I've tinkered with Bash prompts for a lot of years.  That's where this problem originates, but it can be considered just as a thought experiment if you prefer.

We have the directory we're in, for example:

    /Users/gorr/.bashprompt/really/deep/directory/structure/even/deeper

(I keep my prompts in ~/.bashprompt/ , and a very long directory name is often a cause of breakage, so I keep one around to break them ...)

I want to shorten the directory name to just the first letters:

    export newPWD=$(echo ${PWD} | sed -e "s@${HOME}@~@" -e 's@\(/.\)[^/]*@\1@g')

    echo $newPWD

    ~/./r/d/d/s/e/d

This does '~' replacement for $HOME and then substitutes the first letter of each directory for the full directory name.

Here's the question: if the first letter of the directory name is a dot '.', can sed then capture one character more so that the output would become:

    ~/.b/r/d/d/s/e/d

I think this would be pretty easy with Bash and a loop, but that's a lot of processing so I'd rather not go down that road.  I suspect sed is capable of this, but I haven't delved deeply enough into the tool to even know where to start.  This may in fact be a regex problem more than a sed problem - either way I'm kind of stumped.  I'm open to simpler implementations using other (standard system) tools as well.

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