What regex can allow a delimiter, but only if the surrounding groups exist?

I need to design a regex that matches 4 parts separated by a delimiter (: in this case). The 4 parts must be in order, and the delimiter must be present between each part, but the parts are all optional.

These should all match:

A
B
C
D
A:B
A:C
A:D
B:C
B:D
C:D
A:B:C
A:B:D
A:C:D
B:C:D
A:B:C:D

Meanwhile, these should not match.

A:    // cannot end with delimiter
:C    // cannot start with delimiter
ABC   // cannot omit delimiter
A:BC  // cannot omit delimiter
A::B  // cannot duplicate delimiter
...   // many other possible non-matches

My problem is that I don't know how to match the delimiter.

• I tried ^(A)?:?(B)?$, but then it will match AB.
• I tried ^(A:)?(B)?$, but then it will match A:.
• I can manually enumerate all the 15 match types, but that is a big mess (^(A)|(B)|(C)|(D)|(A:B)|(A:C)...........$). In the actual regex, A and B, etc. are more complex.
• I tried a more concise enumeration like this ^((A)|((A:)?((B)|((B:)?((C)|((C:)?(D)))))))$, but it looks like a total mess to me still. I'll go with this option if I can't find a better answer though.

You can make the colon optional but use word boundaries to require it between letters:

^\b(?::?A)?(?:\b:?B)?(?:\b:?C)?(?:\b:?D)?$

See this demo at regex101 - To allow empty matches, move first \b inside leftmost group.

Or you make all characters optional and use a lookahead at start to validate the format:

^(?=\w(?::\w)*$)A?:?B?:?C?:?D?$

Another demo at regex101 - For empty matches change the lookahead to (?=(?:\b:?\w)*$).
Here a \w is used which is a shorthand for word character and usually matches [A-Za-z0-9_].

A more general idea to accept empty matches is simply attaching |^$ at the end of the pattern (regex101 demo) or wrap the whole pattern into an ^(?: optional group )?$ (regex101 demo).