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I have read some papers that give the definition of one-wayness of PKE schemes.

Let $\Pi = (G,E,D)$ be a PKE scheme, and the security game of OW-CPA is defined as follow:

$$\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(k) = \Pr\left[ m = m' \left\vert \begin{gathered} (pk,sk) \gets G(1^k) \\ m \gets \mathcal{M}_{k} \\ c \gets E_{pk}(m) \\ m' \gets \mathcal{A}(pk,c) \\ \end{gathered}\right.\right]$$

We say $\Pi$ is secure in the sense of OW-CPA if $\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(\cdot)$ is nelgibile.

My question is that how does the challenger choose $m$ over $\mathcal{M}_{k}$ randomly? Is $m$ an element picked uniformly from $M_{k}$?

If $\mathcal{M}_{k} = \mathcal{M} = \{0,1\}^*$, $m$ cannot be picked uniformly from $M_{k}$.

If $|\mathcal{M}_{k}| = \mathrm{poly}(k)$, then $\Pi$ cannot be secure in the sense of OW-CPA if $\mathcal{D}_{k}$ is uniform.

Let $F$ be the sample circuit, then $$\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(k) = \Pr\left[ m = m' \left\vert \begin{gathered} (pk,sk) \gets G(1^k) \\ m \gets F(\mathcal{M}_{k}) \\ c \gets E_{pk}(m) \\ m' \gets \mathcal{A}(pk,c) \\ \end{gathered}\right.\right]$$

Assume that the distribution of message is $\mathcal{D}_{k}$, thus $m$ follows $\mathcal{D}_{k}$. We cannot define that $\Pi$ is OW-CPA secure if for every distribution $\mathcal{D}_{k}$, $\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(\cdot)$ is negligible, because it is not OW-CPA secure if $\mathcal{D}_{k}$ is a one-point distribution.

Perhaps, we may define the PKE scheme as $\Pi = (G,E,D,F)$ where the sample circuit $F$ is given at first. But the notion of IND-CPA or SS-CPA does not need $F$.

[FO99] Secure Integration of Asymmetric and Symmetric Encryption Schemes

[BF03] Identity-Based Encryption from the WeilPairing

[GH05] Security Notions for Identity Based Encryption

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  • $\begingroup$ I would suppose that $\mathcal M_k$ is a message sampler, not a set. Usually the text leading up to this formula would introduce $\mathcal M_k$. $\endgroup$ – SEJPM Aug 19 at 18:10
  • $\begingroup$ @SEJPM, It is $c$, I have corrected it. It is OK if you regard $\mathcal{M}_{k}$ as a message sampler. As I know, IND-CPA implies OW-CPA. But the definition of IND-CPA does not need the message sampler. So, if there is an IND-CPA secure PKE scheme with a strange message sampler (e.g. the distribution is almost a one-point distribution), then it cannot be OW-CPA secure. $\endgroup$ – TeamBright Aug 20 at 1:16
  • $\begingroup$ Are the references at the end papers that have this OW-CPA definition as well? $\endgroup$ – SEJPM Aug 21 at 15:37
  • $\begingroup$ @SEJPM [GH05] and [BF03] give the definitions of OW-ATK for IBE. The definitions come from [FO99]. The defintions make me confuse. I do not why does IND implie OW, whatever it is IBE or PKE. $\endgroup$ – TeamBright Aug 22 at 4:51
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Generally in cryptography papers, if a distribution is not specified, either it's uniform, or the results are essentially determined by characteristics of the distribution like its min-entropy.*

  • In the Fujisaki–Okamoto paper, $\mathtt{MSPC}_k$ (corresponding to $\mathcal M_k$) could be taken to be a set of cardinality $2^{\operatorname{poly}(k)}$ sampled uniformly at random, or it could be a more generally distribution itself with min-entropy $H_\infty = \operatorname{poly}(k)$. It's not clear from the paper but the results hold either way.

  • In the Galindo–Hasuo paper, the notation is explicitly defined as such in §2.

Obviously the upper bound on advantage is at least $2^{-H_\infty}$, or $1/\!\left|\mathcal M_k\right|$ in the uniform case.

It is true that OW-CPA (or ‘OW-Passive’ as some authors call it, because there's no oracle involved) does not imply IND-CPA—for example, the RSA trapdoor permutation $x \mapsto x^3 \bmod n$ has OW-CPA security but not IND-CPA security. This is why, e.g., RSAES-OAEP exists: to shoehorn a message $m$ from some nonuniform distribution into a nearly uniform random ‘message representative’ $x$.

(A simpler generic approach, of course, is to choose $x$ uniformly at random and then use $H(x)$ as a secret key for a symmetric authenticated cipher, as RSA-KEM does, and as is the modern paradigm for new cryptosystems like the NIST PQC submissions; this is what the Fujisaki–Okamoto paper is about.)


* This is a rule for readers, not for authors. Authors: Please say ‘uniform’ if that's what you mean!

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  • $\begingroup$ I know that OW-CPA does not imply IND-CPA. I want to know why does IND-CPA imply OW-CPA. I have understood your answer. Can I say that $\text{IND-CPA} \Rightarrow \text{OW-CPA}$ means that for every IND-CPA secure PKE scheme, there exists a distribution of message space such it is OW-CPA secure? Actually, it holds true if the distribution is uniform enough. $\endgroup$ – TeamBright Sep 9 at 16:03
  • $\begingroup$ I think (but haven't worked out and formalized and proven) that you can probably say: for every IND-CPA PKE scheme, for all distributions on messages with min-entropy $\varepsilon$, the OW-CPA advantage is bounded by $2^{-\varepsilon} + \mathit{negl}$. $\endgroup$ – Squeamish Ossifrage Sep 10 at 0:21
  • $\begingroup$ Thanks, this conclusion can be the part of answer of my another question. crypto.stackexchange.com/questions/65905/…. The relation between two security notions depends on the size of plaintext space. $\endgroup$ – TeamBright Sep 10 at 1:20

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