FF-SA: Fragmentation-Free Spatial Allocation

Abstract

Given a grid G containing a set of orthogonal grid cells and a list L of choices on each grid cell (e.g., land use), the fragmentation-free spatial allocation (FF-SA) aims to find a tile-partition of G and choice assignment on each tile so that the overall benefit is maximized under a cost constraint as well as spatial geometric constraints (e.g., minimum tile area, shape). The spatial constraints are necessary to avoid fragmentation and maintain practicality of the spatial allocation result. The application domains include agricultural landscape design (a.k.a., Geodesign), urban land-use planning, building floor zoning, etc. The problem is computationally challenging as an APX-hard problem. Existing spatial allocation techniques either do not consider spatial constraints during space-tiling or are very limited in enumeration space. We propose a Hierarchical Fragmentation Elimination (HFE) algorithm to address the fragmentation issue and significantly increase the enumeration space of tiling schemes. The new algorithm was evaluated through a detailed case study on spatial allocation of agricultural lands in mid-western US, and the results showed improved solution quality compared to the existing work.

Appendix A Proof on APX-hardness

We show FF-SA is more general than the MAX-3SAT problem [8], which is APX-hard.

Definition 1

(3SAT and MAX-3SAT). Given a Boolean formula: $(x_1\vee x_2 \vee x_3)\wedge (\lnot x_2\vee x_4 \vee x_5)\wedge (\lnot x_5\vee x_6 \lnot x_7)\wedge ...$, which is composed by m clauses where each clause has 3 binary variables, 3SAT determines whether there exists a 0-1 assignment of all variables such that the output of the Boolean formula is 1 (satisfiable). MAX-3SAT is a variation of 3SAT. Instead of determining whether the formula is satisfiable, MAX-3SAT finds a 0-1 variable assignment to maximize the number of satisfied clauses.

Definition 2

(BOX-PACK and MAX-BOX-PACK). Given a set S of axis-parallel square boxes with integer height $h_i$, width $w_i$ and coordinates (locations fixed), BOX-PACK determines whether there exists a subset of B non-overlapping boxes in S. MAX-BOX-PACK, instead of determining this existence, finds the subset that has the maximum number of non-overlapping boxes.

The proof in [9] reduces 3SAT to BOX-PACK. It was shown given a 3SAT instance, a specific two-dimensional allocation of boxes can be constructed in polynomial time in BOX-PACK, and 3SAT can be satisfied (return 1) if and only if a subset of B non-overlapping boxes can be found. Every unsatisfied clause in 3SAT will cause a deduction of 1 from B. The proof needs to be extended in two ways to reduce MAX-3SAT to FF-SA: (1) extend from “3SAT$\implies$BOX-PACK” to “MAX-3SAT$\implies$ MAX-BOX-PACK”; and (2) re-formulate the box-allocation in MAX-BOX-PACK to the grid representation of FF-SA by assigning specific grid cell values.

Polynomial-time reduction from BOX-PACK to MAX-BOX-PACK: (1) If BOX-PACK can be determined in polynomial time, then we show there exists a simple algorithm to find the maximum number $B^*$ in MAX-BOX-PACK. Suppose the total number of boxes in the instance is N. Then the algorithm can simply enumerate through all integers in [1, N] using BOX-PACK and find $B^*$ in polynomial time. (2) If MAX-BOX-PACK can be solved in polynomial time, then for any number B in BOX-PACK, we can compare it to $B^*$ and get the decision in polynomial time. Thus, based on (1) and (2), BOX-PACK and MAX-BOX-PACK can be mutually transformed in polynomial time.

MAX-3SAT is a special case of MAX-BOX-PACK: We show MAX-3-SAT can be reduced to MAX-BOX-PACK in polynomial time. So far we have shown 3-SAT can be reduced to MAX-BOX-PACK. Suppose we have a 3SAT instance $I_{3SAT}$ and a corresponding BOX-PACK instance $I_{BP}$, and B is the largest number of non-overlapping boxes achievable in $I_{BP}$ when $I_{3SAT}$ is satisfiable. [9] shows that every unsatisfied clause in $I_{3SAT}$ is equivalent to a deduction of one in B of $I_{BP}$. Since for $I_{3SAT}$ we can create a corresponding MAX-BOX-PACK instance $I_{MBP}$ with exactly the same input (box-allocation) as $I_{BP}$. Let $B^*$ denote the maximum number of non-overlapping boxes achievable in $I_{MBP}$, and $B^* = B$ when $I_{3SAT}$ is satisfiable. Similarly, we construct a MAX-3SAT instance $I_{M3SAT}$ using the same input of $I_{3SAT}$. Given B, $B^*$, and m (total number of clauses in $I_{M3SAT}$), the maximum number of satisfied clauses in $I_{M3SAT}$ can be computed as $M^*=m-(B-B^*)$. Thus, $I_{MBP}$ can be constructed in polynomial time given $I_{M3SAT}$, and the solution $M^*$ of $I_{M3SAT}$ can be achieved in polynomial time given $B^*$.

MAX-3SAT is a special case of FF-SA: To prove FF-SA is APX-hard, we use the above-mentioned special case (specific input box-allocation) of MAX-BOX-PACK to build the corresponding instance of FF-SA. First, we reformulate the special case of box-allocation into a grid representation. Since the coordinates of boxes in MAX-BOX-PACK are all integers, it is straightforward to map the boxes to a grid with cell-size 1. Next, we enrich all the grid cells with values (e.g., benefits) to create the corresponding instance of FF-SA. The goal is to make sure that for the special case of box-allocation, the optimal solution of FF-SA gives the optimal solution of MAX-BOX-PACK in polynomial time.

Overview of the construction of box-allocation [9]: Generally, for each variable in MAX-3SAT, there is a loop of equal-size boxes with side-length 2 and the number of boxes is even. Half of the boxes are labeled “1” and the others are “0”, where the boxes representing “1” and boxes representing “0” are interleaving (Fig. 7(a)), so that each box with “0” (resp. “1”) intersects with two boxes with “1” (resp. “0”). Given this allocation, the largest subset of non-overlapping boxes for each loop is either all boxes with “1” or all with “0”, which corresponds to the binary choice of each variable in MAX-3SAT. For a given clause with three variables, there will be a clause region, where the three box-loops corresponding to the three variables approach each other. The space available in the clause region is determined by the 0-1 choices made at the three loops. The allocation is designed in a way that if any loop chooses “1” boxes, then there will be enough space to include one and only one more box in the clause region into the solution (a subset of all boxes) without overlapping any other boxes. Given a solution containing a subset of boxes in MAX-BOX-PACK, the number of satisfied clauses in MAX-3SAT can be inferred by counting how many boxes are added from all the clause regions. [9] shows it is sufficient to discuss three general cases appearing in the box-allocation. To avoid heavy redundancy, we will focus on the three cases without proving how exactly they translates to MAX-3SAT, as this omitted proof is in [9].

In the following discussion of FF-SA instance $I_{FF}$, (1) we relax the cost constraint by choosing a $+\infty$ budget; (2) there are only two choices for each grid cell, namely “A” and “B”. A’s benefit values are shown as the grid cell values, and B’s benefit values are always 0; and (3) for choice A, the minimum area is 4 and minimum width is 2; for choice B, the minimum area and width are 1 (discussed in scope, Sect. 2).

Case 1: A fragment of a single loop and this fragment does not intersect with any other loop (Fig. 7(a), left). It is easy to see we can either choose all boxes with “1” or all with “0” for a maximum non-overlapping subset. To achieve the same result in FF-SA, a benefit-value assignment is proposed in Fig. 7(a) (right), where $\gamma$ denotes a very large positive value (e.g., 9999). Since tile with choice “A” cannot overlap and must satisfy minimum area 4 and width 2, the optimal solution must be choice “A” on either all $2\times 2$ grid boxes corresponding to “1” or “0”, which is the same as MAX-BOX-PACK. The rest of the cells will just choose choice “B” with 0 benefit.

Fig. 7.

Three cases of box-allocation: from MAX-BOX-PACK to FF-SA. (best in color)

Case 2: An intersection of two different loops (Fig. 7(b), left). An intersection is a special case in this box-allocation. A square region (dashed) of $3\times 3$ is added to the intersection, which contains 4 mutually overlapping $2\times 2$-boxes (not drawn) and the four boxes do not belong to either loop. This special modeling guarantees that whatever combinations of “1” and “0” $x_1$ and $x_2$ (Fig. 7(b)) choose, we can always add one and only one $2\times 2$-box from the dashed region into the optimal non-overlapping subset. Thus, the intersection has no effect on the decision in choosing the optimal subset. A corresponding benefit-value assignment is proposed in Fig. 7(b)(right). For both loops, choosing boxes with either “1” or “0” will include one and only one cell with value of -1 at the intersection. Same as MAX-BOX-PACK, it does not matter whether “0” or “1” boxes are chosen by the two loops to choose choice “A”, since we can always add one and only one $2 \times 2$ box with choice “A” at the intersection and increase benefit by 1. The cells left in the intersecting region will choose choice “B” (0 benefit value).

Case 3: This case is the only one that affects the final optimal solution of MAX-BOX-PACK and FF-SA. As shown in Fig. 7(c), there are three loops of boxes intersecting one dashed region at their turns. This dashed region is called a “clause region”. The dashed region has an up-side-down “L” shape and it has three mutually overlapping $2\times 2$-boxes. Each “clause region” in the box allocation corresponds to a clause in the MAX-3SAT problem. The loops are constructed such that only a “0” box intersects the “clause region”. In the clause region, if there exists at least one loop that chooses boxes with “1”, then we can add one and only one more $2\times 2$-box to the final subset. If none of the three loops choose boxes with “1”, then we cannot add any of the $2\times 2$-box due to overlaps. Thus, for the MAX-BOX-PACK problem, finding a maximum non-overlapping subset is equivalent to finding the maximum number of satisfied clauses in MAX-3SAT. For FF-SA instance $I_{FF}$, a benefit-assignment is given in Fig. 7(c) (right). Similarly, any cell with negative $\gamma$ is prohibitive for choice “A” and needs to be assigned with choice “B”. As with MAX-BOX-PACK, if at least one of the three loops chooses the “1” boxes with choice of “A”, then we can add one and only one more $2\times 2$ “A” box into the FF-SA solution. Due to the spatial constraints, the rest of the cells in the clause region must choose “B”. In addition, we cannot allocate a $3\times 2$ tile in this region because it will cause a decrease in total benefit.

Thus, with the optimal solution of $I_{FF}$, all $2\times 2$ boxes with choice “A” represent exactly the optimal subset of boxes chosen in MAX-BOX-PACK instance $I_{MBP}$, which then gives the optimal solution of the MAX-3SAT instance $I_{M3SAT}$ in polynomial time.