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!
@jasertio
7 жыл бұрын
Johnathon Schultz shor you are
@XxZigiixX
7 жыл бұрын
labobo Do you even know what your saying??
@durnsidh6483
7 жыл бұрын
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.
@durnsidh6483
7 жыл бұрын
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.
@marcelo55869
7 жыл бұрын
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.
@kevinexit9840
6 жыл бұрын
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.
@MD-pg1fh
7 жыл бұрын
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.
Ah, yes good ol python. Get the anaconda bundle for all the imports you'll ever need
@jpkjnn6733
7 жыл бұрын
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!
@flymypg
7 жыл бұрын
I made the comment because it wasn't my usual reaction to a great KZitem video. Normally, my brain goes: "That's cool! What's next?" This time my brain went: "FEED ME NOW!"
@moversti92
5 жыл бұрын
??? Profit
@charliewalker9148
4 жыл бұрын
hahhahahha this is me but 3 years after you said it
@haflam.
7 жыл бұрын
1:33 Actually, testing all the primes smaller then square root of 35 will do.
@YellowToad
4 жыл бұрын
so only 2, 3, and 5
@pierfrancescopeperoni
3 жыл бұрын
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!
@pierfrancescopeperoni
3 жыл бұрын
Unless you divide n by the prime factor you found and repeat the procedure.
@Schnorzel1337
3 жыл бұрын
@@pierfrancescopeperoni No Hans is right. His argument is the following: If a prime N = p * q is present and p
@deannasmith4443
7 жыл бұрын
as a former cryptographer for the army... this was a good video review for RSA. well done.
@Tadesan
6 жыл бұрын
deanna smith '... We aren't as cool as the Marines.'
@rydohg
7 жыл бұрын
I've learned more from this channel than I have so far in pre-calculus
@stephescobar575
6 жыл бұрын
She is easily the best non-fiction presenter on youtube. Keep up the good work, PBS. And of course props to Kelsey.
@andrasfogarasi5014
6 жыл бұрын
How to factor huge numbers: Step 1: Get a computer that will last a trillion years. Step 2: Factor.
@dank6617
5 жыл бұрын
Answer: 42
@carlkai9832
3 жыл бұрын
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!
@erikziak1249
7 жыл бұрын
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.
@13thxenos
7 жыл бұрын
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!
@scheimong
7 жыл бұрын
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.
@keithrobertson7579
7 жыл бұрын
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?
@Reddles37
7 жыл бұрын
Wow, I won! I completely forgot the t-shirts were a thing :). I love the show by the way, keep up the great work!
@xhitiz22
6 жыл бұрын
Congrats mate (Y) Though neither i understood the video nor the question. :(
@Quacky_Batak
7 жыл бұрын
One of the fav episode ever. Quantum Computing maths awaiting...
@blazingkin
7 жыл бұрын
I'm so glad this channel exists to provide such great explanations of things I thought were out of my reach.
@TheLinkr
7 жыл бұрын
Wow ethan, great moves! Keep it up, proud of you.
@anthonyn9096
7 жыл бұрын
papa bless.
@aniruddhdeshpande7319
7 жыл бұрын
Tyler Driscoll what?
@aniruddhdeshpande7319
7 жыл бұрын
Ethan from h3
@aniruddhdeshpande7319
7 жыл бұрын
What has it got to do with this video though
@highlewelt9471
7 жыл бұрын
Hes from the Channel H3H3
@judeleon8485
3 жыл бұрын
Wow! You swept me off my feet! Your teaching is excellent, with unbeatable clarity. Thanks
@numankaraaslan
7 жыл бұрын
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 :)
@Fran7842
7 жыл бұрын
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.
@enzogiannotta
7 жыл бұрын
Nice video number theory is my favourite
@sinithparanga2481
7 жыл бұрын
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
@OGUZHANKOSARMD
6 жыл бұрын
Beyoond shweret. Tackling all of the obstacles. Just as it has been done for the e pie video. Great. 👍🏻👏🏻👏🏻👏🏻
@JorgetePanete
7 жыл бұрын
Her: you should check these steps. Me: **watches screen nervously**
@Tadesan
6 жыл бұрын
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!
@Twisol
7 жыл бұрын
0:44 'q' isn't a digit! (Unless we're using base 27...)
@pbsinfiniteseries
7 жыл бұрын
10 points for attention to detail. Wow!
@MrZyroid
7 жыл бұрын
Lol wow
@undeadreaper3x
7 жыл бұрын
So q is 26?
@stumbling
7 жыл бұрын
Yes, considering that was on screen for roughly q seconds.
@snatchngrab8262
7 жыл бұрын
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.
@PubstarHero
7 жыл бұрын
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.
@Tadesan
6 жыл бұрын
PubstarHero lol, computer people are dumb as rocks.
@aliasgeranees8893
7 жыл бұрын
madly in love with this channel....
@alexanderreusens7633
7 жыл бұрын
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
@recklessroges
7 жыл бұрын
Love this into to shor's algorithm. The music was a little too loud during the comment reply.
@tylershepard4269
6 жыл бұрын
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.
@Youezor
7 жыл бұрын
I like this serie so much. There is nothing similar in my country (as far as I know) so thanks !
@nonameforyouokpeterrodney9051
6 жыл бұрын
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!
@shaylempert9994
6 жыл бұрын
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!
@KaiKunstmann
7 жыл бұрын
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.
@cem_kaya
2 жыл бұрын
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
@Cr42yguy
7 жыл бұрын
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.
@aliasgeranees8893
7 жыл бұрын
plz continue making such good maths videos....really sparks my interest...
@doodelay
7 жыл бұрын
God I love this channel *Shouts to space* I LOVE THIS CHANNEL!
@thisaccountisdead9060
7 жыл бұрын
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.
@vincent-olivierroch8217
7 жыл бұрын
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
@jacoblavin7598
5 жыл бұрын
2:30 : okay... seems a little complicated 2:32 : I'm sorry what?
@youteubakount4449
7 жыл бұрын
An episode about finitist mathematics? It feels much more beautiful and rational (heh) than mathematics introducing infinites
@ilmbrk6570
7 жыл бұрын
Dat moment when you try to do some serius math wilst beeing tired af and english is not your first language.
@carlosjavierpalacios6194
6 жыл бұрын
same here :)
@rickh3714
5 жыл бұрын
That moment when you try to do some serious math(s) whilst being tired as f___ and English IS your first language!
@YouTubist666
6 жыл бұрын
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.
@afsharalithegreatiranian9777
7 жыл бұрын
Great explanation!!!
@pablobragato7164
7 жыл бұрын
amazing! loving u, cant wait for the quantum video, tnkx
@SHA256HASH
7 жыл бұрын
Please make a video on the Basel Problem, Euler's solution, and Apery's constant! 🙏🙏🙏
@reristavi
6 жыл бұрын
Wow.. I really have to update myself with maths. Good work girl.. Keep it up Bravo..
@nonameforyouokpeterrodney9051
6 жыл бұрын
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'.
@yamansanghavi
7 жыл бұрын
I love the background music starting at 0:19. Can anybody tell me its name or where can I get it ?
@deepjoshi356
7 жыл бұрын
Explaining how Diffe-Hellman works is also interesting. RSA looks like a special case of it.
@Thaidory
7 жыл бұрын
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?
@ariel_haymarket
7 жыл бұрын
Where can I get the "Let (epsilon) < 0" shirt?
@jacobsiemons687
7 жыл бұрын
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.
@pairot01
7 жыл бұрын
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
@vorpalneko
7 жыл бұрын
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.
@Twisol
7 жыл бұрын
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".
@Twisol
7 жыл бұрын
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.
@TheManxLoiner
7 жыл бұрын
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.
@unvergebeneid
7 жыл бұрын
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.
7 жыл бұрын
This is awesome that modular arithmetic is just what i need it.
@markoneal9677
7 жыл бұрын
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
@kaustubhken
7 жыл бұрын
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.
@maiprue
Жыл бұрын
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.
@dumbledoor9293
6 жыл бұрын
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?
@deadalnix
7 жыл бұрын
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.
@JM-us3fr
7 жыл бұрын
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
@deadalnix
7 жыл бұрын
:)
@paulthompson9668
7 жыл бұрын
What if I want to spell-check prime numbers?
@QuackersForMath
7 жыл бұрын
I've just compleated a bunch of coursework on this exact topic - ciphers and encryption.
@noone7219
7 жыл бұрын
I was thinking instead of factoring the primes every time can't we hold a list of all the numbers N and their respective primes p and q. And we see the number and look it up in this list to find its primes.
@javahaxxor
Жыл бұрын
Wow. I wish someone showed us real world applications for math when I was a student.
@juliengrijalva8606
6 жыл бұрын
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.
@DavidKennyNZL
7 жыл бұрын
Thought I understood it then blinked. Will have to watch it again with pen and paper.
@bitcoindaddy748
3 жыл бұрын
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?
@iannjan
7 жыл бұрын
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.
@shortcutDJ
7 жыл бұрын
Speaking about encryption makes you look so attractive, and i imagine you must get this alot. But it's true.
@trava4156
3 жыл бұрын
annnndddd my brain melted. still better explanation than my profs.
@xenontesla122
7 жыл бұрын
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.
@SFKelvin
6 жыл бұрын
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?
@georgeandrews2839
5 жыл бұрын
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.
@KnakuanaRka
5 жыл бұрын
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.
@Mach1Greeble
7 жыл бұрын
Yet another Infinite Series that is above my head. Smart people rule.
@deine1mama92
7 жыл бұрын
Why does the chair has a head? 12:35
@ZardoDhieldor
7 жыл бұрын
And two hand to go with it! Spooky!
@michaelyadchuk4989
7 жыл бұрын
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?
@jh-wq5qn
7 жыл бұрын
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 _________________________________________________________549 610 214 910 621 437 189 488 551 962 500 154 576 272 999 772 508 872 170 141 911 603 900 898 852 415 433 977 710 366 103 375 830 102 149 185 577 929 972 357 769 341 511 196 414 304 749 632 357 779 734 355 709 488 612 862 091 194 127 294 880 073 277 699 150 565 429 467 496 996 958 779 743 272 220 076 348 594 180 597 707 318 187 551 446 468 345 234 151 894 294 028 382 273 477 446 771 268 659 926 181 348 263 086 203 148 316 169 606 155 302 598 524 324 237 961 447 139 159 888 291 393 791 764 571 773 165 075 078 716 119 893 422 925 278 853 342 646 888 100 190 544 735 633 787 496 553 695 870 654 215 935 198 140 278 110 380 565 641 135 299 916 018 128 277 271 837 967 431 487 438 014 442 595 159 053 655 885 211 711 006 749 189 589 798 817 021 130 442 604 161 798 061 902 319 215 653 388 577 993 592 902 252 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419 934 115 010 042 876 119 316 900 116 832 811 335 628 179 693 340 009 580 982 478 132 038 430 912 600 376 412 448 843 436 442 391 417 907 739 448 835 179 325 155 219 391 469 060 877 589 696 168 560 556 290 395 976 653 850 142 613 881 670 374 690 755 919 213 908 275 678 497 538 839 566 771 471 243 021 166 867 180 425 892 838 202 920 159 760 038 500 700 696 839 410 774 829 443 373 230 478 692 077 692 328 893 234 692 682 949 900 612 889 244 530 909 600 636 893 342 320 536 408 430 922 106 532 986 990 854 048 741 571 434 829 213 276 712 779 036 042 915 530 565 351 041 560 738 559 880 771 915 047 293 646 117 679 688 640 805 905 672 933 941 049 670 528 937 104 493 833 269 108 557 549 151 042 994 239 868 007 582 474 680 849 642 083 395 371 631 608 273 230 474 915 445 102 522 223 230 676 260 840 328 274 931 926 340 794 835 994 912 299 685 964 562 052 140 652 179 725 701 287 796 274 310 676 216 894 654 746 904 862 659 255 832 527 390 239 036 378 321 555 769 338 347 432 884 522 058 963 140 622 160 275 248 616 217 072 989 397 491 184 873 954 510 696 772 819 831 947 727 992 837 786 214 303 617 460 351 756 434 976 420 202 087 240 939 014 550 752 797 612 133 922 619 294 951 882 180 944 041 434 652 929 592 538 669 504 016 909 081 700 936 533 983 091 586 241 346 685 928 280 612 044 290 578 150 665 837 275 168 811 989 153 670 043 476 132 168 822 101 061 928 078 499 098 602 437 674 059 307 695 610 846 415 461 357 957 759 439 598 404 587 188 466 280 004 374 858 650 460 556 080 050 819 718 638 596 913 659 428 197 968 055 270 523 604 981 406 962 427 248 475 657 022 756 860 203 086 247 248 402 221 934 927 034 202 531 541 120 409 031 493 228 003 655 423 775 810 891 129 085 371 038 189 539 144 797 644 394 150 660 071 644 041 447 300 539 300 443 472 898 574 544 717 613 170 619 256 713 830 078 294 505 047 655 624 582 286 319 558 738 811 984 656 933 997 251 163 154 829 327 778 020 579 557 155 113 795 066 188 005 376 _________________________________________________________ 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 ;)
@izzi532
5 жыл бұрын
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?
@calculon000
7 жыл бұрын
I know this video is supposed to be about math and not programming, but I have to say that this would be much easier for me to understand if you used the notation MOD(a,N)=x rather than a≡x mod N. I use spreadsheets for work on a regular basis using the MOD function and that's one place where I think a much larger number of students will encounter it in their adult lives using that notation rather than what I assume is the classic pure math notation. Of course, this is just my experience and in the end which you use is just a *pure vs applied* stylistic choice on your part on how to teach these concepts. Great choice in how these concepts can be applied to the real world!
@zairaner1489
7 жыл бұрын
But thsat would make it absolutely clunky to use. Also, that is the most commonly used notation
@321Mohan
3 жыл бұрын
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.
@emjackson81989
5 жыл бұрын
So, here's a question that may or may not be covered in the next episode: Does Shor's Factorization Algorithm hold even for large numbers that are multiples of n large numbers where n > 2? That is, does it hold for N of the form p*q*r...*z?
@atrumluminarium
7 жыл бұрын
So when in the future we get quantum computers, RSA will be obsolete. Is this true for other cryptosystems? Will elliptic curve cryptography be still good or is that the same thing? Are there any cryptosystems that are suitable for quantum computes? (future video maybe?) :p
@Twisol
7 жыл бұрын
There are two questions wrapped up in this: cryptography that quantum attacks don't have a significant speed-up on, and cryptography that actually relies on quantum computing. Luckily, both of these are active research subjects: "post-quantum cryptography" for classical algorithms that don't rely on the hard problems that quantum computers can solve so easily, and "quantum cryptography" for utilizing quantum computing as a core piece of a cryptographic system.
@atrumluminarium
7 жыл бұрын
Cool thanks for the reply :)
@doit9854
7 жыл бұрын
a simple "today" solution to p=np issues like shor's algo is as such: encapsulate RSA ciphers within each other - which has been typically used for cold storage of TSCSI information since the early 90's. The idea here is that the cipher length of the keys, encapsulated within each other, becomes another layer of entropy, another "key" per-say. ex: 2^8 within 2^14 within 2^10 within 2^ et cetera et cetera. You will have to know the cipher lengths of the certificates to break confidentiality within a shor's enabled world. You would also have to implement such a solution so it doesn't leak the key "strength". Tomorrow solutions involve recursive inverse square "reaching long tail/almost zero equations" using deep floating point (bigdouble anyone?) decimals &, for now, elliptical curve class algo's. Although its almost zero doesn't mean it is, nor is not, zero... where does your accuracy round? How many qbits do you need to defeat, how many decimal points would that take?
@jakubpekarek6400
7 жыл бұрын
In theory it is possible to base cryptography on (almost) any hard problem. While factoring comes from NP and so can be affected by quantum computation, there are plenty of higher layers of complexity, all the way to absolute unsolvability regardless the technology used. But it is of course practical to use some problems that are just out of reach rather than problems that are hopelessly out of reach as that would make the proper usage explode in complexity as well.
@doit9854
7 жыл бұрын
Pretty sure your comment couldnt be solved with a quantum computer homie. Luckily your point is a mathamatical trap door so your onto something. #jibbajabba
@Compins
6 жыл бұрын
Alright, I'm back. I kinda don't understand the explanation at 9:52 - 10:06. I don't really see the logic there. I see why it has to be like that, so that we don't up with 1*N = p*q. Is it *a^(r/2) =/= 0 mod N* because we can't get *a^(r/2) = 0 mod N* from *a^r = 1 mod N* ?
@Centrinia
7 жыл бұрын
You also want p and q to be distinct primes. You should first compute the square root of N. Applying that algorithm to squares could lead to a situation where you will never find a valid a.
@mrbdzz
7 жыл бұрын
Phase 1. Pick a < N Phase 2. ??? Phase 3-4. Profit
@ve2zzz
5 жыл бұрын
Hi... May someone explain me ? I found this explanation highly interesting... So, i wrote some C for curiosity... Code works, but... ...Finding"r" is waaaaaay longer than brute force factoring... For example: Factoring 21583 into 113 and 191 using 8 for "a" required 2660 iterations to find "r" while BF needs 57 iterations for a complete factor, Another example: factoring 4661 into 59 and 79 using 19 for "a" required 1131 iterations to find "r" while brute force needs only 30. What did i do wrong ???ffff
@mrinmoybanik5598
3 жыл бұрын
At its heart this algorithm exploits the efficiency of Euler's algo to find G.C.D.
@shanelstevens
6 жыл бұрын
Loved this episode so much! 👍 Kesley is brilliant!
@DustinRodriguez1_0
7 жыл бұрын
And if anyone is wondering... no, that's not how hackers break cryptography when it does get broken. There's usually no need at all to figure out what the keys (the primes) used were when there are other ways to figure out what the message is. There are so many ways to leak information, and usually if you can get knowledge of 1 bit out of a system, you can just repeat doing that over and over to get the whole thing.
@catStone92
7 жыл бұрын
also, I had modular mathematics in college (we called it discrete mathematics for some reason, but I guess there was more to the class than just modular maths) but I didn't know about the periodicity of the remainders of the powers thing
@nicktheoregonian
7 жыл бұрын
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?
@Petch85
7 жыл бұрын
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?
@zairaner1489
7 жыл бұрын
Yes we only used numbers with two prime factors. I guess the cryptography uses these?
@tynansigg5472
7 жыл бұрын
I don't see how the period-finding step could have a higher complexity than O(n). The period cannot be more than n, so why isn't it simply a matter of checking a^1 through a^n for a power that results in 1 mod n?
@SSM24_
5 жыл бұрын
Because raising something very large to a very large power is computationally expensive - even more so if you're doing that for _every_ power from 1 to some potentially huge number.
@viniciusdesouzamaia
7 жыл бұрын
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.
@xorenpetrosyan2879
7 жыл бұрын
Very good explanation
@filipsperl
7 жыл бұрын
How does the choice of 'a' affects the calculations? Shouldn't I be luckier with the relatively small numbers or it is safer with big ones? What about primes? We know that 'p' and 'q' are primes, so why don't we just try every prime found instead? Here's my test of this algorithm btw: 91 = p*q a = 4 --> a^2 = 16, a^3 = 64, 256, 1024, ... r1 (remainder) = 4, r2 = 16, r3 = 64, r4 = 75, r5 = 23, r6 = 1, r7 = 4, ... --> period is 6 4^3 +1 = 65 - not a multiple of 91 p = gcd (63;91) = 13 q = 91/13 = 7 I've tried with slightly bigger numbers and it gets really hard to calculate/orientate in.
@stevethecatcouch6532
7 жыл бұрын
It appears that if you use a square for a, there is no need for r to be even. a^(r/2) will be an integer.
@Robertwardrip
7 жыл бұрын
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
@zairaner1489
7 жыл бұрын
No, because we start n with 1. Otherwise, the period would always be 0
@josesaldivar655
6 жыл бұрын
Kelsey, what is the integral of one over the squared root of a log .(a x), DX How do you do that
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