# Random oracle vs implementations like hash function

In this answer, it is stated

It has actually been shown (by Canetti, Goldreich and Halevi) that random oracles cannot exist "in all generality" in the following sense: it is possible to build pathological signature and asymmetric encryption schemes, which are secure when they internally use a random oracle, but which are insecure whenever an actual computable function is used instead of the mythical gnome-in-the-box.

I tried reading the actual paper but I am quite lost. Is there some intuition/simple example that one can provide for a cryptographic task that is secure if random oracles exist but fails when one uses a computable function instead of the random oracle?

I have also not been satisfied with the "classic" counterexamples of random oracle instantiability. I came up with the following for the upcoming version of my book, and am happy to give a sneak preview here.

Consider the following symmetric-key encryption scheme, parameterized by hash function $$H : \{0,1\}^\lambda \to \{0,1\}^\lambda$$:

$$\textsf{Enc}(K,M)$$:

• $$R \gets \{0,1\}^\lambda$$
• interpreting $$M$$ as the encoding of a boolean circuit, test whether $$M(R) = H(R)$$
• if so, set $$S = K$$
• else, set $$S \gets \{0,1\}^\lambda$$
• encrypt $$M$$ under CBC mode using $$S$$ as the IV; output the result.

• If $$H$$ is modeled as a random oracle, then the construction is CPA-secure. With overwhelming probability, $$\textsf{Enc}$$ chooses an $$R$$ that no one has ever queried to the random oracle, so $$H(R)$$ is independent of whatever $$M(R)$$ is. With overwhelming probability, $$\textsf{Enc}$$ behaves just like random-IV CBC mode, which is CPA-secure.
• If $$H$$ is any public, efficiently computable function, then the construction is not CPA-secure (not even one-time secure). The attack is obvious: ask for an encryption where the plaintext is a boolean circuit that implements $$H$$. The resulting ciphertext will contain the encryption scheme's key.