The Byzantine Secretary Problem 30th Nov, 2018 Sahil singla Princeton University Joint work with Anupam gupta, Domagoj Bradac, And Goran Zuzic

Example: Diamond-Selling Optimal Stopping-Time Sell One Diamond: Multiple potential buyers Buyers Arrive and Make a Take-it-or-Leave-it Bid Decide Immediately and Irrevocably When to Accept the Bid? Goal: Maximize value of the accepted bid Cannot go back to a declined bid Similar Examples Selling ad-slots Finding a secretary/marriage partner

Tight approx factor is 1/𝑒 The Secretary Problem 𝑣 1 Problem n unknown adversarial values 𝑣 1 > 𝑣 2 >…> 𝑣 𝑛 IID arrival times t 𝑒 ~ Unif[0,1] Decide immediately and irrevocably Max probability of selecting 𝑣 1 Dynkin’s Algorithm Ignore elements 𝑒 with 𝑡 𝑒 <1/2 Otherwise, select if larger than all previous elements Pr Selecting v 1 ≥ Pr t 1 > 1 2 ⋅Pr t 2 < 1 2 = 1 4 𝑣 2 𝑣 3 𝑣 4 𝑡=0 𝑡=1 𝑡=1/2 Tight approx factor is 1/𝑒

Why not Adversarial Arrival? 𝑡=0 𝑡=1 Don’t know the “scale” of the problem Sequence of increasing elements, and then all zeroes Select at random ⟹ 1 𝑛 success probability Why Random order? Mathematically necessary and makes sense intuitively Weaker assumption than iid values from unknown distrib Related Work: Combinatorial problems, e.g, sum of top k, Matchings, Matroids and their intersections, facility location, max indep set on interval graphs Packing linear programs (large budget assumption) What if only some of the arrivals adversarial? (outlier arrivals)

Remark: Fine if all reds small A Byzantine World Byzantine General’s Problem: Lamport, Shostak, Pease Some arrivals adversarial: Red 𝑅 vs Green 𝐺 elements All 𝑛= 𝑅 +|𝐺| values are adversarial Red elements 𝑅 at adversarial times Green elements 𝐺 at t 𝑒 ~ Unif[0,1] Elements do not reveal their color Dynkin’s algorithm no longer works One large red in the beginning Handle outliers: Need robust algorithms What’s even the right benchmark? Largest in 𝑅∪𝐺? Largest in 𝐺? 𝑡=0 𝑡=1 Remark: Fine if all reds small

Value vs Probability Maximization Dynkin’s Algorithm Needs current time 𝑡. Does not know 𝑛 Only needs relative order (ordinal setting) Maximizes probability of selecting 𝑣 1 Can we do better if given values? Maximize ratio E[Alg] v 1 No! Don’t know the “scale” of the problem For any algo, there exist instances with E Alg v 1 ≤ 1 𝑒 +ϵ 𝑡=0 𝑡=1

OUTLINE Motivation: Secretary Problem in a Byzantine World Benchmark Value-Maximization Probability-Maximization Other Results and Open Problems

Benchmark: Get the Max Overall Max? All green 𝐺 values are zero Back to adversarial arrival on red elements 𝑅 At best probability 1 |𝑅| Max Green? Only one non-zero green Consider increasing sequence of red elements Don’t know which is green ⟹Θ( 1 |𝑅| ) probability Think Ordinal 𝑡=0 𝑡=1 Green What to do next?

Drop the Maximum! Get 2nd Max Green Can we get the 2nd max green? Maximize probability: “Victory” on selecting ≥ 𝑣 ⋆ Maximize value: Ratio of E[Alg] to 𝑣 ⋆ Comparing to 2nd max green helps? Unclear! Adversary cannot control relative order of max and 2nd max greens Why it makes sense? Applications where small gap between max and 2nd max Mathematically this is the “price” that we have to pay Similar benchmarks in other applications, e.g., digital goods auctions 𝑣 ⋆ :=𝑣(2nd max green)

Main results 𝑣 ⋆ :=𝑣(2nd max green) Thm [BGSZ’18]: For Value Maximization we get O lo g ∗ n 2 approximation. Remark: Performance guarantee independent of |𝐺|, e.g., even for 𝐺 =2 Thm [BGSZ’18]: For Probability Maximization we get O log n 2 approximation. Pr[selecting ≥ 𝑣 ⋆ ] Remark: This result is only existential as it uses the minimax principle

OUTLINE Motivation: Secretary Problem in a Byzantine World Benchmark: Drop the Maximum! Value-Maximization: 𝐎 𝐥𝐨𝐠 ∗ 𝐧 𝟐 approx Probability-Maximization: 𝐎 𝐥𝐨𝐠⁡𝐧 𝟐 approx Other Results and Open Problems

Approach: Done or Refine Scale 𝑣 ⋆ :=𝑣(2nd max green) Observ: Given 𝑣 ⋆ ∈[𝑎 , 𝑚⋅𝑎] at 𝑡=0 implies 2⋅log m approx. Define log m levels: 𝑎,2𝑎 , 2𝑎, 2 2 𝑎 , …, 2 log m−1 𝑎, 2 log m 𝑎 Guess ≈𝑣 ⋆ w.p. 1 log m , and select first element above Question: What if good estimate of 𝑣 ⋆ by 𝑡=1/2? Define Checkpoints 𝑇 𝑖 : Partition time into intervals Idea in any Interval: Either a simple algo works (e.g., Dynkin’s, Select a random element) Or keep refining estimate of 𝑣 ⋆ , and set a threshold in the final interval Only works for value-max setting 𝑡=0 𝑡=1 Done Refine How many Checkpoints?

One Checkpoint: 𝐎(𝐥𝐨𝐠⁡𝐧) approximation 𝑣 ⋆ :=𝑣(2nd max green) Case 1 (“small”): Every red before 𝑡= 1 2 is below 𝑣 ⋆ Dynkin’s algorithm works Case 2 (“large”): Exists a red before 𝑡= 1 2 above 𝑛⋅ 𝑣 ⋆ Select a random element: Correct w.p. 1 𝑛 Otherwise (“medium”): Exists a red before 𝑡= 1 2 with value 𝑣 0 ∈[ 𝑣 ⋆ , 𝑛⋅ 𝑣 ⋆ ] Gives a factor m=𝑛 approx to 𝑣 ⋆ Define log 𝑛 levels. Condition on 𝑣 ⋆ at t> 1 2 Guess ≈𝑣 ⋆ w.p. 1/ log 𝑛 , and select first above E Alg ≥ Pr Guess 𝑣 ⋆ ⋅Pr 𝑣 ⋆ at t> 1 2 ⋅ 𝑣 ⋆ 2 ≥ 𝑣 ⋆ 4⋅ log n Done 𝑡=0 𝑡=1 𝑡=1/2 Refine Run one of the three algorithms uniformly at random Q.E.D.

Two Checkpoints: 𝐎(𝐥𝐨𝐠𝐥𝐨𝐠⁡𝐧) approx 1 3 2 3 Last slide implies for 𝑡∈[0, 1 3 ] : Exists a red element with value 𝑣 0 ∈[ 𝑣 ⋆ , 𝑛⋅ 𝑣 ⋆ ] Case 1 (“small”): For 𝑡∈[ 1 3 , 2 3 ], all red below 𝑣 ⋆ Dynkin’s algorithm works Case 2 (“large”): For 𝑡∈[ 1 3 , 2 3 ] there is a red above ( log 𝑛 ⋅ 𝑣 ⋆ ) Random level guessing gets it w.p. 1 log 𝑛 works Otherwise (“medium”): Exists a red for t∈[ 1 3 , 2 3 ] with value 𝑣 1 ∈[ 𝑣 ⋆ , log 𝑛 ⋅ 𝑣 ⋆ ] Gives a factor m=log 𝑛 approx to 𝑣 ⋆ Define log log 𝑛 levels: 𝑣 1 log 𝑛 , 2⋅ 𝑣 1 log 𝑛 ,…, 𝑣 1 4 , 𝑣 1 2 , 𝑣 1 2 , 𝑣 1 𝑡=0 𝑡=1 Done Definition of “large” changes Refine Run one of the Θ(1) algorithms randomly Q.E.D.

Multiple Checkpoints: 𝐎 𝐥𝐨𝐠 ∗ 𝐧 𝟐 approx Use lo g ∗ n checkpoints 𝑇 𝑖 Algorithm Guess a random interval Run a simple algorithm after refining till now Proof Idea Case 1 (“small”): There is an interval with all small reds Dynkin’s algo Case 2 (“large”): There is an interval with “large” red Guess level & Random elem Otherwise (“medium”): Refine till the end to get better scale of 𝑣 ⋆ 𝑇 𝑖 𝑡=0 𝑡=1 Q.E.D.

OUTLINE Motivation: Secretary Problem in a Byzantine World Benchmark: Drop the Maximum! Value-Maximization: 𝐎 𝐥𝐨𝐠 ∗ 𝐧 𝟐 approx Probability-Maximization: 𝐎 𝐥𝐨𝐠⁡𝐧 𝟐 approx Other Results and Open Problems

How to Capture Scale What does “scale” for 2nd max mean? 𝑣 ⋆ :=𝑣(2nd max green) What does “scale” for 2nd max mean? There is no notion of values Define a confusion set 𝑆 𝑡 Elements that are candidates to be 𝑣 ⋆ at 𝑡 Idea: Condition on 𝑣 ⋆ arriving in the first interval ⟹ 𝑆 𝑡 always subset of first interval 𝑡=0 𝑡=1 𝑆 𝑡 Observ: Given 𝑆 𝑡 implies 𝑆 𝑡 /(1−𝑡) approx. Proof: Set a random element of 𝑆 𝑡 as threshold and let largest green arrive after 𝑡.

Application: beating Random guessing 𝑣 ⋆ :=𝑣(2nd max green) Lemma: There exists 𝑂(√𝑛) approximation. Case 1: Number of red elements above 𝑣 ⋆ is Ω(√𝑛) Select a random element Case 2: Consider top √𝑛 elems at 𝑡= 1 2 as confusion set 𝑆 𝑡 Condition on 𝑣 ⋆ before 𝑡= 1 2 Guess a random elem to be 𝑣 ⋆ & set as threshold: Correct w.p. 1 √𝑛 Pr Selecting≥ 𝑣 ⋆ ≥ Pr Guess of 𝑣 ⋆ correct ⋅Pr max green at 𝑡 > 1 2 ≥ 1 2√𝑛 𝑡=0 𝑡=1 𝑆 𝑡 Q.E.D. Thm: For Probability Max, there exists O log n 2 approx. How to Refine Confusion Set?

Approach Using The Minimax Principle WLOG, assume we know the correlated arrival distribution General distribution over permutations of elements Values (order) of all 𝑅∪𝐺 and arrival times of Reds 𝑅 Idea: As time 𝑡 increases, better idea of the permutation Approach: Done or Refine Scale Use log n checkpoints 𝑇 𝑖 Either some simple algorithm works Otherwise: Refine at each checkpoint by discarding half of 𝑆 𝑡 Finally, randomly guess 𝑣 ⋆ when 𝑆 =𝑂(1) For 𝑒∈ 𝑆 𝑡 , maintain 𝑝 𝑒 𝑡 =Pr[𝑒 is 𝑣 ⋆ ] 𝑡=0 𝑡=1 𝑇 𝑖 Done 𝑆 𝑇 𝑖 Refine

How to Refine Let 𝑐 𝑖 ∈ 𝑆 𝑇 𝑖 be the median of 𝑆 𝑇 𝑖 . ∑ 𝑝 𝑒 𝑇 𝑖+1 𝑡=0 𝑡=1 Let 𝑐 𝑖 ∈ 𝑆 𝑇 𝑖 be the median of 𝑆 𝑇 𝑖 . Case 1 (“large”): At 𝑇 𝑖+1 , confusion probability below 𝑐 𝑖 is ≈1 𝑆 𝑇 𝑖+1 = Elements in 𝑆 𝑇 𝑖 below 𝑐 𝑖 Case 2 (“small”): At 𝑇 𝑖+1 , confusion probability below 𝑐 𝑖 is < 1 log 𝑛 𝑆 𝑇 𝑖+1 = Elements in 𝑆 𝑇 𝑖 above 𝑐 𝑖 Otherwise (“medium”): At 𝑇 𝑖+1 , confusion probability below 𝑐 𝑖 is in ( 1 log 𝑛 , 1) Set 𝑐 𝑖 as a threshold and select first element above ≥ 1 log 𝑛 probability that 𝑣 ⋆ is below 𝑐 𝑖 . Moreover, exists a red elem above 𝑐 𝑖 . ∑ 𝑝 𝑒 𝑇 𝑖+1 𝑐 𝑖 Refine 𝑆 𝑇 𝑖 𝑆 𝑇 𝑖+1 Done

Wrapping Up: O log n 2 Approximation Use log n checkpoints 𝑇 𝑖 Algorithm Guess a random interval 𝑖 With prob 1 2 run Dynkin’s algo in ( 𝑇 𝑖−1 , 𝑇 𝑖 ) Else, set 𝑐 𝑖 as threshold after refining till now Proof Idea Done Case: We “correctly” guess 𝑖 w.p. 1 log 𝑛 and 𝑣 ⋆ is below 𝑐 𝑖 w.p. 1 log 𝑛 Refine Till End: We guess final checkpoint where 𝑆 =𝑂(1) and set a random element as threshold 𝑇 𝑖 𝑡=0 𝑡=1 𝑆 𝑡 Q.E.D. 21

OUTLINE Motivation: Secretary Problem in a Byzantine World Benchmark: Drop the Maximum! Value-Maximization: 𝐎 𝐥𝐨 𝐠 ∗ ⁡𝐧 𝟐 approx Probability-Maximization: 𝐎 𝐥𝐨𝐠⁡𝐧 𝟐 approx Other Results and Open Problems

How to Select Multiple Items? Value Maximization (Sum of values): Arrival contains Red R and Green G elems What is the benchmark? Maximum sum of values in G∖{ g max } Thm [BGSZ’18]: For uniform Matroids with k=Ω(log n) we get O(1) approximation. Thm [BGSZ’18]: For Partition Matroids we get O loglog n 2 approximation. Remark: We get max-green element in most parts. Observation: Easy to get O(log n) approx for general matroids.

summary Questions? Byzantine Secretary Model: Compare to 2nd max green O log ∗ n 2 approx for value maximization Done or Refine Scale O log n 2 approx for probability maximization Minimax Principle O(1) for uniform and O loglog n 2 for partition matroids Open Problems Super constant lower bound? What are the optimal approx factors? How to make probability maximization algo constructive? How to extend to general matroids and to general packing LPs? Questions?

Uniform Matroids Thm [BGSZ’18]: For k=Ω(log n) we get O(1) approximation.

Partition Matroids Thm [BGSZ’18]: For Partition matroids we get O loglog n 2 approx.