For now, it is the case. But later, it will be a vanity search on r-values (when grinding s-values alone will be too difficult), and then s-values again, after picking some r-value. Also, reusing two identical r-values will give everyone k-value, which means, that everyone will know, how to make a bit smaller signatures, if any solver will share it at least two times. And also note, that I used public key, equal to the generator. Which means, that it is "s=(z+r)*2". However, if you replace it with a different key, then you will have "s=(z+template)*2". And then, other puzzles can be made, which would allow vanity searches on different s-values, picked by any puzzle maker. Another interesting thing, is that each solver will always hash something different, because everyone will make its own transaction. And also, because TXIDs are unique, no solution can be reused, because "txid:vout" is always covered by all sighashes, even SIGHASH_NONE with SIGHASH_ANYONECANPAY. Which means, that the same Proof of Work cannot be reused, and if you deposit something twice, then each attempt to move it, will require grinding it again from scratch. Also, it is one of those puzzles, which are more similar to collision puzzles, than to famous N-bit key puzzle, because the creator doesn't know the solution upfront. So, after sweeping the first coins, you can be easily convinced, that if my puzzle transaction will be confirmed, then I won't be able to sweep everything at once, and end the challenge, because it would require solving the puzzle by myself.
For now, it is the case. But later, it will be a vanity search on r-values (when grinding s-values alone will be too difficult), and then s-values again, after picking some r-value. Also, reusing two identical r-values will give everyone k-value, which means, that everyone will know, how to make a bit smaller signatures, if any solver will share it at least two times. And also note, that I used public key, equal to the generator. Which means, that it is "s=(z+r)*2". However, if you replace it with a different key, then you will have "s=(z+template)*2". And then, other puzzles can be made, which would allow vanity searches on different s-values, picked by any puzzle maker. Another interesting thing, is that each solver will always hash something different, because everyone will make its own transaction. And also, because TXIDs are unique, no solution can be reused, because "txid:vout" is always covered by all sighashes, even SIGHASH_NONE with SIGHASH_ANYONECANPAY. Which means, that the same Proof of Work cannot be reused, and if you deposit something twice, then each attempt to move it, will require grinding it again from scratch. Also, it is one of those puzzles, which are more similar to collision puzzles, than to famous N-bit key puzzle, because the creator doesn't know the solution upfront. So, after sweeping the first coins, you can be easily convinced, that if my puzzle transaction will be confirmed, then I won't be able to sweep everything at once, and end the challenge, because it would require solving the puzzle by myself.

For now, it is the case. But later, it will be a vanity search on r-values (when grinding s-values alone will be too difficult), and then s-values again, after picking some r-value. Also, reusing two identical r-values will give everyone k-value, which means, that everyone will know, how to make a bit smaller signatures, if any solver will share it at least two times. And also note, that I used public key, equal to the generator. Which means, that it is "s=(z+r)*2". However, if you replace it with a different key, then you will have "s=(z+template)*2". And then, other puzzles can be made, which would allow vanity searches on different s-values, picked by any puzzle maker. Another interesting thing, is that each solver will always hash something different, because everyone will make its own transaction. And also, because TXIDs are unique, no solution can be reused, because "txid:vout" is always covered by all sighashes, even SIGHASH_NONE with SIGHASH_ANYONECANPAY. Which means, that the same Proof of Work cannot be reused, and if you deposit something twice, then each attempt to move it, will require grinding it again from scratch. Also, it is one of those puzzles, which are more similar to collision puzzles, than to famous N-bit key puzzle, because the creator doesn't know the solution upfront. So, after sweeping the first coins, you can be easily convinced, that if my puzzle transaction will be confirmed, then I won't be able to sweep everything at once, and end the challenge, because it would require solving the puzzle by myself.

For now, it is the case. But later, it will be a vanity search on r-values (when grinding s-values alone will be too difficult), and then s-values again, after picking some r-value. Also, reusing two identical r-values will give everyone k-value, which means, that everyone will know, how to make a bit smaller signatures, if any solver will share it at least two times. And also note, that I used public key, equal to the generator. Which means, that it is "s=(z+r)*2". However, if you replace it with a different key, then you will have "s=(z+template)*2". And then, other puzzles can be made, which would allow vanity searches on different s-values, picked by any puzzle maker. Another interesting thing, is that each solver will always hash something different, because everyone will make its own transaction. And also, because TXIDs are unique, no solution can be reused, because "txid:vout" is always covered by all sighashes, even SIGHASH_NONE with SIGHASH_ANYONECANPAY. Which means, that the same Proof of Work cannot be reused, and if you deposit something twice, then each attempt to move it, will require grinding it again from scratch. Also, it is one of those puzzles, which are more similar to collision puzzles, than to famous N-bit key puzzle, because the creator doesn't know the solution upfront. So, after sweeping the first coins, you can be easily convinced, that if my puzzle transaction will be confirmed, then I won't be able to sweep everything at once, and end the challenge, because it would require solving the puzzle by myself.