If we parellelize, should get a speedup of 24, right? AddRange upper ; } if chkDigits. I have this brute force code, where you input a password and runs through combinations of numbers, lowercase and uppercase letters, and special characters until it match the password given. Why don't you go through the characters in the password, and check if every character is in the set of allowed characters? So let's just say that we can find any 8 character password in 2 hours. Ok, let's talk about that. For instance, if you take into account the length, you could restrict the generated passwords to be only of that length. Essentially, this is a dictionary attack too, but you're building the password dictionary incrementally each time you run your program.
Use MathJax to format equations. This is called a , by the way. The algorithm in the page is not implemented via brute force. I don't know how old you are, but suffice it to say that if you plan on obtaining the results before you expire, you're going to need a lot more machines. So the actual speedup will be something less than 24. It is an O n and O 1 space algorithm.
Then you can check your known list of possible passwords before generating more. If you consider the actual characters, then well, you already have the password, so you're done. Provide details and share your research! Changed your sources to compare only 4 bytes, but to my surprise original password wasn't found. I'm sure you can code a bunch of mutators and rainbow table checks in there to make it much smarter. By the way, the maximum password length you examine is 15. They really take the fun out of brute forcing.
I'd say that the algorithm can't be improved. So what now, purchase more machines? If you don't find a match, then you can fall back to the brute force approach. The power of exponentiation will win every time. So, the bottom line is, the only way to brute force something is through brute force. If you do plan on doing this, please let me know what server provider you'll be using. Is there a way I can implement multi-threading to speed up the process? This piece of code is so old I must actually update the article.
Shukhrat Nekbaev 20-Dec-12 2:10 20-Dec-12 2:10 Hi, I needed to bruteforce a hash by 4 first known bytes of it but I knew that password was 4 characters long. So the set of all 9 character passwords will be 68 times larger than the set of all 8 character passwords. So with 24 cores, it now takes 2 hours. After that I could make things work. That's pretty good compared to 2 days! I'd implement this in the following way. This is just the background of a more complex code I'm going to make. Parallelization can help, but only if the problem is tractable to begin with.
And, since you have to keep a running total, at least O 1 additional space is required. But what about 9 character passwords. To learn more, see our. Load the list and cycle through them until you have a match, or have exhausted the list. Background Bruteforcing has been around for some time now, but it is mostly found in a pre-built application that performs only one function.
Exit 0 ; } } } } } Do what a real cracker would do. I also found so many bugs in the algorithm that I changed the way it works. If you get that every character of the password is in the set of allowed characters, then the password is valid, otherwise it isn't. . Instead of generating each combination of characters every time the program is run, cache the calculated strings to disk along with the data that allows you to start where you stopped. My attempt to bruteforcing started when I forgot a password to an archived rar file.
In fact I am not interested in this specific problem. If I counted correctly, there are 68 characters you use to generate passwords. However, it's almost certainly possible to optimize the implementation. To examine this number of passwords will take 6,722,988,818,432 times longer than it takes to examine all 8 character passwords. I think you should look at the problem a bit differently. I'd love to invest in them! No I havent tried anything extra with this yet. AddRange lower ; } if chkUpper.