class Counter:
def __init__(self):
self.n = 0
def increment(self):
self.n += 1
def query(self):
return self.n


# Trivial

Consider the following trivial counter to the right. It keeps a tally $n$ of how many times increment has been called, which can then be queried at any time. Under the hood, since $n$ is an integer it's represented in binary (base-2) which uses $\text{len}(n_2) = \lceil \log_2 (n + 1) \rceil = \lfloor \log_2 (n) + 1 \rfloor$ bits of memory. Thus this counter has a space complexity of $\mathcal{O}(\log (n))$. You can see this in the table below:

 $n = n_{10}$ $n_2$ $\lfloor \log_2 (n) + 1 \rfloor$ 0 1 2 3 4 5 8 16 32 0 1 10 11 100 101 1000 10000 100000 1 1 2 2 3 3 4 5 6

# Morris Counter

Can we do better? That is, can we come up with a better representation (data structure)? Lets say there does in fact exist some schema such that we can represent a $n$umber using less than $\log(n)$ bits. Then, by the Pigeon Hole Principle there exist two distinct integers $a,\ b \in [n]$ corresponding to two different iterations of increments which map to the same representation in memory. Therefore, when queried they both return the same value which is a contradiction! So at least until Quantum Computers aren't a meme, we can't do better.

This lead to an interesting insight: what if we could be 'okay' with $a,\ b$ mapping to the same memory? This would involve losing some information, so we'd only have approximations the true values of $a$ and $b$ represent. Why and when would we want to do this? In 1985 Robert Morris was working at Bell Labs on a spellchecking program and wanted to keep track of trigram counts, requiring $26^3$ simultaneous counters. To reduce the memory needed, he invented the Morris Counter which approximates this count proportional to $\mathcal{O}(\log (\log (n)))$ space. You can read much more of the history here. And even with the cheap and bountiful memory we have now, Redis implements an altered Morris counter in it's key eviction policies' Least Frequently Used algorithm: this requires keeping track how many times each key in the database has been queried to prune the least used.


class MorrisCounter:
def __init__(self):
self.X = 0
def increment(self):
if random.random() < 1/(2**self.X):
self.X +=1
def query(self):
return (2**self.X) - 1

The Morris counter itself works by incrementing the counter probabilistically. That is, after increment is called $n$ times then our representation $X_n$ is increment with a probability of $\frac 1 2^{X_n}$ and query returns the estimate $\hat n = 2^{X_n} - 1$. Notice how this estimate can only take on a subset of values-- this is the tradeoff that allows us to obtain better memory complexity. Intuitively, the chance we increment $X_n$ decreases logarithmically with $n$ so we can speculate that our estimate has space complexity of $\mathcal{O}(\log (\log (n)))$. To prove that that our $\hat n$ is given by the prior formula is accurate, we can prove with induction that $\Bbb{E} \left(2^{X_n}\right) = n+1$: \begin{align*} \mathbb{E}\Big(2^{X_{n+1}}\Big) &= \sum_{i=0}^\infty \Bbb{P} (X_n = i) \cdot \Bbb{E} \Big( 2^{X_{n + 1}} \mid X_n = i\Big) \\ &= \sum_{i=0}^\infty \Bbb{P} (X_n = i) \cdot \left(\left(1- \frac 1 2^i \right) 2^i + \frac 1 2^i 2^{i+1} \right) \\ &= \sum_{i=0}^\infty \Bbb{P} (X_n = i) \cdot \left((2^i - 1) + 2\right) \\ &= 1 \cdot \sum_{i=0}^\infty \Bbb{P} (X_n = i) + 2^i \cdot \sum_{i=0}^\infty \Bbb{P} (X_n = i) \\ &= 1 + \Bbb{E} \Big(2^{X_n}\Big) \\ &= n + 2 \\ \end{align*} The second term expands to the expected value if incremented plus the expected value if not incremented. And after expanding, the first term simplifies to $1$ given by law of total probability and the second is the expected value of $X_n$.

It can also be shown by a similar process that the variance is given by $\text{Var}(\hat n) = \frac{n^2}{2} - \frac n 2 - 1 < \frac{n^2}{2}$. With this, we can use Chebyshev's inequality to set up an equation to bind the variance: \begin{align*} \mathbb{P}(|\hat n - n| \geq \varepsilon n) & \leq \frac{\text{Var}(\hat n)}{\varepsilon^2 n^2} \leq \delta \\ & < \frac{\frac{n^2}{2}}{\varepsilon^2 n^2} = \frac{1}{2\varepsilon^2} \leq \delta \\ \end{align*}

# Generalization & Improving

Instead of incrementing with probability $\frac 1 2 ^X$, we use $\frac 1 {1 + \alpha}^X$, where alpha is some constant ($\alpha = 1$ in the case of OG Morris Counter). We also have return our estimate $\frac {(1 + \alpha)^X}{\alpha} - 1$. This now gives us control over the behavior of our counter: intuitively, as $\alpha \to 1$ we approach a deterministic counter so $\alpha$ is directly related to memory usage (bits) and inversely related to variability. In his original paper, Morris showed that if we set $\alpha = \varepsilon^2 \delta$ leads to a $(1+\varepsilon)$-approximation with $1-\delta$ probability using $\mathcal{O}\left(\log \log n + \log \frac 1 \varepsilon + \log \frac 1 \delta\right)$ bits. Then, in 2022 (40+ years later!) Jelani Nelson and Huacheng Yu proved that there exists an even tighter bound of $\mathcal{\Theta}(\log \log n + \log \frac 1 \varepsilon + \log \log \frac 1 \delta)$.

Another improvement we could implement is to use a deterministic counter for small values and switch over to approximate after a certain threshold lim:


class MorrisAlpha:
def __init__(self, a=.05, lim=8):
self.X = 0
self.a = a
self.lim = lim
def increment(self):
if self.X < self.lim:
self.X += 1
elif random.random() < 1/((1 + self.a)**self.X):
self.X += 1
def query(self):
if self.X <= self.lim:
return self.X
return ((1 + self.a)**self.X)/self.a - 1


# Concurrent

What if we used many concurrent, independent Morris counters? That is, we have $y$ counters $\{\hat n_1 \dots \hat n_y\}$ and we returned the average of their outputs when queried: $\hat n = \frac 1 y \sum_{i=1}^y \hat n_i$. The expectation of a summation of independent random variables is given by $\Bbb{E}(\hat n) = \Bbb{E}\left(\frac 1 y \sum_{i=1}^y \hat n_i\right) = \Bbb{E}(\hat n_i) = n$. When a random variable is multiped by a constant $\gamma = \frac 1 y$ then it's variance changes by $\gamma^2$: $\text{Var}(\hat n) = \text{Var}\left(\frac 1 y \sum_{i=1}^y \hat n_i\right) = \frac {1}{y^2} \sum_{i=1}^y \text{Var}\left(\hat n_i\right) < \frac{n^2}{2y}$. We can then set up Chebyshev's inequality to get the following bound: $$\Bbb{P}(|\hat n - n| \geq \varepsilon n) \leq \frac{1}{2 \varepsilon^2 y} < \delta$$ for $y = \left\lceil \dfrac {1}{2 \varepsilon^2 \delta}\right\rceil$. This has space complexity of $\mathcal{O} (y \cdot \log \log n) = \mathcal{O} \left(\dfrac {\log \log n}{\varepsilon^2 \delta}\right)$.

The main takeaway is that we can now choose how tight we want the variance to be by averaging many copies of Morris, but at the cost of blowing up our memory.


class MorrisConcurrent:
def __init__(self, y):
self.X = 0
self.ys = [MorrisCounter() for _ in range(y)]
def increment(self):
[y.increment() for y in self.ys]
def query(self):
return sum([y.query() for y in self.ys])/len(self.ys)


# Mo(o)re

We can go even further by using another $z$ independent instances of the concurrent counters, each with chance of failure $\delta = \frac 1 3$. Then, we output the median of all $y$ counters where the median is valid iff we have $\frac z 2$ 'successful counters'. Since $\delta = \frac 1 3$ then a deviation of $\frac z 2 - \frac z 3 = \frac z 6$ from the median is valid. We can then set up a Hoeffding bound: $$\Bbb{P}\left(\sum_{i=1}^z Y_i \leq \frac z 2\right) < \Bbb{P}\left( \left| \sum_{i=1}^z Y_i - \Bbb{E}(Y_i) \right| < \frac z 6 \right) \leq e^{-2 (\frac{1}{6})^2 z} \leq \delta %>$$ Thus, we have a total of $yz = \left\lceil\dfrac {\log{\frac 1 \delta}}{\varepsilon^2}\right\rceil$ copies of the original Morris counter. Once any of the counters reach $\log \frac {yzn} \delta$ the chance that it increments again is $\frac \delta {yzn}$ so the chance it increments in the next $n$ increments is $\frac \delta {yz}$. Thus, if we set the bound $\delta$ such that none of the counters go above $\log \frac {yzn} \delta$ (using $\mathcal{O} \left({\log{\log \frac {yzn}\delta}}\right)$ bits) then we have a bounded space complexity of $\mathcal{O} (yz \cdot \log\log n) = \mathcal{O} \left(\dfrac{\log \frac 1 \delta \log\log n}{\varepsilon^2}\right)$.

You can see that the subsequent two algorithms are actually slower than a Generalized Morris counter with optimal $\alpha$ that Morris proved in his original paper. However, this process of taking the median of many means (of a fixed failure rate) exponentially decreases the overall failure rate and can be applied to other algorithms to obtain a better space complexity.