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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
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?