So it's a little after 1:30AM and I've after binge-watching whatever related video popped up next I've been slowly ground by the general terribleness of users on youtube. However, your comment kinda made my day and restored a little faith in humanity. So glad that there are people out there who want to learn, actively seek knowledge, thoughtfully contemplate new ideas, experiment with and apply that knowledge and just generally advance the arts and move the whole world along. Thank you!
I made the comment because it wasn't my usual reaction to a great TH-cam video. Normally, my brain goes: "That's cool! What's next?" This time my brain went: "FEED ME NOW!"
I would just like to express how much I love that this channel is not afraid of challenging its viewers. Thank you for making content for viewers with a nonzero attention span.
I studied discrete math in high school and regretted not being math or compsci major in uni. I understand about 40% of this. Shor’s algorithm was new to me. I knew Euclidean algorithm. Thank you for this! I will now study number theory for RSA to fully solidify my understanding! Thanks so much PBS giving me direction on where to study next.
OMG! I am so excited for the next episode. I am a computer scientist and have always wanted a good explanation of Shor's algorithm and how it runs on a quantum computer! Thanks PBS Infinite Series!
labobo quantum computing is rarely ever mentioned in computer science courses since it deals with quantum mechanics and most computers are built on boolean logic.
Johnathon Schultz What was shown was not Shor's algorithm, that is a quantum algorithm and much more complicated. Here you have to guess numbers for a repeatedly until you find a good result.
Quantum computers are good at fourier transformation . Fast fourier transformation is O (n*2**n). Quantum fourier transformation is O (n**2). if you can reduce your problem to a fourier transformation, then you go from exponential complexity to quadratic. Not all algorithms have a step reduceble to a FT so not all of them will enjoy the minimization on complexility to polinomial time.
That's when you have to check if 35 prime, since you will always find a factor less than square root of n is n is not prime. But that is not enough to factorise the number!
Episode suggestion: In RSA cryptography we create a key based on two large primes. You've discussed the difficulty in finding the factors of a large number; it seems even harder to demonstrate that a large number has no factors, i.e. is prime. Yet millions are produced every month. On my computer I can generate a X.509 certificate with its own key nearly instantly (key-pair actually, but irrelevant here), which means it has to find two large primes to form the key. I believe we start by randomly generating a large number, then advancing until we find one that's prime; do this twice, and we have two primes which are unlikely to match those for any other key generated. But during each progression, how do we determine that a large number is prime? I would love for you to discuss the numerical methods used to determine that a large number is prime. And when the process tells us a number is prime, how certain are we it is right? Are we 100% certain or is it "just" highly probable?
Jonathan Castello Digits are whatever symbol we decide. Q could represent a value in any system, and base 27 (or higher) need not necessarily use Q as a digit. Which makes for a better encryption anyway, if only you and the recipient know what the symbols mean. In simpleat form, it is a basic cypher which spells words by using a matched alphabet. a is b, b ia c, d is f Just like the cryptograms in newspapers next to the crosswords. My only point in posting is to say "never assume", and especially with cryptography.
It is scary how much I forgot, but watching this some remote memories popped up again. I never was really into this stuff back then, considered it not that interesting. Now I see so much more, but at the same time forgot even more.
I dont mean to be offtopic but does any of you know a trick to log back into an Instagram account? I was dumb lost my password. I would appreciate any help you can offer me!
Just trying something to speed up step 2 a tad. You only need to find one to have the other. In order to find the smaller of the 2 prime factors of N, you only have to check all the primes up to sqrt(N). One prime factor of N is less than sqrt(N), and the other one is greater. (Note: there is a possibility that the 2 prime factors are both sqrt(N), however, as we are dealing with encryption, this seems to me a very stupid N to use, so I don't mention it anymore) Now p=gcd(A1,N), q=gcd(A2,N) and p>sqrt(N), qsqrt(N) r>log(N)/log(a) (Note: if p is the smaller prime factor, the result is still valid as than A2>sqrt(N), which comes to the same conclusion) As step 2 is the longest step, probably because the periode r can become very large, it would be wise to take the value of a large. It's not a guaranty that r will be smaller, but now at least it is not *necessary* for r to be huge. Hope I didn't make any mistakes in my reasoning
I can't believe how much I love you, and this channel. More than PBSSpaceTime actually. I've read this attack in my cryptography class 4 years ago, and I was trying to understand Shor's algorithm for the last two years. Yet just watching this video made me understand it! Without even checking the Shor's paper again!
3:00 Actually, the modulo applies to both sides: "A ≡ X (mod N)" reads "A and X are equivalent, modulo N", i.e. "the remainder of A/N is equal to the remainder of X/N" (and not just equal to X), or (likewise and less ambiguous) "the difference A-X of both sides is a multiple of N". For example, "22 ≡ 7 (mod 5)" is true and implies that 5|22-7 (i.e. 22-7=15 is divisible by 5), but it does not imply that the remainder of 22/5 must be equal to 7.
This video travels the path of a rigorous proof with the casual nature of sightseeing. By watching it you are seeing more than you are aware of. Thank you for your thoughtfulness! PS, I love you!
Great Video. I went along just fine doing all the math in excel. The real clue for step two will be to find a where r is not bigger than X. That would be my approach. Example: If N=1537 and a = 30, then r = 4. Which is nice (r
This gave me flashbacks to my college differential equations class. The professor would stand at the front of the room and just make giant leaps in logic with almost no way to follow it
1:32 There's a slightly faster method. You can just check all the prime numbers before the square root of 35, since every divisor above the square root can be paired with one below it.
Great lecture! Reminds me of a class I took 6 years ago. Possible topics could include pseudo random number generators, and mathematical flaws in lotteries exploited. I'm not worrying about quantum computing destroying modern society since it was warned of a decade ago.
A minor correction: at 5:15 you should say "the smallest POSITIVE number r such that..." because when r=0 then x^r is always 1 mod N. This also makes it obvious why r is the period, because it is congruent to 1 at 0 and r, and nowhere in between.
Answer: Use AES... it does not rely on prime numbers at all. It’s quite hard to crack, because it’s addition is mod 2 which ends up being XOR and it’s hard to guess input bits if you know the result of an XOR.
at 3:06 a == x mod n doesn't mean that x is the remainder of a divided by n, but they do have the same remainder. so the right thing to say would be that n divides (a-x). so we could say in that logic that for all integers x y z n, x == y mod n implies that x == y + zn mod n. For example, 5 == 8 mod 3, but 8 is not the remainder of 5 divided by 3
Reddles37's binary approach to the last video is so cool it is almost cheeky. The consequences of quantum computing are fascinating to encryption and financing because prime numbers become exponentially less frequent as you count up, meaning there is less and less of them to use for encryption - it creates some hard real world dilema's that I think eclipses most issues commonly discussed today.
1:27 Actually, you only need to check the primes up to the square root of the number being factored; if there were any factors larger than that, then the other factors must all be smaller, and would have been discovered earlier, so you could divide them out and continue to get the full factorization.
I take a break from studying for security certs (specifically sections on EC Cryptography). Was not expecting Math to show up on random with youtube in the background. I'm done today, my brain just decided to turn off.
Can someone explain where does 3Mod10 sequence (3.46) comes from? What is the full operation? Do we divide 3 to the power of K by 10 and the sequence is the remainder?
Here are 2 mathematical expressions: 1. P= round(.25*(((3+(2*sqrt(2)))^n)+((3-(2*sqrt(2)))^n))^2)-2 (n=positive integer >=1) 2. Factors = sqrt(((1+T(sqrt(P)))^2)+P) +/- T(sqrt(P)) +/- 1 (P=composite integer,T=truncation operator) The 1st expression above generates Composite integers whose size (no. of digits) grows at an exponential rate with increasing n. The amazing thing about the 1st expression above is that it produces composites that is factorable by the 2nd expression above! Try it! but remember to set the precision for calculation high enough so that the 2 expressions above produce integers as output!
9:28 can anybody explain this part? why does p MUST divide one of the factors on the left? and q MUST divide another factor on the left? does this mean that a^(4/2) - 1 should be divisible by p without a reminder? and how do we know it to be true just by looking at the quasion?
I did follow the idea and figured out the arithmetic, but representation at 3:50 to 4:00 (3mod10 and 2mod7) is confusing. I thought the number in front of mod is the remainder, in this case should have been 3 and 2.
The solution to factoring any composite integer is equivalent to determining the non-trivial, positive real-value of a variable (k) I call the 'Coefficient of factorization' used in a unique, integer-factorization expression of mine which I've termed the 'Transformula'.
I wrote an encryption method for messages that quantum computers can’t break. It doesn’t use keys and doesn’t allow you to know exactly when a encoded message stops or starts in the output.
How many large prime numbers are there in the keyspace? With computers all around the world finding them, is a dictionary of large primes being created?
If N = p * q then you could create a rectangle using p as length and q as breadth and N as area using computer, you could extend the rectangle until area of rectangle is N and know p and q.
Great video as always, but sometimes the background music is just extremely loud and distracting. In the after credits it's even louder than Kelsey's voice.
Well yes and no, today most cryptography uses eliptic curves. You can also check all prime number up to sqrt(35) not 35 ;) Overall, good job. You dumbed RSA down enough so it can be understood without oversimplifying.
Kevin Leung Equals means two things are the exact same, while congruent means two things have the same remainder when divided by the modulus. In the relation 12 ≡ 2 (mod 5), 12 and 2 are not equal to each other (since they are different numbers). Rather, they both have the same remainder when divided by 5 (i.e., congruent modulo 5). In computer science, modulo is actually an operator, so the equals sign is used there (which can lead to confusion).
Ahh i remember this video i was at high school back then and now i am studying for my cryptography exam in university and still do not intuitively understand any number theory
Serious question: How can I cite your videos in academic discourse? What is your opinion, I mean. You gave me a -huge- idea... I don't want to not work on it because of this technicality.
This is a great video on fundamentals of math application. I hope they can also show-discuss some other aspects of math like Incompleteness Theorem, Halting problem, Impossibility Theorem, Shannon's Entropy, etc. &/or applications in card counting, poker/casino games, finanicial investments, index movements, voting systems/flaws, etc. so as to clarify the "myths" from the facts😁
This question is not about the video, but I just had a question regarding mathematics. Is there a limit to a number of different songs that can be created? Is there a limit to how many different notes that can be arranged at different frequencies to produce a musical piece?
Finding th eperiod is the key. How many Qbits quantum computer is needed to find factors for SHA256 encryption in six month time?. So we can rotate the keys before it reaches sixmonth and let them try for another six months. Unless new algorithm comes to find the factor faster.
Would a similar solution be possible if you took more than 2 large primes and multiplied them together to form the key? Essentially, if you used more than two large primes, could the hitch be a period-calculation which could leverage quantum computing capabilities?
when you explained modular counting I am confused because in one example you included 0 and in another you didn't. does modular counting include 0 or not?
I saw the "let" word before a variable name followed by a an equal sine and some value (let [variable name] = [some value]) in this video and I have seen it in other places too and my idea is that it is used to declare constants. Can someone please tell me what "let" stands for or/and why it was chosen to declare constants?
Can anyone help me with this…? For the function, y_n = x^(y_{n-1}), as n approaches infinity y_n approaches e for values of x ≤ e^{1/e}, but y_n approaches positive infinity where x > e^{1/e}. Can someone explain what this means? I randomly discovered it messing around with my calculator. I know it is just e saying hello but I don't quite understand it.
Well if x > e^{1/e}, it's pretty obvious you are powering bigger and bigger numbers so it's going to infinity, for second inequality read en.wikipedia.org/wiki/E_(mathematical_constant)#Exponential-like_functions
I am not sure that is correct. Just to check, is the sequence b_1 = a b_2 = a^a b_3 = a^(a^a) b_4 = a^(a^(a^a)), etc. If so, then I tried starting with a=1.4, and it seems to approach 1.8866633062433... Graphically, I think what is happening is as follows: Plot a graph of y=x and of y=a^x on the same axis. If these graphs intersect, then the sequence approaches the point of intersection. Doing some googling reveals that this is indeed the case, and it even has a name: `cobwebbing' or `cobweb diagrams'.
It's confusing because the video oversimplifies. a mod N is the remainder, but a ≡ x (mod N) means that a and x have the *same* remainders, not that x is the remainder (or that N divides a-x). Those are different things, so it makes sense to have different notations for them.
It is a bit confusing, but it's standard mathematical notation. It's better if you read `a = b mod m` as `a (= mod m) b`, that is, the modulus is part of the equals/congruence sign. In English, "a and b are congruent mod m".
Yes, but *relations* are much less common (hello, Prolog!), and what we're dealing with here is actually mathematical notation for a relation. The fact that there's a handy function corresponding to this relation is irrelevant to the fact that this notation has the relation in mind.
An analogy which might be useful: Saying `10 ≡ 19 mod 3' is notationally similar to saying `10.412 = 10.399 to 1 decimal place'. It turns out that the `mod' idea is very useful and has lots of nice properties, unlike the `to 1 decimal place' idea.
I think if the notation had been introduced in the video with one of the great explanations given in this thread, I would have found it much less confusing and distracting. I'd probably have to watch the video again now so I can focus on the steps instead of mentally pausing to unpack the (to me) weird mod notation.
I feel so dumb That's why I love this show. it was the same thing when I started watching space time. But I know, if I put in the effort, I'll be breezing through this soon.
Thank you for making a non-mathematician understand a complicated subject such as breaking cryptography 👍 Also is prime number product the only way to make reliable cryptography?
Grate video can't wait for next week. I might have misunderstood, but ... Are we really using N = p1 * p2 (where p1 and p2 are primes). I through we used N = p1^k1 * p2^k2 *... pn^kn. (all p's prime) (k - you can reuse primes) Can this method be generalized to work for N = p1 * p2 *... pn? even if you do not know n? Can you tell me all the p's that create 315?
Watch straight through.
Start over. Pause, think, pause, think.
Watch again. Pause, Google, wikipedia, pause, Google, mathworld.
Sleep.
Watch again. Write some Python.
Peek ahead at Shor's Algorithm.
Ah, yes good ol python. Get the anaconda bundle for all the imports you'll ever need
So it's a little after 1:30AM and I've after binge-watching whatever related video popped up next I've been slowly ground by the general terribleness of users on youtube. However, your comment kinda made my day and restored a little faith in humanity.
So glad that there are people out there who want to learn, actively seek knowledge, thoughtfully contemplate new ideas, experiment with and apply that knowledge and just generally advance the arts and move the whole world along. Thank you!
I made the comment because it wasn't my usual reaction to a great TH-cam video. Normally, my brain goes: "That's cool! What's next?" This time my brain went: "FEED ME NOW!"
???
Profit
hahhahahha this is me but 3 years after you said it
I would just like to express how much I love that this channel is not afraid of challenging its viewers. Thank you for making content for viewers with a nonzero attention span.
I studied discrete math in high school and regretted not being math or compsci major in uni. I understand about 40% of this. Shor’s algorithm was new to me. I knew Euclidean algorithm. Thank you for this! I will now study number theory for RSA to fully solidify my understanding! Thanks so much PBS giving me direction on where to study next.
OMG! I am so excited for the next episode. I am a computer scientist and have always wanted a good explanation of Shor's algorithm and how it runs on a quantum computer! Thanks PBS Infinite Series!
Johnathon Schultz shor you are
labobo Do you even know what your saying??
labobo quantum computing is rarely ever mentioned in computer science courses since it deals with quantum mechanics and most computers are built on boolean logic.
Johnathon Schultz What was shown was not Shor's algorithm, that is a quantum algorithm and much more complicated. Here you have to guess numbers for a repeatedly until you find a good result.
Quantum computers are good at fourier transformation .
Fast fourier transformation is O (n*2**n). Quantum fourier transformation is O (n**2).
if you can reduce your problem to a fourier transformation, then you go from exponential complexity to quadratic.
Not all algorithms have a step reduceble to a FT so not all of them will enjoy the minimization on complexility to polinomial time.
1:33 Actually, testing all the primes smaller then square root of 35 will do.
so only 2, 3, and 5
That's when you have to check if 35 prime, since you will always find a factor less than square root of n is n is not prime. But that is not enough to factorise the number!
Unless you divide n by the prime factor you found and repeat the procedure.
@@pierfrancescopeperoni No Hans is right. His argument is the following: If a prime N = p * q is present and p
as a former cryptographer for the army...
this was a good video review for RSA.
well done.
deanna smith '... We aren't as cool as the Marines.'
She is easily the best non-fiction presenter on youtube. Keep up the good work, PBS. And of course props to Kelsey.
I've learned more from this channel than I have so far in pre-calculus
Episode suggestion: In RSA cryptography we create a key based on two large primes. You've discussed the difficulty in finding the factors of a large number; it seems even harder to demonstrate that a large number has no factors, i.e. is prime. Yet millions are produced every month. On my computer I can generate a X.509 certificate with its own key nearly instantly (key-pair actually, but irrelevant here), which means it has to find two large primes to form the key. I believe we start by randomly generating a large number, then advancing until we find one that's prime; do this twice, and we have two primes which are unlikely to match those for any other key generated. But during each progression, how do we determine that a large number is prime? I would love for you to discuss the numerical methods used to determine that a large number is prime. And when the process tells us a number is prime, how certain are we it is right? Are we 100% certain or is it "just" highly probable?
0:44 'q' isn't a digit! (Unless we're using base 27...)
10 points for attention to detail. Wow!
Lol wow
So q is 26?
Yes, considering that was on screen for roughly q seconds.
Jonathan Castello Digits are whatever symbol we decide. Q could represent a value in any system, and base 27 (or higher) need not necessarily use Q as a digit.
Which makes for a better encryption anyway, if only you and the recipient know what the symbols mean. In simpleat form, it is a basic cypher which spells words by using a matched alphabet. a is b, b ia c, d is f
Just like the cryptograms in newspapers next to the crosswords.
My only point in posting is to say "never assume", and especially with cryptography.
It is scary how much I forgot, but watching this some remote memories popped up again. I never was really into this stuff back then, considered it not that interesting. Now I see so much more, but at the same time forgot even more.
How to factor huge numbers:
Step 1: Get a computer that will last a trillion years.
Step 2: Factor.
Answer: 42
I dont mean to be offtopic but does any of you know a trick to log back into an Instagram account?
I was dumb lost my password. I would appreciate any help you can offer me!
Just trying something to speed up step 2 a tad.
You only need to find one to have the other.
In order to find the smaller of the 2 prime factors of N, you only have to check all the primes up to sqrt(N).
One prime factor of N is less than sqrt(N), and the other one is greater.
(Note: there is a possibility that the 2 prime factors are both sqrt(N), however, as we are dealing with encryption, this seems to me a very stupid N to use, so I don't mention it anymore)
Now p=gcd(A1,N), q=gcd(A2,N) and p>sqrt(N), qsqrt(N)
r>log(N)/log(a)
(Note: if p is the smaller prime factor, the result is still valid as than A2>sqrt(N), which comes to the same conclusion)
As step 2 is the longest step, probably because the periode r can become very large, it would be wise to take the value of a large. It's not a guaranty that r will be smaller, but now at least it is not *necessary* for r to be huge.
Hope I didn't make any mistakes in my reasoning
I can't believe how much I love you, and this channel. More than PBSSpaceTime actually.
I've read this attack in my cryptography class 4 years ago, and I was trying to understand Shor's algorithm for the last two years. Yet just watching this video made me understand it! Without even checking the Shor's paper again!
Wow, I won! I completely forgot the t-shirts were a thing :).
I love the show by the way, keep up the great work!
Congrats mate (Y)
Though neither i understood the video nor the question. :(
Thanks so much for this video! I'm a first year computer science student and I am really struggling with discrete maths, and this video really helped.
Until 5:15 it was elementary math and i was happy. After that, hell of math broke loose and now i feel stupid :D Thx for awesome videos :)
One of the fav episode ever. Quantum Computing maths awaiting...
3:00 Actually, the modulo applies to both sides: "A ≡ X (mod N)" reads "A and X are equivalent, modulo N", i.e. "the remainder of A/N is equal to the remainder of X/N" (and not just equal to X), or (likewise and less ambiguous) "the difference A-X of both sides is a multiple of N". For example, "22 ≡ 7 (mod 5)" is true and implies that 5|22-7 (i.e. 22-7=15 is divisible by 5), but it does not imply that the remainder of 22/5 must be equal to 7.
Her: you should check these steps.
Me: **watches screen nervously**
Came here a few months later.. I am surprised that I understood it this time..
Now I can truly say that shor is a genius, and this video is great!
This video travels the path of a rigorous proof with the casual nature of sightseeing.
By watching it you are seeing more than you are aware of.
Thank you for your thoughtfulness!
PS, I love you!
I'm so glad this channel exists to provide such great explanations of things I thought were out of my reach.
Great Video. I went along just fine doing all the math in excel. The real clue for step two will be to find a where r is not bigger than X. That would be my approach. Example: If N=1537 and a = 30, then r = 4. Which is nice (r
madly in love with this channel....
Wow ethan, great moves! Keep it up, proud of you.
papa bless.
Tyler Driscoll what?
Ethan from h3
What has it got to do with this video though
Hes from the Channel H3H3
This gave me flashbacks to my college differential equations class. The professor would stand at the front of the room and just make giant leaps in logic with almost no way to follow it
1:32 There's a slightly faster method. You can just check all the prime numbers before the square root of 35, since every divisor above the square root can be paired with one below it.
Great lecture! Reminds me of a class I took 6 years ago. Possible topics could include pseudo random number generators, and mathematical flaws in lotteries exploited. I'm not worrying about quantum computing destroying modern society since it was warned of a decade ago.
Wow! You swept me off my feet! Your teaching is excellent, with unbeatable clarity. Thanks
Nice video number theory is my favourite
A minor correction: at 5:15 you should say "the smallest POSITIVE number r such that..." because when r=0 then x^r is always 1 mod N. This also makes it obvious why r is the period, because it is congruent to 1 at 0 and r, and nowhere in between.
I love the background music starting at 0:19. Can anybody tell me its name or where can I get it ?
Answer: Use AES... it does not rely on prime numbers at all. It’s quite hard to crack, because it’s addition is mod 2 which ends up being XOR and it’s hard to guess input bits if you know the result of an XOR.
Beyoond shweret. Tackling all of the obstacles. Just as it has been done for the e pie video. Great. 👍🏻👏🏻👏🏻👏🏻
at 3:06 a == x mod n doesn't mean that x is the remainder of a divided by n, but they do have the same remainder. so the right thing to say would be that n divides (a-x). so we could say in that logic that for all integers x y z n, x == y mod n implies that x == y + zn mod n.
For example, 5 == 8 mod 3, but 8 is not the remainder of 5 divided by 3
Why does the chair has a head? 12:35
And two hand to go with it! Spooky!
God I love this channel
*Shouts to space*
I LOVE THIS CHANNEL!
Explaining how Diffe-Hellman works is also interesting. RSA looks like a special case of it.
Thought I understood it then blinked. Will have to watch it again with pen and paper.
Yet another Infinite Series that is above my head. Smart people rule.
Can you please do a video with inaccessible cardinals and ordinals.
Wow.. I really have to update myself with maths. Good work girl.. Keep it up Bravo..
Reddles37's binary approach to the last video is so cool it is almost cheeky. The consequences of quantum computing are fascinating to encryption and financing because prime numbers become exponentially less frequent as you count up, meaning there is less and less of them to use for encryption - it creates some hard real world dilema's that I think eclipses most issues commonly discussed today.
I've just compleated a bunch of coursework on this exact topic - ciphers and encryption.
1:27 Actually, you only need to check the primes up to the square root of the number being factored; if there were any factors larger than that, then the other factors must all be smaller, and would have been discovered earlier, so you could divide them out and continue to get the full factorization.
plz continue making such good maths videos....really sparks my interest...
Dat moment when you try to do some serius math wilst beeing tired af and english is not your first language.
same here :)
That moment when you try to do some serious math(s) whilst being tired as f___ and English IS your first language!
2:30 : okay... seems a little complicated
2:32 : I'm sorry what?
amazing! loving u, cant wait for the quantum video, tnkx
I like this serie so much. There is nothing similar in my country (as far as I know) so thanks !
I take a break from studying for security certs (specifically sections on EC Cryptography). Was not expecting Math to show up on random with youtube in the background.
I'm done today, my brain just decided to turn off.
PubstarHero lol, computer people are dumb as rocks.
Love this into to shor's algorithm. The music was a little too loud during the comment reply.
Where can I get the "Let (epsilon) < 0" shirt?
Can someone explain where does 3Mod10 sequence (3.46) comes from? What is the full operation? Do we divide 3 to the power of K by 10 and the sequence is the remainder?
Please make a video on the Basel Problem, Euler's solution, and Apery's constant! 🙏🙏🙏
4:31 Should be "2^k mod 7" not "2 mod 7"
a mod x = b means if you divide a by x you get b. a congruent b mod x means a and b have the same remainder when dividing by x.
Here are 2 mathematical expressions:
1. P= round(.25*(((3+(2*sqrt(2)))^n)+((3-(2*sqrt(2)))^n))^2)-2 (n=positive integer >=1)
2. Factors = sqrt(((1+T(sqrt(P)))^2)+P) +/- T(sqrt(P)) +/- 1 (P=composite integer,T=truncation operator)
The 1st expression above generates Composite integers whose size (no. of digits) grows at an exponential rate with increasing n. The amazing thing about the 1st expression above is that it produces composites that is factorable by the 2nd expression above! Try it! but remember to set the precision for calculation high enough so that the 2 expressions above produce integers as output!
9:28
can anybody explain this part? why does p MUST divide one of the factors on the left? and q MUST divide another factor on the left? does this mean that a^(4/2) - 1 should be divisible by p without a reminder? and how do we know it to be true just by looking at the quasion?
annnndddd my brain melted. still better explanation than my profs.
Great explanation!!!
This is awesome that modular arithmetic is just what i need it.
I did follow the idea and figured out the arithmetic, but representation at 3:50 to 4:00 (3mod10 and 2mod7) is confusing. I thought the number in front of mod is the remainder, in this case should have been 3 and 2.
The solution to factoring any composite integer is equivalent to determining the non-trivial, positive real-value of a variable (k) I call the 'Coefficient of factorization' used in a unique, integer-factorization expression of mine which I've termed the 'Transformula'.
I wrote an encryption method for messages that quantum computers can’t break.
It doesn’t use keys and doesn’t allow you to know exactly when a encoded message stops or starts in the output.
An episode about finitist mathematics? It feels much more beautiful and rational (heh) than mathematics introducing infinites
Loved this episode so much! 👍 Kesley is brilliant!
Wow. I wish someone showed us real world applications for math when I was a student.
Speaking about encryption makes you look so attractive, and i imagine you must get this alot. But it's true.
How many large prime numbers are there in the keyspace? With computers all around the world finding them, is a dictionary of large primes being created?
If N = p * q then you could create a rectangle using p as length and q as breadth and N as area using computer, you could extend the rectangle until area of rectangle is N and know p and q.
Great video as always, but sometimes the background music is just extremely loud and distracting. In the after credits it's even louder than Kelsey's voice.
Any chance for a video on elliptic curves?
looking forward to the next episode! 🙂👍
Well yes and no, today most cryptography uses eliptic curves.
You can also check all prime number up to sqrt(35) not 35 ;)
Overall, good job. You dumbed RSA down enough so it can be understood without oversimplifying.
deadal nix Yes, and technically the general number field sieve takes even less steps than trial division, but we've only got 15 minutes man
:)
What if I want to spell-check prime numbers?
Question, is there a reason to use the congruent comparison when using mod rather than equals?
Kevin Leung Equals means two things are the exact same, while congruent means two things have the same remainder when divided by the modulus. In the relation 12 ≡ 2 (mod 5), 12 and 2 are not equal to each other (since they are different numbers). Rather, they both have the same remainder when divided by 5 (i.e., congruent modulo 5).
In computer science, modulo is actually an operator, so the equals sign is used there (which can lead to confusion).
Great explanation. And what a surprise to learn this is Shor's algorithm, which I first heard of in an episode of Big Bang Theory.
I use mod 86400 for keeping track of time. I'm hoping to go metric sometime soon, though.
Ahh i remember this video i was at high school back then and now i am studying for my cryptography exam in university and still do not intuitively understand any number theory
How do you do this for numbers with multiple prime factors?
Kelsey, what is the integral of one over the squared root of a log .(a x), DX How do you do that
Serious question: How can I cite your videos in academic discourse? What is your opinion, I mean. You gave me a -huge- idea... I don't want to not work on it because of this technicality.
This is a great video on fundamentals of math application. I hope they can also show-discuss some other aspects of math like Incompleteness Theorem, Halting problem, Impossibility Theorem, Shannon's Entropy, etc. &/or applications in card counting, poker/casino games, finanicial investments, index movements, voting systems/flaws, etc. so as to clarify the "myths" from the facts😁
This question is not about the video, but I just had a question regarding mathematics. Is there a limit to a number of different songs that can be created? Is there a limit to how many different notes that can be arranged at different frequencies to produce a musical piece?
If you're following the typical 12 note musical system, and let's say it's limited to 5 octaves, that means 60 options for each note (61 if you consider the lack of a note a note, which for this we will and then 62 for continuation of that note). If you then assume that all music can be cut up with standardized time signatures and what not, let's just simplify it down to to 1/32nd notes (a pretty standard highest-speed note). Let's forgot about odd time signatures, triplets, tempo, and then things line tone and the actual sound of the instrument, and let's even set the standard amount of time to 5:00. At 120 bpm across the board for every song in this scenario, that means 8 beat possibilities per second, or 2400 time slots for each of those 62 options, given our criteria. Using some basic combinatorics, it becomes clear that this is equal to 62^2400 ~=
NOTE:THIS IS LONG SO FEEL FREE TO SKIP TO BOTTOM
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Based on our limited criteria, this is how many options there are, but in reality there are exponentially more factors, leading to a number probably not conceivable by modern home computers. So it is in fact limited as long as there is a finite number of features, and each feature has a finite number of options, but we may as well just call it infinite. We could never understand that large of a number. And, it actually would become infinite if you had no maximum length of a song! As someone could hypothetically add on another 1/32nd note at the end (aka multiply the above number by 62 again) and create a distinctly, measurably different piece of music. And, to top it all of, this is assuming just one instrument! Multiply the end result of whatever equation you get times the number of instruments. So in the end, it comes down to how specific you want to be about infinite and whether you set a limit to the length of the song, but regardless, there is no chance that we will be running out of musical options in any human's lifetime ;)
Finding th eperiod is the key. How many Qbits quantum computer is needed to find factors for SHA256 encryption in six month time?. So we can rotate the keys before it reaches sixmonth and let them try for another six months. Unless new algorithm comes to find the factor faster.
Would a similar solution be possible if you took more than 2 large primes and multiplied them together to form the key? Essentially, if you used more than two large primes, could the hitch be a period-calculation which could leverage quantum computing capabilities?
when you explained modular counting I am confused because in one example you included 0 and in another you didn't. does modular counting include 0 or not?
I saw the "let" word before a variable name followed by a an equal sine and some value (let [variable name] = [some value]) in this video and I have seen it in other places too and my idea is that it is used to declare constants. Can someone please tell me what "let" stands for or/and why it was chosen to declare constants?
Love these videos. You should collaborate with numberphile
Is there a formula to calculate the period of a mod n?
This helped me understand cryptography a bit better!
Can anyone help me with this…? For the function, y_n = x^(y_{n-1}), as n approaches infinity y_n approaches e for values of x ≤ e^{1/e}, but y_n approaches positive infinity where x > e^{1/e}. Can someone explain what this means? I randomly discovered it messing around with my calculator. I know it is just e saying hello but I don't quite understand it.
Well if x > e^{1/e}, it's pretty obvious you are powering bigger and bigger numbers so it's going to infinity, for second inequality read en.wikipedia.org/wiki/E_(mathematical_constant)#Exponential-like_functions
I am not sure that is correct. Just to check, is the sequence
b_1 = a
b_2 = a^a
b_3 = a^(a^a)
b_4 = a^(a^(a^a)), etc.
If so, then I tried starting with a=1.4, and it seems to approach 1.8866633062433...
Graphically, I think what is happening is as follows:
Plot a graph of y=x and of y=a^x on the same axis. If these graphs intersect, then the sequence approaches the point of intersection. Doing some googling reveals that this is indeed the case, and it even has a name: `cobwebbing' or `cobweb diagrams'.
Can't wait to the next episode!
At its heart this algorithm exploits the efficiency of Euler's algo to find G.C.D.
around 4:38 you say the last number in the period in 1, but would that be the first number in the period? shouldn't the sequence start @ 2^0 = 1
No, because we start n with 1. Otherwise, the period would always be 0
This notation is so confusing because in computing 'mod' is a function. So you would say 8 mod 7 = 1, or 5 mod 3 = 2
It's confusing because the video oversimplifies. a mod N is the remainder, but a ≡ x (mod N) means that a and x have the *same* remainders, not that x is the remainder (or that N divides a-x). Those are different things, so it makes sense to have different notations for them.
It is a bit confusing, but it's standard mathematical notation. It's better if you read `a = b mod m` as `a (= mod m) b`, that is, the modulus is part of the equals/congruence sign. In English, "a and b are congruent mod m".
Yes, but *relations* are much less common (hello, Prolog!), and what we're dealing with here is actually mathematical notation for a relation. The fact that there's a handy function corresponding to this relation is irrelevant to the fact that this notation has the relation in mind.
An analogy which might be useful: Saying `10 ≡ 19 mod 3' is notationally similar to saying `10.412 = 10.399 to 1 decimal place'. It turns out that the `mod' idea is very useful and has lots of nice properties, unlike the `to 1 decimal place' idea.
I think if the notation had been introduced in the video with one of the great explanations given in this thread, I would have found it much less confusing and distracting. I'd probably have to watch the video again now so I can focus on the steps instead of mentally pausing to unpack the (to me) weird mod notation.
I feel so dumb
That's why I love this show. it was the same thing when I started watching space time. But I know, if I put in the effort, I'll be breezing through this soon.
Thank you for making a non-mathematician understand a complicated subject such as breaking cryptography 👍 Also is prime number product the only way to make reliable cryptography?
Grate video can't wait for next week.
I might have misunderstood, but ...
Are we really using N = p1 * p2 (where p1 and p2 are primes).
I through we used N = p1^k1 * p2^k2 *... pn^kn. (all p's prime)
(k - you can reuse primes)
Can this method be generalized to work for N = p1 * p2 *... pn?
even if you do not know n?
Can you tell me all the p's that create 315?
Yes we only used numbers with two prime factors. I guess the cryptography uses these?