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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
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Introduction Discrepancy Problem Broadcast Scheduling Problem Our Results and Techniques Negative Results O(log 1.5 n)-Approximation Outline
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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 χ : 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
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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
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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
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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 2 6 3 1 4 2 5 1 5 2 3 4 6 χ : 1 2 3 4 5 6 5 6 3 1 4 2 6 3 1 4 2 5 1 5 2 3 4 6 discrepancy = 2
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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?
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Introduction Discrepancy Problem Broadcast Scheduling Problem Our Results and Techniques Negative Results O(log 1.5 n)-Approximation Outline
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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 2 3 4 5 response time = 3 135342 135135 245245 345345 124124
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Resource Allocation Scheduling Theory
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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
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Introduction Discrepancy Problem Broadcast Scheduling Problem Our Results and Techniques Negative Results O(log 1.5 n)-Approximation Outline
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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
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Introduction Discrepancy Problem Broadcast Scheduling Problem Our Results and Techniques Negative Results O(log 1.5 n)-Approximation Outline
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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.
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Fractional Schedule from LP integral schedule fractional schedule Time response time 0.4 × 1+0.6 × 2=1.6 requests 135135 345345 124124 245245
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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
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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 5431 8627 Req: 5431 8627 6352 7814 6352 7814 7386 1254 7386 1254 m/2
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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 P7P7 5431 8627 Req: 5431 8627 6352 7814 6352 7814 7386 1254 7386 1254 Brd: 3458 1276 8534 6721 4835 7216 3485 3 3 6 6 7 7
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5431 8627 Req: 5431 8627 6352 7814 6352 7814 7386 1254 7386 1254 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
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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 5431 8627 Req: 5431 8627 6352 7814 6352 7814 7386 1254 7386 1254 Brd: 3421 5867 4312 6785 1324 7856 3421
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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 P7P7 5431 8627 Req: 5431 8627 6352 7814 6352 7814 7386 1254 7386 1254 Brd: 3421 5867 4312 6785 1324 7856 3421 3 2 14 4312
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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 = 1 2 8 3 5 4 10 6 7 11 12 9 broadcasts = 9 1 6 3 5 4 broadcast after request : good broadcast before request : bad d = 2
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“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
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Introduction Discrepancy Problem Broadcast Scheduling Problem Our Results and Techniques Negative Results O(log 1.5 n)-Approximation Outline
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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”
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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
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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
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Locally Consistent Distribution t 1025643 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
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Interesting Intervals # time slots ≤ 2Δ × n “error” repeat log n times : backlog = O(log 3/2 n) 64Δ …… λ = 0 λ = 1 λ = 2 …
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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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