Chapter 3. Reasoning about correctness

Although this book focuses primarily on the data side of computation, its study cannot truly be separated from the study of procedure. The next two chapters focus on two procedural concepts that are crucial to our study of data structures: reasoning about a procedure's correctness, and reasoning about a procedure's speed.

3.1. Invariants

For most of the algorithms we've seen thus far, their correctness has been fairly obvious, and we haven't needed to argue why each algorithm provided the correct answer. Some algorithms, though, are so clever that we really need a mathematical proof to be convinced that they work. One of the most useful tools for proving that something works is the notion of an invariant.

An invariant is a condition that is preserved during the course of executing a program. Invariants are particularly useful for reasoning about loops, where the invariant is a condition that holds at the beginning of each iteration. As an example, consider the following simple method; it's obvious enough to require no proof of its correctness, but we do best to begin by examining a simple loop before progressing to less obvious examples.

public static int maxArray(int[] data) {
    int ret = data[0];
    for(int i = 1; i < data.lengthi++) {
        if(data[i] > retret = data[i];
    return ret;

Here, the loop maintains an invariant that ret is the maximum of the first i elements of data, or more succinctly,

ret = max0 ≤ j < i data[j].

To be an invariant for a loop, two properties should hold. (If you've studied mathematical proof techniques, you will recognize the similarity between these two properties and the two parts of a proof by induction (a basis step and a induction step).)

If we have a loop invariant (which by definition must satisfy these two properties), then we can infer that the invariant will still hold just after the loop terminates. Since i will be data.length following the loop, the invariant implies that ret is the maximum value of the first data.length entries — that is, all entries — of the data array.


Technically, the argument is missing something. We know that the loop ends when i is no longer less than data.length, but this doesn't imply that i will equal data.length: It could be greater than data.length. To resolve this, we could add another fact to the invariant: that i is always less than or equal to data.length. This, coupled with the fact that i is no longer less than data.length after the loop completes, will imply that i must equal data.length. We must confirm that this additional assertion adheres to the two invariant properties: First, it is true when the computer first hits the while loop (i starts at 1, and data.length will be at least that); and second, since we know i is less than data.length at the beginning of each iteration, adding 1 to the integer value will not make it exceed data.length.

A complete proof of correctness would also include some argument that the loop will eventually terminate. This is not a technicality that will concern us, though.

Note that a single loop can have many valid invariants satisfying the two properties. For example, another invariant to the above loop is i ≥ 1: It holds in starting the first iteration (when i is 1), and in each subsequent iteration, i only increases. (Because of overflow considerations, a complete argument would depend on the fact that it stops once i reaches data.length.) For that matter, 1 + 1 = 2 is another invariant, as it is something that is true at the beginning of each iteration; of course, this is an invariant for all loops, so it's hardly an interesting one. Because there are many possible invariants for a loop, to refer to the invariant for a loop is technically invalid.

Despite such intricacies, we nonetheless talk about the invariant for a loop. When we do, we are talking about the invariant that encapsulates the important information about what the loop does. In the case of our code for computing the maximum, the first invariant we examined would be this invariant. In short, if you're asked to identify the invariant for a loop, you shouldn't expect anybody to be impressed (or, on a test, to give you any credit) for responding, 1 + 1 = 2.

3.1.1. Case study: Searching an array

Onward to more interesting examples. Consider the problem of determining where within an array of numbers a particular value lies. Our method should return the index of the value, or −1 if the requested value doesn't appear at all in the array. For example, suppose we have the following array.

Array: 2, 3, 5, 7, 11, 13

Given a request to find 5, we want to respond with 2, the index within the array where 5 can be found.

The simple solution, called sequential search, simply starts at the beginning of the array and searches forward until the number is found. If the method reaches the end of the array, it returns −1.

public static int sequentialSearch(int[] dataint value) {
    for(int i = 0; i < data.lengthi++) {
        if(data[i] == valuereturn i;
    return -1;

Here, the invariant to the loop is that value does not appear in the first i elements of the array data. This is trivially true at the beginning, when i is 0, so the invariant satisfies the first required property. For the second property, assuming that the invariant holds at the beginning of one iteration means assuming that value is not among the first i elements. We will reach the next iteration only if data[i] — that is, the (i + 1)st value of i — is also not value. Thus, at the beginning of the next iteration, even though i has gone up by one, value is still not among the first i elements — and the invariant still holds.

That invariant was not a difficult example. Notice how it is similar to the invariant of maxArray: In both cases, the loop has an index (i) that iterates through subsequent integers, and the invariant simply gives information about the cumulative job that the algorithm has completed thus far. Our next example, though, does not follow this model as closely.

Suppose that we want to search an array that we already know is in increasing order. While we could use the sequential search algorithm, the binary search algorithm is a more efficient way to solve the problem. In this algorithm, we keep track of a range of indices (marked by a and c in our program) where value could lie. With each iteration, we take the midpoint of the interval, see how the value in data at that index compares to value, and discard either the lower half or the upper half based on the result.

It's a simple idea, but the binary search algorithm is notoriously difficult to program. Computer scientist Jon Bentley taught programming seminars to many professionals, and he regularly began by giving the participants half an hour to write the binary search algorithm on paper. He found that, after giving the assignment to over a hundred programmers, about 90% of participants discovered a mistake in their logic while participating in his seminar (Jon Bentley, Programming Pearls, Addison-Wesley, 1986, 36). (He was relying on their own reports — he didn't check the other 10% himself.) In another study, Donald Knuth surveyed a broad range of early computing literature and found six different versions published between 1946 and 1958; but he found that the first correct binary search code was not published until 1962 (Donald Knuth, The Art of Computer Programming, vol. 3, Addison-Wesley, 1973, 419). Here's a version written in Java.

// precondition: data array must be in increasing order
public static int binarySearch(int[] dataint value) {
    int a = 0;
    int c = data.length - 1;
    while(a != c) {
        int b = (a + c) / 2;
        if(data[b] >= valuec = b;
        else                 a = b + 1;
    if(data[a] == valuereturn a;
    else                 return -1;

Given the difficulty that others have had writing the program, though, you shouldn't trust this code without a proof. And for that, we need an invariant. Ours will have two parts, which we label (a) and (b).

For this to be correct, it needs to satisfy the two properties of the invariant.

The first property is that the invariant must hold when we first hit the loop. Clause (a) of the invariant holds trivially: In fact, since a starts at 0, there are no indices i where 0 ≤ i < a, so anything is true for all such indices. We hit a snag, though, when we look at clause (b). Since c starts at data.length − 1, there is an index i where c ≤ i < data.length — namely, c itself. And it could be that data[c] is less than value. So the invariant doesn't hold.

Hmm. If this code works, the invariant will be something different. We'll shoehorn another clause in to force our candidate invariant to be true at the beginning.

As we reach the loop, data[c] will either be less than value, or it will be at least value. If data[c] is less than value, then all values in data are less than value (since data must come to us in increasing order), and clause (c) holds true. On the other hand, if data[c] is at least value, then clause (b) holds. In either case, the entire invariant now holds when we first reach the loop.

Now to the second property of invariants. We'll suppose that the invariant holds at the beginning of the iteration; we want to show that it also holds for the values of a and c following the iteration. We'll use a and c to represent the values of the variables' respective values previous to the iteration, and a' and c' to represent the values at the iteration's end. We have two cases to consider.

data[b] ≥ value

In this case, a' is the same as a, and so clause (a) of the invariant will still hold. Note that (c) has nothing to do with a or c, so that won't change either. But the truth value of (b) depends on c, so its truth value may change; we need to show that it won't. Since c' will be b, data[c'] will be at least value. The array is increasing, so all values in data following entry c' will be at least data[c'], which is itself at least value. Thus clause (b) holds, and from that follows the invariant.

data[b] < value

In this case, c' is the same as c, so the meaning of clauses (b) and (c) do not change. One of them held as the iteration started, and the same one will hold for the next iteration too. We have only to handle clause (a). Since a' is b, we know that data[a'] is less than value. Moreover, since the array is increasing, all values in data preceding entry a' must be at most data[a'], which itself is less than value. Thus clause (a) holds, and from that follows the invariant.

With both properties proven, we have confirmed that our invariant is valid.

Since the loop coninues as long as a != c, a and c will be equal once the loop completes. Since they'll be equal then, we can replace c by a in the invariant, and we'll get something that is true once the loop finishes:

We now want to know whether the method returns the proper value. For this, note that the invariant implies that either (b) or (c) holds. If (c) holds, then the correct return value is −1, and this is indeed what the method returns, since data[a] won't match value. The other possibility is that (b) holds. If the entry at a matches value, then returning a (as the method does) is of course correct. But suppose data[a] doesn't equal value. From (a), we know that nothing in data before index a matches value. We already know from (b) that data[a] must be greater than or equal to value, and since we're supposing it's not equal, it must be greater. Since the array is increasing, this means that all the values in data after index a are also greater than value. So value can't appear before index a, at index a, or after index a. None of the entries of data match value, and returning −1 is correct.

3.1.2. Case study: Exponentiation

Consider the problem of taking a value to an integer power. (The Math class includes a pow class method for exponentiating doubles, so you may wonder why we would want to talk about exponentiating numbers. Our algorithms, though, will also apply to exponentiating other values, such as matrices.) We can easily do this the intuitive way.

public static double slowExponentiate(double baseint exp) {
    double ret = 1;
    for(int i = 0; i < expi++) {
        ret *= base;
    return ret;

You wouldn't question that this works. (If you did, you should be able to convince yourself using the invariant that ret = basei.)

But the technique is slower than necessary. If we want to find x100, then it will require 100 multiplications. The following technique would require only 9 multiplications; it is generally a much faster algorithm.

public static double fastExponentiate(double baseint exp) {
    double a = 1;
    double b = base;
    int k = exp;
    while(k != 1) {
        if(k % 2 == 0) {            b = b * bk =    k    / 2; }
        else           { a = a * bb = b * bk = (k - 1) / 2; }
    return a * b;

Of course, being faster is only useful if it also arrives at the correct answer: If a wrong answer were acceptable, then we could simply return 1 without multiplying anything at all. So: Does this algorithm always work?

To analyze the method's correctness, we need to find a loop invariant. Here is the one that works:

a ⋅ bk = baseexp

You're bound to wonder: Where did this come from? The truth is that identifying invariants is not always easy. While a beginning student can be expected to identify invariants for maxArray and slowExponentiate with a bit of practice, identifying the invariant here on one's own takes much more practice than we'll get in this course.

Given a proposal for the invariant, though, we should be prepared to verify that it indeed fulfills both properties. First, the invariant must hold when the computer first reaches the loop, which is trivial to verify, since the method sets up a as 1, b as base, and k as exp. Second, when the invariant holds at the beginning of one iteration, it must hold at the beginning of the next. We'll use a', b', and k' to refer to those variables' values in the next iteration: Knowing that a ⋅ bk = baseexp, we want to show that a' ⋅ (b')k' = baseexp. We will handle the two cases of the if statement separately.

k is even:
a' ⋅ (b')k'
=a ⋅ (b²)k/2
=a ⋅ b2 ⋅ k/2
=a ⋅ bk
k is odd:
a' ⋅ (b')k'
=a ⋅ b ⋅ (b²)(k − 1)/2
=a ⋅ b ⋅ b2 ⋅ (k − 1)/2
=a ⋅ b ⋅ bk − 1
=a ⋅ bk

In either case, then, each iteration preserves the invariant.

Thus, the invariant will still be true once the loop completes. At that time, k will be 1, and so a ⋅ b1 is the desired result. This is precisely what the method returns.

3.2. Data structure invariants

Though we couched our discussion about program correctness in terms of analyzing procedure, in fact the same concepts apply to analyzing data, too.

In the context of data structures, an invariant is a property that will hold after the completion of each operation on the data structure. Our LinkedList implementation of Figure 2.5, for example, maintained several data structure invariants:

We didn't explicitly state these invariants in Chapter 2, but you surely thought about them as invariants, though without using that exact word.

Formally, a data structure invariant is a condition that satisfies two properties. First, it must be true just after the data structure is created. And second, for each possible operation on the data structure, if the condition is true previous to the operation, then it will still be true after the operation completes. (Note the close correspondence to the two properties of a loop invariant.)

With data structure invariants, it is particularly important that the data be accessed only when completing the specified operations. This is another reason why instance variables in Java classes corresponding to data structures should be declared private.