# Building adversary to show a PRF is not secure

Let $F(k,x)$ be a secure PRF over $(\mathcal{K},\mathcal{X},\mathcal{Y})$ where $\mathcal{K} = \mathcal{X} = \mathcal{Y} = \{0,1\}^n$.

Let $F'(k, x) = F(F(k, 0^n), x) \; \Vert \; F(k, x)$.

$a \; \Vert \; b$ means $a$ concatenated to $b$.

How can I show that $F'$ is not a secure PRF? I've been trying to build an adversary that could find some pattern but I couldn't. My best guess was to initially use $x = 0^n$, so the output would be something like $F(k', 0^n) \; \Vert \; k'$ (where $k' = F(k, 0^n)$, which is the key of the left side) and I would be able to use $k'$ for the next calculations, but it didn't seem to be enought to find any pattern in the sequence of outputs.

Query $0^n$ to your oracle, receive $a\Vert b$ as a response. If $F(b,0^n) = a$ output $1$, otherwise output $0$.

In the case that the oracle is $F'$, since $F'(k,x)$ is defined as $F(F(k,0^n),x)\Vert F(k,x)$ your response will be $a\Vert b = F(F(k,0^n),0^n)\Vert F(k,0^n) = F(k',0^n)\Vert k'$, it will always be true that $F(b,0^n) = a$. I.e. $$\Pr_k[\mathcal{A}^{F'(k,\cdot)}(1^n)=1] = 1.$$

In the case that the oracle is a truly random function $g : \{0,1\}^n \times \{0,1\}^n \to \{0,1\}^n$, $a$ is distributed uniformly and independently of $b$. Therefore the probability of $F(b,0^n) = a$ is $2^{-n}$. Thus, $$\Pr_g[\mathcal{A}^{g(\cdot)}(1^n)=1] = 2^{-n}.$$

And therefore

$$\left|\Pr_k[\mathcal{A}^{F'(k,\cdot)}(1^n)=1]-\Pr_g[\mathcal{A}^{g(\cdot)}(1^n)=1]\right|=1-2^{-n},$$ which is clearly non-negligible.

• What's the meaning of this first line? You call the output $a \Vert b$ and then ask if $F(b, 0^n) = a$, but you just defined $F(b, 0^n) = a$ when you said the output was $a \Vert b$, it looks to me like comparing $x$ with $x$. Apr 17, 2018 at 20:53
• I did not define $F(b,0^n)=a$. I defined $a\Vert b$ as the output of the oracle. The oracle is either the PRF with a uniformly chosen random key (in which case it happens to be true that $F(b,0^n)=a$) or it is a truly random function (in which case it is highly unlikely). And your distinguisher needs to distinguish between those two cases. I get the feeling that maybe your problem is a misunderstanding of the definition of a PRF? Apr 17, 2018 at 22:12
• The function $F$ is known and both inputs are known (the key is $b$ and the input is $0^n$). So you just compute the value. Apr 18, 2018 at 1:36
• @Maeher do we have oracle access to $F(k,x)$ by the PRF definition for every $k$ of our choice? It seems to me as if the definition only gives $F(k,x)$ with a fixed $k$ (uniformly random, not of our choice) or $f(x)$: random... Where is my misunderstanding?
– user16972
Nov 27, 2019 at 0:31
• @Jason if you choose the key, you don't need an oracle. You know the description of the algorithm for computing $F$, so just follow it for the key and input of your choice. Nov 27, 2019 at 7:16

Let $F_k(x)=F(k,x)$ and $F'_k(x)=F'(k,x)$.

A PPT distinguisher $\mathcal{D}$ with access to an oracle $\mathcal{O}(x)=F'_k(x)$ can query it for $$F'_k(0^n)=F_{F_k(0^n)}(0^n)||\overbrace{F_k(0^n)}^a \in \{0,1\}^{2n}$$ we ignore the first part and keep part $a$

Then pick random input $x\in \{0,1\}^n$ and compute the value $b=F_{a}(x)=F_{F_k(0^n)}(x)$

Lastly $\mathcal{D}$ can query $\mathcal{O}$ for $F'_k(x)=\overbrace{F_{F_k(0^n)}(x)}^c||F_k(x)$ and output $1$ if $b=c$, 0 otherwise.

Clearly, this distinguisher runs in polynomial time and distinguishes $F'_k(x)$ from a random function $f(x)\in Func_{2n}$ with non-negligable probability.

• didn't make sense for me... in the middle step you pick random input $x$ and say $b = F_{F_k(0^n)}(x)$ and in the last step you pick input $x$ and say $c = F_{F_k(0^n)}(x)$, so what's the meaning at comparing $b$ and $c$ if they are obviously equal? Apr 17, 2018 at 20:47
• $b$ and $c$ are equal when the oracle is equal to $F'_k$ and with negligible probability when the oracle is a uniform function $f$. This allows $\mathcal{D}$ to distinguish $F'_k$ from $f$ and thus $F'_k$ is not a PRF. Apr 18, 2018 at 7:18
• Also we sample $x \in \{0,1\}$ only once. Apr 18, 2018 at 7:20