Generate two different strings with the same hashcode

Question

I want to do some tests which require some strings with the same hash code, but not the same strings. I couldn't find any examples, so I decided to write a simple program to do it for me.

The code below generates two random strings over and over until they generate the same hash code.

    static Random r = new Random();
    static void Main(string[] args)
    {
        string str1, str2;
        do
        {
            str1 = GenerateString();
            str2 = GenerateString();
        } while (str1.GetHashCode() != str2.GetHashCode() && str1 != str2);

        Console.WriteLine("{0}\n{1}", str1, str2);
    }

    static string GenerateString()
    {
        string s = "";
        while (s.Length < 6)
        {
            s += (char)r.Next(char.MaxValue);
        }
        return s;
    }

This code seems to be working (theoretically), but it may take centuries to complete. So I was thinking of doing vice versa and generate two strings from one hash code.

I know it's not possible to retrieve a string from a hash code, but is it possible to generate possible strings from it?

I'm using Visual Studio 2015 Community Edition. Version: 14.0.23107.0D14REL.

.NET Framework: 4.6.00081.

Answer 1

Finding two strings by repeatedly comparing random strings will take practically forever. Instead generate strings and store them in an a dictionary by hashcode. Then look up each. Match found pretty quickly.

MATCH FOUND!! xqzrbn and krumld hash code 80425224

void Main()
{

    var lookup = new Dictionary<int,string>();

    while(true) {
        var s = RandomString();        
        var h = s.GetHashCode();
        string s2;
        if (lookup.TryGetValue(h, out s2) && s2 != s) {
            Console.WriteLine("MATCH FOUND!! {0} and {1} hash code {2}",
                lookup[h],
                s,
                h);
            break;
        }
        lookup[h] = s;

        if (lookup.Count % 1000 == 0) {
            Console.WriteLine(lookup.Count);
        }
    }
}

static Random r = new Random();

// Define other methods and classes here
static string RandomString() {

    var s = ((char)r.Next((int)'a',((int)'z')+1)).ToString() +
            ((char)r.Next((int)'a',((int)'z')+1)).ToString() +
            ((char)r.Next((int)'a',((int)'z')+1)).ToString() +
            ((char)r.Next((int)'a',((int)'z')+1)).ToString() +
            ((char)r.Next((int)'a',((int)'z')+1)).ToString() +
            ((char)r.Next((int)'a',((int)'z')+1)).ToString();

    return s;
}

Answer 2

Take advantage of the Birthday Paradox. Instead of only testing two strings directly, test all strings you have seen before.

using System;
using System.Collections.Generic;

namespace ConsoleApplication2
{
    class Program
    {
        static void Main(string[] args)
        {
            var words = new Dictionary<int, string>();

            int i = 0;
            string teststring;
            while (true)
            {
                i++;
                teststring = i.ToString();
                try
                {
                    words.Add(teststring.GetHashCode(), teststring);
                }
                catch (Exception)
                {
                    break;
                }
            }

            var collisionHash = teststring.GetHashCode();
            Console.WriteLine("\"{0}\" and \"{1}\" have the same hash code {2}", words[collisionHash], teststring, collisionHash);
            Console.ReadLine();
        }
    }
}

For my machine it produces the output

"699391" and "1241308" have the same hash code -1612916492

almost instantly.

Due to how strings are hashed in .NET you may not get the exact same output as me, but it should be just as fast.