TITLE: Minimal blocking objects with shadows like a cube
QUESTION [14 upvotes]: This is a more geometric version of the previous question,
"Lattice-cube minimal blocking sets".  I will first specialize to $\mathbb{R}^3$, $d=3$.
View an $n \times n \times n$ cube $C_3(n)$ as formed of $n^3$ unit cubes glued face-to-face.
I would like to find a minimal blocking object $B$ inside $C_3(n)$,
which I define as a collection of the unit cubes in $C_3(n)$ with the following
two properties:
(a) The shadow of $B$ by parallel light rays in the three orthogonal directions
(parallel to the cube edges) is an $n \times n$ filled square.
(b) $B$ is a connected object, in the sense that its dual graph is connected.
Here the dual graph has a node for each unit cube in $B$ and an edge
between each pair of cubes that share a face.
$B$ is intended to be a minimal volume object that casts shadows like a cube.
The connectedness condition ensures one could build a physical model of $B$.
The previous MO question did not include the connected condition. There it was
shown that blocking sets of size $n^2$ are attainable, $n^{d-1}$ in
dimension $d$.
Certainly that lower bound is no longer achievable in general,
as can be seen with $C_2(2)$: 
a $2 \times 2$ square in dimension $d=2$ needs $3$ rather than $2=2^1$
unit squares to form a connected blocking object.
Exploring $n=3$ in $\mathbb{R}^3$, I have been unable to create a connected blocking
object with fewer than 15 cubes:

 
 
 
 
 
 


\[ \left[\begin{array}{ccc} 0 & 1 & 1 \\\\ 0 & 1 & 0 \\\\ 1 & 1 & 0 \end{array}\right] \hspace{0.25 in} 
\left[\begin{array}{ccc} 0 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 1 & 0 \end{array}\right] \hspace{0.25in}
\left[\begin{array}{ccc} 1 & 0 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 0 & 1 \end{array}\right]
\]
Fifteen seems excessive—more than half!.  Can anyone see a better solution?
In addition, I do not see how to generalize to $C_3(n)$, let alone to
$C_d(n)$, the same question in $d$ dimensions.
Addendum. Here is Joel's 13-cube blocker from his answer below:

REPLY [9 votes]: Lower bound:
Let there be $c$ cubes in a blocking configuration. Consider the graph whose vertices are cubes so that cubes are connected if they share a face. Any connected graph has at least  $c-1$ edges. By the pigeonhole principle, there is at least one direction with at least $(c-1)/d$ edges in that direction. When you project parallel to that axis, the image has size $n^{d-1}$, and the size of the image is also at most $c - (c-1)/d$ since at least one vertex on each of those edges is redundant. So, 
$$ n^{d-1} \le c - \frac{c-1}{d}  = c \frac{d-1}{d} + \frac 1d$$ 
$$ c \ge \bigg(\frac {d}{d-1}\bigg)n^{d-1} - \frac {1}{d-1}. $$
For $d=3$, $c \ge \frac 32 n^2 - \frac12$ (mentioned by Gjergji Zaimi in the comments). This is sharp for $n=3$ by the construction with $13$ cubes shown by Joel David Hamkins.
Upper bound ($d=3$):
Here is a construction of a connected blocking configuration with $\frac32n^2 + O(n)$ cubes related to Zack Wolske's constructions. We'll identify the cubes with lattice points. We start with the points $\lbrace(x,y,z) | x-y \equiv z \mod n, 0 \le x,y,z \lt n \rbrace$, illustrated for $n=7$.
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |.....x.|
|.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|
|....X..|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|
|...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|
|..X....|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|
|.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|
|X......|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

This is made of two triangles of points, in the planes $x-y=z$ (marked X) and $x-y=z-n$ (marked x), which are separated by some distance. We'll first make the connections within the triangles, and then connect the triangles to each other.
To connect the bottom triangle, use $n-1$ extra points (marked A) to connect the base of the triangle in the plane $z=0$ to itself and to the points in the triangle with $z=1$. Then for $i = 2, ..., n-1$, use $\lceil (n-i-1)/2 \rceil$ points (marked O) to connect the points in the plane $z=i$ to each other and to the lower points in the triangle. These points have $y$ odd, and are just above an X in the layer below. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |.....x.|
|.....XA|  |......X|  |x.....O|  |.x.....|  |..x....|  |...x...|  |....x..|
|....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|
|...XA..|  |....X..|  |....OX.|  |.....OX|  |x.....O|  |.x.....|  |..x....|
|..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|
|.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O|
|XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

This has added $n-1$ A's (this is one of the few correct uses of apostrophes to indicate a plural), and $0+1+1+2+2+...+\lfloor (n-1)/2 \rfloor = \lfloor (n-1)^2/4 \rfloor$ O's (see A002620). 
We repeat the process upside down to connect the upper left triangle using $n-2$ A's and $\lfloor (n-2)^2/4 \rfloor$ O's. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |....Ax.|
|.....XA|  |O.....X|  |xO....O|  |.xO....|  |..xO...|  |...x...|  |...Ax..|
|....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |..Ax...|
|...XA..|  |....X..|  |....OX.|  |O....OX|  |xO....O|  |.x.....|  |.Ax....|
|..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |Ax.....|
|.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O|
|XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

Finally, we connect the upper and lower triangles with $n-2$ Z's. There are many choices for how to do this. We'll put them in $z=1$, $y=n-2$, $1 \le x \le n-2$. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |....Ax.|
|.....XA|  |OZZZZZX|  |xO....O|  |.xO....|  |..xO...|  |...x...|  |...Ax..|
|....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |..Ax...|
|...XA..|  |....X..|  |....OX.|  |O....OX|  |xO....O|  |.x.....|  |.Ax....|
|..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |Ax.....|
|.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O|
|XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

In total, this configuration contains $n^2$ X's, $2n-3$ A's, $n-2$ Z's, and $\lfloor (n-1)^2/4\rfloor +\lfloor(n-2)^2/4 \rfloor = {n-1 \choose 2}$ O's, a total of $\frac 32 n^2 + \frac 32 n - 4$. 
Therefore, 
$$ \frac 32 n^2 - \frac 12 \le \min |C_3(n)| \le \frac 32 n^2 + \frac 32 n - 4.$$