01.

Â So the attacker can guess here that the first three bytes of each of these,

Â the underlying plain text corresponding to these two initial ciphertexts.

Â Are letters.

Â But in the fourth position we have a pairing of

Â a space character, and a letter.

Â Right, the, again, the attacker doesn't know whether the first plain-text has

Â the letter and the second plain-text has the space or vice versa.

Â But it can guess, with high confidence, that one of

Â those blue shaded bytes corresponds to a letter in the underlying plain-text.

Â And the other one corresponds to a space.

Â Now given that, the attacker can look at the full byte,

Â the full value of the fourth byte in the XOR of the two cipher texts, and it then

Â knows that that byte is equal to the XOR of a space character with some other.

Â Plain text character with, with the ASCII representation of some letter.

Â You can easily solve this in this particular example to

Â see that the letter is the letter p, the lower case letter p.

Â So again, that means that the attacker has now learned that one of the blue-shaded

Â regions corresponds to a space character in the underlying plain text.

Â And the other blue shaded region corresponds to a lower case letter p in

Â the underlying plain text.

Â And if you extend this, where you can imagine that the attacker can

Â learn lots of information about each of the underlying plain text.

Â Things get even worse if multiple plain texts are all encrypted using the same key

Â because then the attacker can apply the same analysis.

Â To each possible pair, of cipher-texts.

Â Now what we've said is that the one time pad has

Â these two key drawbacks first of all that the key is long as the message, and

Â secondly that the one time pad encryption scheme is only secure,

Â if a particular key is used to encrypt only a single plain text.

Â Now it turns out that these limitations or these drawbacks are not something specific

Â to the one time pad encryption scheme, but they're in fact inherent limitations for

Â any encryption scheme achieving perfect secrecy.

Â Right, we might have thought that the one time pad encryption scheme has these

Â drawbacks, but we can hope to do better by designing some

Â other perfectly secret encryption scheme that doesn't suffer from these drawbacks.

Â But that's not the case.

Â All right. These two drawbacks are inherent for

Â any scheme achieving perfect secrecy.

Â I want to prove this for the particular case of the key links.

Â That is, I want to show that the one-time pad is, in fact, optimal as far as

Â perfectly secret schemes go, when it comes to the length of the key.

Â And the theorem I want to prove is the following.

Â If we have an encryption scheme defined by the algorithm Gen, Enc, and

Â Dec, and with message space M.

Â And if that encryption scheme is perfectly secret, then it must be

Â the case that the size of the key space is at least the size of the message space.

Â And that means that if we represent keys and messages.

Â Using some fixed length bit strings.

Â That the length of the key.

Â Must be at least the length of the messages that can be encrypted.

Â It turns out that the intuition for the theorem is actually very simple.

Â The basic idea is that given any ciphertext,.

Â An attacker can simply try decrypting that ciphertext using every possible key in

Â the key space.

Â We've seen this before for the example of the shift cipher, where we mentioned that

Â given a ciphertext, an attacker can just decrypt using all 26 possible keys.

Â If the attacker does this, then it obtains a list.

Â Of possible messages that can correspond to that cypher text.

Â All right, the only messages that can possibly correspond to that cypher text,

Â are the ones that can be obtained by decrypting that cypher text,

Â under every possible key.

Â Now the point is that that list can contain at most,

Â the size of the key space number of messages.

Â Each time I decrypt with some key.

Â I get one message.

Â It's possible that two different keys map onto the same message, that's fine.

Â But the point is that the list of possible messages can have

Â a size at most the size of the key space.

Â Now if the size of the key space is smaller than the size of

Â the message space.

Â Then that means that there's some message not on that list.

Â If a message is not on the list, it means that that message could not possibly have

Â been the one that the parties encrypted and

Â commu, and wanted to communicate with each other.

Â And that's already a leakage of information to the attacker.

Â Let's turn now to the formal proof.

Â We're given some private key encryption scheme defined by algorithms Gen, Enc and

Â Dec, and with message space M.

Â Let's assume that the key space is smaller than the message space.

Â We'll argue that this implies that the scheme can not be perfectly secret.

Â So what we need to do to show that the scheme is not perfectly secret is to

Â show some distribution over the message base, some message m,

Â and some ciphertext c such that the probability that the message is m,

Â conditioned on observing the ciphertext c.

Â Is not equal to the a priori probability with which the message was equal to M.

Â So let's take the uniform distribution on the message space.

Â In fact the exact distribution is not really important for this proof.

Â What's important is just that every message in the message space can occur

Â with some non zero probability.

Â In any event, we'll take the uniform distribution for completeness.

Â Take any ciphertext c in the ciphertext space.

Â Now, consider the set M(c) which is just the decryption of

Â that ciphertext c using all possible keys in the key space.

Â That is we take the cypher text and we decrypt using every possible key.

Â And each of those goes into our set M(c).

Â The key observation that we had before was that

Â these are the only possible messages that could've yielded the ciphertext c.

Â Now the size of this set.

Â Which corresponds to the list we talked about on the when we

Â discussed the intuition for the proof.

Â The size of the set is at most equal to the size of

Â the key space which by assumption was strictly smaller than the message base.

Â Now, that means that there's some message in the message base.

Â Which is not in this set.

Â And therefore could not have been

Â the message that the parties were communicating.

Â So that means that the probability that the message was equal to

Â that message M which is not on not in that set.

Â Conditioned on observing the ciphertext c is 0.

Â It's simply not, possible.

Â For the parties to have been encrypting that message M and then obtain,

Â to obtain the ciphertext c.

Â Simply because there's no key, for which the encryption of m using that key,

Â would result in ciphertext c.

Â So again, the probability that the message was equal to M,

Â conditioned on observing the ciphertext c, is 0.

Â And that's different from the priority probability that the message was

Â equal to M which we said was non-0.

Â So this concludes the proof that the encryption scheme, having a key

Â space smaller than a message space, could not possibly be perfectly secret.

Â So where do things stand?

Â Well we've defined perfect secrecy,

Â we've also shown that the one-time pad achieves it.

Â And we've just now shown that the one-time pad is optimal,

Â at least as far as its key length is concerned.

Â Does this mean we're done?

Â Well, I hope not because we've got several weeks left in the class.

Â In fact, we're not done at all and things are just starting to get interesting.

Â What we'd like to do is to now relax the definition of perfect secrecy.

Â And obtain a definition which is still meaningful and

Â useful in practice, but allows us to circumvent these inherit draw backs of

Â perfect secrecy that we've discussed in this lecture.

Â And this is what will occupy us for the next few lectures.

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