Development resources at your finger tips
Build with the coolest Web3 projects
Recurring funding for Open Source
Learn about Web3 & earn rewards
Show appreciation for each other
Meet fellow developers, designers, futurists and more. Collaborate and BUIDL awesome projects together.
Discover great web3 organizations, work on meaningful projects and build relationships with like minded people. Browse Tribes
Meet the top hunters and contributors from our community.
Heya Gitcoiners & Gitcoinerettes! They say that July is the warmest month of the year (location depending), and we’re definitely getting ready fo…
Howdy, Gitcoiners! This month we’re running hackathons, we launched Grants Round 6 (which we just tweaked big time btw), we’re building, we’re improv…
Gitcoin is GDPR complaint. Learn more in
Gitcoin's Terms & Conditions.
Check out the Issue Explorer
Looking to fund some work? You can submit a new Funded Issue here.
Edit the .py file mentioned further below that will run if/then loop statements while generating random mnemonics until a valid palindromic one is found. Additional reference:
1) generate a valid mnemonic randomly using the tool (and note the values of bit position 8,9,10 & 11, which are the last four bits of the first word or first group of 11-bits from the left)
2) put the words into reverse order, then translate them back into groups of 11 bits,
3) SHA256 hash the first 128 bits (not 132) as a bytearray and compute the 4-bit checksum,
4) check if the checksum matches the leftover 4 bits from prior step which should match the last four bits of the first group noted in step 1. If they match, a mnemonic is palindromic and should be valid both as a little endian and big endian format. If no match, the program should loop back and repeat steps 1-4.
The file to use as a basis for this experiment is: https://github.com/hatgit/BIP39_mnemonic_creation_light_python/blob/master/Palindromic_Mnemonic_experiment.py
Example of steps that didn't lead to a palindromic mnemonic and should loop over again:
Initial entropy plus checksum:
11-bit groups (note the last 4 bits of first group):
['0110100**0111**', '01101110010', '11110100010', '01110011111', '01100010111', '11010110110', '10111011011', '01101100011', '11100000010', '11001100010', '00010111011', '11101010110']
Index values of words:
[839, 882, 1954, 927, 791, 1718, 1499, 867, 1794, 1634, 187, 1878]
happy hour violin initial glass strategy rocket hockey theory slush blast turn
Putting the above mnemonic into reverse:
turn blast slush theory hockey rocket strategy glass initial violin hour happy
Extract first 128 bits and compute checksum:
checksum didn't match:
['11101010110', '00010111011', '11001100010', '11100000010', '01101100011', '10111011011', '11010110110', '01100010111', '01110011111', '11110100010', '01101110010', '01101001100'] <— checksum here of 1100 would have had to been **0111**, in order for the mnemonic to have been palindromic (not its entropy) where last word would have been "happy" and not "hat".
[1878, 187, 1634, 1794, 867, 1499, 1718, 791, 927, 1954, 882, 844]
turn blast slush theory hockey rocket strategy glass initial violin hour hat
Repeat all steps until a match is found.