Is it possible to match a list of comma-separated decimal integers, where the integers in the list always increment by one?
These should match:
0,1,2,3
8,9,10,11
1999,2000,2001
99,100,101
These should not match (in their entirety - the last two have matching subsequences):
42
3,2,1
1,2,4
10,11,13
Yes, this is possible when using a regex engine that supports backreferences and conditions.
First, the list of consecutive numbers can be decomposed into a list where each pair of numbers are consecutive:
(?=(?&cons))\d+
(?:,(?=(?&cons))\d+)*
,\d+
Here (?=(?&cons)) is a placeholder for a predicate that ensures that two numbers are consecutive. This predicate might look as follows:
(?<cons>\b(?:
(?<x>\d*)
(?:(?<a0>0)|(?<a1>1)|(?<a2>2)|(?<a3>3)|(?<a4>4)
|(?<a5>5)|(?<a6>6)|(?<a7>7)|(?<a8>8))
(?:9(?= 9*,\g{x}\d (?<y>\g{y}?+ 0)))*
,\g{x}
(?(a0)1)(?(a1)2)(?(a2)3)(?(a3)4)(?(a4)5)
(?(a5)6)(?(a6)7)(?(a7)8)(?(a8)9)
(?(y)\g{y})
# handle the 999 => 1000 case separately
| (?:9(?= 9*,1 (?<z>\g{z}?+ 0)))+
,1\g{z}
)\b)
For a brief explanation, the second case handling 999,1000 type pairs is easier to understand -- there is a very detailed description of how it works in this answer concerned with matching a^n b^n. The connection between the two is that in this case we need to match 9^n ,1 0^n.
The first case is slightly more complicated. The largest part of it handles the simple case of incrementing a decimal digit, which is relatively verbose due to the number of said digits:
(?:(?<a0>0)|(?<a1>1)|(?<a2>2)|(?<a3>3)|(?<a4>4)
|(?<a5>5)|(?<a6>6)|(?<a7>7)|(?<a8>8))
(?(a0)1)(?(a1)2)(?(a2)3)(?(a3)4)(?(a4)5)
(?(a5)6)(?(a6)7)(?(a7)8)(?(a8)9)
The first block will capture whether the digit is N into group aN and the second block will then uses conditionals to check which of these groups was used. If group aN is non-empty, the next digit should be N+1.
The remainder of the first case handles cases like 1999,2000. This again falls into the pattern N 9^n, N+1 0^n, so this is a combination of the method for matching a^n b^n and incrementing a decimal digit. The simple case of 1,2 is handled as the limiting case where n=0.
Complete regex: https://regex101.com/r/zG4zV0/1
Alternatively the (?&cons) predicate can be implemented slightly more directly if recursive subpattern references are supported:
(?<cons>\b(?:
(?<x>\d*)
(?:(?<a0>0)|(?<a1>1)|(?<a2>2)|(?<a3>3)|(?<a4>4)
|(?<a5>5)|(?<a6>6)|(?<a7>7)|(?<a8>8))
(?<y>
,\g{x}
(?(a0)1)(?(a1)2)(?(a2)3)(?(a3)4)(?(a4)5)
(?(a5)6)(?(a6)7)(?(a7)8)(?(a8)9)
| 9 (?&y) 0
)
# handle the 999 => 1000 case separately
| (?<z> 9,10 | 9(?&z)0 )
)\b)
In this case the two grammars 9^n ,1 0^n, n>=1 and prefix N 9^n , prefix N+1 0^n, n>=0 are pretty much just written out explicitly.
Complete alternative regex: https://regex101.com/r/zG4zV0/3
