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

2. 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

3. 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 ≥ Pr t 1 > 1 2 ⋅Pr t 2 < =
𝑣 2
𝑣 3
𝑣 4
𝑡=0
𝑡=1
𝑡=1/2
Tight approx factor is 1/𝑒

4. 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 ⟹ 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)

5. 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

6. 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

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

8. 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?

9. 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)

10. 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

11. 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

12. 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?

13. 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 ⋆ ≥ 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.

14. 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 , 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.

15. 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.

16. 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

17. 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.

18. 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√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?

19. 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

20. 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

21. Wrapping Up: O log n 2 Approximation
Use log n checkpoints 𝑇 𝑖
Algorithm
Guess a random interval 𝑖
With prob 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

22. 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

23. 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.

24. 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?

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

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