Protecting Ring Oscillator Physical Unclonable Functions Against Modeling Attacks

Abstract

One of the most common types of Physical Unclonable Functions (PUFs) is the ring oscillator PUF (RO-PUF), a type of PUF in which the output bits are obtained by comparing the oscillation frequencies of different ring oscillators. One application of RO-PUFs is to be used as strong PUFs: a reader sends a challenge to the RO-PUF and the RO-PUF’s response is compared with an expected response to authenticate the PUF. In this work we introduce a method to choose challenge-response pairs so that a high number of challenge-response pairs is provided but the system has a good tolerance to modeling attacks, a type of attacks in which an attacker guesses the response to a new challenge by using his knowledge about the previously-exchanged challenge-response pairs. Our method targets tag-constrained applications, i.e. applications in which there are strong limitations of cost, area and power on the system in which the PUF has to be implemented.

A Appendix

Algorithm 1 can be implemented with $O(n^4)$ complexity.

The algorithm keeps a pool $C$ of available challenges which shrinks as challenges are selected or become unusable, and an $n \times n$ matrix $O$ whose entry at row $i$ and column $j$ (entry $O(i,j)$) is $1$ if it is known that $f_i>f_j$.

At the beginning of the algorithm, the matrix is initialized with all zeroes and the pool of available challenges is initialized with all $\frac{n (n-1)}{2}$ possible challenges.

On the first step of the algorithm, one challenge “$f_i>f_j?$” is chosen randomly among all $\frac{n (n-1)}{2}$ possible challenges in $C$. The response determines the relation between $f_i$ and $f_j$ and a $1$ is inserted in the matrix either at position $O(i,j)$ or at position $O(j,i)$.

In subsequent steps of the algorithm, when a new randomly chosen CRP determines that $f_i > f_j$, a $1$ is written in the matrix at position $O(i, j)$ and also at position $O(i, k)$, with $k$ being all the values for which matrix entry $O(j,k)$ is $1$: for these entries, the attacker knows that $f_i > f_j$ and that $f_j > f_k$, and therefore can deduce that $f_i > f_k$.

Then the algorithm looks at all frequencies $f_l$ which are known to be slower than $f_j$. A $1$ is written in $O(i,l)$: for these entries, the attacker knows that $f_i>f_j$ and that $f_j>f_l$, and therefore can deduce that $f_i>f_l$. A $1$ is also written in $O(m,l)$, with $f_m$ being all the frequencies that are known to be bigger than $f_i$ (for which $O(m,i)$ is one): for these entries, the attacker knows that $f_m>f_i$, that $f_i>f_j$ and that $f_j>f_l$, and therefore can deduce that $f_m>f_l$.

Every time that a $1$ is written in $O(x,y)$, the challenge “$f(x)>f(y)?$” or “$f(y)>f(x)?$” is eliminated from the pool $C$ of available challenges because the attacker can predict the response with $100\,\%$ certitude.