Computational complexity of three-dimensional discrete tomography with missing data

Abstract

Discrete tomography deals with problems of determining shape of a discrete object from a set of projections. In this paper, we deal with a fundamental problem in discreet tomography: reconstructing a discrete object in $\mathbb {R}^3$ from its orthogonal projections, which we call three-dimensional discrete tomography. This problem has been mostly studied under the assumption that complete data of the projections are available. However, in practice, there might be missing data in the projections, which come from, e.g., the lack of precision in the measurements. In this paper, we consider the three-dimensional discrete tomography with missing data. Specifically, we consider the following three fundamental problems in discrete tomography: the consistency, counting, and uniqueness problems, and classify the computational complexities of these problems in terms of the length of one dimension. We also generalize these results to higher-dimensional discrete tomography, which has applications in operations research and statistics.

Omitted proofs

Continuation of the proof of Proposition 13

We show that the construction in the proof of Proposition 13 in Sect. 3.2 is a parsimonious reduction in what follows.

We restate variable gadgets in the construction. Intuitively, $x_k = 1$ holds if and only if $p_{1,k,2k-1} = 1$ holds, and $x_j = 0$ holds if and only if $p_{1,k,2k-1} = 0$ holds, for $1 \le k \le n$. We thus call the variables $p_{1,k,2k-1}$ the (variable) gadget for variable $x_k$.

We first show how assignments to the variable gadgets propagate. For this, we show the following claim.

Claim 22

Let P be a solution of $I_\varphi$. Then for $1 \le k \le n$, we have

$$\begin{aligned} p_{1,k,2k-1}= & {} p_{3,k,2k}, \end{aligned}$$

(52)

$$\begin{aligned} 1- p_{1,k,2k-1}= & {} p_{1,k,2k} = p_{3,k,2k-1}. \end{aligned}$$

(53)

Moreover, for $1 \le j \le m$ and $1 \le k \le n$, we have

$$\begin{aligned} p_{1,k,2k-1}= & {} p_{Z(j),X(j)+2k,Y(j)+2k} \end{aligned}$$

(54)

$$\begin{aligned}= & {} p_{Z(j),X(j)+2k-1,Y(j+1)+2k-1} \end{aligned}$$

(55)

$$\begin{aligned}= & {} p_{3,X(j)+2k-1,Y(j)+2k-1} \end{aligned}$$

(56)

$$\begin{aligned}= & {} p_{3,X(j)+2k,Y(j+1)+2k}, \end{aligned}$$

(57)

$$\begin{aligned} 1- p_{1,k,2k-1}= & {} p_{Z(j),X(j)+2k-1,Y(j)+2k-1} \end{aligned}$$

(58)

$$\begin{aligned}= & {} p_{Z(j),X(j)+2k,Y(j+1)+2k} \end{aligned}$$

(59)

$$\begin{aligned}= & {} p_{3,X(j)+2k,Y(j)+2k} \end{aligned}$$

(60)

$$\begin{aligned}= & {} p_{3,X(j)+2k-1,Y(j+1)+2k-1}. \end{aligned}$$

(61)

Furthermore, for $1 \le j \le m-1$ and $1 \le k \le n$, if $x_k \in C_j$, then

$$\begin{aligned} p_{1,k,2k-1}= & {} p_{Z(j),X(j)+2n+2,Y(j+1)+2k} \end{aligned}$$

(62)

$$\begin{aligned}= & {} p_{Z(j+1),X(j)+2n+1,Y(j+1)+2k-1}, \end{aligned}$$

(63)

$$\begin{aligned} 1- p_{1,k,2k-1}= & {} p_{Z(j),X(j)+2n+1,Y(j+1)+2k-1} \end{aligned}$$

(64)

$$\begin{aligned}= & {} p_{Z(j+1),X(j)+2n+2,Y(j+1)+2k}, \end{aligned}$$

(65)

and if $x_k \not \in C_j$, then

$$\begin{aligned} p_{1,k,2k-1}= & {} p_{Z(j),X(j)+2n+2+\mathrm{IDX}(j,k),Y(j+1)+2k} \end{aligned}$$

(66)

$$\begin{aligned} = & {} p_{Z(j+1),X(j)+2n+2+\mathrm{IDX}(j,k),Y(j+1)+2k-1}, \end{aligned}$$

(67)

$$\begin{aligned} 1- p_{1,k,2k-1}= & {} p_{Z(j),X(j)+2n+2+\mathrm{IDX}(j,k),Y(j+1)+2k-1} \end{aligned}$$

(68)

$$\begin{aligned} = & {} p_{Z(j+1),X(j)+2n+2+\mathrm{IDX}(j,k),Y(j+1)+2k}. \end{aligned}$$

(69)

Finally, if $x_k \in C_m$, then

$$\begin{aligned} p_{1,k,2k-1}= & {} p_{Z(m),X(m)+2n+2,Y(m+1)+2k} \end{aligned}$$

(70)

$$\begin{aligned}= & {} p_{3,X(m)+2n+1,Y(m+1)+2k-1}, \end{aligned}$$

(71)

$$\begin{aligned} 1- p_{1,k,2k-1}= & {} p_{Z(m),X(m)+2n+1,Y(m+1)+2k-1} \end{aligned}$$

(72)

$$\begin{aligned}= & {} p_{3,X(m)+2k,Y(m+1)+2k}, \end{aligned}$$

(73)

and if $x_k \not \in C_m$, then

$$\begin{aligned} p_{1,k,2k-1}= & {} p_{Z(m),X(m)+2n+2+\mathrm{IDX}(m,k),Y(m+1)+2k} \end{aligned}$$

(74)

$$\begin{aligned}= & {} p_{3,X(m)+2n+2+\mathrm{IDX}(m,k),Y(m+1)+2k-1}, \end{aligned}$$

(75)

$$\begin{aligned} 1- p_{1,k,2k-1}= & {} p_{Z(m),X(m)+2n+2+\mathrm{IDX}(m,k),Y(m+1)+2k-1} \end{aligned}$$

(76)

$$\begin{aligned}= & {} p_{3,X(m)+2n+2+\mathrm{IDX}(m,k),Y(m+1)+2k}. \end{aligned}$$

(77)

We can routinely check that the above claim holds. Thus, we omit the proof.

Recall that $C_j = (x_{j_1} \vee x_{j_2} \vee x_{j_3})$ for $j = 1,\dots , m$. For each $1 \le j \le m$, variables $p_{Z(j+1),X(j)+2n+1,Y(j+1)+2j_i-1}$ ($i=1,2,3$) can be seen as a clause gadget. Intuitively, variable $x_{j_i}$ is the unique variable that is true (i.e., one) in clause $C_j$ if and only if $p_{Z(j+1),X(j)+2n+1,Y(j+1)+2j_i-1}=1$ holds, for $1 \le j \le m$ and $i=1,2,3$. We thus call the variables $p_{Z(j+1),X(j)+2n+1,Y(j+1)+2j_i-1}$ ($i=1,2,3$) the (clause) gadget for clause $C_j$ ($1 \le j \le m$).

From Claim 22, assignments to the variable gadgets are propagated through the rectangular parallelepiped. Note that a solution P of $I_\varphi$ is uniquely determined by the assignment to $p_{1,k,2k-1}$ for $1 \le k \le n$, since $p_{ijk} = 0$ for the indices not appearing in equations in Claim 22 by the definitions of F, S and T.

We are now ready to show that the construction is a parsimonious reduction. For this, we define a bijection between the solution sets of $\varphi$ and $I_{\varphi }$. We first define a mapping from the solution set of $\varphi$ to that of $I_{\varphi }$. Let x be a solution of $\varphi$. Define an assignment P of $I_{\varphi }$ as follows. For $1\le k \le n$, set

$$\begin{aligned} p_{1,k,2k-1} = x_k, \end{aligned}$$

(78)

and set variables appearing in Claim 22 so that they satisfies the equations in Claim 22. Set $p_{ijk} = 0$ for all the remaining $p_{ijk}$. We show that this assignment P is a solution of $I_\varphi$ in the following.

Observe that the constraints corresponding to the equations in the definitions of matrices F, S, and T are all satisfied, except ones corresponding to Equations (29), (30), (33), and (34). This is because P satisfies the equations in Claim 22.

Moreover, the constraints corresponding to Equations (29), (30), (33), and (34) are also satisfied by P since x is a solution of $\varphi$. For example, fix $1 \le j \le m-1$ and consider equation $f_{Z(j+1),X(j)+2n+1} = 1$ in Equation (30). By the definition of P and equations in Claim 22, we have $p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_i-1} = 1$ if and only if $x_{j_i} (= p_{1,j_i,2j_i-1}) = 1$. Since x is a solution of $\varphi$, we have $\sum _{i=1}^{3}p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_i-1} = 1$. Moreover, by the definition of P, we have $p_{Z(j+1), X(j)+2n+1, \ell } = 0$ for $\ell \ne p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_i-1}$ ($i=1,2,3$) . Therefore, we have $\sum _{\ell =1}^d p_{Z(j+1), X(j)+2n+1, \ell } = 1$, implying that P satisfies the constraint corresponding to $f_{Z(j+1),X(j)+2n+1} = 1$. Similarly, P satisfies the other constraints corresponding to Equations (29), (30), (33), and (34). Therefore, P is a solution of $I_\varphi$.

We next define a mapping from the solution set of $I_\varphi$ to that of $\varphi$. Let P be a solution of $I_\varphi$. Define an assignment x of $\varphi$ as

$$\begin{aligned} x_k = p_{1,k,2k-1} \end{aligned}$$

for $1 \le k \le n$. We show that x defined as above is a solution of $\varphi$. Fix $1 \le j \le m-1$. Observe that $p_{Z(j+1), X(j)+2n+1, \ell } = 0$ for $\ell \ne p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_i-1}$ ($i=1,2,3$) since $t_{X(j)+2n+1,\ell } = 0$ for such $\ell$’s. Thus, from $f_{Z(j+1), X(j)+2n+1} = 1$, we have

$$\begin{aligned} \begin{array}{l} \sum _{\ell =1}^d p_{Z(j+1), X(j)+2n+1, \ell } \\ \quad =p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_1-1} + p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_2-1} \\ \qquad +p_{Z(j+1), X(j)+2n+1, Y(j+1)+2j_3-1} \\ \quad =1. \end{array} \end{aligned}$$

Since $p_{Z(j+1),X(j)+2n+1,Y(j+1)+2j_{\ell }-1} = p_{1,j_{\ell },2j_{\ell }-1}$ for $\ell = 1,2,3$ from Claim 22, we have $p_{1,j_{\ell },2j_1-1} + p_{1,j_{\ell },2j_2-1} + p_{1,j_{\ell },2j_3-1} =1$. Therefore, $x_{j_1} + x_{j_2} + x_{j_3} = 1$ by the definition of x, implying that exactly one $x_{j_\ell }$ has value one in clause $C_j$. This can be shown for the case of $j=m$ in a similar way. Therefore, x is a solution of $\varphi$.

So far, we have defined two mappings between the solution sets of $\varphi$ and $I_\varphi$. This implies that $\varphi$ is reduced to $I_\varphi$. We finally show that the mappings are inverse to each other, which implies that these mappings are bijective and hence the reduction is parsimonious.

Let x be a solution of $\varphi$ and let P be the solution of $I_{\varphi }$ corresponding to x. Moreover, let $x'$ be the solution constructed from P. We show that $x=x'$ holds. Observe first that $x_k = 1$ if and only if $p_{1,k,2k-1} = 1$ for $1 \le k \le n$. Furthermore, $p_{1,k,2k-1} = 1$ if and only if $x'_k = 1$ by construction for $1 \le k \le n$. Hence, $x_k = x'_k$ for $1 \le k \le n$ and thus $x=x'$.

Conversely, let P be a solution of $I_{\varphi }$ and let x be the solution of $\varphi$ constructed from P. Moreover, let $P'$ be the solution constructed from x. We show that $P=P'$ holds. Firstly, for $1 \le k \le n$, $p_{1,k,2k-1} = 1$ if and only if $x_k = 1$ holds.

Furthermore, $x_k = 1$ holds if and only if $p'_{1,k,2k-1} = 1$ for $1 \le k \le n$. Hence, P and $P'$ coincide in the indices $(1,k,2k-1)$ for $1 \le k \le n$. From Claim 22, the values of P are uniquely determined by the values for these indices. Hence, $P=P'$ holds. Therefore, the two mappings between the solution sets of $\varphi$ and $I_\varphi$ are inverse to each other. This completes the proof. $\square$