A question about the work per recursive call in FPT vertex cover of size k algorithm

Question

I have been looking at the FPT(Fixed Parameter) algorithm for checking if a vertex cover of size k exists.The algorithm goes as follows:

VertexCoverFPT$(G, k)$

if $G$ has no edges then return true

if $k=0$ then return false

let $uw$ be some edge of $G$

if VertexCoverFPT$(G-u, k-1)$ then return true

if VertexCoverFPT$(G-w, k-1)$ then return true

return false

Its said that this algorithm has $2^k$ recursive calls each with $O(n)$ of work in each recursive call. My question is how is it that there is only $O(n)$ work in each recursive call when we need to create the 2 graphs ${G-u}$ and ${G-v}$ (u and v are the vertices we are currently checking) which takes $O(V+E)$ time (Doing a deep copy of graph G excluding the node and the edges connected to it). Is there a more efficient way than constructing ${G-u}$, ${G-v}$ from G or am I missing something?

Answer 1

You do not need to create any new graph.

You can simply maintain an additional array $VT$ of size $|V| = n$ such that $VT[i] = 1$ if the vertex $i$ is deleted from the graph. Otherwise, $VT[i] = 0$.

While accessing an edge $(i,j)$, first the algorithm can check if $VT[i]$ and $VT[j]$ are both $0$ or not. If both are $0$ then the algorithm access that edge. Otherwise, it does not.

In addition to it, maintain the count of edges a vertex $i$ is currently connected to. Suppose the count is stored in an array $C$ of size $n$. In the beginning of the algorithm $C[i]$ is initialized to the $degree(i)$.

---

Accessing an Undeleted Edge: Go to the adjacency list of that vertex which has $VT[i] = 0$ and $C[i] \neq 0$. Since we know that at least one edge in the list is not deleted, the algorithm will find an undelete edge in $O(n)$ time. Updating the count array after deleting a vertex takes $O(n)$ time. Overall operation is $O(n)$ time.