Repairing 3D Binary Images Using the FCC Grid

Abstract

A 3D image I is well-composed if it does not contain critical edges or vertices (where the boundary of I is non-manifold). The process of transforming an image into a well composed one is called repairing. We propose to repair 3D images by associating the face-centered cubic grid (FCC grid) with the cubic grid. We show that the polyhedral complex in the FCC grid, obtained by our repairing algorithm, is well-composed and homotopy equivalent to the complex naturally associated with the given image I with edge-adjacency (18-adjacency). We illustrate an application on two tasks related to the repaired image: boundary reconstruction and computation of its Euler characteristic.

Appendix: Details of the Proof of Homotopy Equivalence

We prove here the lemmas used in the proof of Propositions 2 and 3 (see Sect. 5.2).

Fig. 30

a The only configuration of $2\times 2\times 2$ cubes containing triangles in cases IV, IX and X (colored). Letterings used b in Lemma 2 for case IV, c in Lemma 3 for case IX and d in Lemma 4 for case X (Color figure online)

1.1 Lemma for Proposition 2

Lemma 1

Each triangle $V_1V_2V_3$ in the protruding boundary of X is mapped by h onto $|\varSigma |$.

Proof

The image by h of the triangle $V_1V_2V_3$ depends on the type of its vertices. If $V_1V_2V_3$ comes from triangulating a rhombic face through the longer diagonal, then two vertices are open and in $|\varSigma |$, the third one is closed and protruding (case a. below). If it comes from triangulating a rhombic face through the shorter diagonal, then two vertices are closed and protruding, and the third vertex is open and may be non-protruding (case b. below) or protruding (case c. below).

• a. For a triangle $FF'K$, where $F,F'$ are (non-protruding) open vertices and K is closed (the protruding part of a rhombic face r only halfway outside $\varSigma$), if F is a vertex in $\varSigma$ and $F'$ is the center of a square face of some cube in I, the closed vertex K is mapped to F, and the triangle $FF'K$ is mapped onto the line segment $FF'$ in the boundary of $|\varSigma |$.

• b. For a triangle FKL with one non-protruding open vertex F, we distinguish two cases depending on the position of F.

  • b.1 If F is also an endpoint of an edge of some cube in I, the protruding closed vertices K and L are mapped to F, and the triangle FKL is mapped on the vertex F in the boundary of $|\varSigma |$.

  • b.2 If F is the center of some black face s, the protruding closed vertices K and L are mapped to the endpoints U and V of an edge of s, and the triangle FKL is mapped on the triangle FUV in the boundary of $|\varSigma |$.

• c. For a triangle FKL, where F is a protruding open vertex, we distinguish three cases depending on the number of dodecahedra incident to F.

  • c.1 If F is incident to exactly one dodecahedron E, corresponding to an edge e, the open vertex F is mapped to the midpoint M of e and the closed vertices K and L are mapped either to the same or to the two different endpoints of e, and thus the triangle FKL is mapped onto one half of the edge e or onto e (which is in the boundary of $|\varSigma |$).

  • c.2 If F is incident to two dodecahedra $E_1$ and $E_2$, corresponding to edges $e_1$ and $e_2$ sharing a vertex V, then the open vertex F is mapped to V and the closed vertices K and L of the triangle FKL on $E_1$ (and similarly for $E_2$) are mapped to one of the endpoints V or $V_1$ of the edge $e_1$; thus, the triangle FKL is mapped either onto the edge $VV_1$ or on the vertex V, which are both in the boundary of $|\varSigma |$.

  • c.3 If F is incident to three dodecahedra E, $E_1$ and $E_2$ corresponding to the edges e, $e_1$ and $e_2$, respectively, then the introduction of the new vertices on the edges incident only to the dodecahedron $E_1$ or $E_2$ and to F ensures that the image of each sub-triangle on $E_1$ or $E_2$ is either one of the edges $e_1$ or $e_2$, or one half of the edge e, which are both in the boundary of $|\varSigma |$ (see Fig. 21). If the triangle FKL is on E, it is mapped to the edge e in the boundary of $|\varSigma |$.

$\square$

1.2 Lemmas for Proposition 3

We give here the proof that cases IV, IX and X can only occur in a configuration of $2\times 2\times 2$ cubes where exactly three of them are black, two of them mutually face-adjacent and the third one strictly edge-adjacent to one of them, like in Fig. 26. This situation is shown in Fig. 30a with the positions of triangles FUB lying in the three cases. Note that there are two triangles in each case, triangles in cases IX and X share their edge FU, and triangles in cases IV and X share their edge UB. All triangles share their vertex U.

Lemma 2

(Case IV) If the vertex U of the triangle FBU on the square face s of a black cube C is filled, the edge e on s incident to U is filled, the edge $e'$ on the other square face $s'$ of C incident to the edge UV is not filled and $s'$ is black, then the only black cubes incident to U are C, the cube adjacent to C through black face $s'$, and the cube strictly edge-adjacent to C at filled edge e, and no other edge incident to U is filled.

Proof

Referring to Fig. 30b, let $C'$ be the black cube adjacent to C at $s'$. Then, the other two cubes incident to $e'$ must be both white, or $e'$ would be filled; the other two cubes incident to the edge UV must be both white, or UV would be filled; the cube edge-adjacent to C through the edge e must be black or e would not be filled; the remaining cube incident to U must be white or either e or $e'$ would be filled (configuration with two collinear critical edges). Therefore, the edges incident to U, with the exception of e, are not filled. $\square$

Lemma 3

(Case IX) If the vertex U of the triangle FBU on the square face s of a cube C is filled, no edge of the cube C incident to U is filled, and the other square face $s'$ of C incident to the edge UV is not black, then the filled edge e is collinear with the edge BU.

Proof

As U is filled, let e be a filled edge (one of the three such edges) incident to U and not on C. There are four cubes incident to e: one ($C_1$) is face-adjacent to C, two ($C_2$ and $C_3$) are strictly edge-adjacent to C, and one ($C_4$) is strictly vertex-adjacent to C. By Property 1, at least two of the four cubes incident to e must be black. Cubes $C_2$ and $C_3$ must be both white, or also the shared edge (on C, incident to U) would be filled. Thus, $C_1$ and $C_4$ must be black (see Fig. 30c).

The edge e can be in three different positions with respect to the triangle FBU.

• If e is orthogonal to the plane of face s, then $C_1$ cannot be black, or s would be black; a contradiction.

• If e is orthogonal to BU, in the plane of s, then C and $C_1$ are adjacent through the square face $s'$, which is thus black; a contradiction.

• If e is collinear with the edge BU, then $C_1$ and $C_4$, which are strictly edge-adjacent, share the critical edge e, which is filled. It is possible, depending on the conventional directions, that no other edge incident to U is filled.

$\square$

Lemma 4

(Case X) If the vertex U of the triangle FBU on the square face s of a cube C is filled, no edge of the cube C incident to U is filled, and the other square face $s'$ of C incident to the edge UV is black, then the filled edge e is orthogonal to BU and lies in the plane of the square face s.

Proof

Referring to Fig. 30d, let $C'$ be the cube adjacent to C through $s'$, and let us consider the other six cubes incident to U. Four of them are face-adjacent to either C or $C'$. If any of these four cubes were black, that would make one of the two common edges of C and $C'$ incident to U an L-shape edge, and thus filled; a contradiction. Of the two remaining cubes, the one strictly edge-adjacent to C cannot be black, since that would make their common edge (on C and incident to U) filled. The remaining cube, strictly edge-adjacent to $C'$, can be black, making the edge e, incident to U, orthogonal to BU and in the plane of the square face s, filled. $\square$