Building pathological input for Java HashMaps - hash collisions generator

8 minute read

I’ve recently stumbled upon an article about the Weisterstrass Function, which led me to another article about Patological Objects in Mathematics, meaning, functions, numbers, etc., that have deviant, irregular, or counterintuitive properties.

There was a time when (even) the introduction of \(0\) broke the mathematics of the day. At first, \(0\) was highly pathological. For example, the 7th-century mathematician Brahmagupta said about division by zero:

A positive or negative number when divided by zero is a fraction with the zero as denominator. Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator. Zero divided by zero is zero.

Two hundred years later, another Indian Mathematician, called Mahāvīra, made a different error refuting the assertions of his predecessor:

A number remains unchanged when divided by zero.

In our time, we have no problem bending both \(0\) or \(\infty\) to our mathematical needs, so mathematicians always have found a way to incorporate the seemingly impossible into broader theories. The same story goes for irrational numbers, complex numbers, and all the other peculiarities discovered and accepted through the centuries.

In Computer Science, Pathological Input is a slightly different concept, strongly linked to the study of algorithms and data structures. In this case, the input causes atypical behavior for an algorithm, such as a violation of its average-case complexity.

In this article, we will generate pathological input for Java HashMaps, and see how bad they perform once hit with our malicious set of keys. Compared to a real-world scenario, our pathological set will contain only values that are highly susceptible to collisions. The way to do that is by “reverse engineer” the way the Object::hashCode() method works in Java.

For example, the String::hashCode() looks like this:

// This is the internal representation of a String 
// as an array of bytes.
private final byte[] value;

//... more code here

public int hashCode() {
    int h = hash;
    if (h == 0 && !hashIsZero) {
        h = isLatin1() ? StringLatin1.hashCode(value)
                        : StringUTF16.hashCode(value);
        if (h == 0) {
            hashIsZero = true;
        } else {
            hash = h;
    return h;

Going further into StringLatin1.hashCode(), we will see how simple is the actual hashing function used in Java. It’s a straightforward implementation of a polynomial hash function.

public static int hashCode(byte[] value) {
    int h = 0;
    for (byte v : value) {
        h = 31 * h + (v & 0xff);
    return h;

As you can see, the hashCode() is computed in a series of steps, each result (h) depending on the previous one. So from a mathematical perspective, our code looks like this:

\[\begin{cases} h_{0} = 0 \\ h_{1} = 31*h_{0} + v_{1} = v_{1} \\ h_{2} = 31*h_{1} + v_{2} = 31 * v_{1} + v_{2} \\ h_{3} = 31*h_{2} + v_{3} = 31 * (31*v_{1}+v_{2}) + v_{3} = v_{1}*31^{2} + v_{2}*31 + v_{3} \\ \\ \vdots \\ \\ h_{N} = 31*h_{N-1} + v_{N} = \ldots = v_{1}*31^{N-1}+v_{2}*31^{N-2}+\ldots+v_{N} \\ \end{cases}\]


  • \(N\) represents the size of the byte[] value array. In coding terms, \(N\) is the value.length
  • For simplicity reasons, we also consider \(N\) the size of the String, because we are going to use the US_ASCII charset. In this case, each character will be represented as a single byte.
  • \(v_{i}\) are the actual elements from the byte[] value array, where \(1 \leq i \leq N\). So think our String equivalent to the ordered set \((v_{1}, v_{2}, ... , v{i}, ... v_{N})\).
  • \(h_{0}\) is the initial value of h (initially int h=0).
  • \(h_{i}\) is the intermediary hashCode() value computed in the loop when iterating over \(v_{i}\) (at step \(i\)).
  • \(H=h_{n}\) is the final haschCode() value, the one returned by the Java function (at step n, when the main loop ends).

\(H\) can also be written as the sum: \(H=h_{N}=\sum_{i=1}^{N} v_{i}*31^{N-i}\)

If \(N=1\), the formula becomes: \(H=h_{1}=v_{1}\)

It means that hashCode() for single-character Strings is the actual byte value of the Character (\(v_{1}\)).

If we run the following code, we will see that our “assumption” is correct:

String a = "a";
String b = "b";

System.out.println("The byte representation of 'a' is: " + a.getBytes(StandardCharsets.US_ASCII)[0]);
System.out.println("The hashCode representation of 'a' is:" + a.hashCode());

System.out.println("The byte representation of 'b' is: " + b.getBytes(StandardCharsets.US_ASCII)[0]);
System.out.println("The hashCode representation of 'b' is: " + b.hashCode());

With the output:

The byte representation of 'a' is: 97
The hashCode representation of 'a' is:97
The byte representation of 'b' is: 98
The hashCode representation of 'b' is: 98

Because of this, it’s impossible to generate colliding single-character Strings (when \(N=1\)). All characters are different; thus, they have unique byte[] representations. But the situation becomes much more enjoyable when \(N \geq 2\).

So let’s take the case when \(N=2\). Our formula for obtaining the hashCode() of two-character Strings becomes:

\[H=h_{2}=31*v_{1} + v_{2}\]

In this regard, \(v_{1}\) is the byte value of the first character, and \(v_{2}\) is the byte value of the second character of the String. And here comes the funny part, we can find various values for \(v_{1}\) and \(v_{2}\) so that \(H\) remains the same, thus creating a collision.

There are multiple combinations of numbers \(v_{1}^{i}\) and \(v_{2}^{i}\) so that \(H=h_{2}=v_{1}^{i}+v_{2}^{i}\) is true, because \(H\) can be written as:

\[H=31*(v_{1} + 0) + (v_{2} - 31*0) \\ H=31*(\underbrace{v_{1}+1}_{v_{1}^{1}}) + (\underbrace{v_{2}-31*1}_{v_{2}^{1}}) \\ H=31*(\underbrace{v_{2}+2}_{v_{1}^{2}}) + (\underbrace{v_{2}-31*2}_{v_{2}^{2}}) \\ H=31*(\underbrace{v_{2}+3}_{v_{1}^{3}}) + (\underbrace{v_{2}-32*3}_{v_{2}^{3}}) \\ \vdots \\ H=31*(\underbrace{v_{1}+i}_{v_{1}^{i}}) + (\underbrace{v_{2}-31*i}_{v_{2}^{i}}) \\ \vdots \\ \text{ and so on ...}\]

So theoretically, there is an infinite number of pairs \(v_{1}^{i}\) and \(v_{2}^{i}\) to satisfy the condition: \(H=h_{2}=31*v_{1}^{i}+v_{2}^{i}\). To determine them, we only apply the following formulas on the initial two characters: \(v_{1}\) and \(v_{2}\):

\[v_{1}^{i}=v_{1} + i \\ v_{2}^{i}=v_{2} - 31*i\]

In practice, we will use the US_ASCII encoding, so our characters will be in the interval [32, 127].

Applying the formulas, it becomes pretty straightforward to determine all the possible colliding 2-character Strings for a given String:

public static Set<String> getCollidingStrings(String srcString) {
    if (srcString.getBytes(US_ASCII).length>2) {
        throw new IllegalArgumentException("The string should have two characters only");
    HashSet<String> result = new HashSet<>();
    byte[] crt = srcString.getBytes(US_ASCII);
    while(true) {
        crt[0] += 1; // we increment v_{1} with 1
        crt[1] -= 31; // we decrement v_{2} with 31
        if (crt[0]>127 || crt[1]<32) break; // we exit our bounds (break loop)
        result.add(new String(crt, US_ASCII)); // we add the result
    return result;

Let’s pick "aa" as the srcString:

List<String> result = getCollidingStrings("aa");
result.forEach(s -> {
    System.out.printf("s='%s'\n", s);
    System.out.printf("'%s'.hashCode()=%d\n", s, s.hashCode());



So those 3 Strings are, in a way, pathological. But they are insufficient to prove our point. A HashMap<String, T> can efficiently deal with three elements colliding, we need to generate significantly more collisions by building bigger colliding Strings.

The exciting aspect is that we can reuse "aa", "bB", "c#" to build lengthier inputs by concatenating them into different combinations. The math behind this last assumption is quite simple.

Let’s generate four-character long strings that collide, \(N=4\). The formula for the hashCode will become:

\[H=h_{8}=\sum_{i=1}^{4} v_{i}*31^{4-i}=v_{4}*31^3+v_{3}*31^2+v_{2}*31+v_{1}= \\31^2(\underbrace{v_{4}*31 + v_{3}}_{H_{2}})+(\underbrace{v_{2}*31+v_{1}}_{H_{1}})=31^2*H_{2} + H_{1}\]

So if you look closer, what we have to do now is to find pairs of values (\(v_{3}^i\), \(v_{4}^i\)) and (\(v_{1}^{i}\), \(v_{2}^{i}\)) for which \(H_{2}\) and respectively \(H_{1}\) remain constant. It’s the same exercise we did before when \(N=2\).

We can generate new ones, or we can reuse the ones we’ve discovered so far:

\[(v_{1}^{i}, v_{2}^{i}) \in \{\text{"aa", "bB", "c#"}\} \\ (v_{3}^{i}, v_{4}^{i}) \in \{\text{"aa", "bB", "c#"}\} \\\]

So if we re-arrange the Strings obtained when \(N=2\), we can create longer Strings that, when hashed with Java’s algorithm, will collide:




We can also use other values for our \((v_{3}^{i}, v_{4}^{i})\). For example, let’s find out colliding Strings for "go" by calling getCollidingStrings("go"). The results are: "go", "hP", "i1". So let’s use:

\[(v_{1}^{i}, v_{2}^{i}) \in \{\text{"aa", "bB", "c#"}\} \\ (v_{3}^{i}, v_{4}^{i}) \in \{\text{"go", "hP", "i1"}\} \\\]

By re-combining the Strings obtained with \(N=2\), we will get another set of four-character Strings that collide when hashed:




The number of colliding Strings we can generate using this algorithm is usually \(3^{N/2}\). If \(N=32\) (our Strings are 32 characters long), we can generate \(3^{32/2}=3^{16}=43.046.721\) colliding values, which is more than enough to ruin the performance of a HashMap<K,V>, by degrading it to the performance of a TreeMap<K,V> (for get/put operations).

So let’s define the set \(A=\text{\{"aa", "bB", "c#"\}}\) containing three String that collide.

  • If we compute \(A \times A\) (the Cartesian Product), we will generate all the possible four letters colliding Strings (with their values based on the elements of A).
  • If we compute \(A \times A \times A\), we will generate all the possible six letters colliding Strings (with their values based on A).
  • If we compute \(A \times A \times A \times A\), we will generate all the possible eight letters colliding Strings (with their values based on A).
  • … and so on

If we use guava, there’s a Sets.cartesianProduct(Set...) we can (re)use to implement the algorithm. If not, you can implement the algorithm yourself.

So our code becomes:

public static Set<String>  generateCollisions(Set<String> baseSet, int nTimes) {
    Set<String>[] sets = new Set[nTimes];
    Arrays.fill(sets, baseSet); // fill-up nTimes the array with baseSet
    return Sets.cartesianProduct(sets)
                .map(s -> String.join("",s))


  • baseSet is obtained by calling the previously defined method: getCollidingStrings().
  • nTimes is the number of times we perform the Cartesian Product. If, for example, nTimes=13, we will generate 26-character-long Strings, by doing 13 Cartesian Products between baseSets. The number of elements generated is 3^nTimes=1594323.

To test the code above code, let’s run the following:

public static void main(String[] args) {
    Set<String> baseSet = getCollidingStrings("aa"); // "aa", "bB", "c#"
    Set<String> collidingStrings = generateCollisions(baseSet, 13); 
    System.out.println("Strings generated: " + collidingStrings.size());
    System.out.println("Size of the String: " +   collidingStrings
    System.out.println("Distinct hash values: " + collidingStrings

The output will be:

Strings generated: 1594323
Size of the String: 26
Distinct hash values: 1

The elements from collidingStrings look like this:



  • It’s pretty easy to generate colliding Strings, given the simplicity of the (default) Java hash function;
  • If we were to use these elements as keys for a HashMap<String, T>, the performance of HashMap would degrade to the one of a TreeSet.

Thanks for reading so far!