正则表达式匹配一个数字后跟一个重复多次的符号？

Question

How can I create a RegEx that can match the following:

a3bbb
aaaa3bbb
a4bbbb
aaa5bbbbb

Ie, a (one or more times), then a non-negative number, then b repeated 'that many times' (as many as the number between a and b ).

Is this language regular? If not, can we construct a CFG for this?

Edit: As for whether the number is single digit, I would say no. (also as Daniel Centore and rici point out, the language is not even CF. Then the natural question is, is it context-sensitive or unrestricted?)

Answer 1

Like other answers have said, if the number is unbounded, the language is neither regular (if it's regular, pumping lemma says for a sufficiently long string, the b 's could be extended indefinitely independent of the number) nor context-free (if it's context-free, pumping lemma says for a sufficiently long number, the number and the b 's could be repeated, but not correctly).

But the language is context-sensitive, as it can be generated using the following grammar (I do it for base-3 number for simplicity, you can extend to base 10):

(1) S -> aS | aB
(2) B -> BN | N
(3) aN -> a0 | a1b | a2bb
(4) 0N -> 00 | 01b | 02bb
(5) 1N -> 10 | 11b | 12bb
(6) 2N -> 20 | 21b | 22bb
(7) bN -> WN
(8) WN -> WX
(9) WX -> NX
(10)NX -> Nbbb

Rule (1) is to generate the a 's

Rule (2) is to generate each digit in the number

Rule (3)-(6) is to replace the left-most N with a number and respective number of b 's.

Rule (7)-(10) is to have the N "consume" the b 's to its left, and produce 3 b 's (10 b 's in base-10). Technically (7)-(10) is just bN -> Nbbb .

Example:

To generate: a102bbbbbbbbbbb (102 in base-3 = 11 in base-10)
S
aB (1b)
aBN (2a)
aBNN (2a)
aNNN (2b)
a1bNN (3b)
a1NbbbN (7)-(10)
a1NbbNbbb (7)-(10)
a1NbNbbbbbb (7)-(10)
a1NNbbbbbbbbb (7)-(10)
a10Nbbbbbbbbb (5a)
a102bbbbbbbbbbb (4c)

Answer 2

This language is not regular (and thus cannot be expressed as a RegEx). One test for language regularity is to check if it can be expressed by a Finite Automaton. It can be shown that this language cannot be expressed as an FA because the FA would need at least as many states as the number between a and b , but that number is not bounded. However, if it is bounded ( ex the number can only be from 1-10) then it would be Regular.

The language also cannot be expressed as a CFG, which can probably be shown using the pumping lemma.

Answer 3

If the number is a single digit, then the language is regular (because you can just list the nine possible suffixes). But if the number is not bounded, the language not regular. It is not even context-free. So neither a regular expression nor a CFG are available.