If you can do that, congratulations because you just partially broke elliptic curve.
I don't understand what you are saying, but there is no way of reducing the range for unknown keys, for puzzle 66, you have nothing to work with, do you think the creator just randomly generated a 66 bit key for puzzle 66? No, he generated 256 bit keys and removed some characters/ or some 1s and 0s to make each key to fit in their desired bit range. Unless you share your secret algo, there is nothing here for me. Sorry.
No, i mean I can reduce a generator range to skip not random values, so time to bruteforce reduced too.
For example, 23 bit key to test (python 3.11 + ice_secp256k1.dll).
with secret algo:
GOT: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rVkthFNsQ6i7
10.363348245620728 s
with usual range (2^22 ... 2^23-1)
GOT: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rVkthFNsQ6i7
16.832353353500366 swith big values, like 66 bit, a lot of values just skiped as NOT random binary values, because cant be randomly generated by author (by wallet software).
for example, first value for 66-bit range is 100000100100100101010011001011000111000111001011000111000111001011, all values less is fail.
this value give generator as first value applyed with random's rules
anyway, pure python not a good instrument to get result. wanna use numba cuda.jit, but still learning how to.