Feel free to use it. Just make sure to change lvxferre/Hello+Fediverse+ to something else.
What it does: it generates the SHA256sum for strings starting with whatever you want, and ending in a number, between 0 and 10¹². Then if it finds one starting with “z” zeroes, it prints it alongside the number; then it looks for strings with an additional zero at the start. Each million tests it’ll also print some output so you know that the script didn’t freeze.
OK… here’s some dumb bash shit.
#!/bin/bash i=0; z=0 while [[ $i -le 1000000000000 ]]; do o=$(echo "lvxferre/Hello+Fediverse+$i" | sha256sum) if [[ $o =~ ^($z) ]]; then echo "$i: $o" declare -g z="$z""0" fi if [[ $i == *000000 ]]; then echo "$(expr $i / 1000000)M combinations tried..." fi i=$[$i+1] done
Feel free to use it. Just make sure to change
lvxferre/Hello+Fediverse+
to something else.What it does: it generates the SHA256sum for strings starting with whatever you want, and ending in a number, between 0 and 10¹². Then if it finds one starting with “z” zeroes, it prints it alongside the number; then it looks for strings with an additional zero at the start. Each million tests it’ll also print some output so you know that the script didn’t freeze.
Update: so far my best string was
lvxferre/Hello+Fediverse+2393194
, yielding0000006a 48...
I also did some simple optimisations of the code. Basically “the least you do, the faster it’ll be”.
i=7100000 while true; do o=$(echo "lvxferre/Hello+Fediverse+$i" | sha256sum) if [[ "$o" == 00000* ]]; then echo "$o $i"; fi if [[ "$i" == *00000 ]]; then echo "tried $i combinations..."; fi i=$[$i+1] done
Now it’ll show results with more than five leading zeroes, and print a message every 100k tries (to resume later on).
My machine is a potato, mind you. I don’t expect to get into the leaderboard. Still, I’m doing this as a bash exercise.