> “If you run this on a big repository, it will take quite a lot of time because `git log -n1` takes a long time. I think this is the fastest way to get the most recent commit time on a single file? (That's the assertion that I hope someone can correct me on!) In any case, `bttf tag exec` is using parallelism under the hood to make this even faster.”

Instead of running `git log -n1` on every file, I think you can walk through the commits backwards, skipping any files that have been seen. Something like this (these two commands could be followed by bttf commands):

  git log --pretty=format:"DATE:%aI" --name-only |
  awk '/^DATE:/ {date=substr($0, 6); next} $0!="" && !seen[$0]++ {print date, $0}'

This seems to run much faster. The only problem is it'll include files that have been renamed or removed. I got an AI to fix that too, but it starts getting awkward (still fast though!):

  git ls-files |
  awk '
    # Read all existing files from git ls-files into an array
    NR==FNR { lsfiles[$0]; next }

    # Process the git log stream
    /^DATE:/ { date=substr($0, 6); next }
    $0!="" && ($0 in lsfiles) && !seen[$0]++ { print date, $0 }
  ' - <(git log --pretty=format:"DATE:%aI" --name-only)