Bounds and approximation results for scheduling coupled-tasks with compatibility constraints

TL;DR

This paper establishes bounds and develops an approximation algorithm for scheduling coupled-tasks with compatibility constraints, focusing on cases where the constraints form a quasi split-graph topology, under classical complexity assumptions.

Contribution

It provides new bounds and a polynomial-time approximation algorithm for a specific compatibility graph topology in coupled-tasks scheduling.

Findings

Established lower and upper bounds under complexity hypotheses

Developed an efficient approximation algorithm for quasi split-graph compatibility

Analyzed the problem's complexity in different graph topologies

Abstract

This article is devoted to propose some lower and upper bounds for the coupled-tasks scheduling problem in presence of compatibility constraints according to classical complexity hypothesis ($\mathcal{P} \neq \mathcal{NP}$, $\mathcal{ETH}$). Moreover, we develop an efficient polynomial-time approximation algorithm for the specific case for which the topology describing the compatibility constraints is a quasi split-graph.

Full text

Keywords: coupled-task, compatibility graph, complexity, approximation.

1 Introduction, motivations, model

We consider in this paper the coupled-task scheduling problem subject to compatibility constraints. The motivation of this model is related to data acquisition processes using radar sensors: a sensor emits a radio pulse (first sub-task $a_{i}$), and listen for an echo reply (second sub-task $b_{i}$). To make the notation less cluttered, the processing time of a sub-task will be denoted by $a_{i}$ instead of $p_{a_{i}}$ used in the theory of scheduling. Between these two instants (emission and reception), clearly there is an idle time $L_{i}$ due to the propagation, in both sides, of the radio pulse. A coupled-task $(a_{i},L_{i},b_{i})$, introduced by ?, is a natural way to model such data acquisition. This model has been widely studied in several works, i.e. ?. Other works proposed a generalization of this model by including compatibility constraints: scheduling a sub-task during the idle time of another requires that both tasks are compatible. The relations of compatibility are modeled by a compatibility graph $G$, linking pair of compatible tasks only. This model is detailed in ?. In previous works, we studied the complexity of scheduling coupled-tasks with compatible constraints under several parameters like the size of the sub-tasks or the class of the compatibility graph (?).

In this work, we propose original complexity and approximation results for the problem of scheduling stretched coupled-task with compatibility constraints. A stretched coupled-tasks $i$ is a coupled-task having both sub-tasks processing time and idle time equal to a triplet $(\alpha(i),\alpha(i),\alpha(i))$, where $\alpha(i)$ is the stretch factor of the task $i$ - one can apply a stretch factor $\alpha(i)$ to a reference task $(1,1,1)$ to obtain $i$ -.

The objective is to minimize the makespan $C_{max}$. The input of the problem is a collection of coupled-tasks $\mathcal{T}=\{t_{1},t_{2},\dots t_{n}$}, a stretch factor function $\alpha:\mathcal{T}\rightarrow\rm I\!N$, and a compatibility graph $G_{c}=(\mathcal{T},E)$ where edge from $E$ link pairs of compatible tasks only. When dealing with stretched coupled-tasks only, a edge $\{x,y\}\in E$ exists if $\alpha(x)=\alpha(y)$ (then $x$ and $y$ can be scheduled together without idle time as the idle time of one task is employed to schedule the sub-task of the other, thus we can schedule sequentially $a_{x},a_{y},b_{x},b_{y}$ - or $a_{y},a_{x},b_{y},b_{x}$ - in $\frac{4\alpha(x)}{3}$ time units), or if $3\alpha(x)\leq\alpha(y)$ (then $x$ can be entirely executed during the idle time of $y$ i.e. $a_{y},a_{x},b_{x},b_{y}$ and scheduling both tasks requires $3\alpha(y)$ time units). We note $\#(X)$ the number of different stretch factors in a set of tasks $X$, and we note $d_{G}(X)$ the maximum degree of any vertex $x\in X$ in a graph $G_{c}$.

We use the well-known Graham notation (?) to define the problems presented in this paper. In this work, we propose new complexity and inapproximability results when the compatibility graph is a restricted $1-stage~{}bipartite$ graph $G=(X,Y,E)$, i.e. a bipartite graph where edges are oriented from $X$ to $Y$ only. Then we show the problem is $\mathcal{NP}$-complete on a quasi-split graph $G=(G_{X},G_{Y},E)$111A quasi split graph is a connected graph $G=(G_{X},G_{Y},E)$, with $G_{X}$ a connected non-oriented graph (not complete) and $G_{Y}$ a independent set. The other arcs are oriented from $X$ to $Y$ only. even if $\#(V(G_{X}))=1$ and $\#(V(G_{Y}))=1$, but is $5/4$-approximable.

2 Complexity and approximation results

Theorem 2.1**.**

Deciding whether an instance of $1|\alpha,G_{c}=$ $1-stage-bipartite,$ $\#(X)=2,\#(Y)=1,d_{G_{c}}(X)\in\{1,2\},d_{G_{c}}(Y)\in\{3,4\}|C_{max}$ is a problem hard to approximate within $\frac{21-\rho^{\textsc{Max-3DM-2}}}{20}\leq\rho$, where $\rho^{\textsc{Max-3DM}}$ gives the upper bound for the Max-3DM. Since $\rho^{\textsc{Max-3DM-2}}\leq\frac{140}{141}$, we obtain $1+\frac{1}{2820}$.

Proof 2.2**.**

We prove first that the problem is $\mathcal{NP}$-complete via a polynomial-time reduction. Based on this reduction, we apply the gap-preserving reduction.

The proof is based on a reduction from the maximum $3$ Dimensional Matching (Max-3DM) (?): let $A$, $B$, and $C$ be three disjoint sets of equal size, with $n=|A|=|B|=|C|$, and a set $T\subseteq A\times B\times C$ of triplet, with $|T|=m$. The aim is to find a matching (set of mutually disjoint triplets) $T^{*}\subseteq T$ of maximum size. This problem is well known to be $\mathcal{NP}$-complete. The restricted version of this problem in which each element of $A\cup B\cup C$ appears exactly twice is denoted Max-3DM-2 and remains $\mathcal{NP}$-complete (?). In this restricted version, we have $m=2n$.

We transform the instance of Max-3DM-2 to an instance of $1|\alpha,G_{c}=1-stage~{}bi$ $partite,$ $\#(X)=2,\#(Y)=1,d_{G_{c}}(X)\in\{1,2\},d_{G_{c}}(Y)\in\{3,4\}|C_{max}=63n-3k(1-\epsilon)$ as follows: we define a set of tasks $X\cup Y$ and model the compatibility constraint with a graph $G_{c}=(X,Y,E)$. For each element $x_{i}\in A\cup B\cup C$, we add an item coupled-task $x_{i}$ into $X$ with $\alpha(x_{i})=1$. For each triplet $t_{i}\in T$, we add a box coupled-task $t_{i}$ to $Y$ with $\alpha(t_{i})=9$, and an item coupled-task $t^{\prime}_{i}$ with $\alpha(t^{\prime}_{i})=2+\epsilon$. For each $t_{i}\in T$ and each $x_{i}\in t_{i}$, we add the compatibility arc $(x_{i},t_{i})$ to $E$. We also add the compatibility arc $(t^{\prime}_{i},t_{i})$ to $E$. So, the set of $X$-tasks (resp. $Y$-tasks) are constituted by item coupled-task $x_{i}$ and $t^{\prime}_{i}$ (resp. box coupled-task).

Clearly we have $m$ box coupled-tasks (each with an idle time of $9$ units) of degree $4$ in $G_{c}$, $m$ item coupled-tasks with stretch factor $2+\epsilon$ of degree $1$ in $G_{c}$, and $3n$ item coupled-tasks with stretch factor $1$ of degree $2$ in $G_{c}$. Moreover $G_{c}$ is a bipartite graph. The reduction is constructed in polynomial time.

It exists a schedule of length $63n-3k(1-\epsilon)$ iff it exists a matching of size $k$ for Max-3DM-2 instance.

Hereafter, we propose some negative results concerning the existence of subexponential-time algorithms under the following complexity-theoretic hypothesis that is known as the Exponential-Time Hypothesis (see (?) for a survey on exact algorithms for $\mathcal{NP}$-hard problems) for stretched coupled-tasks, and other ones previously studied.

Recall first the Exponential-Time Hypothesis ((?), and (?)): there exists a constant $c>1$ such that there exists no algorithm for $3-$Satisfiability that uses only $O(c^{l})$ time where $l$ denotes the number of variables.

Corollary 2.3**.**

Assuming the Exponential-Time Hypothesis, there exists no algorithm with a worst-case running time that is subexponential in $n$ (the number of vertices), i.e.:

For the $1|a_{i}=b_{i}=p,L_{i}=2p,G_{c}|C_{max}$ problem in $O(2^{o(n)})$ time 2. 2.

For $1|a_{i}=a,b_{i}=b,L_{i}=a+b,G_{c}|C_{max}$ in $O(2^{o(n)})$ time 3. 3.

$1|\alpha,G_{c}=1-bipartite|C_{max}$* in $O(2^{O(n)})$-time algorithm.*

Proof 2.4**.**

For $1|a_{i}=b_{i}=p,L_{i}=2p,G_{c}|C_{max}$: In (?), the authors proved that for Partition into triangles on graphs of maximum degree four, there is no algorithm with a worst-case running time $O(2^{o(n)})$ that is subexponential in $n$.

Therefore, we transform a Partition into triangles instance with $n$ vertices and $m$ edges into an equivalent instance $G_{c}$ for bounded degree at most four. Since the transformation is linear (see (?)) the result holds. 2. 2.

For the problem $1|a_{i}=a,b_{i}=b,L_{i}=a+b,G_{c}|C_{max}$: In (?) the authors proved that for Hamiltonian path there is no $O(2^{o(n)})$-time algorithm. As the same way as previously the transformation is linear (see (?)). 3. 3.

$1|\alpha,G_{c}=1-bipartite|C_{max}$: In (?), the authors proved that for Max 3DM, there is no $O(2^{O(n)})$-time algorithm, therefore this result is transposed to the scheduling problem using the first part of the proof of Theorem 2.1.

Theorem 2.5**.**

Scheduling stretched coupled task in presence of a quasi split graph is a $\mathcal{NP}-$complete problem even if $\#(V(G_{X}))=1$ and $\#(V(G_{Y}))=1$

Proof 2.6**.**

The proof is based on a reduction from a variant of the well-know $\mathcal{NP}$-complete Partition into triangles. This problem consists to ask if the vertices of a graph $G=(V,E)$, with $|V|=3q,q\in\rm I\!N$, can be partitioned into $q$ disjoints sets $T_{1},T_{2},\ldots,T_{q}$, each containing exactly three vertices, such that for each $T_{i}=\{u_{i},v_{i},w_{i}\},1\leq i\leq q$, all three of the edges $\{u_{i},v_{i}\},\{u_{i},w_{i}\},\{w_{i},v_{i}\}$ belong to $E$.

The problem Partition into triangles remains $\mathcal{NP}$-complete even if the graph $G$ can be partitioned into three sets with the same size, $A,~{}B$ et $C$ such that each set is an independent set (?). The polynomial-time transformation is based on this variant. Let $G=(A\cup B\cup C,E)$ be an instance of the variant of Partition into Triangles. We consider the split-graph $G^{\prime}=(A\cup B,C,E^{\prime})$ obtained as follows:

$\forall v\in A$* (resp. $B$), we create a vertex $A_{v}$ (resp. $B_{v}$) with processing time $(1,1,1)$. Moreover, $\forall v\in C$ we create a task $C_{v}$ with processing time $(4,4,4)$. The edges between $A$ and $B$ remained the same as the $G^{\prime}$ whereas the edge between $A\cup B$ and $C$ are oriented. Finally in order to have a connected graph, we add two news vertices (resp. one) $z_{0}$ and $z_{1}$ (resp. $z_{2}$ with processing time equal to $(1,1,1)$ (resp. $(4,4,4)$). We add edges between $z_{0}$ to $A_{v}$ (resp. $z_{1}$ to $B_{v}$). Lastly, we add the three edges $(z_{0},z_{2})$, $(z_{1},z_{2})$ and $(z_{0},z_{1})$.*

Notice that the graph $A_{v}\cup B_{v}$ form a bipartite graph. The problem is clearly in $\mathcal{NP}$. It exists a positive solution for the variant of Partition into triangles iff a valid schedule of length $12\times(|C|+1)$ exists. It is sufficient to execute the two tasks $A_{v}$ and $B_{v^{\prime}}$ in four units of time into a task $C_{u}$.

Theorem 2.7**.**

The problem is $5/4$-approximable on quasi split-graph where $\#(V(G_{Y}))=1$.

Proof 2.8**.**

W.l.o.g., we suppose that the processing time of $X$-tasks (resp. $Y$-tasks) is $(1,1,1)$ (resp. $\alpha(y_{i})$). Indeed, if $\alpha(x)>1$, we put $\alpha(y_{i})=\lfloor\frac{\alpha(y_{i})}{\alpha(x)}\rfloor$ and $\alpha(x)=1$ .

Algorithm: we transform the problem into an oriented maximum flow-problem between $G_{X}$ and $G_{Y}$ with two sources $s$ and $t$, with $\omega(s,x)=\omega(x,y)=1$ and $\omega(y,t)=\lfloor\frac{\alpha(y_{i})}{3\alpha(x)}\rfloor,\forall y_{i}\in Y,\forall x\in XG_{Y}$ where $\omega(i,j)$ is the capacity of an arc $(i,j)$ . After the computation of a maximum flow $F$ of value $f$, for the uncovered remaining $X$-tasks a maximum $M$-matching ($|M|=m$) is applied. The schedule consists in processing first, the $Y$-tasks with $X$-tasks inside. The $M$-tasks are executed after. Lastly, we schedule $s$ isolated-tasks. The length of schedule given by the algorithm is $C_{max}\leq\sum_{y_{i}\in Y}3\alpha(y_{i})+4m+3s$ with $2m+s+f=n=|X|$ and $\sum_{y_{i}\in Y}3\alpha(y_{i})\geq 9f$. In similar way, the optimal length is $C^{*}_{max}\geq\sum_{y_{i}\in Y}\alpha(y_{i})+4m^{*}+3s^{*}$. We suppose that in $Y$-tasks where are $p^{*}$-edges processed and $r^{*}$ isolated-tasks, then we obtain $2(p^{*}+m^{*})+r^{*}+s^{*}=n$, $p^{*}+r^{*}\leq f$, and $\sum_{y_{i}\in Y}\alpha(y_{i})\geq 12p^{*}+9r^{*}$. In the worst-case, the $p^{*}$-edges are split into two tasks (so $p^{*}$ news tasks are added to $s^{*}$), and also the matched-edges are split (for each $m^{*}$ edges one task is executed into the $Y$-task, instead of one of $r^{*}$-tasks). Therefore, $2m^{*}$ tasks are added to the $s$-value. In the worst case, we have $m^{*}=r^{*}$, $s=s^{*}+p^{*}+2r^{*}$ and $m=0$. In such case, $C_{max}\leq 12p^{*}+9r^{*}+3s^{*}+3p^{*}+6r^{*}$ and $C^{*}_{max}=12p^{*}+9r^{*}+4r^{*}+3s^{*}$. Thus $\rho\leq\frac{15p^{*}+15r^{*}+3s^{*}}{12p^{*}+13r^{*}+3s^{*}}\leq\max(5/4,15/13,1)=5/4$.

Tightness: it exists an example for the $C^{*}_{max}=36$, and for the heuristic $C_{max}=45$. Consider the graph: three triangles $(x_{1},x_{2},y_{1})$, $(x_{3},x_{4},y_{2})$, and $(x_{5},x_{6},y_{1})$. We add the edges $(x_{2},y_{3}$, $(x_{3},y_{1})$ and $(x_{5},y_{2})$. The optimal solution consists in executing the $X$-tasks into the $Y$-tasks; whereas the heuristic leads the solution in which three $X$-tasks are processed after the $Y$-tasks.