Generating First Set from the grammar - algorithm

Question

Algorithm for Finding first set:
Given a grammar with the rules A1 → w1, ..., An → wn, we can compute the Fi(wi) and Fi(Ai) for every rule as follows:
initialize every Fi(Ai) with the empty set
set Fi(wi) to Fi(wi) for every rule Ai → wi, where Fi is defined as follows:
Fi(a w' ) = { a } for every terminal a
Fi(A w' ) = Fi(A) for every nonterminal A with ε not in Fi(A)
Fi(A w' ) = Fi(A) \ { ε } ∪ Fi(w' ) for every nonterminal A with ε in Fi(A)
Fi(ε) = { ε }
add Fi(wi) to Fi(Ai) for every rule Ai → wi
do steps 2 and 3 until all Fi sets stay the same.
Grammar:
A -> B C c
A -> g D B
B -> EPSILON | b C D E
C -> D a B | c a
D -> EPSILON | d D
E -> g A f | c
This website generates the first set as follows:
Non-Terminal Symbol First Set
A g, ε, b, a, c, d
B ε, b
C a, c, ε, d
D ε, d
E g, c
But the algorithm says Fi(A w' ) = Fi(A) for every nonterminal A with ε not in Fi(A) so the First(A) according to this algorithm should not contain ε. First(A) = {g, b, a, c, d}.
Q: First(A) for the above grammar is = First(B) - ε U First(C) U {g} ?
This video also follows the above algorithm and do not choose ε.

Answer 1

First(B) = {ε, b}
First(D) = {ε, d}
First(E) = {g, c}
First(C) = {c, d, a}
First(A) = {b, g, c, d, a}
Example:
X -> Y a | b
Y -> c | ε
First(X) = {c, a, b}
First(Y) = {c, ε}
First(X) doesn't have ε because if you replace Y by ε, then First(Y a) is equal to First(ε a) = {a}
First set implementation on my github.
Edit: Updated link
https://github.com/amirbawab/EasyCC-CPP/blob/master/src/syntax/grammar/Grammar.cpp#L229
Computing the first and follow sets are both available on the new link above.