I wonder... If you don't have access to randomness, e.g., you need to base your choice off some fixed program, and with some limited program length, in theory a perfect observer should be able to guess with certainty your next choices after some time, since there's only a fixed number of possible programs? Of course this isn't of much practical use...
Can you use your "free will"? Try your hand
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Re: Can you use your "free will"? Try your hand
#62I wonder... If you don't have access to randomness, e.g., you need to base your choice off some fixed program, and with some limited program length, in theory a perfect observer should be able to guess with certainty your next choices after some time, since there's only a fixed number of possible programs? Of course this isn't of much practical use...
If these inputs or parts thereof are predictable, randomness decreases, making the cryptography ultimately weaker.
Re: Can you use your "free will"? Try your hand
#63i was pressing f and j instead of f and d, and was always getting 100% and i was freaking out :)
Re: Can you use your "free will"? Try your hand
#64Re: Can you use your "free will"? Try your hand
#65I hammered away for a while and the program was always around 50%. I wanted to be surprised at how accurate the prediction was but I simply felt let down. Funny how that works
Re: Can you use your "free will"? Try your hand
#66Re: Can you use your "free will"? Try your hand
#67Re: Can you use your "free will"? Try your hand
#68The prediction algorithm is actually very straightforward[1]. It's a fun exercise to write a sequence generator specifically to defeat it. Here is a sequence that can get the prediction rate down to 13%: ddddd dfddd dffdd dfdfd ddfff ddfdd fdffd dffdf ddfff fdfdf dfffd ffdff fffff dffff ffddf fffdf dfffd ffdff ffffd ddfff Generated using this Perl script: my %table = (); my $s = "fffff"; for(my $i = 0; $i $table{$s}{…
if you are inspecting the prediction algorithm, and you want to specifically defeat it, doesn't that mean copy the prediction algorithm and throw a "not" on it? It's the old halting problem trick...