Scala Vector operations aren't "Effectively Constant" time

One oft-repeated fact is that in the Scala language, immutable Vectors are implemented using a 32-way trees. This makes many operations take O(log32(n)) time, and given a maximum size of 2^31 elements, it never takes more than 6 steps to perform a given operation. They call this "effectively constant" time. This fact has found its way into books, blog posts, StackOverflow answers, and even the official Scala documentation.

While this logic sounds good on the surface, it is totally incorrect, and taking the logic even one or two steps further illustrates why. This post will walk through why such logic is incorrect, explore some of the absurd conclusions we can reach if the above logic is taken to be true, and demonstrate why Scala's Vector operations are not "effectively constant" time.

About the Author: Haoyi is a software engineer, and the author of many open-source Scala tools such as the Ammonite REPL and the Mill Build Tool. If you enjoyed the contents on this blog, you may also enjoy the Author's book Hands-on Scala Programming

How does a Scala Vector work?

A Scala Vector is implemented as a balanced tree with values only at the leaves. Except instead of being binary, as is commonly seen, the trees used to implement a Vector are 32-ary: each node in the tree contains 32 children instead of just 2. Otherwise, operations on a Vector's 32-ary tree work similar to how those on a binary tree work.

Looking up an element at index i within the Vector, for example, involves starting at the root of the tree, and walking down node by node until you find the element you are interested in (There is also some book-keeping you have to do, but that doesn't change the gist of the approach). A balanced binary tree with n items is log2(n) nodes deep, and so there would be log2(n) nodes you need to walk to get from the root to the element you are interested in. In a Scala Vector, which is a 32-ary tree, there would be log32(n) items to walk. Hence this "lookup at index" operation is O(log2(n)) on a balanced binary tree, and O(log32(n)) on a Scala Vector.

Scala's Vectors are immutable, perhaps unlike many binary tree implementations. This does not affect the lookups described above, since lookups do not mutate the tree at all. It turns out many other operations on immutable binary trees are also O(log2(n)) on the size of tree. Operations like

"create a new tree with one element inserted"

"create a new tree with one element removed"

"create a new tree with one element replaced"

All take O(log2(n)) on an immutable binary tree, e.g. the immutable binary AVL-tree described here. I won't go into detail of why this is the case, but you'll have to take my word on this. There are lots of resources online going into more detail, such as this blog post explaining why inserting into a (simplified) immutable binary tree takes O(log2(n)) time.

Since Scala's Vectors are immutable balanced 32-ary trees rather than binary trees, all these operations (lookup, insert, remove, replace) take O(log32(n)) time instead of O(log2(n)) time. Given the max size of a machine integer on the JVM ( 2^31 , or 2147483648 ) that means it never takes more than 6 steps to perform an operation on the Vector.

So far, all this is true and uncontroversial.

According to the official Scala documentation, this makes those operations take "effectively constant" time. It is that widely repeated claim that is entirely incorrect, and the rest of this blog post illustrates why.

O(log32(n)) is O(log2(n)/5) is O(log2(n))

The most straightforward way of seeing that Scala's O(log32(n)) Vector operations are really "logarithmic", O(log2(n)) by asymptotic Big-O analysis, is to reduce one to the other. It turns out that this does not require any tricks, just high-school math and the very-basics of Big-O notation.

log32(n) is the same as logK(n)/logK(32) , for any arbitrary constant K . This is simple math, and unrelated to Big-O notation, data-structures, or the Scala language.

Given K = 2 , log32(n) becomes log2(n)/log2(32) , which resolves to log2(n)/5 , or 0.2 * log2(n) .

Now we know Scala's O(log32(n)) Vector operations are O(0.2 * log2(n)) Vector operations, we can apply the Multiplication by a Constant rule O(kg)=O(g) if k is non-zero , with k = 0.2, g = log2(n) , and find that Scala's Vector operations are O(log2(n))

That is all there is to it. None of the rules here are particularly novel, complex, or exotic. This is something a first-year college student should be able to figure out as part of their first homework on big-O notation.

And so it turns out, that Scala's O(log32(n)) Vector operations are simply logarithmic: they take O(log(n)) time to perform on a Vector with n items.

Arguments for "Effectively Constant"

Unfortunately, this isn't the end of the discussion: there are many arguments that I have seen justifying why Scala Vector operations are "effectively constant", despite the "obvious" analysis above. I will go into various arguments I have seen and explain why they're wrong.

But the real-world performance is good due to cache effects!

One point that often muddles the discussion of the Scala Vector 's performance is the fact that Scala's Vectors are actually reasonably fast. Now, they are not very fast, as can be seen in actual benchmarks, but overall they're not terrible. Part of this is due to things like cache-locality, where the 32-way branching of the tree means that elements at the leaves are stored in relatively-compact 32-element arrays. This makes traversing or manipulating them substantially faster than working with binary trees, where any traversal has to follow a considerable number of pointers up-and-down in order to advance a single element.

This improvement in cache locality is a real effect, that can be measured.

However, cache effects and similar things are irrelevant when analyzing the "asymptotic" Big-O performance of operations on a data structure. In such a scenario, you only want to analyze the behavior of a data structure in an idealized setting. Similarly, in actual benchmarks, the big-O analysis is irrelevant when compared to the hard numbers of "how fast is my code actually running here?". Neither analysis is all encompassing, and both are useful in their own way.

The question then is: what are we discussing? Idealized, asymptotic, Big-O performance, or the performance of actual benchmarks? In the case of the official Scala documentation, there is not a single concrete number to be found, but lots of terms like "Constant", "Log" and "Linear". This is clearly describing asymptotic Big-O performance, so any cache-effects and similar things that would turn up on actual benchmarks can-and-should be disregarded.

Similarly, when programmers discuss and compare various data-structures in the same programming language, without any concrete application benchmark in mind, most often it's in terms of the asymptotic Big-O performance of their various operations.

Now, we could be using some heretofore unheard of analysis technique that's not one of the two described above, but until someone rigorously describes that technique to me, I have to assume we're using one of the well-known approaches mentioned here.

We know we're talking about the idealized, asymptotic, Big-O performance characteristics of Scala Vectors, both in the official documentation and colloquially. While Big-O analysis is no where near the complete picture in a world with cache-effects and significant constant factors, if we choose it to analyze the performance of our data-structures, we must follow the rules for our analysis to make any sense.

And thus, by the rules of Big-O notation, Scala's O(log32(n)) Vector operations are just O(log(n)) , not "effectively constant".

But log32(Int.MaxValue) cannot go above 6!

Given a Scala Vector operation that takes log32(n) steps to complete, given the max size of a Vector is 2^31 (Scala 32-bit signed integers have 2^32 values, from -2^31-1 to +2^31 ), it will never take more than 6.2 steps to complete the operation.

This is true and uncontroversial.

What is invalid is the claim "6 is a constant, therefore the runtime of the operation is effectively constant".

This reasoning falls apart when you consider the formal definition of Big-O notation:

Let f and g be two functions defined on some subset of the real numbers. One writes f(x) = O(g(x)) as x -> infinity if and only if there is a positive constant M such that for all sufficiently large values of x , the absolute value of f(x) is at most M multiplied by the absolute value of g(x)

Note that it is defined for x -> infinity . Not x -> 2^31 , or x -> 31337 , but x -> infinity . Thus, the fact that a Scala Vector running on the JVM uses machine-integers that can't go above 2^31 , is entirely irrelevant to Big-O notation. And as mentioned above, until someone formally-defines a rigorous new analysis approach to replace Big-O notation or actual-benchmarks, those two are the only tools we have to analyze these things.

Reductio Ad Absurdum

While one way of disproving an argument is to find a flaw in the logic, another common approach is to use the same logic to prove something absurd that cannot possibly be correct. Since we know the end-result is false, and we reached that end-result using our argument, we therefore know that there must be a flaw in our argument even if we do not know exactly what the flaw is.

It turns out, using the same arguments that "show" log32(n) is "effectively constant", we can prove all sorts of absurd results. This section will explore a selection of them.

What if the max size of input mattered?

Earlier on, I had shown that we have to be discussing aymptotic, Big-O complexity when considering Vector operation performance, and showed that the " log32(n) cannot go above 6.2 " logic used to shown they were "effectively constant" was invalid given the definition of asymptotic Big-O complexity. But let's say we ran a thought experiment: imagine if we could perform asymptotic, Big-O analyses with a maximum size of input, e.g. n < 2^31 ? What then?

While it is true that log32(n) will not go above a certain constant given the size limits on a JVM machine integer, there are many other rates of growth which also do not grow above a certain constant given the same size limits!

It turns out that given a limit of n < 2^31 :

Any O(log2(n)) operation will complete in no more than 32 steps

operation will complete in no more than steps Any O(n) operation will complete in no more than 2147483648 steps.

32 and 2147483648 are just as constant as 6.2 is. So why can't all my O(n) and O(log2(n)) operations running on JVM machine-integers be called "effectively constant" too?

Even if you try and limit the reasoning to " 6 is a small constant", what defines "small"? Perhaps 2147483648 doesn't count as "small", but I think 31 is pretty small.

For this argument to be valid, there needs to be some reason why f(x) < 6.2 is enough to call f "effectively constant", but f(x) < 31 isn't. Or, we need to accept that O(log2(x)) is also "effectively constant", a claim I think many would agree is absurd. If we aren't able to provide the former and unwilling to accept the latter, we have to accept that this " f(x) cannot go above 6.2 for n limited to JVM machine-integers" logic is bunk.

What if constant factors mattered?

Earlier on, I had shown that given the definition of Big-O complexity, constant factors in performance are entirely irrelevant. Given that O(log32(n)) is equivalent to O(0.2 * log2(n)) , that is then equivalent to O(log2(n)) by the rules of Big-O analysis.

However, we could run a thought experiment: what if constant factors did matter in Big-O analysis, and could shift the balance between "logarithmic" and "effectively constant"?

It turns out that given:

O(log2(n)) algorithms are "logarithmic"

algorithms are "logarithmic" O(log32(n)) , or O(0.2 * log2(n)) , algorithms are "effectively constant"

You can then "prove" all sorts of crazy things!

Let's consider a single O(0.2 * log2(n)) Vector lookup to be "effectively constant". However what if we then perform five vector lookups?

Given that five Vector lookups takes 5 times as long as a single Vector lookup, the combined Big-O complexity of the five lookups is O(5 * 0.2 * log2(n)) which is O(log2(n)) , which is "logarithmic".

Now, we've shown that a single Vector lookup is "effectively constant", but five consecutive Vector lookups is "logarithmic"! This is clearly absurd.

What if constant factors mattered? Part deux

Taking this logic further, let us imagine we have an O(log2(n)) algorithm, say a binary search, implemented in say Python. At a first approximation, CPU-bound Python code runs about ~40x slower than the equivalent Java or Scala code. If we've decided that constant factors matter, we could say that the same algorithm implemented in Java would be O(0.025 * log2(n)) .

Given that we've already said that O(log2(n)) is "logarithmic" and O(0.2 * log2(n)) is "effectively constant", where does that leave O(0.025 * log2(n)) ? At the very least, we'd have to say that it's better that "effectively constant".

You can count the "time" taken to run an algorithm any number of ways: wall-clock, CPU-time, assembly instructions, etc.. Whatever X you use, you'll still find that the algorithm in Java uses ~40x less Xs than the same algorithm in Python.

And thus, we have shown that a "logarithmic" algorithm implemented in Python, becomes "effectively constant" when implemented in Java. Clearly an absurd conclusion, which illustrates precisely why constant factors are not considered in Big-O analysis.

Conclusion

Scala's immutable Vector operations (as well as the related Map and Set operations) are logarithmic: the time they take to execute grows logarithmically with the size of the Vector. We have seen that this can be proved by high-school maths and the basics of Big-O notation. What's more, if we accept the logic arguing that Scala Vector operations are "effectively constant" time, we could then prove that:

None of which make any sense.

The basic problem here is that "effectively constant" is not a thing with a rigorous definition and broadly-accepted meaning.

"Big-O" notation is a thing, as is Big-O constant time O(1) . Small-O, Big-Omega, Small-Omega, Big-Theta notations, while less common, are well-known and have rigorously defined meanings. "Amortized O(1) ", " O(1) assuming integer operations are all O(1) ", or " O(1) expected but O(n) worst-case" are all well-defined properties that people throughout the computer-science/software-engineering communities can understand.

Coming from an entirely different angle, "100ns per operation with a 1,000,000 item collection" or "Operation takes 10 CPU cycles per input element" are also things: useful, rigorous, and broadly understood.

"effectively constant given a maximum size of input" does not have a rigorous, useful definition. The definitions I have seen ascribed to it are full of holes, and lead to the absurd conclusions shown above. This is not surprising, given the first line of the formal definition states as x -> infinity .

It feels good to come up with novel data-structures that in practice fare better that those that came before. It also feels good to optimize them to work well with the underlying hardware. However, we should not abuse well-established notation just because we think "we know better".

If we think there's a better approach for analyzing algorithm performance, we should rigorously define it so it can be properly discussed, and ensure we have satisfactory answers to all the holes described above. Until then, we should toe the line and accept that Scala's Vector operations are all O(log(n)) : their execution time is logarithmic in the number of items involved.

About the Author: Haoyi is a software engineer, and the author of many open-source Scala tools such as the Ammonite REPL and the Mill Build Tool. If you enjoyed the contents on this blog, you may also enjoy the Author's book Hands-on Scala Programming