Can a non-empty string have a hashcode of zero?

Question

By "non-empty", I mean in this question a string which contains at least one non-zero character.

For reference, here's the hashCode implementation :

1493    public int hashCode() {
1494        int h = hash;
1495        if (h == 0) {
1496            int off = offset;
1497            char val[] = value;
1498            int len = count;
1499
1500            for (int i = 0; i < len; i++) {
1501                h = 31*h + val[off++];
1502            }
1503            hash = h;
1504        }
1505        return h;
1506    }

and the algorithm is specified in the documentation.

Before an integer overflow occurs, the answer is easy: it's no. But what I'd like to know is if, due to integer overflow, it's possible for a non-empty string to have a hashcode of zero? Can you construct one?

What I'm looking for would ideally be a mathematical demonstration (or a link to one) or a construction algorithm.

Answer 1

Sure. The string f5a5a608 for example has a hashcode of zero.

I found that through a simple brute force search:

public static void main(String[] args){
    long i = 0;
    loop: while(true){
        String s = Long.toHexString(i);
        if(s.hashCode() == 0){
            System.out.println("Found: '"+s+"'");
            break loop;
        }
        if(i % 1000000==0){
            System.out.println("checked: "+i);              
        }
        i++;
    }       
}

Edit: Joseph Darcy, who worked on the JVM, even wrote a program that can construct a string with a given hashcode (to test the implementation of Strings in switch/case statements) by basically running the hash algorithm in reverse.

Answer 2

just be care of that int h;. It may overflow, every string that satisfy h % 2^31 == 0 may lead to this.

public class HelloWorld {
    public static void main(String []args) {
       System.out.println("\u0001!qbygvW".hashCode());
        System.out.println("9 $Ql(0".hashCode());
        System.out.println(" #t(}lrl".hashCode());
        System.out.println(" !!#jbw}a".hashCode());
        System.out.println(" !!#jbw|||".hashCode());
        System.out.println(" !!!!Se|aaJ".hashCode());
        System.out.println(" !!!!\"xurlls".hashCode());
    }
}

A lot of strings...

Answer 3

Here's code to find and print strings of any desired hashCode value:

public static int findIntInverse(int x) {
    // find the number y such that as an int (after overflow) x*y = 1
    // assumes x is odd, because without that it isn't possible.
    // works by computing x ** ((2 ** 32) - 1)
    int retval = 1;
    for (int i = 0; i < 31; i++) {
        retval *= retval;
        retval *= x;
    }
    return retval;
}

public static void findStrings(
        int targetHash,
        Iterable<String> firstParts,
        Iterable<String> midParts,
        Iterable<String> lastParts) {
    Map<Integer, String> firstHashes = new HashMap<>();
    for (String firstPart : firstParts) {
        firstHashes.put(firstPart.hashCode(), firstPart);
    }
    int maxlastlen = 0;
    int maxmidlen = 0;
    for (String midPart : midParts) {
        maxmidlen = Math.max(midPart.length(), maxmidlen);
    }
    for (String lastPart : lastParts) {
        maxlastlen = Math.max(lastPart.length(), maxlastlen);
    }
    List<Integer> hashmuls = new ArrayList<>();
    String baseStr = "\u0001";
    for (int i = 0; i <= maxmidlen + maxlastlen; i++) {
        hashmuls.add(baseStr.hashCode());
        baseStr += "\0";
    }
    // now change each hashmuls into its negative "reciprocal"
    for (int i = 0; i < hashmuls.size(); i++) {
        hashmuls.set(i, -findIntInverse(hashmuls.get(i)));
    }
    for (String lastPart : lastParts) {
        for (String midPart : midParts) {
            String tail = midPart + lastPart;
            Integer target = hashmuls.get(tail.length()) * (tail.hashCode() - targetHash);
            if (firstHashes.containsKey(target)) {
                System.out.print(firstHashes.get(target));
                System.out.println(tail);
            }
        }
    }
}

Some interesting finds found by using a list of common English words to seed each part:

sand nearby chair
king concentration feeling
childhood dish tight
war defensive to
ear account virus

Using just Arrays.asList(" ") as midParts and a large English word list for firstParts and lastParts, we find the well-known pollinating sandboxes as well as revolvingly admissable, laccaic dephase, toxity fizzes, etc.

Note that if you give findStrings a large list of size N for both firstParts and lastParts and a short fixed list for midParts, it runs in O(N) time.