N. Bansal 1, M. Charikar 2, R. Krishnaswamy 2, S. Li 3 1 TU Eindhoven 2 Princeton University 3 TTIC Midwest Theory Day, Purdue, May 3, 2014.

Better Algorithms and Hardness for Broadcast Scheduling via a

Discrepancy Approach

N. Bansal1, M. Charikar2, R. Krishnaswamy2, S. Li3

1TU Eindhoven

2Princeton University

3TTIC

Midwest Theory Day, Purdue, May 3, 2014

Introduction

Discrepancy Problem

Broadcast Scheduling Problem

Our Results and Techniques

Negative Results

O(log1.5n)-Approximation

Outline

input

ground set U a family S of subsets of U

output: coloring minimize worst discrepancy:

Discrepancy Problem

U : {1,2,3,4,5,6}

S :

χ : 1 2 3 4 5 6

+1 -1χ : 1 2 3 4 5 6

{1,3,5,6}

{2,3,4,6}

{1,4,5,6}

{1,3,5,6} 0

{2,3,4,6} 0

{1,4,5,6} 0

{1,3,5,6} 0

{2,3,4,6} 0

{1,4,5,6} 2

S contains all subsets discrepancy = n/2 -disc. by randomized coloring -disc. (non-constructive) [Spencer 85] -disc. (constructive) [Bansal 10] [Lovett-

Meka 12] -lower bound

Interesting When |U| = |S| = n

Erdos’s Discrepancy

U = {0, 1, 2, 3, ......} S = {all arithmetic progressions starting at 0} open problem: is discrepancy bounded?

Rectangle Discrepancy U = {n points in a 2-D plane} S = {all axis-parallel rectangles} discrepancy? ( between Ω(log n) and O(log2.5n) )

Special Discrepancy Problems

give 3 permutations of [n] find a coloring χ : [n]{±1} minimize the maximum discrepancy over all

prefixes of the permutations

3-Permutation Discrepancy

5 6 3 1 4 26 3 1 4 2 51 5 2 3 4 6

χ : 1 2 3 4 5 6 5 6 3 1 4 26 3 1 4 2 51 5 2 3 4 6 discrepancy = 2

1 permutation : discrepancy=1, trivial 2 permutations : discrepancy=1, easy exercise 3 permutations?

upper bound : O(log n) lower bound [Newman-Nikolov 11]: Ω(log n)

l ≥ 3 permutations upper bound : O(l1/2 log n) lower bound : max{Ω(l1/2), Ω(log n)}

Why 3 Permutations?

Introduction

Discrepancy Problem

Broadcast Scheduling Problem

Our Results and Techniques

Negative Results

O(log1.5n)-Approximation

Outline

a server holding n pages requests come over time broadcast 1 page per time slot minimize average response time offline version

Broadcast Scheduling Problem

response time = 2

Time

1

23

4

5

response time = 3

1 3 5 34 2

135

245

345

124

Resource Allocation

Scheduling Theory

NP-hard [Erlebach-Hall] (1/α)-speed,1/(1-2α)-approximation, α ≤ 1/3

[Kalyanasundaram et al.] (1/α)-speed: broadcast a page only requires α time slots

(1+ε)-speed, O(1/ε) approximation, ε > 0[Bansal-Charikar-Khanna-Naor 05]

O(n)-approx: trivial, cyclic order O(n1/2)-approx [Bansal-Charikar-Khanna-Naor 05] O(log2n)-approx[Bansal-Coppersmith-Sviridenko 08]

Known Results

Introduction

Discrepancy Problem

Broadcast Scheduling Problem

Our Results and Techniques

Negative Results

O(log1.5n)-Approximation

Outline

previous best our results

approximation O(log2n) O(log3/2n)integrality gap 1 + tiny const Ω(log n)

hardness NP-hard Ω(log1/2 n)

Our Results and Techniques

negative results (integrality gap and hardness) connection to permutation discrepancy

positive result Lovett-Meka algorithmic framework for discrepancy

minimization

Introduction

Discrepancy Problem

Broadcast Scheduling Problem

Our Results and Techniques

Negative Results

O(log1.5n)-Approximation

Outline

Main Lemma

Negative Results

l-permutation instance Π

broadcast scheduling instance I

=“discrepancy” optimal response time

LP(I) = O(1)

Main + Ω(log n)-disc. for 3-perm. Ω(log n)-int. gap Main + Ω(l1/2)-hard. for l-perm.(new) Ω(log1/2 n)-hard.

Fractional Schedule from LP

integral schedule

fractional schedule

Time

response time0.4×1+0.6×2=1.6requests 1

35

345

124

245

Main Lemma

l-permutation instance Π

broadcast scheduling instance I

=“discrepancy” optimal response time

LP(I) = O(1)

proof steps: construction of BS instance from l permutations Θ(1) LP value small discrepancy small response time small response time small discrepancy

given 3 permutations π1 π2 π3 of size m

π1 = (5, 8, 4, 6, 3, 2, 1, 7)

π2 = (6, 7, 3, 8, 5, 1, 2, 4)

π3 = (7, 1, 3, 2, 8, 5, 6, 4)

Construction of BS Instance

π1 π2 π3

forbidden interval

P1 P2 P3 P4 P5 P6 P7

permutation interval

54318627Req: 5431

862763527814

63527814

73861254

73861254

m/2

average response time ≈ # bad requests new goal: minimize #bad requests a request in Pi is good if it is satisfied at Pi or Pi+1

otherwise, the request is bad

Good and Bad Requests

P1 P2 P3 P4 P5 P6 P7

54318627Req: 5431

862763527814

63527814

73861254

73861254

Brd: 3458 1276 8534 6721 4835 7216 34853

36

67

7

54318627Req: 5431

862763527814

63527814

73861254

73861254

LP solution each time slot, broadcast ½ fraction of each page requested P7: broadcast ½ fraction of the m pages arbitrarily

all requests are good: ½ of request in Pi is satisfied immediately

remaining ½ satisfied at Pi+1

Θ(1) LP Value

P1 P2 P3 P4 P5 P6 P7

request½ satisfied½ satisfied

How to Make All Requests Good in an Integral Schedule?

P1 P2 P3 P4 P5 P6 P7

all m pages requested in all intervals(except P7)

each P-interval has m/2 slots solution:

m/2 pages are broadcast in P1, P3, P5, P7

m/2 pages are broadcast in P2, P4, P6

giving a balanced ±1 coloring of the m pages

54318627Req: 5431

862763527814

63527814

73861254

73861254

Brd: 3421 5867 4312 6785 1324 7856 3421

enough to make all requests good? No! Broadcast may be before the request

no bad requests only if two requests at the same time have different colors

discrepancy of 3-permutation system is 1

How to Make All Requests Good in an Integral Schedule?

P1 P2 P3 P4 P5 P6 P7

54318627Req: 5431

862763527814

63527814

73861254

73861254

Brd: 3421 5867 4312 6785 1324 7856 3421

3 2 144312

suppose discχ(πi) = d

πi =(1, 10, 2, 6, 8, 7, 3, 11, 5, 12, 4, 9)

χ =(1, 10, 2, 6, 8, 7, 3, 11, 5, 12, 4, 9) order of red elements (1,6,3,5,4,9) right rotate by d-1=1 positions: (9,1,6,3,5,4) broadcast according to this ordering in P2i-1

#bad quests = d-1

Small DiscrepancyFew Bad Requests

requests = 1 2 8 3 5 4 10 6 7 11 12 9 broadcasts = 9 1 6 3 5 4

broadcast after request : goodbroadcast before request : bad

d = 2

“discrepancy” = average discrepancy of l

permutations size of BS instance is exponential in l

lengths of forbidden intervals grow exponentially

Remarks

P1 P2 P3 P4 P5 P6 P7

request good bad

Introduction

Discrepancy Problem

Broadcast Scheduling Problem

Our Results and Techniques

Negative Results

O(log1.5n)-Approximation

Outline

A Rm×n, x [0,1]n, b=Ax,

λ1, λ2, …, λm s.t.

output: y [0,1]n, s.t. ½ fraction of coordinates in y are integral

Lovett-Meka Framework

A x b× =m

n

A y b× =m

n±λ1||A1||

±λ2||A2||

±λ3||A3||

...±λm||Am||

“error”

we may broadcast more than 1 page at a time slot

tentative schedule of backlog b valid schedule, with additive b loss in the average response time

backlog discrepancy

Tentative Scheduling

6 time slots, 11 broadcast, backlog = 5

assumptions:

fractional schedule is ½-intergal every page is broadcast ≤ Δ = O(log n) times # timeslots ≤ 2Δ × n

locally consistent distributions

Goal

with probability 1/2

with probability 1/2

Locally Consistent Distribution

t10 2 5 643

f(t) = # broadcasts of p by time t

1

3

4

2

s

1+s

2+s

3+s

broadcast p at time 0, 1, 4, 5

randomly select a s (0,1) broadcast at time f -1(s), f -1(1+s), f -1(2+s),……

call (0,1,4,5) a shift for page p

Interesting Intervals

# time slots ≤ 2Δ × n

“error”

repeat log n times : backlog = O(log3/2n)

64Δ

……

λ= 0λ= 1λ= 2…

previous best our results

approximation O(log2n) O(log3/2n)integrality gap 1 + tiny const Ω(log n)

hardness NP-hard Ω(log1/2 n)

Summary

Open problems hardness for 3-permutation(implying the same

hardness for broadcast scheduling) discrepancy of l-permutation?

Thank you!

Questions?