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@Murch
stacking since: #127838
200 sats \ 0 replies \ @Murch 14 Dec \ on: South Korea's president impeached by parliament news
Impeached and suspended!
Transactions are chained together by the second transaction spending an output of the first transaction. Anyone that gets paid by the first transaction can use CPFP, so if there is no change output, the sender cannot use CPFP, but the other recipients can.
Using a VPN will help obfuscate your actual IP address, but the information that you are leaking, i.e. which addresses belong your wallet, gets transmitted independently of that to the Electrum server.
The problem with using an Electrum server is that you leak your entire wallet composition: you literally give the Electrum server the list of addresses you are interested in. It’s that you tell a server that has a high likelihood of being run by surveillants that all your addresses belong to the same wallet which they then can use to tie your wallet to your activity on exchanges and other KYCed businesses.
Regarding the Ledger it depends on the software wallet that you use with the Ledger.
Many guides focus on installing Bitcoin Core from the source code, but getting the binary directly from the Bitcoin Core website is much easier: https://bitcoincore.org/en/download/
It is also directly packaged for some Linux distributions, e.g. if you are on Ubuntu, you can get it via snap like this:
sudo snap install bitcoin-core
I’ve made some progress on catching up on BIP reviews. I’ve already reviewed the following five this week:
- Asynchronous Payjoin
- OP_PAIRCOMMIT
- Discrete Log Equality Proots
- QuBit—P2QRH spending rules
- Sending Silent Payments in PSBTs
I also hosted a very short Bitcoin Optech Recap yesterday. The rest of the week, I will work on some Bitcoin Core reviews, preparation for a talk and a workshop session next week, and touching up my own pull request.
This feels a bit outdated. More than 95% of the hashrate has adopted a full-RBF policy. Full-RBF in this context means that even transaction replacements are accepted for all transactions whether they signaled replaceability or not. More crucially, a replacement transaction has to both pay a higher feerate and a higher absolute fee than the original transaction to get accepted, which might be an interesting point to cover in a future explainer.
I already provided my opinion: I have yet to hear or read anything from Lowery that I find remotely interesting.
At best those 400 pages should have been a long blog post.
Jameson’s review covers it more eloquently: https://blog.lopp.net/softwar-thesis-review/
I have been traveling a lot in the past few weeks and a lot happened. I was on vacation, went to the Core Dev meeting and attended TABConf. We announced Localhost Research, a new Bitcoin Core contributor office that we are launching in the Bay Area last weekend.
Work piled up a bit. I already went through new questions on Bitcoin Stack Exchange, co-hosted the Bitcoin Optech Recap and BitDevs yesterday, agreed to a podcast appearance about Cluster Mempool in a couple weeks, went through a bunch of emails. I still need to (not sorted by priority):
- Take some notes about the conversations I had last week at TABConf
- Review a paper on consensus changes
- Review the new compatibility matrix on Optech
- Do about 20 reviews on Bitcoin Core pull requests
- Address review on my own Bitcoin Core pull request
- Address review on my BIP-2 successor draft
- Go over a ton of activity in the BIPs repository from the last weeks
So, I will work on some of that.
Every person has only 24h per day. We cannot be involved in every decision that affects our lives, nor would we as a society be able to decide anything if every single person were to speak on every single topic. You can of course pretend that decisions made by people outside your direct vicinity don’t affect your life if that’s your preference, but if other people decide to weigh in on societal processes by electing someone to represent their interests that’s outside of your self-sovereignty. A representative probably won’t always be exactly in line with your position on everything, and some representatives often fall short of their mandate, but in a functioning democracy they may still help achieve an outcome that benefits overall. Democracy is a messy process of compromises, but the alternative of sticking your head in the sand doesn’t seem all that appealing either.
I spent some time on flights in the past few weeks, so I got around to watching
- Please Stand By
- The Fall Guy
- Furiosa
The Fall Guy was great. The other two were okay.
As one of the co-hosts, I’m obviously biased, but if you are trying to keep up with Bitcoin development, Mike Schmidt and I put out the weekly Bitcoin Optech Recap, where we and our guests talk about the content of the weekly Bitcoin Optech Newsletter.
red: A+B+C+D+E = x
green: B+C+D+E+G+H+I = x
yellow: C+D+G+H+F = x
blue: D+E+H+I+J = x
red - green: A - (G+H+I) = 0
A = G+H+I
green - yellow: B+E+I - F = 0
F = B+E+I
green - blue: B+C+G - J = 0
J = B+C+G
We notice that there are three symmetric groups of letters. As starting points for any solution, A, F, and J are interchangeable, B, G, and I are interchangeable, and C, E, and H are interchangeable.
Since A…J are each unique values and integers from [1..10], the minimum value for three letters to be summed is 1+2+3 = 6. A, F, J must be 6 or bigger.
None of B, C, E, G, H, I can be bigger than 7, as 7 + 2 + 1 = 10
We suspect that D is 7. Proof by contradiction:
Let’s assume A = 6. If A is six, G, H, and I must be 1, 2, and 3 in some configuration.
If that’s the case, for F = B+E+I to be valid, I must be 1 and B and E must be 4 and 5. Otherwise, F would be bigger than 10. However, for the same reason, G would have to be one and B+C would be 9 which is in contradiction to the requirement that each number only appears once. Due to the symmetry, it follows that none of A, F, and J can be 6.
Let’s assume A = 7. This means that G, H, and I must be 1, 2, 4 in some order. It would follow that I is either 1 or 2.
- If I were 4, F would necessarily be greater than 10 as the next two lowest numbers we can use for B and E are 3 and 5.
- If I is 2, B and E must be 3 and 5, which makes F 10. This leaves C to be at least 6, in which case J would be at least 10 as well 3+6+1 = 10. F and J cannot both be 10.
- If I is 1, B and E could be 3 and 5 or 3 and 6. 3a. Let’s assume they are 3 and 6, which makes F = 10. Then the lowest remaining number for C is 5. This would make J at least ten via J = 3+C+2. Again, both J and F would be 10. We follow that due to the symmetry, neither A, F, or J can be 7. A, F, and J must be 8, 9, and 10.
Among B, C, E, G, H, and I three characters appear twice in our formulas for A, F, and J and three appear only once.
Let’s assume one of B, G, and I were 7, for example B. This leads to F = 7+E+I and J = 7+C+G. Let’s assume E and I are 1 and 2, which makes F 10. J ends up being bigger than 10. It follows that B, G, and I cannot be 7.
Let’s assume one of C, E, or H is 7, e.g. H. If H is 7, G and I have to be 1 and 2 for A to only be 10, let’s say G is 1 and I is 2 (the other pick is symmetric).
It follows that F and J have to be either 8 or 9 each.
- Let’s say F is 8, it follows that B + E = F - I = 6. This would only be possible if B and E could either 1 and 5, 2 and 4, or 3 and 3. 1 and 2 are already assigned, and 3 and 3 is not permitted.
- Let’s say F is 9. Because I is already 2, B and E must be 3 and 4. This leaves J to be 8. G is 1 and B is either 3 or 4. C would have to be 4 or 3 respectively, but both are taken.
Because neither the three letters A, F, and J, nor the three letters B, G, and I, nor the three letters C, E, or H can be 7, given all the restrictions, only D can be 7.
Per a close look, we guess that the three letters that only appear once should be 4, 5, and 6, and see that the following configuration would be viable:
A = G+H+I = 1 + 6 + 3 = 10
F = B+E+I = B + E + 3 = 2 + 4 + 3 = 9
J = B+C+G = 2 + 5 + 1 = 8
I.e. A = 10, B = 2, C = 5, D = 7, E = 4, F = 9, G = 1, H = 6, I = 3, J = 8
Checking back in the original formulation of the problem:
red: A+B+C+D+E = 10 + 2 + 5 + 7 + 4 = 28
green: B+C+D+E+G+H+I = 2 + 5 + 7 + 4 + 1 + 6 + 3 = 28
yellow: C+D+G+H+F = 5 + 7 + 1 + 6 + 9 = 28
blue: D+E+H+I+J = 7 + 4 + 6 + 3 + 8 = 28
Q.E.D.