I'll illustrate the RANK measure of a lattice by a simple case: 6-CDL, the complemented distributive lattice on factors of 6. (Two elements of a lattice are mutually complementary if their only "connections" are at the "bottom" & "top" of the lattice, as is the case here with 2 & 3. 6 & 1 are considered mutually complementary by definition. A lattice is COMPLEMENTED DISTRIBUTIVE if DISTRIBUTIVE AND IF EVERY ELEMENT HAS A UNIQUE COMPLEMENT, as below.

                6 (RANK=2)
               / \
              /   \
             2     3 (RANK=1)
              \   /
               \ /
                1 (RANK=0)

Elements of RANK 1 (2 & 3, above) are known as "atoms" of a lattice. (In a FACTOR LATTICE, as above, the atoms are PRIME NUMBERS.) The RANK OF THE LATTICE is that of the "top" or MAXIMAL ELEMENT (RANK 2, above). (As another example, 30-L, the FACTOR LATTICE ON 30=2*3*5 HAS RANK 3, & 3 ATOMS (PRIMES).

It can easily be proven that THE RANK OF A COMPLEMENTED DISTRIBUTIVE LATTICE EQUALS THE NUMBER OF ITS ATOMS (as above). And the number of its elements equals 2^a, a = # atoms. (Above, CD 6-L has 2² = 4 elements.)

Elsewhere, I show how to GENERATE THE FREE LATTICE EXTENSION of 30-L. CD 30-L has 2³=8 elements. Its FREE EXTENSION has 18!

I haven't found, in THE LITERATURE, AN ALGORITHM PRESCRIBING THE RANK (& NUMBER OF ELEMENTS) OF THE FREE EXTENSION OF A CD n-L. Can you discover it, in THE LITERATURE, or by original research?