Using randomized approach we can guarantee that the equality problem has O(1) complexity (in communication). With other definitions of of equality (not strictly equal), is there a general approach to design a randomized algorithm with bounded error? Or how to determine if there is no way such an algorithm can be designed?

  • $\begingroup$ You ask for the case of 1-bit difference, but I suspect once we answer this you will ask about some other case. Please fix your question in advance and stick to it. $\endgroup$ Sep 19, 2017 at 6:07
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1 Answer 1


Suppose that the inputs are $x,y \in \{0,1\}^n$.

Consider the following protocol:

  • The parties decide on a random string $z \in \{1,2\}^n$.
  • Alice sends Bob $\alpha = \sum_{i=1}^n x_i z_i \pmod{3}$, and Bob sends Alice $\beta = \sum_{i=1}^n y_i z_i \pmod{3}$.
  • The parties compute $\gamma = \alpha - \beta \pmod{3}$.

Suppose that the Hamming distance between $x$ and $y$ is $d$. Then $\gamma$ is the sum of $d$ random values in $\{1,2\}$. Using linear algebra, we can calculate $$ \Pr[\gamma = 0] = \frac{1}{3} + \frac{2}{3} \left(-\frac{1}{2}\right)^d.$$ (You can also prove this formula by induction.)

In particular, when $d=0$ the probability is $1$, when $d=1$ the probability is $0$, and otherwise the probability is in the range $[1/4,1/2]$ (corresponding to $n=3$ and $n=2$, respectively).

The suggests a protocol of the following form:

  1. The parties decide on $N$ random strings $z_1,\ldots,z_N \in \{1,2\}^n$.
  2. Alice sends Bob $\alpha_j = \sum_{i=1}^n x_i z_{j,i} \pmod{3}$ and Bob sends Alice $\beta_j = \sum_{i=1}^n y_i z_{j,i} \pmod{3}$, for $1 \leq j \leq N$.
  3. The parties compute $\gamma_j = \alpha_j - \beta_j \pmod{3}$ for $1 \leq j \leq N$.
  4. If all $\gamma_j$ are zero or all are non-zero, output "Hamming distance is probably at most 1", otherwise output "Hamming distance is definitely more than 1".

As in the case of equality, this protocol has only one-sided error: whenever the algorithm outputs "Hamming distance is more than 1", it is always correct. It remains to determine the connection between $N$ and the error probability $\epsilon$. Let $p = \Pr[\gamma = 0]$. If $p \neq 0,1$, the probability that the algorithm outputs the incorrect answer is $$ p^N + (1-p)^N \leq (1/2)^N + (3/4)^N. $$ This shows that $\epsilon$ is exponentially small in $N$, and so in order to guarantee an error probability of $\epsilon > 0$, it suffices to take $N = O(\log(1/\epsilon))$.


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