I'm not a programmer at all — the AI writes all the code for me. But it still couldn't give me a clear answer to my question: is it possible to iterate through private keys within a given range while instantly filtering out "unreliable" keys, without affecting the speed of the iteration itself?
Here's the non-AI (correct) answer.
Once you filter out a single key, then computing the very next key would require one of these 2 options:
1. Multiplication with G.
2. Addition with a precomputed delta*G public key.
2 is faster, but you need to know how many precomputed deltas you'll ever need, or come up with a complicated strategy to optimize the "jump" using your precomputed delta keys, without resorting to option #1.
Also, the CUDA kernel would need to know what private key is computing, which, without filtering concept, is totally not needed at all. The fastest kernels simply go from pubKey A to pubkey B without ever caring what the private key is associated with each result. The CPU handles this association (and also maybe the initial starting point multiplications, as its only needed one time only), freeing up computational time on the GPU.
Oh, and also it's a fallacy to ever think some whatever pattern can't result in the solution. There are multiple puzzles where we had sequences of long 1s or 0s in a row, which would have been skipped if this pattern filtering thing would have been used.