Back Diffie-Hellman is a standard method of Alice and Bob being able to communicate, and end up with the same secret encryption key. It is used in many applications, and uses two numbers (G and N) for the first part of the calculation (of which N must be a prime number). The Elliptic-Curve Diffie–Hellman (ECDH) is an anonymous key agreement protocol that allows two parties, each having an elliptic-curve public–private key pair, to establish a shared secret over an insecure channel. Diffie Hellman calculator in Python and my first story. Paper key lists transported by a trusted courier. The Diffie - Hellman key exchange method allows two parties that have no prior knowledge of each other to jointly establish a shared secret key over an insecure channel. This key can then be used to encrypt subsequent communications using.

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A. The idea

Suppose two people, Alice and Bob [traditional names], want to use insecureemail to agree on a secret 'shared key' that they can use to dofurther encryption for a long message. How is that possible?The so-called Diffie-Hellman method provides a way. This methodis one of the ingredients of SSL, the encryption package that ispart of the Netscape browser.

These notes are a little more detailed than in class, for clarity,but on the exam you are responsible only for doing exercises like Problem 8.2.

B. The mod function

The main ingredient is the 'remainder' or 'modulo' or 'mod' function,denoted % in Perl. For example, 25%10 is 5 (say '25 mod 10is 5') and 25%16 is 9 ('25 mod 16 is 9'). For n%10,the result will always be one of 0,1,..,9.

As you can see, any positive integer modulo 10 is just the lastdigit in base 10: 1537%10 is 7, etc. You can think of'modulo 10' for positive integers as meaning 'ignore all decimaldigits except the last one'.

Doing 'modular arithmetic' with 'modulus' 10 means doingaddition, subtraction, and multiplication (including powers)where you only care about the remainder modulo 10. You can usesome other modulus m instead of 10, as long as it's the samethrough the whole problem. It works very smoothly.

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The 'as often as you want' principle: If you are doingmodular arithmetic to find an the answer modulo m, you cantake the remainder modulo m as often as you want during thecalculations, without changing the answer.

Example 1. To find 1537 x 4248 modulo 10, you couldmultiply out and take the last digit, but a better way would beto replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 =56, and then take 56 mod 10 to get 6 as the answer.

A handy standard notation is to write a b (mod m) ifa and b have the same remainder modulo m. This is read'a is congruent to b modulo m'. In this notation theexample just mentioned looks like this: 1537 x 4248 7 x 8 = 56 6 (mod 10).

Example 2. Find 2⁸(mod 11).

One solution. 2⁸= 256; 11 goes into 256 with quotient 23and remainder 3.

Another solution. Find 2², 2⁴, 2⁸by squaring repeatedly,but take remainders mod 11 each chance you get: 2²= 4,2⁴= 4²= 16 5, 2⁸ 5²= 25 3.

Example 3. Find all the powers of 2 up to 2¹⁰ Synthesia unlock code list. , each modulo 11.

Solution. Keep doubling, taking remainders modulo 11 wheneverpossible:

2, 4, 8, 16 5, 10, 20 9, 18 7, 14 3, 6, 12 1 (mod 11). So the answer is 2, 4, 8, 5, 10,9, 7, 3, 6, 1.

Notice that the powers of 2 run through all possible remaindersmodulo 11, except 0. We say 2 is a 'generator' modulo 11.There is a theorem that if you take a prime modulus, then thereis always some generator, and in fact 2 often works. If 2 doesn't,maybe 3 will.

C. The Diffie-Hellman method

The idea of Diffie and Hellman is that it's easy to computepowers modulo a prime but hard to reverse the process: Ifsomeone asks which power of 2 modulo 11 is 7, you'd have toexperiment a bit to answer, even though 11 is a small prime. Ifyou use a huge prime istead, then this becomes a very difficultproblem even on a computer. Steps:

Alice and Bob, using insecure communication, agree on ahuge prime p and a generator g. They don't care if someonelistens in.
Alice chooses some large random integer x_A < p andkeeps it secret. Likewise Bob chooses x_B < p and keeps itsecret. These are their 'private keys'.
Alice computes her 'public key' y_A g^x_A(mod p)and sends it to Bob using insecure communication. Bobcomputes his public key y_B g^x_Band sends it toAlice. Here 0 < y_A < p, 0 < y_B < p.
As already mentioned, sending these public keys with insecurecommunication is safe because it would be too hard for someoneto compute x_A from y_A or x_B from y_B, just like thepowers of 2 above.
Alice computes z_A y_B^x_A(mod p) and Bob computes z_B y_A^x_B(mod p). Here z_A < p, z_B < p.
But z_A = z_B, since z_A y_B^x_A (g^x_B)^x_A= g^{(x_A x_B)}(mod p) and similarly z_B (g^x_A)^x_B= g^{(x_A x_B)}(mod p). Sothis value is their shared secret key. They can use it toencrypt and decrypt the rest of their communication by somefaster method.
In this calculation, notice that the step y_B^x_A (g^x_B)^x_Ainvolved replacing g^x_Bby its remainder y_B, (in the reversedirection) so we were really using the 'as often as you want'principle.

D. Notes (not on final exam)

It's easy to see why the 'as often as you want' principle works formodular arithmetic with positive integers in base 10. In Example 1,imagine doing the multiplication with paper-and-pencil arithmetic, butignoring everything except the last digit. You getIn other words, you can multiply and then take the last digit,or you can take remainders early, by saving just the 7 and 8,taking their product, and saving its last digit.
Congruences work fine for negative numbers if you alwaysuse a remainder that is positive or 0; for example, -13 7 (mod 10)because -20 is a multiple of 10 and -13 is 7 larger. The% operation in Perl works this way but the same operation inC and C++ does not.
The notation is meant to suggest =, becauseseveral properties of are similar to those of =.For example, a b and b c give a c (all mod m).
Another interesting fact is that modulo 11, we have2¹⁰ 1, 3¹⁰ 1, 4¹⁰1,..,10¹⁰ 1, and of course 1¹⁰=1 to startwith. More generally, there is a theorem saying that for anyprime p and for any a from 1 to p-1 we get a^p-1 1(mod p).
The Diffie-Hellman method works best ifp = 2q+1 where q is also a

Diffie Hellman Calculator