Find the last three words of each paragraph if they are shorter than 20 characters in total

Question

Based on my SO question, but more specifically:

Find a RegEx such that, given a text where paragraphs end with an empty line (or $ at the very end), match the up to twenty last characters of a paragraph excluding the \r or $ but no more than three complete "words" including the "separators" in between and the punctuation following it. The following constraints apply:

• "words" also include abbreviations and the like (and thus punctuation), i.e. "i.e." and "non-trivial" count as one word each
• whitespaces are "separators"
• isolated dashes do not count as "words" but as "separators"
• "separators" before the first word must not be included
• trailing punctuation counts to that limit, but not the leading one nor the linebreaks and $ that must not be in the matching group
• whether trailing punctuation is included in the match or not is up to you

Some examples:

The early bird catches the worm. Two words.

But only if they got up - in time.

Here we have a supercalifragilisticexpialidocious etc. sentence.

Short words.

Try accessing the .html now, please.

Vielen Dank und viele Grüße.

Did any-one try this?

Score is given in bytesteps: bytes multiplied by amount of steps according to https://regex101.com, but there are three penalties:

• if the upper limit of words cannot trivially⁺ be modified: +10%
• if the upper limit of character count cannot trivially⁺ be modified: +10%
• failure to handle unicode used in actual words: add the percentage of world population speaking the language according to Wikipedia the failing word stems from. Example: Failing to match the German word "Grüße" as a word => +1.39%. Whether you consider a language's letters as single characters or count the bytes to utf8-encode them is up to you. The penalty does not have to be applied before someone actually provides a counter-example, so feel free to demote your competitors ;)

Since this is my first challenge, please suggest any clarifications. I assume regular-expression means there are no "loopholes" to exclude.

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⁺ With trivially modified I mean expressions such as {0,2} and {21} are valid but repeating something three times (which would have to be re-repeated to increase the amount words) is not.

Answer 1

37 bytes * 1403 steps = 51911 bytesteps

34 bytes * 1036 steps = 35224 bytesteps

37 bytes * 666 steps = 24642 bytesteps

(?<!\S)(?!.{21})\S+(( | - )\S+){0,2}$

Verify it here!