Odd closed neighborhood Let G be a given graph. prove that you can color some vertices of $G$ red, so that each vertex has odd red vertices in its closed neighborhood.
closed neighbor hood of a vertex $u$, is union of the set of all vertices adjacent to $u$ and the set:$\{u\}$ itself.
the question in other words is: find a induced sub-graph ,$S$, of a given graph $G$, which all vertices in $S$ is of even degree and all vertices in $V(G)\setminus S$ has odd number of neighbors in $S$.
WHAT I HAVE TRIED: I have tried induction on the number of vertices of $G$ so far, but it didn't work.
There is a theorem in this book: The vertex set of every graph can be partitioned into two sets that induce even sub-graphs. I tried same kind of induction that is used to prove this theorem. Choose a vertex $v$ of odd degree and toggle pairs of vertices adjacent to $v$, then omit $v$ from vertex set of graph $G$. by induction hypothesis we can color a set of vertices so that every vertex has odd colored vertices in its closed neighborhood. now if vertex $v$ has odd number of red vertex adjacent to it, do not color $v$ other wise color $v$. by doing this all vertices has odd number of red vertices in their closed neighborhood except those are adjacent to $v$ and aren't colored red. I couldn't do much better for those vertices which after induction may have problems!
 A: at last I found the solution!
I want to generalize the problem.
First I define open vertex and close vertex. 
We call a vertex "open" if it has even number of red vertices in its open neighborhood, we call a vertex "close" if it has odd number of red vertices in its closed neighborhood.
The generalized problem is: given a graph $G$ and a sequence $a(i)$ of size $|V(G)|$  of entries {close, open}. prove that you can color some vertices red so for each $i$ if $a(i)$ is open then $v_{i}$ is open vertex and it is closed otherwise.
(Actually in main problem, the sequence is $a(i)$=close for each i.)
now use induction on $|V(G)|$. suppose in sequence $a$ all entries are open, then if you color no vertex obviously all vertices will be open.
so there will be a close entry in sequence $a$. without loss of generality name it $v_{1}$.
Let U={all vertices adjacent to $v_{1}$}. remove $v_{1}$ from $G$ and toggle all edges in U. (I mean for each pair $x$ and $y$ in $U$ if $xy$ was in $E(G)$ remove this edge and other wise add new edge $xy$). also in sequence $a$, if $v_{i}$ is in $U$ then toggle $a(i)$ from close to open and open to close.
use induction hypothesis in new graph. by a few observation you can finish the solution.
A: See under the term : odd-domination sets. Sutner proved it first in 1988 and I gave a linear algebraic proof in 1996.
