Has anyone studied the consequences of the Kleene star in regular expressions to always be possessive?In other words, if * would always match as much as possible and no backtracking is allowed, would I still be able to express any regular language?Let's say that @ is the possessive * operator. There are cases where it doesn't matter: a*b and a@b both will match any string with 0 or more $a$ followed by a $b$.However there are cases where being possessive is relevant: a.*b will match any string that starts with $a$ and ends with $b$ but a.@b will never match any string as the greedy @ will eat any character, including the ending $b$.  The equivalent expression would be a([^b]@b)+.One may be tempted to think that for any non-possesive expression there is an equivalent possessive expression and viceversa but I wasn't able to find any proof of this.I'm limiting myself to * considering a+ equivalent to aa*.As D.W. suggested in the comments below, I tried to start from the DFA. I didn't go much far, as this seems to have more to do with the way the non-determinism is handled, rather than with the structure of the automaton.Could anyone point me in the right direction?(It has pointed out to me that the right term to use would be possessive (http://www.regular-expressions.info/possessive.html) rather than greedy. Thanks.)
Possessive Kleene star operator
formal languages;regular languages;regular expressions