I am interested in showing that the halving number of the d-dimensional cube is $2^{d-1}$ for $d = 1, 2$ but not for $d \ge 3$.
I tried to find the meaning of $\textit{halving number}$ but I couldn't find. I read from wikipedia about halved cube graph and it appears to not have given an explicit definition of $\textit{halving number}$ but it stated that halved cube graphs have vertices of $2^{n-1}, n\in \mathbb{N}$. In my case, I want to show this for d-dimensional cube for $n=1,2$ but not $n\ge 3$.
I also found a definition which says that: $\textit{halving number}$ is the minimum number of vertices to remove such that every remaining component has order at most half that of the original graph (I believe this might be the correct definition).
Now, using this definition (iff it is correct) how would I do the prove of the question?
$\textbf{update 1:}$
For $d=1,2$ hypercubes, the number of vertices will be $2, 4$ with corresponding halving numbers as $1,2$ respectively. Removing (subtracting) the halving numbers from the number of vertices will lead to the remaining components having an order (number of vertices) of at most half that of the original graphs. That is $2^{d-1}$ holds for $d=1,2.$ For $d=3,$ the number of vertices is $8$ and the halving number is $3 \not= 2^{3-1} =4.$ This is just one case for $d=3$. How do I deal with $d=4, \ldots$ cases if my idea presented is correct.
$\textbf{update 2:}$
Observe that $d=1,2$ contain no cube, but $d\ge3$ contains at least one $2^{n-4}n(n-1)(n-2)/3$ cube with $d=3$ having the minimum cube. This means that the halving number for $d=3$ (which is 3) is a product multiples of the halving numbers for $d\ge3$ with halving numbers $3,6,9, \ldots$ and vertices $8,16,32, \ldots$ for $d=3,4,5.$ respectively. Clearly, the halving numbers $3,6,9, \ldots \not= 2^{d-1} = 4, 8, 16 \ldots$ for $d\ge3.$
$\textbf{update 3:}$
If $d=1$ then a line graph- disconnect either vertex leaves a singleton. If $d=2$ then a square graph- disconnect any adjacent vertices leaves a line graph, disconnect opposite vertices leave a two singletons. In both cases, the remaining components are at most half of the original graph.