A solution to this problem follows from establishing the following result--which is actually a decent amount stronger, we don't have to settle for merely 200 vertices, we can remove 250 vertices:
Theorem A: Let $G$ be a 3-regular graph on $n$ vertices. Then there is an independent set $S$ of vertices in $G$ satisfying $|S|= \frac{n}{8}$ so that $G \setminus S$ is connected.
Idea of proof: The idea that gives a bound of a bit better than
$\frac{n}{10}$ is that you end up with a connected graph $G'$ where the cycles are all vertex-disjoint and so the upper bound on the number of edges is $\frac{4n(G')}{3}$ where $n(G')$ is the number of vertices in $G'$, and this upper-bound is achieved when all the cycles have length 3. [The smaller you can bound the average degree on $G'$,
the larger you can guarantee the size of the set of vertices that you can remove.] In fact, we establish that the connected graph $G'$ that we achieve the cycles remain vertex disjoint, and $G'$ is guaranteed to have either many cycles of length 4 or larger, or many vertices not in any cycle.
The proof I gave last would establish a bound of $\frac{n}{10}$
but not $\frac{n}{8}$. The problem was as this: If you contract all degree-2 vertices
as in Lemma 3 to an edge then what may result is a multigraph with parallel edges
between two vertices and so you cannot assume that the length of the smallest cycle
is 3 anymore.
We first make the following observation:
Claim 1: If $v$ is a cut-vertex in a graph of maximum degree-3, then at
least one edge in that graph incident to $v$ is a cut-edge.
Proof of Claim 1: Removing $v$ makes at least 2 connected components. And so as
$v$ has degree-3, there is at least one such component where exactly
one edge $e$ incident to $v$ has its endpoint in that component. Then
$e$ is a cut edge. $\surd$
We now establish this theorem. Let us suppose that we have already removed
an independent set from $G$ so that the resulting graph is still
connected. Now let $v$ be a vertex in the remaining graph where $v$
still has degree-3 in the remaining graph and $v$ is not a
cut-vertex in the remaining graph. Then if we remove $v$,
then the resulting set of vertices that we have removed from $G$ remains
independent, and the resulting graph is still connected. Thus we do
the following:
(I) If there is an available vertex $v$ that
is in a triangle [on the graph that is left], pick a vertex $v'$ that
is in a triangle to remove. Otherwise
pick another available vertex to remove. STOP when there are no more
available vertices to remove. Write as $G'_{int}$ the resulting graph.
Then the set of vertices in $G \setminus G'_{int}$ is an independent set
of vertices in $G$. Furthermore, $G$ and $G'_{int}$ together have
the property:
(A) Let $C$ be a triangle in $G'_{int}$ and let $w'$ be a vertex
not in $G \setminus G'_{int}$ such that $w'$ is adjacent in $G$ to a
vertex in $C$. Then $w'$ is in a triangle in the original graph $G$.
Indeed, for $w'$ to have been taken out in the first place,
by Claim 1 above $w'$ is not incident to any cut edges at the time. So let
$v$ be the vertex in $C$ adjacent to $w'$ in $G$. Then $vw'$ was
not a cut-edge at the time. As $v$ is in a cycle $C$, neither
are the other 2 edges incident to $v$. So $v$ was available.
As $v$ is in a triangle, by the rule (I) $w'$ had to have been in a
triangle as well.
Then, from $G'_{int}$, do the following if and so long as
there is a triangle $C$ in $G'_{int}$ that
satisfies the following: There is a vertex $w'$ in $G \setminus G'_{int}$ such
that both:
(a) The vertex $w'$ is adjacent in $G$ to a vertex $v$ in $C$, and $w'$
is in a triangle in $G$.
(b) A triangle $w'x'y'$ that $w'$ is in, in $G$, satisfies
the additional property that $x'y'$ is in a cycle in $G'_{int}$.
Then do the following:
- Add back in $w'$ and so the edges $vw'$, $w'x',w'y'$.
- Then check that both $v$ and $x'$ together are available, are not
adjacent to each other, and are each in triangles, so remove
both $x'$ and $v$.
- Then resetting the resulting graph to $G'_{int}$, note that $G'_{int}$
remains connected, and has one fewer triangle than
before.
Then $G'_{int}$ satisfies the invariant (A) after each of the above
iterations as well. Indeed, the set of vertices in a triangle after each
iteration is a subset of the set of vertices that were in a triangle
before. And every vertex $v$ taken out of $G'_{int}$
was in a triangle in $G'_{int}$ before so $v$ is also in a triangle in
$G$.
When there are no more vertices satisfying (a) and (b) above,
the resulting graph $G'$ also satisfies (B):
(B) Let $C$ be a triangle in $G'$ and let $w'$ be a vertex
not in $G \setminus G'$ such that $w'$ is adjacent in $G$ to a
vertex in $C$. Then $w'$ is in a triangle $w'x'y'$
in the original graph $G$, where the additional property is satisfied
that both $x',y'$ are not in any cycles in $G'$.
Claim 2: The cycles in $G'$ are vertex-disjoint.
Proof of Claim 2: In $G'$ there are no more available vertices. So all vertices of
degree-3 are cut-vertices. Thus by Claim 1,
every vertex $v$ of degree-3 is incident to a cut edge. So let $M$ be
the set of cut edges. Each such $e \in M$ is not in a cycle, and as every
vertex of degree-3 in $G'$ is incident to a cut edge, it follows that
$G \setminus M$ has maximum degree-2 and is a collection of
vertex-disjoint cycles. As every edge in $M$ is a cut-edge and is in no cycle in $G'$, it follows
that the set of cycles in $G' \setminus M$ is the same as the set
of cycles in $G' \setminus M$, and these are vertex-disjoint. So the
claim follows. $\surd$
Now let $\cal{C}$ be the collection of cycles in $G'$ and for each integer $k$
let us write as $\cal{C}_k$ the set of cycles of length $k$.
We now make some observations:
As the cycles are vertex-disjoint,
let us collapse each cycle $C \in \cal{C}$ in $G'$ to a vertex $v_C$;
the resulting graph $G"_1$ is a tree. Furthermore, every vertex in $G''_1$
is either in $\{v_C; C \in \cal{C}\}$ or in $A$ where
$A$ $\doteq \{v \in G'$; $v$ is not in any cycle in $G'\}$.
Now let $W'$ be the set of vertices $W'\doteq \{w';$
$w' \in G\setminus G';$ $w'$ is adjacent in $G$ to a vertex in
a triangle $C \in \cal{C}_3\}$. By the fact that $G''_1$ is a
tree gives the following:
Claim 3:
$$|{\cal{C}}_4|+|A| \ge 1+\left[\sum_{C \in \cal{C}_3}d_{G''_1}(v_C)\right]
-2|\cal{C}_3|.$$
Furthermore, the following is also true:
Claim 4:
$$W' \ge \sum_{C \in \cal{C}_3} [3-d_{G''_1}(v_C)].$$
Proof: As Invariant (A) is maintained,
for each cycle $C \in \cal{C}_3$ there are
$3-d_{G''_1}(v_C)$ vertices $w'$ in $G \setminus G'$
such that $w'$ is in a triangle in $G$, and $w'$ is adjacent in
$G$ to a vertex in $C$. So $w'$ is in $W'$.
Thus, to establish this, it suffices to show that each $w' \in W$
is adjacent in $G$ to exactly one vertex $v$ that is any
$C \in \cal{C}_3$. Let $w'$ be a vertex in $W'$, and let
$w'x'y'$ be a triangle
that $w'$ is in, in the original graph $G$, and let $v$ be the
vertices adjacent in $G$ to $w'$ so that $v$ is in a triangle in $G'$.
Then $v,x',y'$ are all of $w$'s neighbors in $G$ because $w'$ has
degree-3, and furthermore, as (B) is satisfied, $x'$ and $y'$ cannot
be in any triangles in $G'$. So the only vertex that $w'$ can be
adjacent to that is in a triangle in $G'$, is $v$. Thus indeed,
each $w' \in W'$ is adjacent in $G$ to exactly one vertex $v$ that is any
$C \in \cal{C}_3$, and so $|W'|$ must be at least
$\sum_{C \in \cal{C}_3} [3-d_{G''_1}(v)]$.
So putting Claims 3 and Claims 4 together yields:
Claim 5: $$|{\cal{C}}_4|+|A|+|W'| \ge |\cal{C}_3|+1$$
We now partition the set $A$ of vertices not in a cycle
into 2 sets $A_1$ and $A_{2}$.
The set $A_1$ is defined to be the set of vertices
$v \in A$ such that either $v$ has degree-3 in $G''$,
or $v$ has degree 2 or 1 but $v$ has no neighbor
in $G'' \cap A$ that has degree 2 or 1.
The set $A_{2}$ is the set of vertices $v \in A \setminus A_1$;
equivalently, the set of vertices $v \in A$ such that both $v$ has
degree-2 or 1, and $v$ is adjacent in $G'$ to another vertex in $A$
that has degree-2 or 1.
Claim 6: Then the inequality
$$|{\cal{C}}_4|+|A|+|W'| \ge |\cal{C}_3|+1$$
from Claim 5 can be strengthened to
$$|{\cal{C}}_4|+|A_1|+\frac{|A_2|}{2}+|W'| \ge |\cal{C}_3|+1.$$
Proof of Claim 6: From $G''_1$ contract each path of 2 or more vertices
in $A$ of degree 2 or less to a single vertex and let
$G''_2$. Then let $A'$ be the set of vertices in $G''_2$
that correspond to vertices in $A$ in $G''_1$. Then
$|A_1| + \frac{|A_2|}{2} \ge |A'|$ on the one hand, and on the
other hand, $G''_2$ remains a tree, the line of reasoning used to
establish Claim 3 above gives:
$$|{\cal{C}}_4|+|A'| \ge 1+
\left[\sum_{C \in \cal{C}_3}d_{G''_2}(v_C)\right]-2|\cal{C}_3|.$$
However, $d_{G''_1}(v_C)=d_{G''_2}(v_C)$
for each $C \in \cal{C}$. So this gives
$$|{\cal{C}}_4|+|A'| \ge 1+
\left[\sum_{C \in \cal{C}_3}d_{G''_1}(v_C)\right]-2|\cal{C}_3|.$$
Plugging this together with Claim 4, and then
noting $|A_1|+\frac{|A|}{2} \ge |A'|$ yields Claim 6.
$\surd$
We also make the following claim:
Claim 7: $|A| \ge \max\{2|W'|+|A_1|, |A_2|\}$.
Proof of Claim 7: We first show the following: Let $w'$ be a vertex
in $W'$ and let $w'x'y'$ be a triangle. We claim that the only
vertex in $W'$ that $x'$ and $y'$ can be adjacent to is $w'$.
[Indeed, then $x'$ and $y'$ are both in
$G'$ and so is the edge $x'y'$. If on the one hand there is
another vertex $w'' \in W'$ such
that $w''x'y'$ is a triangle then
that leaves $D_{G'}(x')=d_{G'}(y')=1$ with the edge $x'y'$ in $G'$
so $x'$ and $y'$ form their own connected component, a
contradiction. If on the other hand there is another
vertex $w'' \in W'$ such that
$w''x'x''$ form a triangle for a vertex $x'' \not = y'$, then
$x''$ and $y'$ are 2 distinct vertices in $G'$ and $w'$ and $w''$ are 2
distinct vertices in $G \setminus G'$, all adjacent to $x'$
which gives $d_G(x') \ge 4$.]
We next claim that $w'$ is not adjacent in $G$ to any vertex in
$A_1$. Indeed, both $x'$ and $y'$ as above
are in $A$ because invariant (B) is satisfied, but both $x'$ and $y'$
have degree no greater than 2 and are adjacent to each other so $x'$ and
$y'$ are both not in $A_1$. So Claim 7 follows. $\surd$
Claim 6 and Claim 7 together imply Claim 8:
Claim 8: $$|A| \ge |A|_1+\frac{|A_2|}{2}+|W'|$$, and so
$$|{\cal{C}}_4|+|A| \ge |\cal{C}_3|+1.$$
Claim 8 implies Claim 9:
Claim 9: Let $n$ be the number of vertices in the original graph $G$ and
let $R$ be the number of vertices in $G \setminus G'$ and let
$n(G')=n-R$ be the number of vertices in $G'$ and let
$m(G')$ be the number of edges in $G'$. Then
$m(G') \le \frac{9n(G')}{7}$, and this is when every vertex is in either a 3-cycle
in $G'$ or a 4-cycle.
Then putting this together and doing algebra on the inequality
$$\frac{9(n-R)}{7} \ge \frac{3n}{2}-3R$$
yields $R \ge \frac{n}{8}$
and thus the theorem. $\surd$
And this is essentially the best you can do, you cannot guarantee a set larger than $\frac{n}{8}+O(1)$: First let $T_1$ and $T_2$ be
2 complete binary trees of the same size where the roots of $T_1$ and $T_2$ have degree-2 and every other interior vertex of each of $T_1$ and $T_2$ has degree-3. Then
Put an edge between the roots of $T_1$ and $T_2$ and call the resulting graph $T$.
Replace each vertex $v$ of $T$ with a cycle $c_v$, where $c_v$ is a triangle if $v$ is an interior vertex, and $c_v$ is a 4-cycle if $v$ is a leaf.
Then connect the $c_v$s as follows: (a) If $v$ and $w$ are adjacent vertices in $T$ then there is exactly one edge from a vertex in $c_v$ to a vertex in $c_w$; if $v$ and $w$ are non-adjacent distinct vertices in $T$ then there are no edges from $c_v$ to $c_w$. (b) If $v$ is an interior vertex of $T$, then each of the exactly 3 vertices in $c_v$ is adjacent to exactly 1 vertex outside of $c_v$ each. If $v$ is a leaf vertex, then exactly 1 of the exactly 4 vertices in $c_v$ is adjacent to exactly one vertex outside of $c_v$.
Then, for each leaf vertex $v$ of $T$, add an additional vertex $i_v$ and put an edge from $i_v$ to the 3 of the vertices in $c_v$ that have degree exactly 2.
Call the resulting graph $G$. Then if $T$ has $n$ vertices, then $G$ has
$5 \times \left(\frac{n}{2}\right)+ 3 \times \left(\frac{n}{2}\right)$ vertices, plus or minus $O(1)$. The largest independent set of vertices that we can remove from $G$ so the resulting graph is connected is $\{i_v; v$ and leaf vertex in $T\}$, and this set has cardinality $\frac{n}{2} = \frac{|V(G)|}{8}+O(1)$. $\surd$
