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

  Introduction  Discrepancy Problem  Broadcast Scheduling Problem  Our Results and Techniques  Negative Results  O(log 1.5 n)-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 : χ : {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(log 2.5 n) ) 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 χ : 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(l 1/2 log n)  lower bound : max{Ω(l 1/2 ), Ω(log n)} Why 3 Permutations?

  Introduction  Discrepancy Problem  Broadcast Scheduling Problem  Our Results and Techniques  Negative Results  O(log 1.5 n)-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 response time =

 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(n 1/2 )-approx [Bansal-Charikar-Khanna-Naor 05]  O(log 2 n)-approx[Bansal-Coppersmith-Sviridenko 08] Known Results

  Introduction  Discrepancy Problem  Broadcast Scheduling Problem  Our Results and Techniques  Negative Results  O(log 1.5 n)-Approximation Outline

 previous bestour results approximationO(log 2 n)O(log 3/2 n) integrality gap1 + tiny constΩ(log n) hardnessNP-hardΩ(log 1/2 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(log 1.5 n)-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(l 1/2 )-hard. for l-perm.(new)  Ω(log 1/2 n)-hard.

 Fractional Schedule from LP integral schedule fractional schedule Time response time 0.4 × × 2=1.6 requests

 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 π 1 π 2 π 3 of size m  π 1 = (5, 8, 4, 6, 3, 2, 1, 7)  π 2 = (6, 7, 3, 8, 5, 1, 2, 4)  π 3 = (7, 1, 3, 2, 8, 5, 6, 4) Construction of BS Instance π1π1 π2π2 π3π3 forbidden interval P1P1 P1P1 P2P2 P2P2 P3P3 P3P3 P4P4 P4P4 P5P5 P5P5 P6P6 P6P6 P7P7 P7P7 permutation interval Req: m/2

  average response time ≈ # bad requests  new goal: minimize #bad requests  a request in P i is good if it is satisfied at P i or P i+1  otherwise, the request is bad Good and Bad Requests P1P1 P1P1 P2P2 P2P2 P3P3 P3P3 P4P4 P4P4 P5P5 P5P5 P6P6 P6P6 P7P7 P7P Req: Brd:

 Req:  LP solution  each time slot, broadcast ½ fraction of each page requested  P 7 : broadcast ½ fraction of the m pages arbitrarily  all requests are good:  ½ of request in P i is satisfied immediately  remaining ½ satisfied at P i+1 Θ(1) LP Value P1P1 P1P1 P2P2 P2P2 P3P3 P3P3 P4P4 P4P4 P5P5 P5P5 P6P6 P6P6 P7P7 P7P7 request ½ satisfied

 How to Make All Requests Good in an Integral Schedule? P1P1 P1P1 P2P2 P2P2 P3P3 P3P3 P4P4 P4P4 P5P5 P5P5 P6P6 P6P6 P7P7 P7P7  all m pages requested in all intervals(except P 7 )  each P-interval has m/2 slots  solution:  m/2 pages are broadcast in P 1, P 3, P 5, P 7  m/2 pages are broadcast in P 2, P 4, P 6  giving a balanced ±1 coloring of the m pages Req: Brd:

  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? P1P1 P1P1 P2P2 P2P2 P3P3 P3P3 P4P4 P4P4 P5P5 P5P5 P6P6 P6P6 P7P7 P7P Req: Brd:

  suppose disc χ (π i ) = d  π i =(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 P 2i-1  #bad quests = d-1 Small Discrepancy  Few Bad Requests requests = broadcasts = broadcast after request : good broadcast 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 P1P1 P1P1 P2P2 P2P2 P3P3 P3P3 P4P4 P4P4 P5P5 P5P5 P6P6 P6P6 P7P7 P7P7 request good bad

  Introduction  Discrepancy Problem  Broadcast Scheduling Problem  Our Results and Techniques  Negative Results  O(log 1.5 n)-Approximation Outline

  A  R m×n, x  [0,1] n, b=Ax,  λ 1, λ 2, …, λ m s.t.  output: y  [0,1] n, s.t. ½ fraction of coordinates in y are integral Lovett-Meka Framework A A x x b b ×= m n A A y y b b ×= m n ±λ 1 ||A 1 || ±λ 2 ||A 2 || ±λ 3 ||A 3 ||... ±λ m ||A m || “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

 Locally Consistent Distribution t f(t) = # broadcasts of p by time t 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(log 3/2 n) 64Δ …… λ = 0 λ = 1 λ = 2 …

 previous bestour results approximationO(log 2 n)O(log 3/2 n) integrality gap1 + tiny constΩ(log n) hardnessNP-hardΩ(log 1/2 n) Summary  Open problems  hardness for 3-permutation(implying the same hardness for broadcast scheduling)  discrepancy of l-permutation?

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