Given that:

$$ SD\bigg( (r, \langle r, s \rangle),(r, b) \bigg) < \mathrm{negl}(n)$$

where $SD$ stands for statistical distance, $r$ is random uniform in $\{0,1\}^n$, $s$ is random uniform in $S \subseteq \{0,1\}^n$ and $b$ is a uniformly distributed bit.

It seems intuitive that, given a collision resistant hash function $h$, it should also hold that:

$$ SD\bigg( \big(h(s), r, \langle r, s \rangle \oplus 0\big),\big(h(s), r, \langle r, s \rangle \oplus 1\big) \bigg) < \mathrm{negl}(n)$$

but I cannot seem to prove this. I've tried using the formal definition of $SD$, but I don't know how to handle the fact there are tuples so I haven't even reached a point where I could incorporate the first claim.

Is there a way to show this from the first claim? Or am I wrong and there's a way to refute this? (for context -- I'm trying to show that $\big(h(s), r, \langle r, s \rangle \oplus b\big)$ is a statistically hiding commitment scheme).


  • $\begingroup$ May I ask how the question arises? $\endgroup$
    – Daniel S
    Commented Aug 11, 2022 at 17:10
  • $\begingroup$ @DanielS Yes - it's in the context of commitment schemes. I'm working through some exercises and this is part of one of them. Note that my construction may be wrong. In any case, would appreciate gaining a general understanding of how such SD claims work. If you have a different, but similar, example in mind - that could also be helpful (i.e. of SD claims involving tuples of distributions rather than just one). $\endgroup$
    – Anon
    Commented Aug 11, 2022 at 17:15

1 Answer 1


I'm not sure whether this answers your question, but here is a proof that it should be computationally infeasible to show a statistical distance between the two distributions for a preimage resistant hash function. Let $h$ and $r_0$ be two values such that $$\left|\mathbb P(h(s)=h, r=r_0, \langle r,s\rangle=0)- \mathbb P(h(s)=h, r=r_0, \langle r,s\rangle=1)\right|>\mathrm{negl}$$ which is equivalent to the distinguishability criterion. This tells us that $$\left|\mathbb P(\langle r_0,s\rangle=0|h(s)=h)-\mathbb(\langle r_0,s\rangle=1|h(s)=h)\right|>\mathrm{negl}$$ note that the law of total probability the sum of the above two probabilities is 1. Let us assume then that $$\mathbb P(\langle r_0,s\rangle=0|h(s)=h)>1/2+\delta$$ for some non-negligible $\delta$ (the other case being very similar). We will use this $h$ and $r_0$ to subexhaustively construct a pre-image $s$ such that $h(s)=h$.

We restrict ourselves to evaluating $h(s)$ on inputs that statisfy $\langle r_0,s\rangle =0$, then by Bayes theorem $$\mathbb P(h(s)=h|\langle r_0,s\rangle=0)=\frac{\mathbb P(\langle r_0,s\rangle=0|h(s)=h)\mathbb P(h(s)=h)}{\mathbb P(\langle r_0,s\rangle=0}=(1+ \frac 2\delta)\mathbb P(h(s)=h)$$ and we have improved our probability of success by a factor of $(1+2 delta)$ over exhaustion. If multiple $r_i$ values deviate non-negligibly and independently, we can improve the advantage by choosing to exhaust inputs that satisfy the multiple linear constraints.


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