qlrd

"""
This python script is aimed to help and teach how Krux can be used to sign files and create PEM public keys so openssl can be used to verify

Requirements:
    - opencv, qrcode
    pip install opencv-python qrcode

    - This script also calls a openssl bash command, so it is required to have verification functionality
 """
import argparse

qlrd

import json
import requests
import argparse

HEADERS = {
    "Accept": 'application/vnd.github+json',
    "X-GitHub-Api-Version": "2022-11-28",
}

parser = argparse.ArgumentParser(

qlrd

Miniscript challenge

Build a simple inheritance scheme on Liana and try to spend a ~170000 sats.

Prelude

During the workshop at BTC++ in 2025, the speaker created a simple inheritance scheme and put you as his heir. He can spend at any time, but you only can spend after some time.

Preliminary instructions

qlrd

A password to decrypt a qrcode

Scan this qrcode and type a sequence of characters that decrypt the QR code.

5ef3b16a pic.twitter.com/2P3G7ux3KE

— qlrd (@qlrddev) February 20, 2025

To find out the "password", you will need to edit solution.py. It have a line where you'll find a certain variable that require a byte string as input to the solution function. The output of solution will be the key to decrypt the mnemonic.

For example:

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Get the correct mnemonic

To solve this puzzle, it is useful to know that krux supports:

• BIP39 with and without passphrases;
• DoubleMnemonic (a valid 24w BIP39 mnemonic where the first and second halves are both valid mnemonics);
• BIP85 derived 12w and 24w BIP39 mnemonics

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Find a password to decrypt a qrcode;

After finding the mnemonic, you must find a 13th/25th word for this mnemonic. To find that, you will do the same thing as in the first part of this challenge:

derive_this_to_generate_the_passphrase = b" ".join([
    b"word 1",
    b"word 2",
    b"word 3",
 b"word 4",

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a derivation path m/48'/0'/0'/<index>'

Once you have entered the passphrase, you will need to select a miniscript policy with a specific type of descriptor (either P2WSH or P2TR), and a custom derivation path m/48'/0'/0'/<index>' to spend the funds.

Basically, the index is a number that you will need to figure out to construct the given descriptor:

# <descriptor> : wsh || tr
#  -&gt; 0, 1, 2, 3 ... who knows?

qlrd

"""
solution.py

Read:

- https://gist.github.com/qlrd/0eed0c8a56aeab3f8894c6d67694d690
- https://gist.github.com/qlrd/d1696dd63e15d89ecbc191fa2d734b36
- https://gist.github.com/qlrd/1b6f6edc99915acee72f7c73b3e649f0
- https://gist.github.com/qlrd/aca897acb42f0b6cbd07d7337f16c790