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Suppose $g$ is a generator of an order $p$ cyclic group in which discrete logarithm is hard and $p$ is a prime (i.e., given $g^x$ for a random $x \in \{0,1,\ldots, p-1\}$, it is hard to recover $x$ except with $negl(\lambda)$, where $negl(\lambda)$ is a negligible function of the security parameter $\lambda$).

My question is as follows: what happens if I am given $g^y$ such that $y$ is chosen randomly from $\{0, 1, \ldots, q-1\}$ for some $q < p$, is there a way to quantify the hardness of discrete logarithm in terms of $q$? perhaps, something like: probability that one recovers $y$ is at most $\frac{1}{q} + negl(\lambda)$ and the probability is taken over the choice of $y$, or is there no way to quantify this?

I am mainly interested in knowing what happens in practice when one does this. For instance, in most libraries, the exponents are chosen to be 160 bits even though $p$ is 1024 bits.

(I posted this on math.stackexchange, but no responses. So, I am moving the question here by deleting it from there.)

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The exponents are 160 bits in integer DL systems with 1024 bit $p$, because there we are actually working in a subgroup of size $q$, with $q$ a 160 bit prime. –  Paŭlo Ebermann Dec 13 '12 at 21:10
    
so, does it mean if I take q to be 160 bits integer in my example above, I can assume the attacker's success probability to be negligible (i.e., can be ignored in practice)? –  user52914 Dec 14 '12 at 5:59
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Actually, the integers mod p are (multiplicatively) not a cyclic group (not even a group at all, but excluding the zero we get a (normally non-cyclic) group). Then we chose a generator which generates a cyclic subgroup of size $q$ (where $q$ is a divisor of $p-1$). And yes, we normally assume that the discrete logarithm in this subgroup is hard. –  Paŭlo Ebermann Dec 14 '12 at 12:48
    
By the way, welcome to Crypto.SE! I hope you'll stay in the community. May I encourage you to read the faq, to learn more about this site? –  D.W. Dec 18 '12 at 22:42
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2 Answers

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The answer depends upon the factorization of $p-1$. Let me explain the situation for two example cases:

  • Suppose $p-1=2p'$, where $p'$ is also prime. This is the best case (for the defender): it minimizes the information leakage. Then the best known attack that takes advantage of the reduced range of the exponents will solve the discrete logarithm in about $\sqrt{q/2}$ time. For instance, if $q$ is 160 bits, then the best attack known will take about $2^{80}$ time (assuming this is less than the time to solve the discrete logarithm modulo $p$ for an arbitrary exponent).

  • Suppose $p-1=2p'p''$, where $p',p''$ are prime and $p'$ is pretty small and $p''$ is pretty big. This situation is bad for the defender and good for the attacker. An attacker can first recover the value of the exponent $y$ modulo $2p'$ in $\sqrt{p'}$ time (using Pohlig-Hellman). Then, the attacker can learn the full value of $y$ in $\sqrt{q/(2p')}$ time. The total running time will be $\sqrt{p'}+\sqrt{q/(2p')}$. If $p' \approx \sqrt{q}$, then the running time of the attack is $O(q^{1/4})$, which is significantly faster than the above -- yet the general discrete logarithm problem (for unrestricted exponents) might still be very hard.

What's going on here? Two things. First, if we know that the exponent comes from a consecutive range of $n$ possible values, then it's possible to recover the exponent in $\sqrt{n}$ time. Second, for any divisor $d$ of $p-1$, one can recover the value of the exponent $y$ modulo $d$ in $\sqrt{d}$ time (raise the group element to the $(p-1)/d$ power, then use Pohlig-Hellman to solve the resulting discrete log problem). You can combine both of these ideas. However, nobody knows how to do any better than that.

So, if you pick the parameters right, using an exponent from a restricted range can be secure. However, you have to pick them carefully. A good choice is to select $p$ so that $(p-1)/2$ is prime, and select $q$ to be at least 160 bits in length.

Even better yet, follow Paŭlo Ebermann's advice: select a prime $q$ that is 160 bits long or longer and a prime $p$ so that $q$ divides $p-1$, and then choose the generator $g$ to be a generator of the cyclic subgroup of order $q$ (e.g., let $g_0$ be a generator for the group of integers modulo $p$, then let $g=g_0^{(p-1)/q}$).

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@user52914, as far as I can see, nothing terribly bad should happen. Releasing $y_1+y_3$ and $y_2+y_3$ doesn't reveal much more information than releasing $y_1-y_2$. And releasing $y_1-y_2$ does not help find $y_1$ or $y_2$ (though if you somehow manage to learn one of $y_1$ or $y_2$, obviously knowledge of $y_1-y_2$ reveals the other). –  D.W. Dec 17 '12 at 22:36
    
@user52914, hire a cryptographer to prove it for you? At a meta level: this has now deviated pretty far from the original topic of this question, so I don't think it's on-topic on this question. FYI, Crypto.SE is not a discussion forum: it's a question-and-answer site. One question per question. If you have a question about the design of a particular system, I think you'll need to post a separate question. If you do that, make sure you post detailed enough information about the design of the system that analysis is possible. –  D.W. Dec 18 '12 at 7:04
    
@user52914, again, that is off-topic. If you want to discuss it, please post a separate question. The more you discuss it here, the more sorry I am that I responded to your second question at all. As far as explaining the site -- well, it sounds like you are not clear on how to use it properly and so I do need to explain the site policies. And no, it is not this way to maximize reputation of users -- it is this way to maximize the utility of the site as an archive of quality answers to relevant questions. Off-topic comments reduce the ability of others to find this information. –  D.W. Dec 18 '12 at 22:38
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Assume there exists an adversary for your q-subset DL problem that will be successful with probability $\mathsf{succ(\lambda)}$. If you just use this adversary to attack the original DL problem, he will get an input that comes from the set it expects with probability $q/p$ and will therefore be succesfull with probability at least $q/p \cdot \mathsf{succ(\lambda)}$. However we know that the DL problem is hard. Thus, we can bound the success probability of the adversary: $$\mathsf{negl}(\lambda) \ge q/p \cdot \mathsf{succ(\lambda)}$$ $$p/q\cdot\mathsf{negl}(\lambda) \ge \mathsf{succ(\lambda)}$$

Thus we get a bound for the success probability against your q-subset DL problem.

Whether this bound is helpful, depends on $q$. As long as $p/q$ is polynomial, the upper bound is still a negligible function. So at least in theory this is still a hard problem.

In practice and for a fixed group size that is of course a completely different story.

Of course, if $p/q$ is superpolynomial, then the all bets are off and this upper bound does not help you in any way.

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There's even a tradeoff between the running time of your algorithm and the success probability since instead of failing if the input doesn't fall in the acceptable subset you can randomize your input via a multiplication by $g^r$ (with random $r$) and run it again. I think that's called Random Self Reducibility –  Alexandre Yamajako Dec 13 '12 at 17:22
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