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Approximating the size of a mesh network


$$ \newcommand{\E}{\mathrm{E}} \newcommand{\Var}{\mathrm{Var}} \newcommand{\Cov}{\mathrm{Cov}} $$


We present here a method for approximating the amount of nodes in a mesh network. We also introduce an experiment to test our approximation method. The size approximation is done by taking the minimum hashes of various node ids and performing a transformation over those values.

The experiments are written in Rust and could be found here [github].

Hashing node ids

Let $G$ be a connected network of nodes. Every node is connected to a few other nodes that we call his neighbors. How can a node get an approximation of the amount of nodes in the network?

We will discuss here a distributed algorithm that allows all the nodes of the network to calculate an approximation.

Assume that every node has a unique identifier. For example, a public key or a hash of a public key. Let $I$ be the set of all possible node ids. Assume that we have a hash functions $h: I \rightarrow [0,1]$. This function takes a node id as input, and returns a value between 0 and 1 as output.

How to practically obtain this kind of function? We could use some cryptographic hash function, like SHA256, as follows:

$$h(x) = sha256("myhash" || x) / 2^{256}$$

Where || means concatenation. We divide by $2^{256}$ to normalize, so that we get a value between $0$ and $1$.

Observing the minimum hash value

Consider the minimum value of the function $h$ over the set $V(G)$. (We implicitly assume that $V(G) \subseteq I$). In other words, we apply $h$ over all the node ids in the network, and pick the minimum value of $h$. We call this value $min$.

$min$ has some relation to the amount of nodes in the network. If there are more nodes in the network, we expect $min$ to be smaller. This is because it is more likely that some node $v \in V(G)$ will get a small value for $h(v)$.

If there is only one node in the network, we expect $min$ to be about $1/2$, on average. Extending this idea, if there are two nodes in the network, we expect $min$ to be about $1/3$. Generally for $n$ nodes, we expect $min$ to have the value $1/(n+1)$.

We can prove this idea more rigorously. Consider $n$ random node ids $v_1, \dots v_{n}$. What is the probability that $min$ is going to be larger than $t$? This requires all of the values $h(v_j)$ to be larger than $t$. The probability of this event is $\Pr[min > t] = (1-t)^n$, because $\Pr[h(v_j) > t] = 1-t$, and the values of $h$ for different nodes are independent. Therefore $F_{min}(t) = \Pr[min \leq t] = 1 - (1-t)^n$. This is also known as the cumulative probability function of $min$.

$F(t)$ is differentiable, and so we can obtain the density function: $f(t) = F'(t) = n(1-t)^{n-1}$. To obtain the mean of $min$ we can then calculate: $$\E[min] = \int_{0}^{1} tf(t)dt = \int_{0}^{1} nt(1-t)^{n-1}dt$$

To solve this we first calculate:

$$\begin{split} \int_{0}^{1}t(1-t)^{n-1}dt = & \left[-\frac{t(1-t)^n}{n}\right]{0}^{1} - \int{0}^{1} 1 \cdot \frac{(1-t)^n}{-n}dt \ & = \int_{0}^{1}\frac{(1-t)^n}{n}dt = \left[\frac{(1-t)^{n+1}}{-n(n+1)}\right]_{0}^{1} \ & = \frac{1}{n(n+1)} \end{split}$$

Hence $\E[min] = n\cdot\frac{1}{n(n+1)} = \frac{1}{n+1}$. This result is similar to our intuitive idea earlier.

For the sake of completeness, we are adding here the computation for $Var[min]$. $Var[min] = \E[min^2] - \E[min]^2$. We already know $\E[min]$, we are now left to calculate $\E[min^2]$.

$$\begin{split} \E[min^2] = & \int_{0}^{1} nt^2(1-t)^{n-1}dt = \left[\frac{nt^2(1-t)^n}{-n}\right]{0}^{1} - \int{0}^{1} n\cdot 2\cdot t\cdot\frac{(1-t)^n}{-n}dt \ & = 2\int_{0}^{1}t(1-t)^n dt = \left[\frac{2t(1-t)^{n+1}}{-(n+1)}\right]{0}^{1} - 2\int{0}^{1} \frac{(1-t)^{n+1}}{-(n+1)}dt \ & = 2\int_{0}^{1} (1-t)^{n+1} dt = \left[2\cdot\frac{(1-t)^{n+2}}{-(n+1)(n+2)}\right]_{0}^{1} \ & = \frac{2}{(n+1)(n+2)} \end{split}$$

Therefore $$\begin{split} \Var[min] = & \E[min]^2 - \E[min^2] = \frac{2}{(n+1)(n+2)} - \left(\frac{1}{n+1}\right)^2 \ & = \frac{2(n+1)}{(n+1)^2(n+2)} - \frac{n+2}{(n+1)^2(n+2)} = \frac{n}{(n+1)^2(n+2)} \end{split}$$

We can also obtain that $\Var[min] \leq \frac{n+1}{(n+1)^2(n+2)} = \frac{1}{(n+1)(n+2)} \leq \frac{1}{(n+1)^2}$.

Using multiple hash functions

Having the minimum of the hash function $h$ over all the node ids in the network gives us some idea about the amount of nodes in the network. $min$ is about $1 / (n+1)$, so we approximate $n$, the amount of nodes in the network, to be $(1 / min) - 1$.

As we have seen above, the variance of $min$ is pretty large. This means that there is a possibility for large errors in our approximation method. One idea to deal with this problem is to add more hash functions. Instead of just one hash function $h$, we will have $r$ hash functions, defined for example as:

$$h_i(x) = sha256("myhash" || x || i) / 2^{256}$$

Where $0 \leq i < r$. With $r$ hash functions we can obtain $r$ different min values: $min_i$ for $0 \leq i < r$. Those $r$ minimum values hold more information about the network size than just one minimum value, however we still have to figure out how to combine all those minimum values into one approximation of the network size.

Experimenting with combining multiple minimums

We wrote a basic experiment (In Rust) to check the various methods for combining minimum values for approximating the amount of nodes. It can be found here [github].

We have tried various methods:

  • Averaging all the minimum values and then calculating $(1 / avg) - 1$.
  • Calculating $(1 / min_i) - 1$ for every $0 \leq i < r$ and then averaging the values.
  • Calculating harmonic average over the minimum values and then calculating $(1 / avg) - 1$.
  • Calculating $(1 / min_i) - 1$ for every $0 \leq i < r$ and then calculating harmonic average over the values.

This is how the various combination schemes look like on code:

pub fn approx_size_harmonic_before(mins: &[u64]) -> usize {

    let fmeans = mins.iter()
        .map(|&m| m as f64)
    let hmean_min = harmonic_mean(&fmeans);
    (((u64::max_value() as f64)/ hmean_min) as usize) - 1

pub fn approx_size_harmonic_after(mins: &[u64]) -> usize {
    let trans = mins.iter()
        .map(|&m| (u64::max_value() / m) - 1)
        .map(|x| x as f64)

    harmonic_mean(&trans) as usize

pub fn approx_size_mean_before(mins: &[u64]) -> usize {
    let fmeans = mins.iter()
        .map(|&m| m as f64)
    let mean_min = mean(&fmeans);
    (((u64::max_value() as f64)/ mean_min) as usize) - 1


pub fn approx_size_mean_after(mins: &[u64]) -> usize {
    let trans = mins.iter()
        .map(|&m| (u64::max_value() / m) - 1)
        .map(|x| x as f64)

    mean(&trans) as usize

These are the results:

$ cargo run --release
   Compiling approximate_net v0.1.0 
    Finished release [optimized] target(s) in 1.44 secs
     Running `target/release/approximate_net`
Calculating error ratios for approximation functions...
num_iters = 100
num_mins  = 40
num_elems = 1000000

err_ratio for approximation functions:
approx_size_harmonic_before    : 10.208276536189299
approx_size_harmonic_after     : 0.14301947584857105
approx_size_mean_before        : 0.14301974775883225
approx_size_mean_after         : 10.208276278935292

Explaining the experiment:

We generate a set of size num_elems. Each element is of type u64: a 64 bit number. We then use num_mins different hash functions to find the minimums over all the elements. We then obtain num_mins different minimums for all the different hash functions. Finally we use one of 4 methods to approximate the amount of elements in the set.

We do this whole process num_iters times, to get a general idea of how much deviation each of those methods have from the original amount of elements in the set. To check if an approximation method works well, we calculate the error ratio. It is calculated as the standard deviation divided by num_elems, the amount of elements in the set. We approximate the standard deviation using this calculation:

$$\sigma \approx \sqrt{\frac{1}{numIters} \sum_{0 \leq j < numIters}(approx_j - numElems)^2}$$

And we then calculate $errRatio = \frac{\sigma}{numElems}$.

Looking at the results, approx_size_harmonic_after and approx_size_mean_before give very similar results, and they have the best results in this group. The other two methods (approx_size_harmonic_before and approx_size_mean_after) have very bad results.

A friend noted that approx_size_mean_after and approx_size_harmonic_before should be identical (up to floating point math errors). This is because: $$\begin{split} mean_i{ \left(\frac{1}{a_i} - 1 \right) } & = \frac{\sum_{i}{\left(\frac{1}{a_i} - 1\right)}}{n} \ & = \frac{\sum_{i}{\frac{1}{a_i}}}{n} - 1 = \frac{1}{\frac{n}{\sum_{i}{\frac{1}{a_i}}} } - 1 \ & = \frac{1}{harmonic_{i}{(a_i)}} - 1 \end{split}$$

For similar reasons, approx_size_mean_before and approx_size_harmonic_after should be pretty close.

Our choice for combining the min values is going to be the approx_size_mean_before, as it is the simplest to reason about.

Distributed algorithm for calculating the hashes minimums

One of the reasons we chose the minimum hashes method for approximating the amount of nodes in the network is that it is easy to calculate in a mesh network.

Every node in the network maintains an array of the nodes with lowest hashes that he ever seen, as follows: $minHash[i] = (nodeId, timestamp, signnodeId)$ for $0 \leq i < r$.

$minHash[i]$ corresponds to the node with lowest hash value for the function $h_i$. $signnodeId$ is a digital signature by the node $nodeId$ over some timestamp. This proves that $nodeId$ was recently alive.

Given that every node knows the set of minimums for all the $r$ different hash functions, every node in the network can calculate an approximation for the amount of nodes in the network.

We now describe the distributed algorithm for a given node $x$:


When a node $x$ enters the network, he first fills: $minHash_x[i] = (x, signx)$ for $0 \leq i < r$. In other words, $x$ fills himself as the node with lowest hash for all the hash functions.

Periodic cleaning

For every node $x$, every constant period of time the following happens:

  1. $x$ checks his list of $minHash$ nodes. If any timestamp is too old, the entry $minHash[i] = (nodeId, timestamp, signnodeId)$ is removed from the list and $x$ puts himself instead: $minHash_x[i] = (x , currentTimestamp, signx)$.

  2. $x$ sends all his direct network neighbors a message the contains the updated $minHash$ entry: $UpdateMinNodeId(minHash_x[i])$

Receipt of UpdateMinNodeId message

Assume that a node $x$ receives an UpdateMinNodeId message with $minHash[i] = (nodeId, timestamp, signnodeId)$. Also assume that $x$ currently have a current entry: $minHash_x[i]$ as the node with lowest value for $h_i$.

  1. If the timestamp is too old, abort.
  2. If the signature by $nodeId$ over the timestamp is not valid, abort.
  3. If $h_i(nodeId) > h_i(minHash_x[i].nodeId)$ then abort. ($x$'s node with lowest $h_i$ value has lower value than the candidate).
  4. If $h_i(nodeId) = h_i(minHash_x[i].nodeId)$ and the timestamp of the candidate is older then abort.
  5. Update $minHash_x[i] = minHash[i]$
  6. $x$ sends all his direct network neighbors a message that contains the updated $minHash$ entry: $UpdateMinNodeId(minHash_x[i])$

Further research


It is known that there are more accurate methods to approximate the unique amount of elements in a set. However, not all of those methods are adequate to work as a distributed algorithm.

We know that the arithmetic mean gives good enough results for approximating the network size for our purposes, however it is possible that better methods exist to approximate the network size only given various minimum values of random hash functions over the node ids.


If a node can choose his own node id, he may generate many node ids until his node id gives a low value for one of the hashes. This can allow one participant in the network to greatly affect the approximation for the amount of nodes in the network. Currently the only way we know of to avoid this problem is to not let nodes pick their own node ids, and to invalidate node ids from time to time. See the article "How to spread Adversarial Nodes? Rotate!" by Christian Scheideler for more about this idea.

Further reading