Question

I am writing a java program which would act upon the input which would be in scheme kind

of language.

Something like

(+ (+ a b))

Say I now want to check the syntax like if there are 2 brackets which have been opened, there have to be 2 other closing brackets. I am not sure how to achieve this with a regex. Can you help me out

Answer 1

Regular expressions can’t pair parentheses to an arbitrary depth. Scheme syntax is not regular.

http://en.wikipedia.org/wiki/Regular_language#The_number_of_words_in_a_regular_language

Thus, a non-regularity of some language L’ can be proved by counting the words in L’. Consider, for example, the Dyck language of strings of balanced parentheses. The number of words of length 2n in the Dyck language is equal to the Catalan number …, which is not of the form p(n)λn, witnessing the non-regularity of the Dyck language.

You’re going to have to tokenize it and then walk over tokens counting paren depth and make sure the depth is zero at the end and never goes negative.

For a simple language that has parentheses spaces and identifiers made from repetitions of the letter ‘a’, you might do

Patter token = Pattern.compile("[() ]|a+|.", Pattern.DOT_ALL);
Matcher m = token.matcher(sourceCode);
int parenDepth = 0;
while (m.find()) {
  char ch = m.group().charAt(0);
  switch (ch) {
    case '(':
      ++parenDepth;
      break;
    case ')':
      if (parenDepth == 0) {
        fail("Too many close parens");
      }
      --parenDepth;
      break;
  }
}
if (parenDepth != 0) {
  fail(parenDepth + " unclosed lists");
}