*This is the latest in a series of blog posts to address the list of '52 Things Every PhD Student Should Know To Do Cryptography': a set of questions compiled to give PhD candidates a sense of what they should know by the end of their first year. This blog is the second in the section on Mathematical Background and concerns how group operations can be used to design cryptographic primitives.*

*As you probably know by now, cryptography often relies on 'hard problems'. That is, we design cryptographic protocols whose security can be proven if we assume that the adversary is unable to solve a certain (mathematical) problem in a reasonable amount of time. This blog post introduces three such problems which are widely used in security proofs. Luckily for me, a) this is really just Group Theory, not Computer Science and b) just two days before writing this I attended a very decent guest lecture by fellow Bristol Crypto researcher Susan Thomson on this exact topic. (That is to say, any inaccuracies in the following can and should be blamed on her!)*

__The Discrete Logarithm Problem (DLP)__

Okay, let $G$ be an abelian group. First we write the operation in $G$ multiplicatively. For any $g \in G$ and any integer $a>1$ let $g^a$ denote $g * g * ... * g$ with $a$ occurrences of $g$. The

**discrete logarithm problem**(DLP) is:

*Given $G$, $g$ and $h=g^a$, find $a$.*

Here $a$ is called the

**discrete logarithm**of $h$ with base $g$, hence the name of the problem.

Is the discrete logarithm problem hard?

**Sometimes, maybe.**As a counter-example, let $G$ be the integers under addition. So now it makes sense to write the group operation additively, not multiplicatively. So the same process of repeating the group operation with the same element $g$ is now written $g + g + ... + g$ with, say, $a$ summands and, since we're working in the integers, this sum, call it $h$, is just equal to the integer $ag$. Therefore $a$, the discrete logarithm of $h$ to the base $g$, can be found by just dividing $h$ by $g$. For example, if I say to you, "find the discrete logarithm to the base 3 of 18 in the integers under addition", you'd just write $3 + 3 + ... + 3 = 18$ with $a$ summands on the left, simplify this to $3a = 18$ and find that $a = 6$. I could change the group to integers modulo $N$ for some integer $N>1$ (still under addition) but the problem wouldn't be much harder: one has to solve an equation like $ag \equiv h \: (\mathrm{mod} \: N)$ which is solved by performing the Extended Euclidean Algorithm to find $g^{-1}\: (\mathrm{mod} \: N)$ (if it exists), multiplying this by $h$ and reducing modulo $N$ to obtain $a$. All of this is can be done in polynomial time - no good for building secure cryptographic primitives.

On the other hand, the DLP in a finite field of prime order viewed as a group under

*multiplication*(after throwing out the zero element) or in elliptic curve groups (more on those next week) is

**believed**to be hard. That is, we do not yet know of any polynomial time algorithms for finding discrete logarithms in these groups. As a concrete example, suppose I ask you, "find the discrete logarithm to the base 3 of 5 in the multiplicative group of the integers modulo 7". This means find an integer $a$ such that $3^a \equiv 5\: (\mathrm{mod} \: 7)$. Now that we are in the multiplicative group, not the additive one and so we really

*do*have to 'take logarithms' somehow, not just multiply by the inverse of 3. In this case, since 7 is fairly small, we can find the answer by just trying all the possibilities one at a time until we find a solution:

- $3^1 = 3 \not\equiv 5\: (\mathrm{mod} \: 7)$
- $3^2 = 9 \equiv 2 \not\equiv 5 \: (\mathrm{mod} \: 7)$
- $3^3 = (3^2)\times3 \equiv 2\times3 = 6 \not\equiv 5\: (\mathrm{mod} \: 7)$
- $3^4 = (3^3)\times3 \equiv 6\times3 = 18 \equiv 4 \not\equiv 5\: (\mathrm{mod} \: 7)$
- $3^5 = (3^4)\times3 \equiv 4\times 3 = 12 \equiv 5\: (\mathrm{mod} \: 7)$

__The Computational Diffie-Hellman Problem (CDH)__

A problem related to DLP is named after Whit Diffie and Martin Hellman who devised a way of two parties agreeing on a secret key over a public channel without revealing it:

- Alice and Bob publicly agree on a cyclic group $G$ and generator $g$.
- Alice chooses a random secret integer $a$ and Bob chooses a random secret integer $b$.
- Alice computes $g^a$ and publicly sends this to Bob. Bob computes $g^b$ and publicly sends this to Alice.
- Alice and Bob both compute $g^{ab}=(g^a)^b=(g^b)^a$ by raising what they received from the other party to power of their own secret integer.

**Computational Diffie-Hellman Problem**(CDH):

*Given $G$, $g$, $g^a$ and $g^b$, find $g^{ab}$.*

CDH is clearly related to DLP, but which is harder? Well, if I can solve DLP then I can efficiently compute the secret integer $a$ from $g^a$ and then find $g^{ab}$ by raising $g^{b}$ to the power $a$ in the same way Alice does, therefore solving CDH. So anyone who can solve DLP can also solve CDH, meaning

**DLP is at least as hard as CDH.**

__The Decisional Diffie-Hellman Problem (DDH)__

This is another 'discrete logarithm' style problem used to prove indistinguishability properties. Say Alice and Bob perform the Diffie-Hellman key agreement protocol as above so that $G$, $g$, $g^a$ and $g^b$ are all public and $g^{ab}$ is the shared secret key. Intuitively, the

**Decisional Diffie-Hellman Problem**(DDH) asks whether an adversary can distinguish Alice and Bob's secret key $g^{ab}$ from a random group element of $G$. Formally:

*Given $G$, $g$, $g^a$, $g^b$ and $T_x$ such that $T_0$ is a random element of $G$, $T_1 = g^{ab}$ and $x$ is chosen uniformly at random from $ \lbrace 0,1 \rbrace $, find $x$.*

*some*information about the secret key $g^{ab}$ that distinguishes it from a random group element, even if it can't be computed directly. What should be clear is that if the adversary can solve the

*computational*Diffie-Hellman problem, then they can actually

*compute*$g^{ab}$ and hence trivially distinguish this element from a random group element, thereby solving the

*decisional*Diffie-Hellman problem. So anyone who can solve CDH can also solve DDH, meaning

**CDH is at least as hard as DDH**.

These are the three problems we wanted to talk about and we've given a sketch proof of their ordering in terms of hardness:

**DLP is the most hard, then CDH and then DDH.**As we've seen, DLP is sometimes easy, making CDH and DDH easy. So the choice of group $G$ and generator $g$ is very important when doing cryptography!