Until very recently, the major commercial programming languages were based on the idea of update in place. C, C++, Pascal, Java etc presuppose that the way to solve a programming problem is to have procedures that read and write values to memory locations, be it directly visible to the programmer (by means of pointers) or not.
Other languages, like Lisp, are based on a different assumption: that a computation is best expressed as a set of functions over immutable values. So, in order to produce any kind of repetitive computation, recursion is essential. This assumption has strong roots in lambda calculus and related mathematical formalisms.
Probably due to the popularity of the procedural paradigm, most programmers find it easier to reason about loops than about recursion. But in order to reason effectively about a program, the functional model is far superior. This is what I intend to show in this post.
The programming model behind procedural code is that of a machine that crunches data for some time and eventually stops. When it stops, we take a look at whatever came up from all this crunching and accept it as the result of the procedure. In some sense, this is the most natural model, since the underlying implementation of the algorithm runs on a real machine that does exactly that.
The problem is that it would be desirable to conduct some kind of formal reasoning about the code at hand. The same kind we use in mathematics, for example. But the machine model relies heavily on mutation of the contents of chunks of memory over time, which is not exactly amenable to a mathematical analysis.
A neat solution to this problem is the use of loop invariants (cf. “Introduction to Algorithms”). A loop invariant is a logical predicate that we ascribe to an algorithm. That predicate remains true from the beginning to the end of the algorithm execution, including all the loops that are run, despite the fact that those loops might be changing things. Loop invariants have to take time into account (not clock time, but the discrete time defined by the sequence of operations made by the algorithm) and the values of the variables at each point in time.
Let’s take a simple and common example to illustrate the point: binary search. Given a sorted array and an element, the binary search algorithm returns the index of the element in the array, or -1 if the element is not found. Intuitively, what the algorithm does is: at each iteration, it cuts down the search space to a half, always looking at the element in the middle. The size of the search space tends to zero as the algorithm runs. So, either the element is eventually found or the search space becomes empty and the element can be declared not found. Here is an implementation of the algorithm in Scala:
def search[T <% Ordered[T]](a: Array[T], elem: T): Int = {
var start = 0
var end = a.length - 1
while (start <= end) {
val mid = (start + end) / 2
if (elem == a(mid)) return mid
else if (elem > a(mid)) start = mid + 1
else end = mid - 1
}
-1
}
We can prove the intuition behind binary search using loop invariants. The property that remains true throughout its execution is:
The subarray a(start..end) contains elem if, and only if, a contains elem. (P1)
Note that we are abusing the Scala notation a bit here, when we use a(i..j) to indicate a subarray of a from i to j, inclusive on both ends.
Loop invariants have to be proved in three main sections: initiation (the initial values assigned to the variables satisfy the property), maintenance (at the beginning of each loop, the property is satisfied) and termination (when the algorithm stops, what it returns matches up with the expected result). So, here are the three sections for the iterative version of the binary search algorithm:
Initiation: lines 2 and 3 define an interval that comprises the whole array, which is equivalent to say that a contains elem if, and only if, a contains elem. So, (P1) is trivially satisfied.
Maintenance: to prove the maintenance, we have to prove two things: a) that if a contains elem then a(start..end) also contains elem; and b) that if a does not contain elem, neither does a(start..end). It is easy to see that part b) is true, because no matter what values we assign to start or end, they will define a subarray of a, even if that subarray is empty. So if elem is not in a, it cannot be in any subarray of a.
Part a) is slightly more complex. For there to be a loop in the first place, the condition in line 6 has to be false (more on this in the next paragraph). In other words, elem is either greater than, or less than, the element in the middle (recall that the array is sorted). If it is greater, start will be updated to mid + 1, effectively dropping the first half of the array, so that, when line 4 is evaluated again, the interval will consist of the second half of the array. And it is at this point that the loop invariant is maintained. The first half of the array could not possibly contain elem, so if it is in the original array, it will also be in the second half, defined by the new values of start and end, satisfying (P1). For the case where elem is less than a(mid), the proof is analogous.
Termination: there are two possible situations in which the procedure stops: when condition in line 4 is false or condition in line 6 is true. Let’s start with the former. start <= end being false implies an empty array. By definition, empty arrays do not contain any element and, in particular, do not contain elem. So it will return -1, which is the correct result in this case. In the latter case, elem == a(mid) means that we have actually found the element we were looking for, and the result in this case is mid.
Even though loop invariant proofs are mathematically precise techniques to show the correctness of an algorithm, they are somewhat cumbersome to work with, for two main reasons:
Now let’s look at a recursive version of the same algorithm:
def search[T <% Ordered[T]](a: Array[T], elem: T): Int = {
def doSearch(start: Int, end: Int): Int =
if (start > end) -1
else {
val mid = (start + end) / 2
if (elem == a(mid)) mid
else if (elem > a(mid)) doSearch(mid + 1, end)
else doSearch(start, mid - 1)
}
doSearch(0, a.length - 1)
}
It is the same algorithm, with the same runtime complexity and even the same actual running time. What is different is the programming model. Instead of a machine that changes data in place, we can think of this function as a kind of generator of an immutable sequence of immutable states. The states, in this case, are the tuples (not in the sense of a Scala syntax construct, but in the more general sense of an ordered list of elements) that make up the function’s list of parameters. The function, in this case, is doSearch().
For example, suppose you call the recursive function with these parameters:
val a = Array(4, 9, 28, 37, 40, 50, 52, 57, 60, 61, 68, 71, 74, 76, 82, 87, 92, 98)
val elem = 61
search(a, elem)
The sequence of states that comprise the execution of the algorithm for the input above is:
(0, 17) → (9, 17) → (9, 12) → (9, 9)
The arrow represents a recursive call, that yields the next state. And every state in that sequence satisfies the property (P1). What we have here, then, is a recursion invariant, that is, an invariant property that is guaranteed to be preserved between recursive calls. The way to prove a recursion invariant is basically the same as we would a loop invariant: initiation, maintenance and termination. But instead of thinking in terms of how a loop changes certain variables, we think of states and the relationship between consecutive states.
In particular, we are interested in the first state (the initiation), the last state (the termination, which can be straightforwardly transformed to the final result) and the inductive step of generating a new state from the current state, assuming that the current one satisfies the invariant. No mutation of variables and no notion of time to worry about; just the sequence of states.
And it gets even better: we don’t even need to think about the sequence itself. It suffices to establish the relationship between input and output. In terms of code, we have to establish the relationship between the first and second parameters in line 2 with the first and second parameters in lines 7 and 8.
To sum up, the next time you write code to solve some problem, try to think about what property your algorithm keeps throughout its execution. And, if possible, try to develop a (tail) recursive version of it, so that you can prove that it works with much more elegance and simplicity. The key to understand how an algorithm changes things is to observe what it does not change (more on this on a future post).