Algorithm SiR is an open-loop and memoryless algorithm in which we constrain the vehicle to move in an angular motion, i.e., either clockwise or anti-clockwise. This can be achieved by moving the vehicle with unit speed in the direction perpendicular to its position vector (see Figure 3A). By doing so, we aim to understand the worst-case scenarios for heterogeneous defenders and gain insights into the effect of the heterogeneity that arises due to the capture range of the defenders, i.e., the capture circle and the engagement range of the turret. We say that a defender sweeps in its dominance region if it turns, from its starting location, either clockwise or anti-clockwise to a specified location and then turns back to its initial location. Algorithm SiR is formally defined in Algorithm 1 and is summarized as follows.

Algorithm SiR first partitions the environment E ( θ ) into two dominance regions and assigns a single defender to a dominance region. Let 2α denote the angle of W _Veh. Then, the vehicle takes exactly 4αz _v to sweep within its dominance region, where z _v denotes the radial location of the vehicle and will be determined shortly (see Case 1 below). Similarly, as the turret can only turn either clockwise or anti-clockwise with at most angular speed ω, the turret takes exactly 4 ( θ − α ) ω to sweep in its respective dominance region W _Tur. Observe that the environment must be partitioned such that the time taken by the defenders to complete their motion in their respective dominance region is equal. Otherwise, in the worst-case, all of the intruders will be concentrated in the dominance region of that defender that takes more time to sweep its dominance region. Mathematically, this means 4 α z v = 4 ( θ − α ) ω must hold which yields that α = θ 1 + ω z v . Observe that as ω → ∞, α → 0. This means that the turret sweeps the entire environment, in time 4 θ ω , if ω is sufficiently high. Recall that the (v, ρ) parameter space is characterized by the time taken by the defenders to complete their motion and can be improved by reducing the time taken by the turret, for high values of ω. In case of very high ω, this can be achieved by having the vehicle remain static at a specific location while the turret sweeps the remaining environment as opposed to the entire environment. This means that although angle α = θ 1 + ω z v characterizes the vehicle’s dominance region, there exists an angle α ̂ ≥ α for some values of problem parameters for which we can obtain an improved parameter regime by assigning a dominance region of angle 2 α ̂ to the vehicle. Thus, in Algorithm SiR, there are two cases based on the values of the problem parameters. First, as described above, the defenders sweep the environment in their respective dominance regions and second, the vehicle remains static at a specific location while the turret sweeps its dominance region. In what follows, we determine the location at which the vehicle must remain static for the second case, followed by formally describing the two cases.

The vehicle’s location must be such that its capture circle covers the perimeter contained in its dominance region entirely, ensuring that any intruder that arrives in that dominance region is guaranteed to be captured. To achieve this, the boundary of the dominance region assigned to the vehicle must be tangent to its capture circle (see Figure 3B) which, through geometry, yields that α ̂ = arctan r c ρ and the location for the vehicle as ( ρ cos ( α ̂ ) , θ − α ̂ ) . Therefore, the angle of the vehicle’s dominance region is defined as 2 α = 2 max { θ 1 + ω z v , α ̂ } , where α ̂ = arctan r c ρ , and angle α determines if the vehicle sweeps in its dominance region or remains stationary. We first describe the motion of the turret followed by formally describing the motion of the vehicle in the two cases.

At time instant 0, the turret is at an angle −θ. The turret turns clockwise, with angular speed ω towards angle θ − 2α. Upon reaching angle θ − 2α, the turret turns anti-clockwise towards angle −θ. Note that the turret takes exactly 4 ( θ − α ) ω time to complete its motion in a particular epoch. We now describe the motion of the vehicle which can be summarized in two cases described as follows:

Case 1 ( α = θ 1 + ω z v ) : At time instant 0, the vehicle is located at (z _v , θ), where z _v = min{ρ + r _c , 1 − r _c } and was determined in (7) and was proved to be optimal in (4). The vehicle then moves anti-clockwise with unit speed in the direction perpendicular to its position vector until it reaches the location (z _v , θ − 2α). Then, the vehicle moves clockwise, with direction perpendicular to its position vector, until it reaches location (z _v , θ). Note that the vehicle takes exactly 4αz _v time to complete its motion in a particular epoch. Since α is chosen so that 4 α z v = 4 ( θ − α ) ω , the vehicle and the turret return to their respective initial locations at the same time instant, at which the next epoch begins.

Case 2 ( α = arctan ( r c ρ ) ) : At time instant 0, the vehicle is located at ( z v = ρ cos ( α ) , θ − α ) and remains stationary at this location for the entire duration. In this case, the next epoch begins once the turret turns back to angle −θ.

The following result characterizes the parameter regime in which Algorithm SiR is 1-competitive.

Theorem 4.1: Algorithm SiR is 1-competitive for a set of problem parameters that satisfy v ≤ min r t − ρ 1 + ω z v 4 θ z v , z v + r c − ρ 1 + ω z v 4 θ z v i f θ 1 + ω z v > arctan r c / ρ r t − ρ ω 4 θ − arctan r c / ρ , o t h e r w i s e . Otherwise, Algorithm SiR is not c -competitive for any constant c .

Proof

Suppose that α = θ 1 + ω z v . At the start of any epoch k, i.e., at time instant k _s , we assume that, in the worst-case, intruders i ₁ and i ₂ are located at (z _v + r _c + ϵ₁, θ) and (r _t + ϵ₂, − θ), respectively, where ϵ₁ and ϵ₂ are arbitrary small positive numbers (see Figure 3A). To ensure that the vehicle (resp. turret) does not lose any intruder, we require that the time taken by the vehicle (resp. turret) to return to location (resp. angle) (z _v , θ) (resp. −θ) must be less than the time taken by intruder i ₁ (resp. i ₂) to reach the perimeter. Formally, r t + ϵ 2 − ρ v ≥ 4 ( θ − α ) ω and z v + r c + ϵ 1 − ρ v ≥ 4 α z v must hold. Given the first condition on v, these two conditions always hold, so any intruder that arrives in the environment is guaranteed to be captured.

If v > min { ( r t − ρ ) ( 1 + ω z v ) 4 θ z v , ( z v + r c − ρ ) ( 1 + ω z v ) 4 θ z v } , then there exists an input instance with intruders arriving only at (1, − θ) such that these intruders are located at (r _t + ϵ, − θ) at the time instant the turret turns from angle −θ. As v > min { ( r t − ρ ) ( 1 + ω z v ) 4 θ z v , ( z v + r c − ρ ) ( 1 + ω z v ) 4 θ z v } all of these intruders will be lost and thus, from Definition 1, Algorithm SiR will not be c-competitive.

Now consider that α = arctan ( r c ρ ) . As the vehicle remains stationary in its dominance region and the location of the vehicle is such that no intruder that is released in that dominance region can reach the perimeter, we only focus on the turret. Assume that, in the worst-case, intruder i ₁ is located at (r _t + ϵ, − θ) where ϵ is an arbitrary small positive numbers. To ensure that the turret does not lose any intruder, we require that the time taken by the turret to return to angle −θ must be less than the time taken by intruder i ₁ to reach the perimeter. Given the second condition on v, i.e., v ≤ ( r t − ρ ) ω 4 ( θ − arctan ( r c / ρ ) ) holds, it is ensured that intruder i ₁ will be captured. The proof for Algorithm SiR not being c-competitive when v > ( r t − ρ ) ω 4 ( θ − arctan ( r c / ρ ) ) is analogous to the previous case and has been omitted for brevity. This concludes the proof.

Although Algorithm SiR is 1-competitive, note that for r _t = ρ, the algorithm is not effective as Theorem 4.1 yields v ≤ 0. However, by allowing the vehicle to sweep the entire environment, it is still possible to capture all intruders for some small v > 0. This is addressed in a similar algorithm below.

At time instant 0, the turret is at angle θ and the vehicle is located at location (min{r _t + r _c , 1}, θ). The idea is to move the two defenders together in angular motion. Thus, the vehicle moves anti-clockwise with unit speed in the direction perpendicular to its position vector until it reaches the location (min{r _t + r _c , 1}, − θ). Similarly, the turret turns anti-clockwise, in conjunction with the vehicle, to angle −θ. Upon reaching −θ, the vehicle and the turret move clockwise until they reach angle θ. The defenders then begin the next epoch. As the two defenders move in conjunction, 2 θ ω = 2 θ min { r t + r c , 1 } ⇒ ω = 1 min { r t + r c , 1 } must hold. Thus, this algorithm is effective for ω ≥ 1 min { r t + r c , 1 } by turning the turret exactly with angular speed 1 min { r t + r c , 1 } .

The following result establishes that Algorithm SiCon is 1-competitive for specific parameter regimes.

Theorem 4.2: Algorithm SiCon is 1-competitive for a set of problem parameters which satisfy ω ≥ 1 min { r t + r c , 1 } and v ≤ min { r t + 2 r c , 1 } − ρ 4 θ min { r t + r c , 1 } . Otherwise, it is not c -competitive for any constant c .

Proof

As the proof is analogous to the proof of Theorem 4.1, we only provide an outline of this proof for brevity. In the worst-case, an intruder requires exactly min { r t + 2 r c , 1 } − ρ v time to reach the perimeter whereas, the defenders synchronously require 4θ min{r _t + r _c , 1} time to complete their motion in any epoch. Thus, as the time taken by the defenders must be at most the time taken by the intruders we obtain the competitive ratio. If the condition on v does not hold, then by constructing an input analogous to the input in the proof of Theorem 4.1, it can be shown that Algorithm SiCon is not c-competitive.

Remark 4.3: (Maneuvering Intruders). As the analysis of the fundamental limit (Theorem 3.1), Algorithm SiR, and Algorithm SiCon are independent of the nature of motion of the intruders, the results of Theorem 3.1, Algorithm SiR, and Algorithm SiCon apply directly to the case of maneuvering or evading intruders.

Recall that in Algorithm SiR, the idea was to partition the environment and assign a single defender in each dominance region. By doing so, we obtain valuable insight into the parameter regime wherein we are guaranteed to capture all intruders. However, we refrain from designing such algorithms in this work due to the following two reasons. First, such an algorithm requires that ratio of intruders captured by a defender to the total number of intruders that arrived in that corresponding defender’s dominance region is equal for both defenders. Otherwise, since the adversary has the information of the entire algorithm, it will release more intruders in the dominance region of the defender that has the lower ratio, which determines the competitive ratio of such an algorithm. Second, such algorithms are not cooperative and thus fall out of the scope of this paper. The objective of this work is to study how heterogeneous defenders can be used to improve the competitive ratio of a single defender. Thus, in the next algorithm, although we partition the environment, the defenders are not restricted to remain within their own dominance region.

Algorithm 2

Split and Capture Algorithm

1 Turret is at angle 0

The motivation for this algorithm is to utilize the vehicle’s ability to move in any direction while the turret rotates either clockwise or anti-clockwise. Since the turret can only turn either clockwise or anti-clockwise, the idea is to first partition the environment into two-halves and turn the turret towards the side which has higher number of intruders. By doing so, we hope to capture at least half of the intruders by the turret, assuming they are sufficiently slow, that arrive in every epoch. Further, while the turret moves to capture intruders on one side, the vehicle moves to the other side to capture intruders, ensuring that the defenders jointly capture more than half of the intruders that arrive in the environment in every epoch. Algorithm Split is formally defined in Algorithm 2 and is summarized as follows where we first describe the motion of the turret in every epoch followed by that of the vehicle.

The heading angle of the turret is always γ k s = 0 at the start of every epoch k. To determine whether the turret turns clockwise or anti-clockwise at time instant k _s , we first describe two sets P _left and P _right. These sets characterize a region on the left and right side of the y-axis, respectively, and are determined once at time instant 0. P r i g h t ρ , v ≔ y , β : ρ + β v ω < y ≤ min 1 , r t + 2 θ − β v ω , ∀ β ∈ 0 , θ , P l e f t ρ , v ≔ y , β : ρ − β v ω < y ≤ min 1 , r t + 2 θ + β v ω , ∀ β ∈ 0 , − θ .

Let P r i g h t k and P l e f t k denote the set of intruders contained in P _right and P _left (see Figure 4A), respectively, at the start of an epoch k and let |S| denote the cardinality of a set S of intruders. Then, at time instant k _s , the turret compares the total number of intruders in P l e f t k to the total number of intruders in P r i g h t k and turns in the direction of the set which has higher number of intruders. More formally, if | P r i g h t k | < | P l e f t k | holds at time instant k _s , then the turret turns anti-clockwise towards angle −θ. Upon turning to angle −θ, the turret turns to angle 0. Otherwise, i.e., if | P r i g h t k | ≥ | P l e f t k | holds at time instant k _s , then the turret turns clockwise towards angle θ. Upon turning to angle θ, the turret turns to angle 0. As the turret’s dominance region is determined at the start of every epoch k, we denote the turret’s dominance region as W T u r k , and consequently, the other dominance region as the vehicle’s dominance region denoted as W V e h k . We now characterize the motion for the vehicle which builds upon the SNP algorithm designed in (Bajaj et al., 2022).

Algorithm Split further divides the environment E ( θ ) into N = ⌈ θ θ s ⌉ sectors, where 2θ _s = 2 arctan(r _c /ρ) denotes the angle of each sector (see Figure 4B). The value 2 arctan(r _c /ρ) of the angle of each of the sectors is to ensure that the portion of the perimeter in a sector can be completely contained in the capture circle of the vehicle. This can be achieved by positioning the vehicle at resting points (see Figure 4B), which is a specific location in every sector and is formally defined as follows.

Definition 2 (Resting points). Let N _l denote the l th sector, for every l ∈ {1, …, N} where N ₁ corresponds to the leftmost sector. Then, a resting point ( x l , ϕ l ) ∈ E ( θ ) for sector N _l , is the location for the vehicle such that when positioned at (x _l , ϕ _l ), the portion of the perimeter inside sector N _l is contained completely within the capture circle of the vehicle. Mathematically, this is equivalent to x l , ϕ l = ρ cos θ s , l − N + 1 2 2 θ s .

Now, let D denote the distance between the two resting points that are furthest apart in the environment (see Figure 4B). Formally, D = 2 ρ cos θ s sin N − 1 θ s ,  if  N − 1 θ s < π 2 , 2 ρ cos θ s ,  otherwise. (3) Observe that when N = 1, D = 0. This means that the vehicle captures all intruders that arrive in the environment by positioning itself to the unique resting point of the single sector.

Next, Algorithm Split radially divides the environment E ( θ ) into three intervals of length Dv, corresponding to time intervals of time length D each (see Figure 4C). Specifically, the jth time interval for any j > 0 is defined as the time interval [(j − 1)D, jD]. Note that this time interval is different than the epoch of the algorithm. Let S l j denote the set of intruders that are contained in the lth, l ∈ {1, …, N}, sector and were released in the jth interval. Then, at the start of each epoch, the motion of the vehicle is based on the following two steps: First, select a sector with the maximum number of intruders. Second, determine if it is beneficial to switch over to that sector. These two steps are achieved by two simple comparisons; C1 and C2 detailed below.

Suppose that the vehicle is located at the resting point of sector N _i at the start of the jth epoch and let N ̃ denote the set of sectors in the vehicle’s dominance region. Then, to specify the first comparison C1, we associate each sector N l ∈ N ̃ with the quantity η i l ≜ | S l j + 2 | + | S l j + 3 | , if  l ≠ i , | S i j + 1 | + | S i j + 2 | + | S i j + 3 | , if  l = i .

Note that η i l is not defined for every sector in the environment. Instead, it is only defined for the sectors in W V e h j , which may contain N _i . Then, as the outcome of C1, Algorithm Split selects the sector N p * , where p * = a r g m a x p ∈ N ̃ η i p . In case there are multiple sectors with same number of intruders, then Algorithm Split breaks the tie as follows. If the tie includes the current sector N _i (which is only possible if N i ∈ N ̃ holds ³ ), then Algorithm Split selects N _i . Otherwise, Algorithm Split selects the sector contained in N ̃ with the maximum number of intruders in the interval j + 2, i.e., p * = a r g m a x p ∈ N ̂ | S p j + 2 | , where N ̂ denotes the set of sectors that have the same number of intruders. If this results in another tie, then this second tie is resolved by selecting the sector with the least index. Let the sector chosen as the outcome of C1 be N _o .

We now describe comparison C2 jointly with the motion of the vehicle in the following two points:

Note that the vehicle takes at most 2D time units in every case above. The vehicle then re-evaluates after 2D time. At time instant 0, the turret’s heading angle is 0 and the vehicle is located at (x ₁, ϕ ₁). The first epoch begins when the first intruder arrives in the environment.

We now describe two key requirements for the algorithm. The first requirement is to ensure that the defenders start their individual motion in an epoch at the same time instant. Recall that the turret requires exactly 2 θ ω time to turn from its initial heading angle to either θ or −θ, at time instant k _s , and turn back to its initial heading angle. On the other hand, the vehicle requires 2D time units to capture intruders in at least one interval. Thus, to ensure that the defenders begin their motion at the same time instant, 2 θ ω = 2 D must hold, i.e., the speed of the turret must be at least θ D . The second requirement is to ensure that the algorithm has a finite competitive ratio. This is achieved by ensuring that any intruder that was not accounted for comparison by the defenders (for instance intruders that are not in P l e f t k or P r i g h t k ) in an epoch k, are accounted in epoch k + 1. Our next result formally characterizes this requirement for the turret.

Lemma 4.4. Any intruder with radial coordinate greater than min { 1 , r t + ( 2 θ − β ) v ω } , ∀ β ∈ [ 0 , θ ] (resp. min { 1 , r t + ( 2 θ + β ) v ω } , ∀ β ∈ ( 0 , − θ ) at time instant k _s will be contained in the set P r i g h t k + 1 (resp. P l e f t k + 1 ) at time instant (k + 1) _s if v ≤ ω ( r t − ρ ) 2 θ holds.

Proof

Without loss of generality, suppose that | P l e f t k | ≤ | P r i g h t k | holds at time instant k _s . Then, the total time taken by the turret to move towards θ and turn back to angle 0 is 2 θ ω . In order for any intruder i to not be captured in epoch k, in the worst-case, the intruder i must be located at (min{1, r _t } + ϵ, θ), where ϵ is a very small positive number, by the time the turret reaches angle θ. Note that 1 + ϵ here means that the intruder is released after ϵ time at location (1, θ) after the turret’s heading angle is θ. Thus, in order to ensure that i can be captured in epoch k + 1, the condition 2 θ ω ≤ r t − ρ v must hold, where we used the fact that r _t ≤ 1 and ϵ is a very small positive number. From the definition of P r i g h t k , if the intruder i can be captured in epoch k + 1, then it follows that the intruder i was contained in the set P r i g h t k + 1 at the start of epoch (k + 1). This concludes the proof.

The proof of Lemma 4.4 is established in the worst-case scenario which is that an intruder, with angular coordinate θ (resp. −θ), is located just above the range of the turret at the time instant when the turret’s heading angle is θ (resp. −θ) in an epoch k. This is because in an epoch k, the angle θ is the only heading angle that is visited by the turret once as the turret turns to all angles β ∀ β ∈ 0 , θ (resp. 0 , − θ ) twice; once when the turret turns to angle θ or −θ and second, when turret turns back to angle 0. This results in the following corollary, the proof of which is analogous to the proof of Lemma 4.4.

Corollary 4.5: Suppose that the turret moves to capture intruders in P l e f t k (resp. P r i g h t k ) in epoch k . Then, the intruders contained in P r i g h t k (resp. P l e f t k ) with radial coordinate strictly greater than r t + v θ ω at time instant k _s will be considered for comparison at the start of epoch (k + 1) if v ≤ ω ( r t − ρ ) 2 θ .

Recall that the adversary selects the release times and the locations of the intruders in our setup. Thus, with the information of the online algorithm, the adversary can release intruders such that all the intruders have their radial coordinates at most r t + θ v ω and at angular location θ of −θ at the start of every epoch k, which is considered to be the worst-case scenario. This ensures that if the turret selects to turn towards −θ (resp. θ) at time k _s , then the turret cannot capture any intruder that was contained in P r i g h t k (resp. P l e f t k ) in epoch k + 1. As the idea is to have the vehicle capture these intruders, we require that the intruders must be sufficiently slow. This is explained in greater detail as follows.

For the vehicle, the requirement is that the intruders take at least 3D time units to reach the perimeter. This is to ensure that the vehicle can account for intruders that are very close to the perimeter at the start of an epoch. From Corollary 4.5, as the intruders with radial location greater than r t + θ v ω are counted for comparison in next epoch by the turret, we require that these intruders must also be counted by the vehicle in the next epoch. This yields that 3 D ≤ min { 1 , r t + θ v ω } − ρ v which implies that either v ≤ 1 − ρ 3 D or v ≤ r t − ρ 2 D must hold, where we used the fact that 2 θ ω = 2 D . Finally, as Lemma 4.4 requires that v ≤ r t − ρ 2 D must hold, the second requirement for Algorithm Split is that v ≤ min { 1 − ρ 3 D , r t − ρ 2 D } . We now establish the competitive ratio of Algorithm Split.

Theorem 4.6: Let θ _s = arctan(r _c /ρ) and N = ⌈ θ θ s ⌉ . Then, for any problem instance P with the turret’s angular velocity ω ≥ θ D , where D is defined in equation 3, Algorithm Split is 3 N − 1 3 ⌊ 0.5 N ⌋ + 2 -competitive if v ≤ min { 1 − ρ 3 D , r t − ρ 2 D } .

Proof

First observe that although the turret can capture intruders from one-half of the environment, the vehicle only captures at most two intervals out of all intervals that are in W V e h k (the total number of intervals in W V e h k will be determined shortly). Thus, in the worst-case, the intruders are released in the environment such that there are as many intruders possible in the vehicle’s dominance region. Since W V e h k is selected based on W T u r k , there cannot be more number of intruders in the vehicle’s dominance region as than those in the turret’s dominance region. This implies that there are equal number of intruders in each dominance region in every epoch in the worst-case. We now characterize the total number of intervals in the vehicle’s dominance region.

If N is even then, the vehicle’s dominance region contains N 2 sectors and 3 N 2 intervals due to the three intervals of length Dv each. Otherwise, the total number of intervals in the vehicle’s dominance region is 3 ⌈ N 2 ⌉ . The explanation is as follows. Observe that, for odd N, the sector in the middle is contained in the turret’s as well as the vehicle’s dominance region. As the portion of the middle sector which is contained in the vehicle’s dominance region may contain intruders and from the fact that the number of intervals must be an integer, we obtain that there are 3 ⌈ N 2 ⌉ intervals in the vehicle’s dominance region. Since the total number of intervals in the environment is 3N, this implies that the turret’s dominance region has 3 N − 3 ⌈ N 2 ⌉ = 3 ⌊ N 2 ⌋ intervals and not 3 ⌈ N 2 ⌉ intervals as we already accounted for the portion of the middle sector contained in the turret’s dominance region in the vehicle’s dominance region by using the ceil function. Intuitively, this means that there is no benefit for the adversary to release intruders in the portion of the middle sector contained in the turret’s dominance region as the turret captures all intruders in its dominance region in an epoch. Thus, the adversary can have all intruders in a single interval within the turret’s dominance region and the number of intruders that the turret capture remain the same, which is not the case in the vehicle’s dominance region. We now account for the number of intruders jointly captured by the defenders in any epoch k.

Since at the start of every epoch k, the turret selects a dominance region based on the number of intruders on either side of the turret, it follows that the turret captures at least half of the total number of intruders that arrive in epoch k. This means that the turret captures intruders in all 3 ⌊ N 2 ⌋ intervals. The number of intruders captured by the vehicle in an epoch k is determined as follows. Recall that in Algorithm Split, the vehicle’s motion is independent of the turret’s motion. The only information exchange that is required is the dominance region selected by the turret at the start of each epoch, which governs the number of sectors that the vehicle must account intruders in. Hence, this part of the analysis of accounting the number of intruders captured by the vehicle is identical to the proof of Lemma IV.5 in (Bajaj et al., 2022), so we only give an outline of the proof. From the fact that the vehicle’s dominance region can have at most 3 ⌈ N 2 ⌉ intervals and by following similar steps as in proof of Lemma IV.5 from (Bajaj et al., 2022), it follows that for every two consecutively captured intervals, the vehicle loses at most 3⌈0.5 N⌉ − 3 intervals. Further, from Lemma 4.4 and by following similar steps as in the proof of Lemma IV.6 from (Bajaj et al., 2022), it follows that every lost interval is accounted for by the captured intervals of the turret and the vehicle. Thus, we obtain that the turret and the vehicle jointly capture at least 2 + 3⌊0.5 N⌋ intervals of intruders and lose at most 3⌈0.5N⌉ − 3 intruders in every epoch of Algorithm Split. Therefore, by assuming that there exists an optimal offline algorithm that can capture all 2 + 3⌊0.5 N⌋ + 3⌈0.5N⌉ − 3 = 3N − 1 intruder intervals in every epoch establishes that Algorithm Split is 3 N − 1 3 ⌊ 0.5 N ⌋ + 2 -competitive. This concludes the proof.

Recall that the motion of the vehicle in Algorithm Split builds upon Algorithm SNP designed in (Bajaj et al., 2022), which was shown to be 3 N − 1 2 -competitive. A major drawback of Algorithm SNP was that its competitive ratio increases linearly with the number of sectors N. The following remark highlights that Algorithm Split does not suffer from this drawback and is effective under the same parameter regime as Algorithm SNP.

Algorithm 3

Partition and Capture Algorithm

1 Turret’s heading angle is θ/3.

Remark 4.7 (Heterogeneity improves competitive ratio of Algorithm Split). The competitive ratio of Algorithm Split is at most 2, achieved when N → ∞ . Further, for r _t = 1, the parameter regime that required by Algorithm Split ( v ≤ 1 − ρ 3 D ) ) is the same as that of Algorithm SNP in (Bajaj et al., 2022).

Further note that if N is odd and N ≠ 1, then the competitive ratio of Algorithm Split is higher than that for N + 1. This is because when N is odd, the adversary can exploit the fact that there are higher number of intervals that the vehicle can lose as compared to that in the turret’s dominance region. Finally, for N = 2, Algorithm Split is 1-competitive. The explanation is as follows. For N = 2, the two sectors of the environment overlap the two dominance region. Thus, in this case, the turret captures all intruders in one dominance region while the vehicle remains stationary at the resting point of the second dominance region, ensuring that all intruders that are released in the environment are captured.

Given that the turret can only move clockwise or anti-clockwise and from the requirement that the defenders must start their motions at the same time instant, the parameter regime of Algorithm Split is primarily defined by the time taken by the turret to sweep its dominance region. This means that by reducing the time taken by the turret to complete its motion, it is possible to achieve an algorithm with higher parameter regime. This is exploited in our next algorithm which is provably 1.5-competitive.

Algorithm Part, formally defined in Algorithm 3, partitions the environment into three equal dominance region, each of angle 2 θ 3 . We denote these dominance regions as W ₁, W ₂, and W ₃, where W ₁ denotes the leftmost dominance region. The idea is to move the vehicle and the turret similar to the motion of the turret in Algorithm Split and capture all intruders from two out of the three total dominance region in each epoch. The dominance region are determined as follows.

At the start of every epoch k, the turret’s heading angle, measured from the y-axis, is set to θ 3 . Similar to Algorithm Split, we describe two sets for the turret that characterize specific regions in the two dominance regions that surround the turret, i.e., W ₂ and W ₃. Intuitively, these sets corresponds to the locations in the environment that the turret can capture intruders at during its sweep motion. T r i g h t ρ , v ≔ y , β : ρ + β − θ 3 v ω ≤ y ≤ min 1 , r t + 5 θ 3 − β v ω ∀ β ∈ θ 3 , θ , T l e f t ρ , v ≔ y , β : ρ − β − θ 3 v ω ≤ y ≤ min 1 , r t + θ + β v ω ∀ β ∈ − θ 3 , θ 3 . Similarly, at the start of every epoch k, the vehicle is assumed to be located at ( z v , − θ 3 ) , where the angle is measured from the y-axis and z _v is as defined for Algorithm SiR. Next, we define two sets that characterize a specific region in W ₁ and W ₂, respectively. V r i g h t ρ , v ≔ { y , β : ρ + θ 3 + β z v v ≤ y ≤ min { 1 , z v + r c + θ − β z v v } ∀ β ∈ − θ 3 , θ 3 } , V l e f t ρ , v ≔ { y , β : ρ − β + θ 3 z v v ≤ y ≤ min 1 , z v + r c + 5 θ 3 + β v z v ∀ β ∈ − θ , − θ 3 } . Let T r i g h t k , T l e f t k , V r i g h t k , and V l e f t k denote the set of intruders contained in T _right, T _left, V _right, and V _left, respectively, at the start of an epoch k. Finally, denote I ^k as the set of intruders contained in V r i g h t k ∩ T l e f t k (see Figure 5).

We now describe the motion of the defenders. The objective is to move the defenders such that intruders from any two sets out of V l e f t k , T r i g h t k , and I ^k are captured. This requires assigning the defenders to the sets containing maximum number of intruders, which can be summarized into two cases.

Case 1: The set I ^k contains maximum number of intruders, i.e., | I k | ≥ V l e f t k and | I k | ≥ T r i g h t k hold at the start of epoch k. This means that one of the defenders must be assigned to the set I ^k . By determining which set has more intruders out of V l e f t k and T r i g h t k , Algorithm Part performs an assignment of the sets to the defenders. Mathematically, if | I k | ≥ | V l e f t k | and | I k | ≥ | T r i g h t k | , then

• If | V l e f t k | ≥ | T r i g h t k | , then the vehicle is assigned the set V l e f t k and the turret is assigned the set I ^k .

Case 2: | V l e f t k | < | I k | or | T r i g h t k | < | I k | holds at the start of epoch k. This implies that at least one set out of V l e f t k and T r i g h t k has the maximum number of intruders out of the three V l e f t k , I k , and T r i g h t k sets. Then, the sets are assigned as follows:

• If | V l e f t k | < | I k | , then the vehicle is assigned the set I ^k . Otherwise, the vehicle is assigned the set V l e f t k .

Once the sets are assigned, the vehicle turns as follows. If the set assigned to the vehicle is I ^k , then the vehicle moves clockwise with unit speed in the direction perpendicular to its position vector until it reaches location ( z v , θ 3 ) . Upon reaching the location, the vehicle moves anti-clockwise with unit speed in the direction perpendicular to its position vector until it returns to location ( z v , − θ 3 ) . Otherwise (if the vehicle is assigned the set V l e f t k ), the vehicle moves anti-clockwise with unit speed in the direction perpendicular to its position vector until it reaches location (z _v , − θ). Upon reaching that location, the vehicle moves clockwise with unit speed in the direction perpendicular to its position vector until it returns to location ( z v , − θ 3 ) .

Before we describe the motion of the turret, we determine its angular speed to ensure that the defenders start an epoch at the same time instant. As we require that the defenders take the same amount of time to return to their starting locations in an epoch, we require that 4 θ 3 ω = 4 θ 3 z v ⇒ ω = 1 z v , which means that the angular speed of the turret must be at least 1 z v .

We now describe the turret’s motion in an epoch. Similar to the motion of the vehicle, if the set assigned to the turret is I ^k , then the turret turns to angle − θ 3 and then turns back to the initial heading angle θ 3 with angular speed 1 z v . Otherwise, the turret turns to angle θ and then back to angle θ 3 with angular speed 1 z v .