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## Concept of Counter Automata, Finite Automata with External Storage

Counter automata:The counter automaton comprises of an fsm M augmented by the register (or counter) C which can hold a single integer of random size. C is initialized to zero. M can decrement and increment the counter and then test it for 0. Therefore, M’s memory is un-bounded, and arguments of the type ‘FAs can’t count’, as formalized in pumping lemma. In principle, any amount of data can be coded to a single integer. For illustration, a sequence of 3 integers a, b and c can encoded reversibly by the single integer 2a 3b 5c (Goedel numbering). Therefore, a CA consists of all the storage capacity one might require. Its computational power is restricted by the access operations limited to increment, decrement and test for 0. Due to such, counter automata ‘can’t do much more than count’.

As an illustration, each of the given counter automata accepts the language of all the strings over A = {a, b} with an equivalent number of a’s and b’s:

The transition is activated by a pair (that is, input symbol, state of counter) and triggers an action on counter.

Testable states of counter are ‘=0’ and ‘≠0’, and the counter actions are decrement ‘dec’ and increment ‘inc’.

Rather than reading an input symbol, M can as well act based on the counter state alone that we represent as ‘reading the null string ε’. M might ignore the state of counter, which we represent by ‘-’ = ‘don’t care’. Finally, M might select to execute no action on the counter C, that we denote by ‘-’ = ‘don’t act’.

Example: Parenthesis expressions. The polish or parenthesis-free notation is for arithmetic expressions.

i) Let consider the language L1 of accurate parenthesis expressions over the alphabet A = {(, )}. Illustrations of correct expressions: (( ) ), ( ) ( ), ( ( ) ) ( ), and the null string ε. Either: exhibit a counter automaton M1 which accepts the language L1 , or exhibit that no such counter automaton exists.

ii) Let consider the language L2 of accurate parenthesis expressions over alphabet A = {(, ), [, ]}, including two pairs of parentheses. Illustrations of accurate expressions: ([ ]), ( ) [ ] ( ), ( ( ) [] ) ( ), and the null string ε. Illustration of a wrong expression: ( [ ) ]. Either: Show a counter automaton M2 which accepts the language L2, or argue that no such counter automaton is exists.

iii) Polish or parenthesis-free notation for an arithmetic expressions come in two versions: suffix and prefix notation.

Let consider operands designated by the single letter, say x, y or z, and four binary operators +, -, *, /. In the prefix notation, the operator is written before its 2 operands, and in suffix notation after x + y becomes +xy or xy+.

Design a counter automaton P which recognizes accurate prefix expressions and a counter automaton S which recognizes accurate suffix expressions.

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