$k$-Vertex Cover:
Given a graph $G = (V, E)$ where $V$ is a set of vertices and $E$ a set of edges, and an integer $k$, the $k$-Vertex Cover problem determines if there exists a subset of vertices $V'$ of $V$ of size at most $k$, such that every edge of $E$ has at least one vertex in $V'$.
Can we solve $k$-Vertex Cover problem in $f(k)+c\log|V|$ space where $c>0$ is some constant?
yes. (proof given in link page number 126 and Theorem 2.3). Can you give Simple proof?
Can we get all mimimum vertex covers of size $k$ from this proof with in
$f(k)+c\log|V|$ space?
There is an other method (bounded search tree method proof given in [link] (http://www.mrfellows.net/papers/CCDF97_AdviceClasses.pdf) page number 127 proof(2)) to show that k-Vertex Cover problem is in parameterized Log space.
Other method goes as follows: There are 2^k possible vertex covers. Think $P$ is at most $k$ length 0's and 1's bit string. For each value of $P$, we can generate the vertex set of size at most $k$ and it may become potential vertex cover. Clearly we can say $P$ is the path in bounded search tree. For generating for all $P$'s need 2^k space. Now we need to find vertex cover for given $P$ and $P$ is at most $k$ length 0's and 1's bit string.
Function $F(P,1)$ returns us first vertex by choosing the lex least edge and then the lex least vertex and this function will take constant space.
Function $F(P,2)$ returns us vertex of lex least edge not connected to the first one ($F(P,1)$).
$F(P,2)$ needs output of $F(P,1)$ and enumerate edges to find the lex least edge not connected to the first one. $F(P,2)$ needs one logspace
Function $F(P,3)$ returns us vertex of lex least edge not connected to $F(P,1)$ and $F(P,2)$. But we are not store the output of $F(P,1)$ and $F(P,2)$. We need to enumerate the edges and find the lex least edge not connected to $F(P,1)$ and $F(P,2)$. For this for each edge, we call $F(P,1)$ and $F(P,2)$ sequentially and check whether edge connected to $F(P,1)$ and $F(P,2)$. For each edge we need one logspace and $F(P,2)$ one logspace $F(P,3)$ needs two logspace
Similarly $F(P,i)$ returns us vertex least edge not connected to $F(P,1) \cdots F(P,i-1)$. we are not store the outputs of $F(P,1) \cdots F(P,i-1)$. We need to call sequentially. We need to enumerate the edges and find the lex least edge not connected to $F(P,1) \cdots F(P,i-1)$. For this for each edge, we call $F(P,1) \cdots F(P,i-1)$ sequentially and check whether edge connected to $F(P,1) \cdots F(P,2)$. So $F(P,i)$ needs (i-1) logspace
Clearly $F(P,k)$ needs k.logspace. But algorithms should takes only $f(k)+c\log n$ space?.
Am I missing something?. Please help out this process where it went wrong?