Precedence Scheduling with Unit Execution Time is Equivalent to Parametrized Biclique

Abstract

We consider the following scheduling problem. Given m machines and n jobs with unit execution times and a precedence relation between the jobs, the problem is to assign each job to one of the machines. The objective is to find a schedule that minimizes the makespan (i.e. the length of the schedule).

We reduce $3\text {-CNF-SAT}$ to this problem and obtain a new lower bound for the running time of $2^{o(\sqrt{n\log n})}$ assuming the Expontential Time Hypothesis ($\text {ETH}$). This improves the previous lower bound of $2^{o(\sqrt{n})}$ also due to the $\text {ETH}$ and a reduction by Ullman [13] or, alternatively, a reduction from the k-Clique problem by Lenstra and Rinnooy Kan [10].

For the corresponding decision problem of whether there is a schedule with target makespan $\mathbf{T }=3$ or not, we further show the equivalence to a classical graph problem, the parametrized Biclique problem. The equivalence also holds for the same scheduling problem with the additional restriction that no job has both a predecessor and a successor. By this we show that an improved lower bound for the running time for the Biclique problem will lead to an improved lower bound for the running time for our scheduling problem and vice versa. Moreover a transfered lower bound for the running time from the Biclique problem would also hold for the running time of approximation algorithms with ratio better than $\frac{4}{3}\text {OPT}$. That is, if for example there was no algorithm solving Biclique in $2^{o(n)}$ and U was the set of vertices in the Biclique problem, then there would be no approximation algorithm finding a solution for the introduced scheduling problem with $\varTheta (|U|)$ jobs, that finds a solution with a target makespan smaller than $\frac{4}{3}$ times the optimal makespan in time $2^{o(n)}$.

Appendices

A Example for the Modified Ullman Reduction

Example 1

Figure 5 visualizes the reduction using an example where $\varphi =(x_1\vee x_2\vee \lnot x_3)\wedge (x_1\vee \lnot x_2 \vee x_4)$. This leads to gadgets of length 9, $m=13$ and $\mathbf{T }=11$. We choose $x_1=1$, $x_2=0$, $x_3=0$ and $x_4=1$ as the truth assignment that makes $\varphi$ true.

The truth assignments for the clauses, $a_{j,d}$, are numbered in order to the binary output of the clauses: $a_{1,1}$ makes $x_1$, $x_2$ and $x_3$ false so the output of the first clause would be $0\vee 0\vee 1$. $a_{1,2}$ makes $x_1$ false but $x_2$ and $x_3$ true and so on.

Fig. 5.

An example for the modified Ulman reduction. In this example the formula is $\varphi =(x_1\vee x_2\vee \lnot x_3 )\wedge (x_1\vee \lnot x_2\vee x_4)$. For reasons of clarity only a few precedences are presented. Jobs of the same color are corresponding to the same variable. Jobs of the same color and pattern are assigned to the same gadget. The dottet gadgets represent the negative assignments and the gadgets without pattern represent the positive assignments.

B Modifiing the Biclique Instance to One with the Needed Property

Given a $k_1\times k_2$-Biclique instance $G=(V\dot{\cup }W,E)$ we want to obtain an equivalent $\widehat{k_1}\times \widehat{k_2}$-Biclique instance $\widehat{G}=(\widehat{(}V)\dot{\cup }\widehat{W},\widehat{E})$ with the property, that there is a $v_d\in \widehat{V}$ and a $w_d\in \widehat{W}$ without any edges and $|\widehat{V}|-\widehat{k_1}=\widehat{k_1}+\widehat{k_2}=|\widehat{W}|-\widehat{k_2}$:

First we add a dummy node $v_d$ to V and a dummy node $w_d$ to W without any edges. Then we take the maximum of $k_1+k_2$, $|V\cup \{v_d\}|-k_1$ and $|W\cup \{w_d\}|-k_2$ and take this number as number of machines m. We modify each instance in a way that $\widehat{k_1}+\widehat{k_2}=|\widehat{V}|-\widehat{k_1}=|\widehat{W}-\widehat{k_2}|=m$:

We obtain a set $\overline{V}$ by adding $v_d$ and other nodes without any edges to V until $|\overline{V}|-k_1=m$ and we do the same with W until $|\overline{W}|-k_2=m$.

Then we add vertices to $\overline{V}$ with edges to every node in $\overline{W}$ and at the same time increase $k_1$ by one for every vertex we add, until we get a set $\widehat{V}$ and a parameter $\widehat{k_1}$ where now $\widehat{k_1}+k_2=m$ and $|\widehat{V}-|\widehat{k_1}=m$ still holds and define $\widehat{k_2}=k_2$ and $\widehat{W}=\overline{W}$.

In this modification we find a $\widehat{k_1}\times \widehat{k_2}$-Biclique if and only if we found a $k_1\times k_2$-Biclique in the old version: The added nodes without any edges do not touch Bicliques in the instance and those with edges to every node increase all Bicliques by the number of added nodes of that kind.

So without loss of generality we can assume a $k_1\times k_2$-Biclique instances $G=(V\dot{\cup }W)$ where $|V|-k_1=k_1+k_2=|W|-k_2$ and dummy nodes $v_d\in V$, $w_d\in W$ with no edges.

C Alternative Reduction from Biclique to the Scheduling Problem Without Chains

There is an alternative for a reduction from $k_1\times k_2$-Biclique to $\mathrm{P}\,|\,\mathrm{prec,}\,p_j=1\,|\,C_\mathrm{max}$ with target makespan $\mathbf{T }=3$, where the Biclique instance is not manipulated in the beginning. Further this reduction provides a possibility to reduce the instance without being important from which set, V or W, the $k_1$ nodes come from and from which the $k_2$ come from. This could also be done with the old reduction from the Biclique problem to $\mathrm{P}\,|\,\mathrm{prec,}\,p_j=1\,\mathrm{cl}\le 2 |\,C_\mathrm{max}$ by testing both. This reduction however may reveal some new ideas for techniques.

Lemma 1

$\mathrm{P}\,|\,\mathrm{prec,}\,p_j=1\,|\,C_\mathrm{max}$ with target makespan $\mathbf{T }=3$ can be reduced from the $k_1\times k_2$-Biclique Problem.

Proof

Let us assume a $k_1\times k_2$-Biclique instance $G=(V\dot{\cup }W)$. Let $U=V\dot{\cup }W$. For each node $v\in U$ we create two jobs $j_v^-$ and $j_v^+$. For each $v,w\in U$ let $j_v^-\prec j_w^+$ iff $\{v,w\}\not \in E$. Let $m=\max \{|U|-k_1+1,k_1+k_2+1,|U|-k_2+1\}$ and let there be sets of dummy jobs $J_{\tiny {\text {bottom}}}$, $J_{\tiny {\text {middle}}}$ and $J_{\tiny {\text {top}}}$, where $|J_{\tiny {\text {bottom}}}|=m-(|U|-k_1)$, $|J_{\tiny {\text {top}}}|=m-(|U|-k_2)$ and $|J_{\tiny {\text {middle}}}|=m-(k_1+k_2)$. Let $j_{\tiny {\text {bottom}}}\prec j_{\tiny {\text {middle}}}\prec j_{\tiny {\text {top}}}$ for all $j_{\tiny {\text {bottom}}}\in J_{\tiny {\text {bottom}}}$, $j_{\tiny {\text {middle}}}\in J_{\tiny {\text {middle}}}$, $j_{\tiny {\text {top}}}\in J_{\tiny {\text {top}}}.$

Let further $j_v^-\prec j_{\tiny {\text {top}}}$ and $j_{\tiny {\text {bottom}}}\prec j_v^+$ for all $v\in U$, and at least one $j_{\tiny {\text {bottom}}}\in J_{\tiny {\text {bottom}}}$, $j_{\tiny {\text {top}}}\in J_{\tiny {\text {top}}}$.

The dummy job structure is to assure if the schedule has length 3, then all jobs from $J_{\tiny {\text {bottom}}}$ are scheduled in the first (bottom) time slot, all jobs $J_{\tiny {\text {middle}}}$ in the second (middle) and all jobs from $J_{\tiny {\text {top}}}$ in the third (top) time slot. Further all $j_v^-$ are scheduled in the first two time slots, $|U|-k_1$ of them in the first one, $k_1$ in the second. Likewise all $j_v^+$ jobs are scheduled in the second and third time slot, $k_2$ of them in the second and $|U|-k_2$ of them in the third.

Let us denote $J^-_{A}=\{j_v^-\ |\ v\in A\}$ and $J^+_{A}=\{j_v^+\ |\ v\in A\}$ for all $A\subseteq U$.

Note that there are no precedences between two jobs $j_v^-$ and $j_w^-$ aswell as no precedences between $j_v^+$ and $j_w^+$ for any $v,w\in U$.

If it is part of the instance, that the $k_1$ vertices are supposed to come from V and the $k_2$ vertices from W, we add the condition $j_v^-\prec j_{\tiny {\text {top}}}$ for all $v\in V$, and at least one $j_{\tiny {\text {top}}}\in J_{\tiny {\text {top}}}$.

If there is a solution of this instance with target makespan $\mathbf{T }=3$, then there are subsets $U^-,U^+\subseteq U$, $|U^-|=k_1$ and $|U^+|=k_2$ and the corresponding jobs $j_v^-$, $j_w^+$, $v\in U^-$, $w\in U^+$ are scheduled in the middle time slot together while all other jobs $j_v^-,j_w^+$, $v,w\in U\setminus (U^-\cup U^+)$ are in the bottom or top time slot. So for all $v\in U^-$ and $w\in U^+$ it is $j_v^-\not \prec j_w^+$, that means $\{v,w\}\in E$.

Fig. 6.

Visualization of a reduced schedule from the Biclique Instance. The arrows denote that there possibly are precedences between the jobs in the particular sets. All dark gray areas represent the jobs from the set $J^-$ and the light gray area represent the jobs from $J^+$.

First we know by this, v and w are not both in V or both in W, because otherwise $\{v,w\}\not \in E$. So we know $U^-\subseteq V$ and $U^+\subseteq W$ or vice versa.

With the extra condition we would know $U^-\subseteq V$, which induces $U^+\subseteq W$.

Second we know that for each $v\in U^-$ and each $w\in U^+$ it is $\{v,w\}\in E$ and so $U^-$ and $U^+$ form a $k_1\times k_2$-Biclique (Fig. 6).

Now assume there is a $k_1\times k_2$-Biclique. Let again $U^-,U^+\subseteq U$ be the sets of nodes, that form the biclique, where $|U^-|=k_1$ and $|U^+|=k_2$ and $U^-\subseteq V$ and $U^+\subseteq W$ or vice versa. For all $v\in U$ we now schedule all corresponding jobs $j_v^-$ in the bottom time slot exept the $j_v^-$ with $v\in U^-$, which we schedule in the middle time slot. For all $v\in U$ we schedule all corresponding jobs $j_v^+$ in the top time slot except the $j_v^+$ where $v\in U^+$, which we also schedule in the middle time slot. The schedule is feasable, because there are no precedences in between the jobs corresponding to $U^-$ or in between those of $U^+$ and there are none between the jobs $j_v^-$ and $j_w^+$ in the middle time slot, because their corresponding vertices form a biclique by assumption.