Need help understanding how x1 in the image has just 11 surfaces and not 12, i cannot visualize it :(
I cannot understand what's going on in the image, can someone please explain in easy terms (dummy alert!), i would greatly appreciate it.
Or maybe any useful links/resources could also be of great help! :)
It seems the unknown 3D operation is the "extrude" operation appears in various 3D CAD/modelling softwares (AutoCAD,SketchUp, ...) or 3D graphics libraries (Three.js, X3DOM).
Different softwares/libaries may use different name but the basic idea is the same. It is a way to create a solid from a cross-section (a geometric shape lies on a plane).
Geometrically, one starts with a 2D planar shape $S$ and a continuous curve $\gamma : [0,1] \to \mathbb{R}^3$ with starting point $\gamma(0)$ belongs to $S$. One then parallel translate $S$ so that the image of $\gamma(0)$ always lies on top of $\gamma(t)$. The resulting solid is the union of the image of $S$ at different $t$. More precisely, it is the 3D solid:
$$V = \bigg\{ p + \gamma(t) - \gamma(0) : p \in S, t \in [0,1] \bigg\}$$
If $\gamma(0)$ is the origin, the solid $V$ is known among geometers as the Minkowski sum of the point set $\gamma([0,1])$ and $S$.
For practical application, one typcially impose some extra constraint to limit what $\gamma$ could be. One choice is require $\gamma$ to be at least piecewise $C^1$, i.e continuous differentiable up to $1^{st}$ order. Aside from finitely many $t$, the tangent vector $\gamma'(t)$ is defined, non-zero and never parallel to the plane holding $S$.
Consider the special case where the cross-section is a planar $n$-gon and $\gamma$ is a polyline consists of $m$ line segments. When one increases $t$ from $0$ to $1$, each edge of $S$ will trace out $m$ quadrilaterals, one for each segment of $\gamma$. In the absense of any relations among directions of segments on $\gamma$ and edges on $S$, solid $V$ will have $mn + 2$ faces.
The actual number of faces of $V$ can be less than this. At those $t_0$ where $\gamma'(t_0)$ is discontinuous. If the two limiting tangent vector $\gamma'(t_0\pm) \stackrel{def}{=}\lim\limits_{\epsilon \to 0\pm}\gamma'(t_0+\epsilon)$ is coplanar with the tangent vector of an edge $e$ of $S$. The two quadrilaterals traced out by $e$ at $t \ge t_0$ and $t \le t_0$ will share the same normal vector. This mean they will merge into a single face. Let $N_\parallel$ be the number of this sort of events. The number of faces of $V$ is $mn + 2 - N_\parallel$.
Let's take the solid $X_1$ as an example. One can generate $X_1$ by taking the pentagon $X$ as $S$ and the union of two leftmost edges of $X_1$ as $\gamma$. Notice at the back of $X$, the tangent vector of the edge there seems to be coplanar with the two tangent vectors on $\gamma$. This means $N_\parallel = 1$ the number of faces of $X_1$ becomes $2(2)(5) + 2 - 1 = 11$.
One can generalize this to other planar shape. In particular, those $S$ whose boundary consists of $n_c$ circular arcs and $n_\ell$ line segments. If $N_\parallel$ is the number of instances where the two limiting tangent vectors $\gamma'(t_0\pm)$ is coplanar with the tangent vector of a straight edge of $S$, the number of faces of $V$ with be $m(n_c + n_\ell) + 2 - N_\parallel$.
Let's take the $Y$ as another example of $S$. Its boundary $\partial Y$ consists of $n_c = 2$ circular arcs and $n_\ell = 10$ line segments. $4$ of the line segment seems to be parallel to the reference direction indicated by the orange dashed line. Since $m = 2$ and there is only one discontinuity in $\gamma'(t)$, $N_\parallel = 4$. The number of faces of corresponding 3D solid $Y_1$ should be $2(2 + 10) + 2 - 4 = 22$.
Notes
I first learned this operation in SketchUp but never know what it is. I finally know its name when looking up documents on x3dom.js many years later. Neither of them explain the concept well. A web search returns this article on the extrude command in AutoCAD. It is quite readable and explain the operation clearly. Take a look if you are interested.