Ambiguous grammar:
E -> UV | EBE | V | [E]
V -> a | b
U -> < | >
B -> ? | ! | @
Some information:
Order of precedence: ? < ! < @, with unary operators (<,>) being the highest
Binary operators ?, !, @ are right associative.
My attempt:
E -> UV | EBT | V | [E]
T -> E
V -> a | b
U -> < | >
B -> ? | B1
B1 -> ! | B2
B2 -> @
I'm not sure if I left out some corner cases during my conversion. Appreciate if you guys can point some mistakes out and offer some hints.
E -> UV | EBE | V | [E]
V -> a | b
U -> < | >
B -> ? | ! | @
Order of precedence: ? < ! < @, with unary operators (<,>) being the highest.
Binary operators ?, !, @ are right associative.
I was confused about your order of precedence, because it implies ! is higher precedence than >.
So assuming this order of precedence:
a,b,<,>,?,!,@
I would note that unambiguous grammars are all about using intermediary characters, normally more than the ambiguous grammars, to ensure that there is a standard procedure of steps to reach a certain terminal symbol, i.e. each string will have the same parse tree. Highest precedence means that it is the closest conversion to the final non-terminal to terminal symbol conversion.
My answer would be:
E -> V?E | V!E | V@E | [E] //Right associativity = right recursive
V-> <V | >V | E | T | a | b
T-> a | b | E
However very difficult to know without knowing your target string and accepted/not accepted.
