Congratulations, you’ve just invented digital images.

Dictionary attacks are definitely not a magic bullet, they require a lot of processing power, just like any other brute-force attack, but not more because of their longer length, as has been implied.

True, there are a lot of english words, but the amount of common words is relatively small. Most people aren’t going to choose a password like “MachicolationRemonstranceCircumambulationSchadenfreude”, even if it were generated for them (which is unlikely).

Sure, it is comparable to a standard 8 characters passward, but even that kind of password is verging on the insecure (it is the absolute minimum, which should be avoided when possible).

There are also a lot of symbols when you count emojies and the entire Unicode standard.

Like someone else said on this thread; that’s just security by obscurity, which is bad. Dictionary attacks will be one of the first (brute force related) attacks attackers will use because word passwords are incredibly popular (though admittedly of fewer words: VeryBigDog34 etc…), and relatively easy to do. I agree that having the password across different devices is somewhat of a challenge with a password manager, but not impossible. My very long and complex password is all down to muscle memory by this point, I couldn’t tell you what it is from memory.

Also you shouldn’t use the same password on multiple things and if you don’t use a password manager you will need to memorize a lot of different passwords.

You can’t compare a 46 random character password to a password composed out of words, the entropy of each is very different. Your kind of password is vulnerable to dictionary attacks which are way more common and easy than brute forcing every possibility. A 50+ characters unique random password for each service that is stored in a password manager which is encrypted with a 20+ characters random password is the most secure and future proof (for now).

In the loop where you iterate over the positions, you translate the matrix model with the position correctly, but when you rotate it in the next line, you rotate the identity matrix, not the translated matrix, and put the result into model, which resets the position. And so all the cubes get rendered at the same place which results in it looking like only one cube is being rendered.

The final threshold.