This series is so cool- Khan Academy really is revolutionary, and it astounds me that, especially with the quality and quantity of content produced, it's all free.
honestly this videos explanation is a little bit wrong. not wrong but its confusing, the equation supposed to be e*d = 1 mod phi(n) , in this equation we dont care what "k" is, thats the "mod"less version in the video. you just find the d with extended euclidean algorithm. to find the k, you also should do this extended euclidean algorithm but if you solve that algorithm, you ll be already found the "d" that you were looking for.
FYI: The youtube title doesn't match the title on the site. It's step 4 on the main site. Awesome videos on RSA BTW. End of semester and I actually understand this now... :)
This might be the worst video of Khan Academy. It makes no sense at all and that background noise is so distracting. How was k calculated? It is assumed that k is 2. e is chosen at random such that (e,phi(n)) = 1. This was not told in any video before this one. What is known and what is unknown is really confusing. Is it the case that for each pair of p1 and p2, there must exist a pair of e and d? Is it e and d that decide the value of n? I chose the value of 307 and 101 as p1 and p2, but i am not able to find any e,d pairs for that.
(This is for C#. If you are using another programming language, please write.) Do you using Math.Pow ? Math.Pow is double. If you are using BigInteger (for big integers) and do with for, then you are not see "Infinity". (Sorry my English, I am Turkish.) (BigInteger's reference is System.Numerics) Example for C#: int taban00 = 1394; int us00 = 2011; BigInteger bigInt = new BigInteger (1); for(int i = 0; i
The complexity of your solution is linear. It is able to do O(lgN) computation You can use divide and conquer to calculate A^m % n for any large A and m If m is even => return (A^(m/2) mod N)^2 mod N if m is odd => return (A^(m-1) mod N * A mod N) mod N
easily the best explanation of rsa
This series is so cool- Khan Academy really is revolutionary, and it astounds me that, especially with the quality and quantity of content produced, it's all free.
What is K in d at 2:47 ? where does it come from?
honestly this videos explanation is a little bit wrong. not wrong but its confusing, the equation supposed to be e*d = 1 mod phi(n) , in this equation we dont care what "k" is, thats the "mod"less version in the video. you just find the d with extended euclidean algorithm. to find the k, you also should do this extended euclidean algorithm but if you solve that algorithm, you ll be already found the "d" that you were looking for.
The background bell sound in the video is really, really distracting. I cannot imagine why it was put in the video.
with the bell, I can imagine what he is saying as a story and that's fantastic
What is the website or the program he uses to create the graph at 3:51 in the video? I could really use it in one of my works!
FYI: The youtube title doesn't match the title on the site. It's step 4 on the main site. Awesome videos on RSA BTW. End of semester and I actually understand this now... :)
awesome explanation of RSA, cool that prime factorization is the heart of all of it
Great video
thank you so much this was really helpful
This might be the worst video of Khan Academy. It makes no sense at all and that background noise is so distracting. How was k calculated? It is assumed that k is 2. e is chosen at random such that (e,phi(n)) = 1. This was not told in any video before this one. What is known and what is unknown is really confusing. Is it the case that for each pair of p1 and p2, there must exist a pair of e and d? Is it e and d that decide the value of n? I chose the value of 307 and 101 as p1 and p2, but i am not able to find any e,d pairs for that.
there is 3 other parts of rsa encryption series;
/watch?v=IY8BXNFgnyI
part 2 there^ says e and d are encryption and decryption I believe iirc
what is the link if i want to calculate time rquired to multiply 2 number
I'm confused with the formula at 0:53... how does the left side equal the right side in that example?
It's not equal. It's "congruent" that's why it's 3 bars.
what if 'm' is greater than 'n'?
Can't make the decryption work.
C^d = 1394^2011 = Infinity
Please help.
(This is for C#. If you are using another programming language, please write.) Do you using Math.Pow ? Math.Pow is double. If you are using BigInteger (for big integers) and do with for, then you are not see "Infinity".
(Sorry my English, I am Turkish.)
(BigInteger's reference is System.Numerics)
Example for C#:
int taban00 = 1394;
int us00 = 2011;
BigInteger bigInt = new BigInteger (1);
for(int i = 0; i
The complexity of your solution is linear. It is able to do O(lgN) computation
You can use divide and conquer to calculate A^m % n for any large A and m
If m is even => return (A^(m/2) mod N)^2 mod N
if m is odd => return (A^(m-1) mod N * A mod N) mod N
it feels like a vsause1 video
Pretty sure the equation for d is bullshit.