On Length-Sensitive Fréchet Similarity

Abstract

Taking length into consideration while comparing 1D shapes is a challenging task. In particular, matching equal-length portions of such shapes regardless of their combinatorial features, and only based on proximity, is often required in biomedical and geospatial applications. In this work, we define the length-sensitive partial Fréchet similarity (LSFS) between curves (or graphs), which maximizes the length of matched portions that are close to each other and of equal length. We present an exact polynomial-time algorithm to compute LSFS between curves under $L_1$ and $L_\infty$. For geometric graphs, we show that the decision problem is NP-hard even if one of the graphs consists of one edge.

Supported by the National Science Foundation grant CCF 2046730.

Supported by the National Science Foundation grant CCF 2107434.

Appendices

A LSFS is Length-Preserving on Connected Components

By definition, $C_h^{\varepsilon }$ only requires the restriction of h to small balls to be length-preserving, we show that indeed all connected components are length-preserving.

Lemma 4 (Path-Connected Components Are Length-Preserving)

Each restriction of h to the preimage of a path-connected component of $C_h^{\varepsilon }$ is a length-preserving map.

Proof

Let $\widetilde{C}$ be a path-connected component of $C_h^{\varepsilon }$. Let $x,y \in \widetilde{C}$. Since $\widetilde{C}$ is path-connected, let $\gamma :[0,1] \rightarrow \widetilde{C}$ be a path that starts at x and ends at y. Then, by the definition of $C_h^{\epsilon }$, for each $t \in [0,1]$, there exists a $\delta _t>0$ such that h restricted to $B_t:={\mathbb {B}_{p}(h^{-1}(\gamma (t)),\delta _t)}$ is length-preserving. Let $\mathcal {U} := \{ \gamma \left( B_t \right) \}_{t \in [0,1]}$. Since ${{\,\textrm{Im}\,}}(\gamma )$ is a compact subspace of $\mathbb {R}^2$ and since $\mathcal {U}$ covers ${{\,\textrm{Im}\,}}(\gamma )$, there exists a finite sub-cover $\widehat{\mathcal {U}}$ of $\mathcal {U}$. Then, there exists a decomposition of $\gamma$ into sub-paths such that each sub-path lies entirely in at least one open set in $\widehat{\mathcal {U}}$. Let $\{ \gamma _i \}_{i=1}^n$ be one such decomposition. Then, we know that ${{\,\textrm{len}\,}}(\gamma )=\sum _i {{\,\textrm{len}\,}}(p_i)$.

Since each ${{\,\textrm{Im}\,}}p_i$ is contained in some $U \in \widehat{\mathcal {U}}$ and since h restricted to U is length-preserving, we know that ${{\,\textrm{len}\,}}(\gamma _i) = {{\,\textrm{len}\,}}(h(\gamma _i))$. Taking the sum over all sub-paths, we find ${{\,\textrm{len}\,}}(\gamma )={{\,\textrm{len}\,}}(h(\gamma ))$. Thus, we have shown that h restricted to $\widetilde{C}$ is length-preserving.

Noting that a curve $f :[0,L_f] \rightarrow \mathbb {R}^d$ is simply a directed graph in $\mathbb {R}^d$, we obtain:

Corollary 1

Each restriction of h to the preimage of a path-connected component of $H^{\varepsilon }$ is a length-preserving map.

B Proof of Lemmas 2 and 3

In this section we discuss the proofs for Lemmas 2 and 3.

1.1 B.1 Proof of Lemma 2

Proof

First we show that for any two points $(x_1,y_1)$ and $(x_2,y_2)$ on the top or right boundary, there exist optimal paths to $(x_1,y_1)$ and $(x_2,y_2)$ that might overlap but do not cross each other. Assuming that the optimal paths to $(x_1,y_1)$ and $(x_2,y_2)$ cross at a point $a_0$, there is another optimal path to $(x_1,y_1)$ that goes through $a_0$ and overlaps with the optimal path to $(x_2,y_2)$ before reaching $a_0$ (as shown in Fig. 11b).

Fig. 11.

A refined cell $\mathcal {C}$ in the free-space diagram $\mathcal {D}_{\varepsilon }(f,g)$. In each sub-figure, we see two optimal paths to $(x_1,y_1) \in T$ and to $(x_2,y_2) \in R$. (a) The two paths intersect transversally, at $a_0$. (b) The two paths first overlap, then separate at $a_0$.

Therefore, the optimal path to any point on the top or right boundary of a cell cuts the cell into two parts such that for all points on the top or right boundary in one part there is an optimal path going through them within that part. Considering the example in Fig. 11b with the purple path being an optimal path to $(x_2,y_2)$, without loss of generality, we can say any point (x, y) such that $y > y_{2}$ has an optimal path that comes from the left boundary (more specifically, above the purple path) which means all points on the top boundary have an optimal path coming from the left boundary, hence (4) applies. Similarly, consider $(x_1,y_1)$ and the orange path in Fig. 11a to be optimal. In this case, all the points (x, y) such that $x > x_{1}$ have an optimal path that comes from the bottom boundary (the right side of the orange path) which results in (5).

1.2 B.2 Proof of Lemma 3

Proof

Let $\mathcal {C}$ be a refined cell in $\mathcal {D}_{\varepsilon }(f,g)$. Let $(x_1,y_B+y_T), (x_2,y_B+y_T)$ be two points on T of $\mathcal {C}$. We want to show: $\mathcal {S}_T(x_2)-\mathcal {S}_T(x_1)=\mathcal {S}(x_2,y_B+y_T)-\mathcal {S}(x_1,y_B+y_T) \le x_2 - x_1$. Consider the vertical line $\ell =\{(x,y)\in D\;|\; x=x_1\}$. The optimal monotone path from (0, 0) to $(x_2,y_B+y_T)$ has to cross $\ell$ at a point $(x_1,y_0)$. Note that $(x_1,y_0)$ does not have to be inside $\mathcal {C}$, see Fig. 12. We know $\mathcal {S}(x_1,y_B+y_T) \ge \mathcal {S}(x_1,y_0)$ and from Observation 3 follows that $\mathcal {S}(x_2,y_B+y_T)-\mathcal {S}(x_1,y_0) \le \min (y_B+y_T-y_0,x_2-x_1)$. Thus we can conclude: $\mathcal {S}_T(x_2)-\mathcal {S}_T(x_1)=\mathcal {S}(x_2,y_B+y_T)-\mathcal {S}(x_1,y_B+y_T) \le \min (y_B+y_T-y_0,x_2-x_1) \le x_2 - x_1$. Similarly we can show the equation above for other boundaries of $\mathcal {C}$ as well.

Fig. 12.

A cell $\mathcal {C}$ in $\mathcal {D}_{\varepsilon }(f,g)$. $(x_1,y_0)$ is on the optimal path to $(x_2,y_T)$.

C Score Function Propagation Within a Cell, Continued

We discuss the remainder of the cases and score functions. Due to symmetric nature of Case (b),Case (c), Case (d), and Case (e) with regard to $\mathcal {S}_{L \rightarrow T}$ and $\mathcal {S}_{B \rightarrow R}$ (likewise, for Case (f) and Case (g)), we do not explain $\mathcal {S}_{L \rightarrow T}$ for these cases to avoid repetition.

Case (b): Consider Fig. 13 and the cell $\mathcal {C}$ that is divided into two propagatable polygons by a line with a positive slope. The maximum value for $\mathcal {L}((0,l),(x_R,r))$ in this case is $x_R-j$ if $x_R<y_T$, and $y_T$ if $x_R>y_T$. Such value can be achieved by drawing a slope-one line from (j, 0) and if the divider’s slope is less than one (see Fig. 13, right), by drawing a slope-one line from $(x_R,y_T)$. For $\mathcal {S}_{L \rightarrow R}$, for any point $(x_R,r_1)$ on R, below the maximum line, the optimal path goes from (0, 0) to $(x_R-r_1,0)$ horizontally and then to $(x_R,r_1)$ via a slope-one segment (red dashed line).

Fig. 13.

Case (b) when $\frac{y_T}{i-j}>1$ on the left, and when $\frac{y_T}{i-j}<1$ on the right. The blue line shows the maximum slope-one length possible. Maximum slope-one segments to $r_1$ and $r_2$ are shown in magenta and orange, respectively. (Color figure online)

For all points $(x_R,r_2)$ above the maximum, the start point of the optimal path depends on $\mathcal {S}_L(l)$. In other words, taking the longest slope-one path to $(x_R,r_2)$ in $\mathcal {C}$ does not necessary yield the maximum score for $(x_R,r_2)$ in this case. This is due to the fact that $\mathcal {L}((0,l),(x_R,r))$ over l (and r) has negative slope. In Theorem 4, we demonstrate how this decision is made and why the number of start-point candidates is at most the number of breakpoints on $\mathcal {S}_L(l)$.

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le x_R-j \\ \max \limits _{l \in [0,y_T(\frac{x_R-r-j}{i-j-y_T})]} \mathcal {S}_L(l)+\mathcal {L}((\frac{i-j}{y_T}(l+j),l),(x_R,r)) &{} \text{ for } &{} r \ge x_R-j\end{array}\right. \end{aligned} \end{aligned}$$

Propagation from B to R can be done similar to Case (a), yielding:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B(x_R-r) + r &{} \text{ for } &{} r \le x_R-j \\ \mathcal {S}_B(j) + x_R-j &{} \text{ for } &{} r \ge x_R-j\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow T}(t)=\left\{ \begin{array}{lcl} \mathcal {S}_B(t) &{} \text{ for } &{} t \le j \\ \mathcal {S}_B(j)+(t-j) &{} \text{ for } &{} t \ge j \end{array}\right. \end{aligned}\end{aligned}$$

Theorem 4

The propagation within Case (b) adds O(1) complexity.

Proof

Assuming $\mathcal {S}_{L}$ has n breakpoints, we want to determine the number of breakpoints on $\mathcal {S}_{L \rightarrow R}$, which consists of two pieces. The first piece, $\mathcal {S}_L(0) + r$ is a linear function. The second piece, is the upper-envelope of the summation of a non-decreasing piecewise linear and a decreasing linear function. See Fig. 14 for an example. To compute $\mathcal {S}_{L \rightarrow R}$, one can find the summation on the breakpoints of $\mathcal {S}_{L}$ and connect them, which results in at most n breakpoints. Taking the upper-envelope makes the resulting function non-decreasing and can only reduce the number of breakpoints. Therefore, there are at most $n+1$ breakpoints on $\mathcal {S}_{L \rightarrow R}$, and the complexity of $\mathcal {S}_{L \rightarrow R}$ is equal to the complexity of $\mathcal {S}_{L}$. The complexity for $\mathcal {S}_{B \rightarrow R}$, $\mathcal {S}_{B \rightarrow T}$, and $\mathcal {S}_{L \rightarrow T}$ can be shown similar to Theorem 1.

Fig. 14.

The score function for $S_{L \rightarrow R} (r)$ when $r \ge x_R -j$.

Case (c): Case (c) represents a cell divided by a positive-slope line with free space on the left, therefore, the maximum length slope-one segment goes through $(i,y_T)$, if the divider’s slope is greater than one ($\frac{y_T}{i-j}>1$). Otherwise the maximum segment intersects B at (j, 0).

The Divider’s Slope is Greater than One: Consider Fig. 15, we first explore the paths from L to R. We draw a slope-one segment starting at (0, 0) to find the score function’s breakpoint. For any point below $(x_R,\frac{j}{\frac{y_T}{i-j}-1})$ the optimal path starts from (0, 0). For all points $(x_R,r)$ above the breakpoint, the optimal path starts from $(0,(1-\frac{i-j}{y_T})r-j)$.

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le \frac{j}{\frac{y_T}{i-j}-1} \\ \mathcal {S}_L((1-\frac{i-j}{y_T})r-j) + \frac{i-j}{y_T}r+j &{} \text{ for } &{} r \ge \frac{j}{\frac{y_T}{i-j}-1} \end{array}\right. \end{aligned} \end{aligned}$$

Similarly for paths from B to R we have:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B((\frac{i-j}{y_T}-1)r+j) + r &{} \text{ for } &{} r \le \frac{j}{\frac{y_T}{i-j}-1} \\ \mathcal {S}_B(0) + \frac{i-j}{y_T}r+j &{} \text{ for } &{} r \ge \frac{j}{\frac{y_T}{i-j}-1} \end{array}\right. \end{aligned} \end{aligned}$$

Fig. 15.

Case (c), when $\frac{y_T}{i-j}>1$ on the left, and when $\frac{y_T}{i-j}<1$ on the right. The blue line shows the maximum slope-one length possible. Maximal slope-one segments to $r_1$ and $r_2$ are shown in magenta and orange, respectively. (Color figure online)

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow T}(t)=\left\{ \begin{array}{lcl} \mathcal {S}_B(0)+t &{} \text{ for } &{} t \le i \\ \max (\mathcal {S}_B(0)+t,\mathcal {S}_B(t)) &{} \text{ for } &{} t \ge i \end{array}\right. \end{aligned}\end{aligned}$$

The Divider’s Slope is Less than One: Consider the right picture in Fig. 15. Since the maximum of $\mathcal {L}$ can be achieved from (j, 0), we do not have a breakpoint in the score functions:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \mathcal {S}_L(0) + r\quad \text{ and }\quad \mathcal {S}_{B \rightarrow R}(r) = \mathcal {S}_B(j) + r \end{aligned} \end{aligned}$$

However for $\mathcal {S}_{B \rightarrow T}(t)$ we have:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow T}(t)=\left\{ \begin{array}{lcl} \mathcal {S}_B(0)+t &{} \text{ for } &{} t \le y_T \\ \mathcal {S}_B(t-y_T)+y_T &{} \text{ for } &{} y_T \le t \le j+y_T \\ \max (\mathcal {S}_B(t-y_T)+y_T,\mathcal {S}_B(t)) &{} \text{ for } &{} t \ge j+y_T \end{array}\right. \end{aligned}\end{aligned}$$

Theorem 5

The propagation within Case (c) adds O(1) complexity.

Proof

The O(1) complexity follows in a similar way to Theorem 1. Note that for $\mathcal {S}_{B \rightarrow T}$ the max function adds at most one breakpoint (Fig. 16).

Case ( d ):

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le \frac{j}{\frac{y_T}{i-j}-1} \\ \max \limits _{l \in [0,(1-\frac{i-j}{y_T})r-j]} \mathcal {S}_L(l)+\mathcal {L}((0,l),(\frac{i-j}{y_T}r+j,r)) &{} \text{ for } &{} r \ge \frac{j}{\frac{y_T}{i-j}-1}\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B((\frac{i-j}{y_T}-1)r+j) + r &{} \text{ for } &{} r \le \frac{j}{\frac{y_T}{i-j}-1} \\ \mathcal {S}_B(0) + \frac{j}{\frac{y_T}{i-j}-1} &{} \text{ for } &{} r \ge \frac{j}{\frac{y_T}{i-j}-1}\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow T}(t) = \left\{ \begin{array}{lcl} \mathcal {S}_B(0) + t &{} \text{ for } &{} t \le i \\ \max \limits _{b \in [0,t-(\frac{y_T}{i-j}(t-j))]} \mathcal {S}_B(b)+\mathcal {B}((b,0),(t,\frac{y_T}{i-j}(t-j))) &{} \text{ for } &{} t \ge i\end{array}\right. \end{aligned} \end{aligned}$$

Fig. 16.

Case (d), when $\frac{y_T}{i-j}<-1$ on the left, and when $\frac{y_T}{i-j}>-1$ on the right. The blue line shows the maximum slope-one length possible. Maximal slope-one segments to $r_1$ and $r_2$ are shown in magenta and orange, respectively. (Color figure online)

Theorem 6

The propagation within Case (d) adds O(1) complexity.

Proof

The O(1) complexity can be discussed in a similar way to Theorem 4 (Fig. 17)

Case ( e ):

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le x_R \\ \mathcal {S}_L(y_T(\frac{x_R-r-j}{i-j-y_T})) + r-y_T(\frac{x_R-r-j}{i-j-y_T}) &{} \text{ for } &{} r \ge x_R\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B(x_R-r) + r &{} \text{ for } &{} r \le x_R-j \\ \mathcal {S}_B(\frac{x_R-r-\frac{jy_T}{i-j}}{1-\frac{y_T}{i-j}}) + x_R-\frac{x_R-r-\frac{jy_T}{i-j}}{1-\frac{y_T}{i-j}} &{} \text{ for } &{} r \ge x_R-j\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow T}(t) = \left\{ \begin{array}{lcl} \mathcal {S}_B(t) &{} \text{ for } &{} r \le t \le i \\ \mathcal {S}_B(\frac{y_T}{i-j}+t-j) + t-\frac{y_T}{i-j}+t-j &{} \text{ for } &{} t \ge i \end{array}\right. \end{aligned} \end{aligned}$$

Fig. 17.

Case (e), when $\frac{y_T}{i-j}<-1$ on the left, and when $\frac{y_T}{i-j}>-1$ on the right. The blue line shows the maximum slope-one length possible. Maximal slope-one segments to $r_1$ and $r_2$ are shown in magenta and orange, respectively. (Color figure online)

Theorem 7

The propagation within Case (e) adds O(1) complexity.

Proof

The O(1) complexity can be discussed in a similar way to Theorem 1

Case (f): As mentioned in Case (c), when a divider has a positive slope and is located on the right side of a free space, extra care is required. Suppose a divider’s slope is less than one, a slope-one segment can be drawn above it, starting at (k, 0). This segment can be a part of the optimal path from B to R. Therefore, we have four possibilities for this case:

Both Dividers’ Slopes are Greater Than One: When both slopes are greater than one, the optimal path between two points does not necessarily have at most three segments. Consider Fig. 18 (right) and point $(x_R,r'_2)$. The longest slope-one path to $(x_R,r'_2)$ is shown in orange, and has multiple slope-one segments. By Lemma 3, this path is always maximal when going through B. The same cannot be said for paths going through L in this case, since there are vertical segments in such a path. Therefore, for a point $(x_R,r)$ above $(x_R,\frac{j+k}{1-\frac{w-k}{y_T}})$, $\mathcal {S}_{L \rightarrow R}(r)$ is the maximum over l of $\mathcal {S}_L(l)$ and the length of slope-one segments in the feasible space between l and r.

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le \frac{j+k}{1-\frac{w-k}{y_T}} \\ \max \limits _{l \in [0,\frac{(1-\frac{w-k}{y_T})r_2-k+j}{1-\frac{i-j}{y_T}}]} \mathcal {S}_L(l)+\mathcal {L}((\frac{(i-j)}{y_T}l+(i-j)j,l),(\frac{w-k}{y_T}r+k,r)) &{} \text{ for } &{} r \ge \frac{j+k}{1-\frac{w-k}{y_T}}\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B((\frac{i-j}{y_T}-1)r+j) + r &{} \text{ for } &{} r \le \frac{j+k}{1-\frac{w-k}{y_T}} \\ \mathcal {S}_B(j)+\frac{w-k}{y_T}r+k-j &{} \text{ for } &{} r \ge \frac{j+k}{1-\frac{w-k}{y_T}}\end{array}\right. \end{aligned} \end{aligned}$$

Fig. 18.

Case (f) when both dividers have slope greater than one. The blue line shows the maximum slope-one length possible. Maximal slope-one segments to $r_1$, $r_2$ and $r'_2$ are shown in magenta and orange. (Color figure online)

Theorem 8

The propagation within Case (f) adds O(1) complexity.

The proof is omitted, as it is similar to Theorem 4 and Theorem 5.

Left Divider’s Slope is Greater than One: Since a divider can only have slope less than one if $x_R>y_T$, the maximum length slope-one segment that can obtained in this case is $y_T$. This amount can be achieved by simply drawing a slope-one line from (K, 0) (see Fig. 19) because the right divider’s slope is less than one. We have:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \mathcal {S}_L(0) + r \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \mathcal {S}_B(k) + r \end{aligned} \end{aligned}$$

Right divider’s slope is greater than one: Similar to the previous sub-case, the maximum length slope-one segment is achievable. However, since the right divider’s slope is greater than one, the maximum can be gained by drawing a slope-one line from $(w,y_T)$ (Fig. 19).

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \mathcal {S}_L(0) + r \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \mathcal {S}_B((\frac{w-k}{y_T}-1)r+k) + r \end{aligned} \end{aligned}$$

Fig. 19.

Case (f) when one divider’s slope is less than one and the other’s is greater than one.

Both Dividers’ Slopes are Less Than One:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(x_R,r) = \mathcal {S}_L(0) + r \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned}&\mathcal {S}_{B \rightarrow R}(r)=\\&\left\{ \begin{array}{lcl} \mathcal {S}_B(k) + r &{} \text{ for } &{} r \le \frac{(i-j)(j-k)}{(i-j)-y_T} \\ \max \limits _{b \in [k,\frac{(w-k)((1-\frac{i-j}{y_T})r+ky_T-j(i-j))}{y_T-w+k}]} \mathcal {S}_B(b)+\mathcal {L}((b,\frac{y_T}{w-k}b-ky_T),(\frac{i-j}{y_T}r+j(i-j),r)) &{} \text{ for } &{} r \ge \frac{(i-j)(j-k)}{(i-j)-y_T}\end{array}\right. \end{aligned} \end{aligned}$$

Case (g): Case (g) consists of two dividers with negative slope. Naturally, there are three possibilities for this case:

Two Parallel Dividers ( II ):

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le \frac{j+k}{1-\frac{w-k}{y_T}} \\ \mathcal {S}_L(\frac{(1-\frac{w-k}{y_T})r-k+j}{1-\frac{i-j}{y_T}}) + \frac{(\frac{w-k-i+j}{y_T})r+k-j}{1-\frac{i-j}{y_T}} &{} \text{ for } &{} r \ge \frac{j+k}{1-\frac{w-k}{y_T}}\end{array}\right. \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B((\frac{w-k}{y_T})r+k) + r &{} \text{ for } &{} r \le \frac{j+k}{1-\frac{w-k}{y_T}} \\ \mathcal {S}_B(j) + \frac{j+k}{1-\frac{w-k}{y_T}} &{} \text{ for } &{} r \ge \frac{j+k}{1-\frac{w-k}{y_T}}\end{array}\right. \end{aligned} \end{aligned}$$

Theorem 9

The propagation within Case (g) adds O(1) complexity.

Proof

The O(1) complexity for sub-case II follows in a way similar to Theorem 1. Likewise, for V and $\varLambda$ we can use the same approach as in Theorem 4.

V-shaped and $\varLambda$-shaped cases share some functions and properties with the parallel case II. We discuss their corresponding functions below:

The Left Divider’s Slope is Greater than the Right Divider’s (V): The score function on R coming from L is identical to the parallel case (II), since $\mathcal {L}((0,l),(x_R,r))$ is non-decreasing over l and r (see Fig. 20). However, $\mathcal {L}((0,b),(x_R,r))$ is decreasing over b, so for $\mathcal {S}_{B \rightarrow R}(r)$ we have:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{B \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_B(j) + r &{} \text{ for } &{} r \le \frac{j+k}{1-\frac{w-k}{y_T}} \\ \max \limits _{B \in [0,(\frac{w-k}{y_T}-1)r+k]} \mathcal {S}_B(b)+\mathcal {L}((b,\frac{y_T}{i-j}b-jy_T),(\frac{w-k}{y_T}r+k,r)) &{} \text{ for } &{} r \ge \frac{j+k}{1-\frac{w-k}{y_T}}\end{array}\right. \end{aligned} \end{aligned}$$

The Right Divider’s Slope is Greater than the Left Divider’s ($\varLambda$): The score function on R coming from B is identical to the parallel case (II) because $\mathcal {L}((0,b),(x_R,r))$ is non-decreasing over b.⁴ However, that is not the case with $\mathcal {L}((0,l),(x_R,r))$:

$$\begin{aligned} \begin{aligned} \mathcal {S}_{L \rightarrow R}(r) = \left\{ \begin{array}{lcl} \mathcal {S}_L(0) + r &{} \text{ for } &{} r \le \frac{j+k}{1-\frac{w-k}{y_T}} \\ \max \limits _{l \in [0,\frac{(1-\frac{w-k}{y_T})r-k+j}{1-\frac{i-j}{y_T}}]} \mathcal {S}_L(l)+\mathcal {L}((\frac{(i-j)}{y_T}l+(i-j)j,l),(\frac{w-k}{y_T}r+k,r)) &{} \text{ for } &{} r \ge \frac{j+k}{1-\frac{w-k}{y_T}}\end{array}\right. \end{aligned} \end{aligned}$$

Fig. 20.

Case (g) with V on the left and $\varLambda$ on the right side. The blue line shows the maximum slope-one length possible. Maximal slope-one segments to $r_1$ and $r_2$ are shown in magenta and orange. (Color figure online)

Case (h). Here, the cell is filled with infeasible space; therefore, for any pair of points $p_1$ and $p_2$ on the boundaries, $\mathcal {L}(p_1,p_2)=0$. Hence, going from L to R (or B to T) is simply: $\mathcal {S}_{L \rightarrow R}(r) = \mathcal {S}_L(r)$ (and $\mathcal {S}_{B \rightarrow T}(t) = \mathcal {S}_B(t)$). Since the score functions on the boundaries are non-decreasing (by Observation 3), we have: $\mathcal {S}_{B \rightarrow R}(r) = \mathcal {S}_B(x_R)$ (and $\mathcal {S}_{L \rightarrow T}(t) = \mathcal {S}_L(y_T)$).

Theorem 10

The propagation within Case (h) adds no complexity.

Proof

Since $\mathcal {S}_{L \rightarrow R}(r)=\mathcal {S}_L(l)$ directly, there is no additional breaking point on $\mathcal {S}_{L \rightarrow R}$. Furthermore, $\mathcal {S}_{B \rightarrow R}$ is a constant function, which means it does not transfer any complexity from $\mathcal {S}_B$. We have similar explanations for $\mathcal {S}_{B \rightarrow T}$ and $\mathcal {S}_{L \rightarrow T}$.