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.
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;
}