Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
mahmood1356
on 02/08/2025, 00:24:06 UTC
Example of How a Filter Works:
We want to exclude private keys that contain triple repeating characters (e.g., “aaa”, “111”, etc.), because no valid Bitcoin private key from Puzzles 1-70 has had this pattern.

Without Filter:
• Generate Private Key
• Convert Private Key to Public Key (elliptic curve multiplication).
• Check Public Key: Convert to Bitcoin address, compare with target.
• If Invalid: Discard the result.

With Filter (Efficiency Gain):
• Generate Private Key
• Filter Check: Check if the private key contains any triple repeating characters.
• If the key fails the filter (e.g., contains “aaa” or “111”), skip the public key calculation.
• If the key passes the filter, proceed to the next step.
• Convert Private Key to Public Key (elliptic curve multiplication).
• Check Public Key: Convert to Bitcoin address, compare with target.
• If Invalid: Discard the result.

We need to update filter with every new private keys…

You only save the cost of EC multiplication and address conversion for filtered keys, but you still have to generate and scan every private key.
The global search space and thus the brute-force difficulty remains the same. Filtering helps a bit, but doesn’t make miracle.

If the filter (no consecutive three same hex digits like aaa) holds true for 71 puzzle:
Why not exclude key generation for entire sub range:
The range we're talking about, 444000000000000000 to 44500000000000000, the first three digits are always "444" (e.g., 444000000000000000, 444100000000000000, ..., 445000000000000000).
Every key in this range starts with the triplet "444", which holds true the filter we're using, as any key containing triple consecutive digits (e.g., "aaa", "111", "444", etc.) is invalid.

It will save computing for quadrillions of keys!


Sure, you’re right if your filter is absolutely reliable, you can skip entire ranges like 444... that violate the rule, which does save a lot of computation.
But remember, this only works if you’re certain the filter will always hold true for any future puzzle. Otherwise, you risk missing possible solutions.

I also stated the same theory in previous posts, but it was useless and not accepted by users.