Minimizing weighted completion time with precedence constraints

1. Minimizing weighted completion time with precedence constraints Nikhil Bansal (IBM) Subhash Khot (NYU)

2. The Problem Given n jobs, arbitrary weights and sizes (job j: wt. wj, size pj) Precedence constraints: DAG (edge (i,j) implies j cannot started before i finishes) Goal: Schedule jobs to minimize weighted completion time. Precedence Graph A valid schedule

3. Some previous work Strongly NP-Hard (Lawler 78, Lenstra-Kan 78) Several 2 approximations Potts linear ordering formulation Completion time Formation Time indexed Formulation Sidney’s decomposition Special case of vertex cover problem < 2 approximations for special cases of precedence graphs [Ambuhl et al 07, Papadimitriou-Yannakakis 79, Woeginger 03, …] Various Researchers

4. Hardness [Ambuhl-Mastrolilli-Svensson 07] Thm 1: No (1+) approximation, for some >0 (assuming SAT cannot be solved in sub-exponential time) Thm 2: General vertex Cover ! Scheduling Instance Objective function = Fixed cost + Vertex Cover cost. Current vertex cover hardness: 1.36 [Dinur-Safra 02], (2-) assuming Unique Games Conjecture (UGC) [Khot-Regev 03] In Thm2 proof instance, Fixed cost >> Vertex Cover cost. Fixed cost comes from precedence constraints. (If hard instances always have large fixed cost, implies <2 approx.)

5. Our results Thm: Assuming a variant of UGC, it is NP-Hard to obtain 2- approximation for any  > 0. Key idea: New dictatorship test Implies a PCP with 1 free bit assuming UGC. Implies VC hard to approximate within 2- even if Independent Independent (1/2-) n (1/2-) n Equivalently, hard to find independent set of size  n even if graph has two disjoint independent sets of size (½-) n

6. Unique Games (Basics)

7. Unique Games Problem Unique games: Equations mod p, with two terms. 12 x1 + 7 x2 = 43 (mod p) 3 x1 + 5 x2 = 11 (mod p) ….. 7 x1 + 4 x2 = 15 (mod p) If there is a valid solution, can find it by Gaussian elimination. (or just propagation). But what if, say, 99% equations are satisfiable? Can we find an assignment that satisfies even 1% of them? Current algorithms do not preclude this (provided p large enough).

8. Unique Games Conjecture Conjecture (Khot 02): For any ,  >0, there is a UG instance, such that it is NP-hard to distinguish if at least 1- fraction of equations are satisfiable, or at most fraction are satisfiable. Extremely successful at showing (tight) new hardness results. Max-Cut .878 Hardness (KKMO), .94 assuming P  NP Vertex Cover 2 (KR 03), 1.36 assuming P P Non-constant hardness for multi-cut, sparsest cut …

9. Unique Games Conjecture Conjecture (Khot 02): For any ,  >0, there is a UG instance, such that it is NP-hard to distinguish if at least 1- fraction of equations are satisfiable, or at most  fraction are satisfiable. Powerful, but also use is quite general and simple. Very Roughly, reduces proving hardness to constructing a type of “integrality gap” instance on boolean hypercube.

10. 1 0 0 1 0 1 0 1 Power of UGC 1. Can you cast problem as 0-1 labeling problem? 2. Create your problem instance on hypercube where: a. Any dictatorship labeling is good. b. Any labeling “far from dictator” bad. (think random labeling) 1,1,0 1,1,1 Vertices are x = (x1,x2,x3) 2 {0,1}3 Labeling: f ! {0,1}3! {0,1} Dictator (co-ordinate) labeling: f(x) = xi 1,0,1 1,0,0 0,1,0 0,1,1 x1 x2 x3 0,0,0 0,0,1 Gap instance (usually) implies hardness of same value

11. Back to 1|prec| wj Cj

12. 1|prec| wj Cj: Simpler instance Bipartite Instances: Thm (Woeginger): Bipartite instances are as hard to approximate as general problem Size=1 Wt =0 Size=0 Wt =1 Set system: (elements, sets) Set is completed when all its elements are scheduled Problem:Order elements, to minimize avg. completion time of sets

13. Good and bad instances elements Good ordering: Bad ordering: set completions Set completion times: “almost” uniformly distributed Average completion time ¼ n/2 Canonical example: n elements, Sets: all sets of size k = 1/2 Easy claim: In any ordering, most sets complete at time > n(1- ) Our Thm: There are instances, s.t. cannot (efficiently) determine 1. If there is some good ordering. 2. Or, if every ordering is bad.

14. Set system (gadget) construction 1,1,0 1,1,1 1,0,1 Elements: vertices of a d dim. boolean hypercube (d is large constant) Vertex = (x1,x2,…,xd) each xi2 {0,1} Sets =  d dimensional subcubes. Subcube = ( x1,*,x3,*,…,xd) One dimensional subcube = edge (line) Two dimensional subcube = faces (square) 1,0,0 0,1,0 0,1,1 0,0,0 0,0,1

15. Mapping to 0-1 problem Schedule = Assign labels 1,…,2d to elements of hypercube. Completion time of a set (subcube) = max label of elements in it. Suppose only care about whether an element placed in first half or second half (0-1 labeling) (Note: For 2- hardness we actually, need O(1/) buckets) A subcube has high completion time if some element in it labeled 1. Goal: Good case: About half subcubes consist only of 0’s. Bad case : Almost all subcubes have 1 in them. 0 1

16. 1 0 0 1 0 1 0 1 Dictatorship test & Result Any Dictatorship function: f(x1,…,xd) = xi Main observation: • Half points 0, half are 1 • A random subcube on d co-ordinates is mono-chromatic with probability 1- Thm (non-trivial): Given any 0-1 labelling of hypercube, if Pr[ random sub-cube is all 0] > , then “close” to a dictator. (labeling must have some non-zero fraction of 1’s.) (Close to a dictator: Some variable with large low degree influence.) Corollary: If “far” from dictator, then most sub-cubes contain a 1.

17. Dictatorship test to Hardness UG: n variables, m equations mod p. Graph view: n vertices, m edges. Need to choose 1 out of p labels for each vertex . Replace each vertex by hypercube on p co-ordinates. Label corresponds to dictatorship in that co-ordinate. Uniqueness property allows to map dictatorship test within a hypercube, to across hypercubes. Our Variant: 1. Mild expansion property. To force enough hypercubes to take both 0’s and 1’s. (usually need for all ordering type problems). 2. In good case, most vertices have all adjacent constraints true. (like in vertex cover [Khot Regev 03] )

18. Open Problems 1. Show hardness of 2-, assuming P  NP (will already imply vertex cover is factor 2 hard) (currently, we cannot rule out a PTAS, assuming P  NP) 2. Is our UG variant equivalent to (standard) UGC. 3. Other applications of our instance? Parallel machine case? 4. UseUGC for other outstanding scheduling problems.

19. Thanks!

20. Our Variant of UG instance 1. In good case, there is subset of 1-\eps fraction of variables, s.t. all equations involving them are satisfied. 2. UG graph has (mild) expansion properties For all S, s.t. |S| = \delta |V|, |N(S)| \geq (1-\delta) |V| With either just 1. or just 2. can show equivalent to UGC, but not clear if this variant is equivalent also. Need expansion to ensure that a non-trivial fraction of hypercubes have non-trivial fraction of both 0’s and 1’s.

21. Other implications Free bit complexity of a probabilistically checkable proof (PCP) = Log2 (accepting answers) (can query arbitrarily many bits) Example: Hastad’s test f(x)+f(y)+f(z)=0 Bellare, Goldreich, Sudan: PCP with 0-free bits, completeness c, soundness s is equivalent to (1-s)/(1-c) Our test: Choose random sub-cube. Query all locations, accept if mono-chromatic. Assuming UGC, this gives a PCP with completeness 1-, soundness , Free bit =1

22. Outline • Introduction • Bipartite precedence graph • Dictators and influences • UGC and its power, examples • Our UGC variant, why is it needed • Free bit, BGS result