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 1

Introduction to Classical Cryptography

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

Maryland Cybersecurity Center

[SOUND] In this lecture, I wanted to go into

Â a little bit more detail about how it's possible to break the Vigenere cipher.

Â This lecture is really intended for

Â those of you who are going to be doing the first programming assignment, and so

Â I've actually tried to present exactly the necessary background you need.

Â In order to complete that assignment here.

Â Those of you who are not planning to do that assignment can feel free to skip

Â this video, although you may find it interesting as well,

Â even if you're not planning on doing that assignment.

Â Now when we introduced the VigenÃ¨re cipher,

Â we presented it in the following way.

Â We said that the key was a string of letters.

Â From the English alphabet and

Â the plaintext will also consist of letters from the English alphabet.

Â To encrypt a given plaintext with a key, what you do is simply shift every

Â character in the plaintext by the amount dictated by the next character of the key.

Â You can view this simple as addition modular 26,

Â because there are 26 letters in the English alphabet.

Â And if the key is sorted then the plaintext you simply wrap around in

Â the key as needed, and decryption just reverses the process.

Â So here was the example we gave, where we're encrypting the plain text,

Â tell him about me, using the key cafe, and so

Â you can view this as simply replicating the key cafe, as many times as necessary,

Â in order to complete to fill out the length of plaintext.

Â And then doing the appropriate addition, ie shifting,

Â modular 26 in order to get the cipher text as displayed her on the bottom line.

Â Now, I wanna present a variant of the Vigenere cipher, which is sort of

Â mathematically the same, although technically a little bit different.

Â And what we are gonna do here is we are gonna work with both plaintext and

Â keys viewed in terms of bytes rather than of letters over the English alphabet.

Â And so in particular what we are gonna do is we are gonna resume,

Â we start with an ASCII plaintext and we are gonna have a cipher text.

Â Which we're gonna represent in hex.

Â So ultimately, of course,

Â the plain text key and cipher text are all going to consist of bytes.

Â It's just that we're gonna view the plain text as nasty string and

Â it's easiest to view the key and cipher text as hex, digits or hex strings.

Â Now why would you want to do this?

Â Well, first of all, it turns out to be easier to implement.

Â So if you look at the code, actually for programming assignment one,

Â you'll see that it's fairly easy to implement this.

Â It's not that it's too much more difficult to implement the VigenÃ¨re cipher the way

Â we had just presented it, but it's slightly easier to do it this way.

Â I think more to the point, is that it's easier to use and the reason for

Â that is the VigenÃ¨re cipher as originally presented

Â was limited to plaintext that only consisted of lowercase English letters.

Â It did not support capital letters, it did not support punctuation,

Â it did not support spaces.

Â And if you were ever gonna do encryption and practice, you'd, of course,

Â like your encryption scheme to be able to handle arbitrary ASCII characters.

Â It turns out that what we're also gonna change is rather than doing addition,

Â modular 26, we're going to do byte-wise XOR.

Â So we're gonna take a byte of plaintext and a byte of the key, and

Â XOR them together to get the next byte of the ciphertext.

Â Now we could have done modular addition here too.

Â The number of possible bytes is 256, because a byte is of course eight bits,

Â and so we could have done addition, modular 256, and that would have been

Â maybe morally closer to the VigenÃ¨re cipher as we had originally presented it.

Â In fact, it doesn't really matter much from a mathematical point of view, and

Â it's easier to implement XOR,

Â it's faster to implement XOR than to implement modular 26 edition I suppose.

Â So let me give an example here.

Â This is just to specify exactly what's going on and

Â you can look in the code actually for programming assignment one.

Â I've given their code for encryption and

Â you can see how that lines up exactly with what I'm saying here.

Â So again we're gonna view the key as a string of bytes.

Â We are gonna view the plaintext as a string of ASCII characters, and

Â to encrypt, what we do is simply XOR, as I said,

Â the next character of the plaintext with the next character of the key.

Â And we wrap around in the key as needed, just like in the original Vigenere cipher.

Â And the decryption can reverse the process because XOR,

Â if invertible in the sense that if we take a plain text character,

Â XOR the next byte of key to get a cipher text character and then XOR the key

Â into the cipher text character, we'll recover the original plaintext character.

Â So here's a worked example.

Â Let's say our plaintext is the string "Hello!" so

Â this is a six character ASCII string.

Â Notice we have a capital letter here.

Â We have lower case letters here and we also have the exclamation point.

Â The key let's just take as the hex string.

Â A1 2F, so that's going to be a two byte key.

Â If we represent the plaintext "Hello!" as a sequence of hex characters and

Â you can do this just by looking in an ASCII table,

Â then you'll see that if we write it out in hex,

Â the plaintext is equal to hex 45 65 6C 6C 6F 21.

Â We're going to then XOR that with the key were again, the key is replicated as many

Â times as needed to fill out the length of the plain text.

Â So that means that in this particular case, we're going to XOR

Â the hex string on the previous line with the hex string A1 2F A1 2F.

Â Now it's easiest to see what's going on if you write everything out in terms of its

Â underlying binary representation.

Â So let's take the first pair of characters.

Â So we have the plain text character hex 48 and

Â the first character of the key, hex A1.

Â If we write those out in binary,

Â hex 48 is equal to 0100 1000 hex A1 is 10100001.

Â If you XOR those bit by bit you get the result 11101001.

Â And then if you convert that back into hex you get the hex string E9.

Â And if you do that for each character of the plaintext,

Â throwing it with the correspondent character of the key, you should find that

Â you get the cipher text, the six byte cipher text E9 4A CD 43 CE 0E.

Â Now how can we attack this variant Vigenere cipher?

Â And from now on, for the rest of this lecture,

Â we're going to assume that we're attacking this variant Rather than the original.

Â Well there are two steps involved.

Â So the first is determining the key length.

Â So, I didn't specify exactly how the key was chosen.

Â And in fact there are different ways you can imagine choosing the key.

Â But what I'm thinking of here, is that the key length is chosen uniformly in some

Â known interval and then after you fix the key length,

Â you then chose a uniform sequence of bytes of the specified length.

Â So when we're attacking the scheme,

Â we're gonna be given some cipher attacks, and we're gonna try to recover the key.

Â And we're gonna do that in two steps by A, first determining the length of the key,

Â how many bytes are in the key.

Â And then determining every byte of the key until we have the whole thing.

Â So let's go through these steps one by one.

Â When attacking the Vigenere cipher,

Â what we're going to make use of known plaintext letter frequencies for

Â the English language alphabet, we're gonna assume here that we're gonna

Â be given cipher text that results from encryption of English language test.

Â We've said already that if it's not the case,

Â if we're working with some other language.

Â Where the plaintext is from some other language, we can, of course,

Â do a corresponding attack, but just using different frequencies.

Â But here I'm assuming English,

Â though it won't matter in any of the mathematical equations that I show.

Â You can get these letter frequencies from Wikipedia.

Â You can also get them from calculating them yourself over some corpus.

Â But for our purposes it suffices to take the values that are given

Â on Wikipedia or in this chart.

Â So how do we determine the key length?

Â Well, first of all, let's let p i, for i ranging from 0 to 255,

Â be the frequency of the byte i in some English language text.

Â That is, p0 represents the frequency

Â of having the byte 0 in some English language plaintext.

Â Now of course, if you look at the ASCII table, you'll see that ASCII

Â characters corresponding to values less than 32 are non-printable characters.

Â And so if you're encrypting English language text,

Â you would have that p i is equal to 0 when i is less than 32.

Â And similarly, the value of p i for i greater than 127 is also 0.

Â The ASCII character with the value 127 is some other non-printable character,

Â and ASCII values are not even defined once you get above 128.

Â So the values of p i are going to be 0, the frequencies

Â of those bytes are going to be 0, when i is less than 32 or i is greater than 127.

Â As another example, if you look at the ASCII value corresponding

Â to the lowercase letter a, that's a 97.

Â And so p 97 is the frequency of the lowercase character a in

Â standard English texts, let's say.

Â And the main point here is, first of all, that this distribution can be calculated.

Â Here it wouldn't suffice to look at the distribution on Wikipedia

Â that only has English letters.

Â You would need also to take into account punctuation and

Â spaces, perhaps numerals, perhaps the difference between upper and lowercase.

Â But nevertheless you could calculate these from a known corpus.

Â And more to the point is that this distribution is very far from uniform.

Â As we've already seen, many of the p i's are 0.

Â Even the p i's that are non-0 are not uniformly distributed among those values.

Â An e is the most common letter, for example, and

Â a space would be relatively common as well.

Â And the key point we're gonna use in our attack is that if the length of the key

Â that was used to generate some ciphertext is N, then every Nth

Â character of the plaintext was encrypted using the same shift, or XOR in this case.

Â And so if we take every Nth character of the ciphertext and

Â then calculate frequencies of different bytes in every Nth character.

Â So we just imagine taking out every Nth character or

Â selecting out every Nth character of the ciphertext, and

Â then running your frequencies, your statistics over that.

Â Then what you expect to get are exactly the p i values, but

Â in some permuted order.

Â Because every time you have an e, let's say a lowercase e in the plaintext,

Â that's gonna get shifted by the same amount if you look at every Nth character

Â to some ciphertext character, some ciphertext byte.

Â And then you expect to see that corresponding byte appear with the same

Â frequency in the ciphertext, again, if you look at every Nth character.

Â So what you're doing is simply shuffling around the p i's, but

Â the values themselves should be replicated.

Â On the other hand, if you take every Mth character, where M is not a multiple of N,

Â then you're picking out characters that are all shifted by different key bytes.

Â And in that case, you would not expect the frequencies of those characters

Â to match up with the frequencies of the underlying plaintext.

Â And so, in particular, if you calculate the resulting frequencies

Â that you get from doing that, heuristically speaking, you

Â expect to get something that will be close to uniform or at least closer to uniform.

Â Because again, you're shifting your plaintext by different amounts

Â in this particular selection of ciphertext characters.

Â So the question of determining the key length boils down to distinguishing

Â the case where you have a bunch of frequencies that are permutations of

Â these known p i values,

Â versus the case where you've got a bunch of frequencies that are close to uniform.

Â There are various ways you could imagine distinguishing these two possibilities,

Â but the easiest is to do something like the following.

Â So say you've computed some candidate, some observed set of frequencies,

Â q 0 through q 255, where, again, q i here represents

Â the observed frequency of byte i in your selection of ciphertext characters.

Â Say, again, by picking out every Nth character of the ciphertext.

Â Calculate those frequencies and then just compute the summation of q i squared.

Â And let's see what we expect.

Â So if your q i values are close to uniform, i.e.,

Â each q i is about 1/256, well,

Â then the summation of q i squared is going to be the summation of 256 values that

Â are all approximately 1/256 squared, and that equals 1/256.

Â On the other hand, if the q i values are a permutation of the p i values,

Â then the summation of the q i squared will be about equal to, or exactly equal to

Â if they're exactly a permutation, the summation of the p i squared.

Â As I said earlier, you could compute the p i's from a known corpus, and

Â then compute what you expect to get for the sum of the p i's squared.

Â This would be somewhat annoying and time consuming to do, but

Â you certainly could do it.

Â But the key point, actually, is that this value,

Â the sum of the p i squared, will be significantly larger than 1/256.

Â And so this gives a way to determine the key length.

Â What you do is you simply try all possibilities for the key length, and

Â for every candidate value N, you pick out every Nth character of the ciphertext.

Â Compute the frequencies of each byte in that selection of characters.

Â Compute the summation of the frequency squared.

Â And then just look for

Â the maximum value among all the different choices of the key length.

Â And again, what you expect is that when you try the wrong value for the key

Â length, you should get approximately the summation of q i squared being 1/256.

Â And when you hit upon the right key length,

Â you expect to get something that's much larger than 1/256.

Â It will be close to the sum of p i squared, though again,

Â you don't really need to know what that value is.

Â You just need to look for the maximum.

Â So now that we have a way to get the length of the key,

Â we're left with determining each of the bytes of the key.

Â So let's assume we've done the first part correctly and

Â the key length N is now known.

Â Well, what we're gonna do, again,

Â is look at every Nth character of the ciphertext, pluck out every Nth character.

Â Note that we can do this starting with any character we want.

Â And in particular if we wanna figure out what the ith character or

Â byte of the key is, what we'll do is simply start at the ith character

Â of the ciphertext and then count every N characters from that and

Â pick out every Nth character starting from the ith character.

Â If we start from the first character and then get the first,

Â n plus first, two one plus first,etc., that will all correspond to a bunch of

Â ciphertext characters generated by shifting using the first byte of the key.

Â And so on, if we had started with the second character,

Â that would correspond to shifting or XORing with the second byte of the key.

Â And I'll call this the ith ciphertext stream.

Â So the sequence of every Nth character starting with the first will be the first

Â ciphertext stream, etc.

Â And again, the observation is that every byte in the stream was generated by XORing

Â some underlying plaintext with the same byte of the key.

Â What we'll then do is we'll take this resulting stream and

Â we'll try decrypting it using every possible byte value B.

Â So we'll try every possible value of that byte of the key ranging from 0 to 255.

Â And that will give us a candidate plaintext stream for

Â each possible value of that byte.

Â So we can imagine if we did them all at once, we would get 256 possible plaintext

Â streams that could result from decrypting the corresponding ciphertext stream.

Â Now these are all going to be gibberish because they don't correspond to adjacent

Â characters, they correspond to taking every nth character and if you took every

Â nth character of a plain text, you would get something which is gibberish.

Â However, you do still expect that the plain text frequencies are observed.

Â That is, even if you take every eleventh character of your plain text, you would

Â expect that the letter E will occur most frequently in that plaintext stream.

Â And so, when the guess B is correct, then you can deduce a couple of things.

Â So when you guess B for the shift value,

Â the value of that ith byte of the key is correct, then first one all,

Â every byte in the plaintext stream, that you've obtained,

Â should be a valid character that would appear in English text.

Â And, in particular, that means every byte in the plaintext stream

Â should take on a value between 32 and 127.

Â If it's outside that range,

Â then unless the plaintext came from something else that you didn't expect,

Â that would mean that your guess B could not possibly be correct,

Â because the English language text does not have any ASCII characters lower than 32 or

Â greater than 127.

Â Moreover, as we said earlier, the frequencies

Â of each of the letters would correspond to their frequencies in regular English text.

Â And so, in particular, if you just look at the frequencies of all the lower

Â case letters, say right as a fraction of all lower case letters over all so

Â the number of times that the lowercase E appears as a fraction of all lowercase

Â letters that should be close to known English letter frequencies.

Â And so what you can do is you can then try your guess B for the next byte of the key,

Â pluck out ever end character from your decipher text, everything by B

Â to get a candidate value for every nth character of plain text.

Â And then look and tabulate the observed frequencies for

Â each of the lowercase letters, say.

Â And we'll call these values qa through qv, so they just correspond,

Â those are the observed frequencies for the letter A among all lowercase letters,

Â the letter B among all lowercase letters,

Â up to the letter z over all lowercase letters.

Â And again when you guess B is correct, then you should find that

Â the summation of q i p i where PI are the known letter frequencies of the lower

Â case English letters should be equal to the sum of the PI squared.

Â And this is because you expect in this case QI to in fact be equal to PI right?

Â QI here is observed frequencies in a candid of plain text not in

Â the cipher text itself.

Â And that value is a known value you can calculate it for yourself or

Â you can take my word for it and use the value display here.

Â So again, what you expect is that when you hit upon the correct guess B for

Â the next byte of the key and then decrypt the Ith ciphertext

Â stream using that byte B, you should get a plain text stream, a sequence of bytes,

Â where first of all every byte in the stream is between 32 and 127.

Â Furthermore, the calculated frequencies of lower case letters in that stream

Â should give you some values, q a through q z,

Â such that the sum of q i p i is about .065.

Â In practice, you're not going to get exactly .065 and what you can do is simply

Â take the value of b that maximizes the value of summation q a p i.

Â Subject to this caveat of everything lying between 32 and 127 and

Â possibly others as well, other restrictions I mean as well.

Â So for example, if you happen to fee a huge amount of commas, that would be

Â a sign that you probably that you're guess for B is probably not correct.

Â And then you would try another one.

Â What's the time that we need to implement these attacks?

Â Well, let's say that we know that the key length is in the range of 1 to L.

Â So the determining the key length takes about time 256 times L.

Â These numbers are very roughly, they're just meant to be indicative.

Â There are different operations going on and so

Â this is not any kind of very specific measure.

Â But just in general, it's about 256 times L, because, for

Â each candidate value of the key length,

Â you're doing about 256 works to figure out all the observed frequencies, and

Â do the calculations, and then figure out which one is the maximum.

Â Once you've determined N, the actual length of the key.

Â Then, determining every byte of the key is done by

Â repeating N times the process of finding the next byte of the key.

Â And finding the next byte of the key takes about time 256 square because you're

Â enumerating overall 256 possible values for that byte.

Â And then, for each candidate value of a byte, you're doing about 256 work.

Â You're doing that.

Â That 256 squared work a total of N times.

Â N is at most L, so the work here is at most 256 squared times L,

Â and the point here is that it's linear in L.

Â So this really is not only polynomial time, but it's really quite efficient.

Â And it's instructive to compare this to the time it would require

Â to carry out a brute force key search.

Â So a brute force key search would require time about 256 to the L.

Â So the number of possible keys of length exactly L is 256 to the L,

Â if the key has L bytes in it and each byte takes on 256 values.

Â So we've gone here from the trivial brute-force key search, which takes

Â exponential time in L down to linear in L, just by being a bit clever in our attack.

Â Now I do want to point out a couple of things in practice,

Â now the first thing is that the attacks that I have described will become

Â more reliable as the ciphertext length grows larger, and the reason for

Â that is because when your ciphertext is larger, your plaintext is larger, and

Â when your plaintext is larger, you expect that the observed

Â letter frequencies in the plaintext will more closely correspond to

Â the letter frequencies tabulated on say Wikipedia.

Â And for very short ciphertexts and correspondingly short plain text, it's

Â possible that you have large deviations from those average letter frequencies.

Â But if you have a very short playing text of about 30 characters,

Â you might have no E's at all.

Â And then that would completely throw everything off.

Â But as you have hundreds of characters and

Â even thousands of characters, the attacks get better and better.

Â Now the attacks do still work for relatively short ciphertexts and

Â in particular I did it myself and the attacks I described do work

Â on the ciphertexts I've given you for homework assignment one.

Â However, the attacks don't quite work or may not quite work out of the box as I've

Â described them, and you may need to do a little bit of tweaking in the algorithms,

Â and a little bit of manual involvement might be needed.

Â You might have to even potentially guess.

Â You might get, for example, two different candidate bites for

Â the first bite of the key.

Â And then you have to look yourself and

Â see which one give you something which looks meaningful.

Â And if you have to do this in order to solve homework one,

Â you shouldn't feel you're doing something wrong,

Â it's perfectly fine to do that as long as you get the right answer in the end.

Â So, with that in mind, I'll just leave you to take

Â a look at the programming assignment and try your luck with it.

Â Please do feel free to use the discussion boards to ask questions and

Â to try to get help from both myself and other students in the class.

Â You shouldn't feel if you're stuck you should feel free to ask and

Â get yourself unstuck as quickly as you can.

Â And with that I just wish you good luck and

Â I do hope that you have fun with this assignment.

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