Excellent..I liked they way the professor explained equivalence classes passionately. My target is to finish this course for the pleasure it gives. I have done my under graduation 20 years before
Hi, my recommendation for you is to NOT start with cryptography, but start with mathematics for computer science, networking or a low level language course (learn C++ 😅).
Loving the lecture and I really appreciate, that you explain what is meant by the special mathematical notations. In a lot of books an lectures it is just anticipated, that you know it, but how? Thank you for making learning easy.
This professor is on fire! I never had a professor with this much fervor and energy and brilliance, and I have well over 300 semester credits, mainly math, science, and computer science. This course is freakin' amazin' and free!@#$%
my professor uses mr. paar's book "Understanding Cryptography - A Textbook for Students and Practitioners" for our "Network Security and Cryptography" lesson, so I'm very much glad to watch him lessons on youtube, on the other hand, him explanation is far better than others did. thank you for sharing these series.
I'm watching this lectures and studying from your awesome book for my cybersecurity exam and i'm loving it. I just wanted to thank yo for making this fantastic lectures available on RU-vid. If you will ever read this comment i'd like to learn modular arithmetic deeply. Can you recommend some books for beginners? Thanks a lot!
Thank you, thank you, thank you!!! I was struggling so much with the modular arithmetic and why the number of keys in the affine cipher is 312. Now I understand. THANK YOU!
Clocks are actually infinite sets that just increment the day by 1 every time the base of 24 is satisfied. And we are just truncating the incrementing day value. A better example of a modular set would be the alphabet, since once you get to the end of the set, it doesn't increment a higher place value. If a clock is a modular set then it follows that the decimal system's 1s place is a modular set if you don't show the tens place.
This was great. Looking forward to watching the remainder (pun intended) of the series soon. Based on where I am now in my career, this is what I wish I'd studied at university.
A finite set is one which resets to 0 or some other start-value when a certain end-value is overreached. Programmers call it "overrun". In other words, it's a fixed number range. The complete range of ASCII codes 0 to 255, for instance, is a finite set too. If you add 1 + 255 you get 0 again. Same with time. After 2359 hrs comes 0000 hrs again and not 2400 or 2401. So "finite set" basically means "fixed number range". That's how a crypto developer like me calls it. And yes, you can envision it as a circle, a loop or even a ring.
@1:08:35 how did he get 2^-1 ≡ 5 mod 9? And @1:13:18 he uses German to describe a structure that uses all four operations + - * /. What is the English? Field Structure?
Quick tip, using the negative mod can be very useful, what time is 23:00? Notice 23 is one less than a multiple of 12, thus 23 ≡ -1 (mod 12) so you know 23 ≡ 11 (mod 12) thus it's 11pm. (Of course 23 - 12 is just as easy, but when numbers get larger negative modulo equivalents can make life easier)
"Since there are only 26 different keys (shift positions), one can easily launch a brute-force attack by trying to decrypt a given ciphertext with all possible 26 keys. If the resulting plaintext is readable text, you have found the key." @1:20:40 Sure, from an attacker point of view there is a 26 sized keyspace, only after 26 iterations of shifting will you have %100 certainty to see the decrypted version of the ciphertext But is this true for choosing a key? Because you have 26 minus 1, so 25 possible 'shifts'. Maybe purely from a mathematical point of view shifting by 26 would be "correct", but in practise the result would be the same plaintext again, so is the 'keyspace' always defined from an attacker point of view?
If the identity key doesn't count for choosing a key, then why would it count for an attacker? The attacker only needs to perform the 25 shifts that would actually be chosen to get a 100% certainty.
I love this. I've always wanted a more in depth look into cyphers. Is this a university where students to learn english while learning other subjects? Ich moechte wissen wo und was diese ist.
Thanks for your interest. We are just a plain German university (and are, of course, public :)). Teaching is mostly in German, at least the B.Sc. courses. We have every year several foreign exchange students, that's why I had taught this one year in English. The approximately 150 students in the room are 1st year students and since most of them are German, I explain some tricky questions in German. Regards, christof
Danke für ihre beantwörte! Ich studierte Computer Ingenieurwesen als hauptfach und natürlich Deutsch als nebenfach an der Uni aber möchte immer lerner. Ich wünsche ihnen nur das beste.
are there learning algorithms associated with cryptography? ie can a crypher be taught to defend it self from attacks?? an AI crypher would be a great thing.
Note that for the affine cipher all arithmetic is done mod 26. That means that 27 == 1 mod 26 and, thus, the value of "a" must be from (0,1, ..., 25). hope this helps, christof
if A = m*q + r, and we have 15*17 = X mod 6, we're just replacing (m1*q1+r1)*(m2*q2+r2) but every term having m is divisible by "6", then you're left with the remainders. In this case it could be (2*6+3)(2*6+5) and the only trouble is 3*5/6. so it's 3 mod 6 I guess the corollary is we can find remainders easier. It was a bit overly complicated from the video to me.
In your lecture you said that the key space for shift cipher is 26. But actually, it is 25 only. Because shift by 0 is the plain text itself. Your lectures are useful for me and I am trying to listen all the lectures by you. Thanks and regards.
Hannah JP, it's true that a shift of 0 results in plaintext, but it still in the key space. bartłomiej Jakubowski Yup. With the modulo operator, you can't get a result the same as the modulus. It *has* to be 26.
Hannah Jp Modulo is not for shifting here, only to stay within the range (alphabet). In this Cipher the shift comes from "K". PS: Now i got it, you are right.
1:19:35 (You subtract modulo 26). I am here to try to understand the modulo operator. So if you have the encrypted letter "b" which is 1, and you subtract the key 3, you will obviously get a negative number here: -2. Try dividing -2 with 26 to get the remainder and you will be surprised it is outside the alphabet set. The result is -2 for 1 - 3 mod 26. -2 is an integer still. It is just not the result you would expect. I am here to try to solve this by mathematics, and not by special exceptions such as, if the value before division is less than 26, add 26, just so we'll never go below 0...... I am not expecting this to be solved without special exceptions.
Hi Professor Christof May I ask please why the remainder was taken in to account on 43:00? The equation before that was totally understandable but why did we enquire about the remainder of diving 55 to 5? Many Thanks and thank you for continuously supporting us
+YOLOBOSS In this example we always look at results that we obtain if we compute modulo 5. Please have also a look at the first equation (2 lines above the 55 = 0 mod5 line). There we consider 200 modulo 5. Hope this helps. regards, christof
Thank you so much, that was great. How did you even decide to record the lessons? I thought it was a kind of secret. Or else one won't need to enter the university :D But don't stop, if you occasionally see my comment, please!
Hi Professor, Thank you for your video lectures. Can you please tell from where we can get access to the homework. Also I want to ask do you have lectures on other subjects too like discrete maths?
For problem sets, please visit www.crypto-textbook.com and go to online course -> videos. The solutions to the ODD NUMBERED problems are also on the website, look at "book". Sorry, but we are not allowed to release the other solutions. cheers, christof
Thanks for th lectures! I'm leraning a lot thatnks to you. Out f curiosity, what happens in a shift cipher when you chose a key that varies, for example 3+position of the letter in the alphabet?
+Elitios In principle, a varying key is a good idea. However, your proposal will not work. In your example A and N would both be mapped to the same ciphertext with the shift cipher. Assume k=1, the "A" would be encrypted as e(A) = 0 + 0 + 1 = 1 =B but also e(N) = 13 +13 + 1 = 27 = 1 = B mod 26 But again, a varying key is exactly what we need for a strong cipher, but it must be constructed wiith a more sophisticated rule. Cheers, christof
Maybe I missed it, but for those confused at 28:33 its because A=B(modC) is equivalent to AmodC = BmodC. Thats how he got 7 and -3 ...... 12= 7(mod5) --> 12mod5 = 7mod5 the remainder is 2 not 12
can pick any number basically to sqaure beacuse the set is inifinite so even if 4 was from the 5 equivalence class it is also form the 7 equivalence class so I can square it to get a smaller number which is easier to divide by 7?
41:26 - i'll be honest, that *did* actually blow my mind a little bit. Dr. Paar wasn't lying haha. [edit] okay the part with the negative numbers *definitely* blew my mind
i have a stupid question so would we replace 3 to the expondent of 8 with any number from the 7 equivalence classs and then see how many time 7 goes into that number to find the remainder
+Youcef Mahdadi Sure, we are using "Understanding Cryptography" by Jan Pelzl and myself. The video lectures follow the book very closely. You find more information about the book and sample chapters at www.crypto-textbook.com. You may also want to check out the reviews at Amazon.com. People seem to really like the book. Regards, Christof
also im confused with equaivalence class if I create an equivalence class of ( 7, 15,22, 29,3 etc) and of these numbers if i substitute them instead of 81 I will get a remainder of of then i would get 1 mode 7? someone please explain.
The lecture closely follows my textbook "Understanding Cryptography": www.amazon.de/Understanding-Cryptography-Textbook-Students-Practitioners/dp/3642041000 Please note that the 2nd edition should become available in appr. 2 months.
yes it require to do so , the final answer will be 16 ≡2 mod 7 . i think he didn't do the final step but in his book he show up the final answer as the answer which i have mentioned .
Hi Sir , I have one question. Let's consider ( 13 * 16 ) mod 5 Here i replace one with +ve and one with -ve number ( 8 * -4 ) mode 5 = -32 mod 5 Does this equal to 3 or 2 ?
