SODA07 - Microsoft Research
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Transcript SODA07 - Microsoft Research
Estimating the Sortedness of a
Data Stream
Parikshit Gopalan
U T Austin
T. S. Jayram
IBM Almaden
Robert Krauthgamer
IBM Almaden
Ravi Kumar
Yahoo! Research
Data Stream Model of Computation
Input
X1 X2 X 3
…
Storage
• Computing with Massive data sets.
• Sequential access.
•Small storage space, update time.
[Alon-Matias-Szegedy, …]
Xn
Sorting on Data-Streams
Cannot sort efficiently.
Can we tell if the data needs to be sorted?
[Ergun-Kannan-Kumar-Rubinfeld-Vishwanathan,
Ajtai-Jayram-Kumar-Sivakumar, Gupta-Zane,
Cormode-Muthukrishnan-Sahinalp, LibenNowell-Vee-Zhu,
Ailon-Chazelle-Commandur-Liu]
Sorting on Data-Streams
•
Cannot sort efficiently on a data-stream.
•
Can we tell if the data needs to be sorted?
[Ergun-Kannan-Kumar-Rubinfeld-Vishwanathan,
Ajtai-Jayram-Kumar-Sivakumar, Gupta-Zane,
Cormode-Muthukrishnan-Sahinalp, LibenNowell-Vee-Zhu,
Ailon-Chazelle-Commandur-Liu]
•
Measuring distance from Sortedness:
Kendall Tau distance
Spearman Footrule distance
Ulam distance
Candidate metrics
1. Spearman’s footrule [ℓ1 distance] :
3 5 7 9 10 4 1 2 6 8
e
1 2 3 4 5 6 7 8 9 10
Easy to compute.
Candidate metrics
2. Kendall Tau distance [No. of Inversions]
Inversions: Positions i < j where (i) > (j)
3 5 7 9 10 4 1 2 6 8
Candidate metrics
2. Kendall Tau distance [No. of Inversions]
Inversions: Positions i < j where (i) > (j)
3 5 7 9 10 4 1 2 6 8
Candidate metrics
2. Kendall Tau distance [No. of Inversions]
Within a factor-2 of Spearman’s footrule. [DiaconisGraham]
An O(log n) space, 1-pass (1 + ) algorithm. [AjtaiJayram-Kumar-Sivakumar]
Candidate metrics
3. Ulam distance [Edit Distance]:
Ed(): Number of deletions needed to sort.
Ulam: Fastest way to sort a bridge hand.
Edit Distance and the LIS
Ed(): Number of deletions needed to sort.
5 7 8 1 10 4 2 3 6 9
Edit Distance and the LIS
Ed(): Number of deletions needed to sort.
5 7 8 1 10 4 2 3 6 9
Delete
5 7 8 10
Insert
1 2 3 4 5 6 7 8 9 10
Edit Distance and the LIS
Ed() : Number of deletions needed to sort .
LIS() : Length of the longest increasing sequence.
Ed() + LIS() = n
Studied in statistics, biology, computer science …
Both take a global view of the sequence.
Hard for models like streaming, sketching,
property-testing.
151 … 190
51 … 80
81 … 100
Prior Work
• Exact Computation of Ed() and LIS() :
– Patience Sorting [Ross,Mallows]
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
5
7
8
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
1
5
7
8
10
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
1
4
5
7
8
10
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
2
1
4
5
7
8
10
Number in place i: Earliest end to IS of length i.
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
2
1
4
3
5
7
8
10
Number in place i: Earliest end to IS of length i.
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
2
1
4
3
6
5
7
8
10
9
Number in place i: Earliest end to IS of length i.
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
0
2
1
4
3
6
5
7
8
10
9
Number in place i: Earliest end to IS of length i.
Patience Sorting
5 7 8 1 10 4 2 3 6 9 0
LIS
0
2
1
4
3
6
5
7
8
10
Length of LIS
9
Prior Work
• Exact Computation of Ed() and LIS() :
– Patience Sorting [Ross,Mallows]
– O(n) space, 1-pass streaming algorithm.
– (√n) space lower bound. [LibenNowell-Vee-Zhu]
• Approximating Ed() and LIS() :
– No sub-linear space algorithms, no lower bounds.
[Ajtai et al, Cormode et al, LibenNowell et al]
• LIS Algorithms parametrized by length of LIS :
[LibenNowell-Vee-Zhu, Sun-Woodruff]
• Computing Ed() in other models:
– Property Testing [Ergun et al, Ailon et al]
– Sketching [Charikar-Krauthgamer]
Our Results
• Approximating Ed() :
– An O(log2 n) space, randomized 4-approximation for Ed().
– A O(√n) space, deterministic (1 + ε)-approximation for Ed().
• Approximating the LIS:
– A O(√n) space, deterministic (1 + ε)-approximation for LIS().
• Exact Computation of Ed() and LIS():
– An (n) space lower bound for randomized algorithms.
– Independently proved by [Sun-Woodruff].
• Lower bounds for approximating the LIS:
– Conjecture: Deterministic algorithms require (√n) space for
(1 + ε)-approximation
Computing the Edit Distance
Thm: For any ε > 0,there is a one-pass randomized
algorithm using O(ε-2log2 n) space and update time, that
gives a (4 + ε) approximation to Ed().
1.
Combinatorial measure that approximates Ulam
distance. Builds on [Ergun et al, Ailon et al].
2.
Sampling scheme to compute this measure in one pass.
A Voting Scheme [Ergun et al.]
Combinatorial measure called Unpopularity.
Neighborhoods of (i) : Intervals starting or ending at i.
1
3
7
8
6
5
9
2
A Voting Scheme [Ergun et al.]
Combinatorial measure called Unpopularity.
Neighborhoods of (i) : Intervals starting or ending at i.
Deciding if (i) is unpopular:
For every neighborhood of (i)
Every number in the neighborhood votes on “Is (i)
out of order?”
If majority in some neighborhood vote against (i), it is
marked unpopular.
Let U() denote no. of unpopular numbers.
[Ergun et al]:
Ed() ≤ U()
[Ailon et al]:
U() ≤ 2 Ed()
A Voting Scheme [Ergun et al.]
Can we estimate U() using a streaming algorithm?
4 5 3 7 1 2
A Voting Scheme [Ergun et al.]
Can we estimate U() using a streaming algorithm?
4 5 3 7 1 2
Impossible to decide if (i) is unpopular before
seeing the entire input.
A New Voting Scheme
• Neighborhoods of (i) : Intervals ending at i.
• If majority in some neighborhood vote against (i), it is
marked unpopular.
• Unpopularity based only on past, not the future.
Thm: Let V() denote no. of unpopular numbers. Then
Ed()/2 ≤ V() ≤ 2 Ed()
A Voting Scheme
• Let Ed() = k. Then V() ≤ 2k.
• Fix an optimal Bad set of size k to delete.
How many numbers can be Unpopular ?
Partition Unpopular into Good and Bad.
Good numbers form an increasing sequence.
Good never votes against Good.
Good + Unpopular ≡ Bad neighborhood !
A Voting Scheme
• Let Ed() = k. Then V() ≤ 2k.
• Fix an optimal Bad set of size k to delete.
Good + Unpopular ≡ Bad neighborhood !
If k numbers are Bad,
At most k are Good + Unpopular.
Bad numbers might all be Unpopular.
Hence V() ≤ 2k.
A Voting Scheme
• Let Ed() = k. Then V() ≤ 2k.
• Bound can be tight.
100 99 98 … 91 1 2 3 … 10 11 12 …
90
100 99 98 … 91 1 2 3 … 10 11 12 …
90
100 99 98 … 91 1 2 3 … 10 11 12 …
90
A Voting Scheme
• Let V() = k. Then Ed() ≤ 2k.
• Fix the set of k Unpopular elements.
Algorithm to produce an increasing sequence:
1. Scan right to left.
2. Delete Unpopular elements + Inversions w.r.t last
number in sequence.
At least half of deletions are Unpopular numbers.
What remains is an increasing sequence.
A Voting Scheme
• Let V() = k. Then Ed() ≤ 2k.
• Bound can be tight.
11 … 50 91 92 93 … 100 1 2 3 … 10 51 … 90
11 … 50 91 92 93 … 100 1 2 3 … 10 51 … 90
11 … 50 91 92 93 … 100 1 2 3 … 10 51 … 90
A New Voting Scheme
• Neighborhoods of (i) : Intervals ending at i.
• If majority in some neighborhood vote against (i), it is
marked unpopular.
• Unpopularity based only on past, not the future.
Thm: Let V() denote no. of unpopular numbers. Then
Ed()/2 ≤ V() ≤ 2 Ed()
Can we estimate V() efficiently?
Outline of Sampling Scheme
Taking a vote in one neighborhood:
– Take O(log n) samples, take the (approx)
majority.
Reservoir Sampling [Vitter].
1
3
7
8
6
5
9
2
Computing V() :
Need O(log n) samples from every neighborhood.
1
3
7
8
6
5
9
2
Outline of Sampling Scheme
Computing V() :
Need O(log n) samples from every neighborhood.
1
3
7
8
6
5
9
2
Key observation: Don’t need samples across intervals
to be independent!
Roughly O(log2 n) samples suffice.
Deterministic Algorithm for LIS
Thm: For any ε > 0,there is a one-pass deterministic
algorithm using O(n/ε)1/2 space and update time, that
gives a (1 - ε) approximation to LIS().
Based on multiplayer communication protocol for LIS:
10 51 … 19
32 … 80
15 … 50
• Algorithm simulates protocol for √n players.
Two-Player Protocol for LIS
1000 5123
… 1319
n/2
Patience
Sorting
6 24 … 1000
k
Multiples
of εk
6…1000
1/ε
3245 4582 …
8021
Approximating the LIS
Consider k-player communication protocol for LIS:
10 51 … 19
32 … 80
15 … 50
• As k increases, maximum message size increases.
Conjecture: For some ε0 > 0, every 1-pass deterministic
algorithm that gives a (1 + ε0) approximation to LIS()
requires (√n) space.
Proving the conjecture requires analyzing k ≥ √n
Lower Bounds for approximating the LIS
Conjecture: For some ε0 > 0, every 1-pass deterministic
algorithm that gives a (1 + ε0) approximation to LIS()
requires (√n) space.
Candidate Hard Instances?
1.8
2.9
3.7 4.9
1.6
2.8 3.5 4.6
1.3
2.5
3.3 4.5
1
2
3.2 4.2
Lower Bounds for approximating the LIS
Conjecture: For some ε0 > 0, every 1-pass deterministic
algorithm that gives a (1 + ε0) approximation to LIS()
requires (√n) space.
Candidate Hard Instances?
Yes
No
1.8
2.9 3.7 4.9
1.7
2.8 3.4 4.8
1.6
2.8 3.5 4.6
1.6
2.6 3.5 4.6
1.3
2.5 3.3 4.5
1.3
2.5 3.6 4.5
1.1
2.1
1
2
3.2 4.2
3.9 4.2
Lower Bounds for approximating the LIS
Conjecture: For some ε0 > 0, every 1-pass deterministic
algorithm that gives a (1 + ε0) approximation to LIS()
requires (√n) space.
Candidate Hard Instances?
Yes
No
1.8 2.9 3.7 4.9
1.7
2.8 3.4 4.8
1.6
2.8 3.5 4.6
1.6
2.6 3.5 4.6
1.3
2.5 3.3 4.5
1.3
2.5 3.6 4.5
1.1
2.1
1
2
3.2 4.2
3.9 4.2
Lower Bounds for approximating the LIS
Conjecture: For some ε0 > 0, every 1-pass deterministic
algorithm that gives a (1 + ε0) approximation to LIS()
requires (√n) space.
Candidate Hard Instances?
Yes
No
1.8 2.9 3.7 4.9
1.7 2.8 3.4 4.8
1.6
2.8 3.5 4.6
1.6
2.6 3.5 4.6
1.3
2.5 3.3 4.5
1.3
2.5 3.6 4.5
1.1
2.1 3.9 4.2
1
2
3.2 4.2
Open Problems
Estimate the Edit distance between two permutations.
Tight bounds for approximation:
Show (√n) lower bound for deterministic algorithms.
Randomized algorithm for LIS ?