I recently stumbled across Thomas Pornin's old answer about deterministic (EC)DSA again. There he states the following:

Note that $k$ must be generated uniformly in the $[1, q-1]$ range (where $q$ is the subgroup order). Any information on $k$, even partial (such as: values between $1$ and $2^{160}-q$ are twice as probable than values between $2^{160}-q$ and $q$), can be exploited by the attacker.

When asked for clarification on this, he responded with a link to "The Security of DSA and ECDSA Bypassing the Standard Elliptic Curve Certification Scheme" by Vaudenay (PS) which mentions that Bleichenbacher discovered the attack and mentioned it in private communication with the author.

Now the paper has a short paragraph (section 2.2), which basically states an approximate equation for the bias and says that

Bleichenbacher actually used it in order to approximate the secret key more and more precisely with signatures.

Now my question:
How does this attack work in a more detailed description? or formulated differently: How does one exploit this tiny bias to recover the secret key and how many signatures (ie oracle queries) are required for this along with what approximate computational effort (if it is non-negligible)?

The relevant section from the paper is (for your convenience):

The initial standard pseudorandom generator in DSA for $k$ was simply a 160-bit pseudorandom number reduced modulo $q$. Bleichenbacher observed that the probability of $k$ in the $[0, 2^{160} − q]$ range have probability which is twice of the others. This leads to a bias

$$E\left(e^{\frac{2i\pi k}q}\right)\approx \frac{q e^{i\pi\frac{N-1}q}}{\pi N} \times \sin\left(\frac{\pi N}q\right)$$

where $N=2^{160}$. Since $q \approx N$, this may be large depending on the $\frac{\pi N}q$ angle. Bleichenbacher actually used it in order to approximate the secret key moreand more precisely with signatures.

  • $\begingroup$ If you also wonder how one could possibly arrive at the approximating equation, I have posted an according question on Math.SE. $\endgroup$
    – SEJPM
    Commented Mar 11, 2017 at 19:45
  • $\begingroup$ The attack is described with much more detail in two recent papers: Mulder et al. and Aranha et al.. $\endgroup$ Commented Mar 11, 2017 at 23:34

1 Answer 1


Nguyen and Shparlinski detail the method in The Insecurity of the Elliptic Curve Digital Signature Algorithm with Partially Known Nonces. I'll largely follow the notation in that paper.

The idea is to convert the determination of a private key from biased $k$ nonces in several ECDSA signatures into instances of the hidden number problem (HNP), and then solve the HNP as a reduction to the closest vector problem.

So you collect a lot of signatures and construct a lattice out of them, and reduce that lattice using LLL or BKZ, and then you can extract the private key from the reduced basis vectors.

Estimated computational effort required

I tested this with a hand-crafted LLL Python implementation (it's so slow it surely must constitute an upper bound in performance...) running on a Ubuntu 16.04 VM with 2GB RAM, and I have the following sample runs:

  • 96-bit bias, 5 signatures: (failed) 31 seconds
  • 64-bit bias, 10 signatures: (success) 4 minutes
  • 32-bit bias, 12 signatures: (success) 12 minutes
  • 16-bit bias, 17 signatures: (success) 49 minutes
  • 8-bit bias, 19 signatures: (failed) 79 minutes
  • 8-bit bias, 21 signatures: (success) 244 minutes
  • 8-bit bias, 20 signatures: (success) 177 minutes

I stopped there. Nguyen and Shparlinski were apparently able to recover the secret key with only 3 bits of bias with 100 signatures and think that recovery is possible with only 2 bits being known.

More problem details and setup

So let's say you know the low $\ell$ bits of $k$. (This paper describes one way of recovering those bits.) Then you can write $k$ as

$$k = a + 2^\ell b$$

In other words, you know $a \in [0, 2^\ell -1]$. For simplicity, let $a = 0$. Then the equation for $s$ in the ECDSA signature $(r,s)$ becomes

$$s = (h + rx) \cdot (2^\ell b)^{-1}$$

where $h$ is your hashed message and $x$ is your private key, and everything is modulo $q$, the order of your base point. Rewrite this as

$$xr \cdot (2^\ell s)^{-1} = -h \cdot (2^\ell s)^{-1} + b$$

Define $t \equiv r\cdot (2^\ell s)^{-1}$ and $u \equiv -h \cdot (2^\ell s)^{-1}$ and you have

$$xt = u + b$$

Remembering that $0 \lt b \lt q/2^\ell$, you have

$$xt - u \lt q/2^\ell$$

So this is basically the HNP. $b$ is smaller by a factor $1/2^\ell$ than $x$,$t$, and $u$, so approximate this equation:

$$xt - u \approx 0 \longrightarrow xt - u - jq \approx 0$$

since everything is mod $q$. So first collect $n$ signatures, giving you several tuples of $t_i$, $u_i$, and $j_i$, and you can construct a matrix out of basis vectors:

$$ \begin{pmatrix} q & 0 & 0 & \cdots & 0 \\ 0 & q & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & q \\ t_0 & t_1 & t_2 & \cdots & t_n \\ u_0 & u_1 & u_2 & \cdots & u_n \\ \end{pmatrix} $$

(The first $n$ rows contain all zeroes and one $q$, one for each of the unknown $j_i$.) This is an $(n+2) \times n$ matrix. Now let $T \equiv (t_0, t_1, \cdots, t_n)$ and $U \equiv (u_0, u_1, \cdots, u_n)$. Then the short vector of interest, $X$ is going to be something like

$$U - xT + \text{(other stuff)}$$

You can modify your matrix to include "sentinel values" $s_T$ and $s_U$ so that you can identify $X$; if you see a reduced vector $T'$ with $s_T$ as its last slot, most likely you'll see the next-to-last entry containing $-x\cdot s_U$ from which you can recover $x$. Disclaimer: I saw this trick in a cryptopals problem first and makes things way easier than just guessing. The modified matrix looks like:

$$ \begin{pmatrix} q & 0 & 0 & \cdots & 0 & 0 & 0\\ 0 & q & 0 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & 0 & 0\\ 0 & 0 & 0 & \cdots & q & 0 & 0\\ t_0 & t_1 & t_2 & \cdots & t_n & s_T & 0 \\ u_0 & u_1 & u_2 & \cdots & u_n & 0 & s_U\\ \end{pmatrix} $$

Now you can use an implementation of LLL from NTL or fplll, find the sentinel value, and extract $x$.


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