DZone
Thanks for visiting DZone today,
Edit Profile
  • Manage Email Subscriptions
  • How to Post to DZone
  • Article Submission Guidelines
Sign Out View Profile
  • Post an Article
  • Manage My Drafts
Over 2 million developers have joined DZone.
Log In / Join
Refcards Trend Reports
Events Video Library
Refcards
Trend Reports

Events

View Events Video Library

Related

  • An Introduction to Bloom Filters

Trending

  • Run Gemma 4 on Your Laptop: A Hands-On Guide to Google's Latest Open Multimodal LLM
  • Your AI Agent Tests Are Passing, But Your Agent Is Still Broken
  • Bringing Intelligence Closer to the Source: Why Real-Time Processing is the Heart of Edge AI
  • Feature Flag Debt: Performance Impact in Enterprise Applications

Unique hashCodes is Not Enough to Avoid Collisions

By 
Peter Lawrey user avatar
Peter Lawrey
·
Oct. 15, 13 · Interview
Likes (0)
Comment
Save
Tweet
Share
12.1K Views

Join the DZone community and get the full member experience.

Join For Free

There is a common misconception that if you have unique hashCode() you won't have collisions.  While unique, or almost unique, hashCodes are good, this is not the end of the story.

The problem is that the size of a HashMap is not unlimited (or at least 2^32 in size)  This means the hashCode() number has to be reduced to a smaller number of bits.

The way HashMap, and thus HashSet and LinkedHashMap ,work is to mutate the bits in the following manner

    h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);


and then apply a mask for the lowest bits to select a bucket.  The problem is that even with unique hashCode()s as Integer does, there will be values with different hash code map to the same bucket. You can research how Integer.hashCode() works ;) 

public static void main(String[] args) {
    Set integers = new HashSet<>();
    for (int i = 0; i <= 400; i++)
        if ((hash(i) & 0x1f) == 0)
            integers.add(i);
    Set integers2 = new HashSet<>();
    for (int i = 400; i >= 0; i--)
        if ((hash(i) & 0x1f) == 0)
            integers2.add(i);
    System.out.println(integers);
    System.out.println(integers2);
}
static int hash(int h) {
    // This function ensures that hashCodes that differ only by
    // constant multiples at each bit position have a bounded
    // number of collisions (approximately 8 at default load factor).
    h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);
}
this prints
[373, 343, 305, 275, 239, 205, 171, 137, 102, 68, 34, 0]
[0, 34, 68, 102, 137, 171, 205, 239, 275, 305, 343, 373]

The entries as in the reverse order they were added as the HashMap is acting as a linked list, placing all entries into the same bucket.

Solutions?

A simple solution is to have a bucket turn into a tree instead of a linked list.  In Java 8, it will do this for String keys, but this could be done for all Comparable types AFAIK.
Another approach is to allow custom hashing strategies to allow the developer to avoid such problems, or to randomize the mutation on a per collection basis, amortizing the cost to the application.

Other notes

I would favour supporting 64-bit hash codes, esp for complex objects.  This has a very low chance of collision in the hash code itself and supports very large data structures well. e.g. into the billions.
Collision (computer science)

Published at DZone with permission of Peter Lawrey. See the original article here.

Opinions expressed by DZone contributors are their own.

Related

  • An Introduction to Bloom Filters

Partner Resources

×

Comments

The likes didn't load as expected. Please refresh the page and try again.

  • RSS
  • X
  • Facebook

ABOUT US

  • About DZone
  • Support and feedback
  • Community research

ADVERTISE

  • Advertise with DZone

CONTRIBUTE ON DZONE

  • Article Submission Guidelines
  • Become a Contributor
  • Core Program
  • Visit the Writers' Zone

LEGAL

  • Terms of Service
  • Privacy Policy

CONTACT US

  • 3343 Perimeter Hill Drive
  • Suite 215
  • Nashville, TN 37211
  • [email protected]

Let's be friends:

  • RSS
  • X
  • Facebook