Estimating the longest increasing sequence in polylogarithmic time

1
Estimating the longest increasing sequence in
polylogarithmic time

• C. Seshadhri (Sandia National Labs)
• Joint work with Michael Saks
• (Rutgers University)

Sandia National Laboratories is a multi-program
laboratory managed and operated by Sandia
Corporation, a wholly owned subsidiary of
Lockheed Martin Corporation, for the U.S.
Department of Energy's National Nuclear Security
Administration under contract DE-AC04-94AL85000
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The problem
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24
10
9
15
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20
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4
10
15
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• Given array fn ? N, find (length of)
• Longest Increasing Subsequence (LIS)
• Rather self-explanatory
• By now, textbook dynamic programming problem
• CLRS 01 Chapter 15.4 (Longest Common
  Subsequence), Starred Problem 15.4-6
• Schensted 61, Fredman 75 O(n log n) algorithm

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Too much to read
LIS is in range 0.4n, 0.6n
Algorithm
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7
4
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• Array f is extremely large, so cant read all of
  it
• What can we say about LIS length, if we see very
  little?
• LIS LIS length
• Read only poly(log n) positions
• Obviously randomized

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Uniform sample says nothing
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10
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4
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• Choose uniform random sample of poly(log n) size
• LIS n/2, but random sample always increasing
• So not really that easy to learn about LIS

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Our result
LIS in this range
Algorithm
1
n
LIS

• We want range to be small

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Our result
LIS in this range
dn
Algorithm
1
n
LIS

• We want range to be small
• This work For any (constant) d gt 0
• Algorithm gives additive dn approximation to
  LIS
• Running time is (1/d)1/d(log n)c

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Our result
Ad alert!
dn
1
n
n/2
LIS

• We want range to be small
• This work For any (constant) d gt 0
• Algorithm gives additive dn approximation to
  LIS
• Running time is (1/d)1/d(log n)c
• Ailon Chazelle Liu S 03 Parnas Ron Rubinfeld
  03
• Previous best d ½

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Our result
Ad alert!
dn
1
n
n/2
LIS

• We want range to be small
• This work For any (constant) d gt 0
• Algorithm gives additive dn approximation to
  LIS
• Running time is (1/d)1/d(log n)c
• We get (1 d)-approx to distance to monotonicity
• Previously best was factor 2

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Prelims the array in space
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Prelims the array in space
Violation
Increasing sequence

• Input is points in plane, given as array
• (LIS is longest chain in partial order)

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A hard example
k
k
10 points in each
k
k
LIS 4k
LIS 2k
3k
k
k
3k

• The decision for a point depends on small scale
  properties of far away portions

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A hard example
k
k
k
k
3k
k
k
3k

• Random samples in neighborhoods of points are
  identical!
• Can we really estimate LIS in polylog time?
• Is it time for some heavy work?
• I mean, time for lbs (lower bounds).

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Outline (or lack thereof)

• Will I show proofs?
• No
• Will I show the algorithm?
• Maybe
• I will try to demonstrate the main insight
• By a series of thought experiments

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The dynamic program
Closest LIS point to left
Splitter
n/2

• Closest LIS point to left gives splitter
• Find LIS is each blue region. Piece together!
• So we break up original problem into subproblems

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The dynamic program
S
n/2

• But we dont know right splitter.
• So try all possible! Only n different choices
• Choose the one that gives the largest sum of
  LISs
• MaxS (LIS-below-S LIS-above-S)

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The dynamic program
n/2

• If you LIS in all small boxes, you can build LIS
  for bigger boxes
• Not the most efficient DP
• So our sublinear algo will mimic this process

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The IP
Is this point on LIS?
LIS is in blue region
Splitter
n/2
Where is the splitter?
It is there.
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The IP
This point NOT on LIS
LIS is in blue region
n/2
Where is the splitter?
It is there.
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The IP
n/2
3n/4
I wish we knew the splitter in that region
It is there.
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The IP
n/2
3n/4
5n/8
I think I know what will happen next
Youre lucky Im here
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The IP
n/2
3n/4
5n/8
I think I know what will happen next
Youre lucky Im here
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The IP
n/2
3n/4
5n/8
I think I know what will happen next
Youre lucky Im here
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The interactive protocol

• If point stays in blue region till very end, then
  it is good (on LIS). Otherwise, bad.
• This takes (log n) steps, with the help of the
  wizard

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The interactive protocol

• If point stays in blue region till very end, then
  it is good (on LIS). Otherwise, bad.
• This takes (log n) steps, with the help of the
  wizard
• If we could simulate the wizard

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The interactive protocol

• If point stays in blue region till very end, then
  it is good (on LIS). Otherwise, bad.
• This takes (log n) steps, with the help of the
  wizard
• If we could simulate the wizard

What?? If you could simulate the wizard, you know
the LIS!
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Find a splitter
If very few LIS points outside blue, this is not
a bad splitter
n/2

• Finding splitter may be hard, so try for
  approximate versions?
• But how do we determine the number of LIS points?

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Find a splitter
Total no. of points outside bluelt µn
Conservative splitter
n/2

• If µ lt 1/(100 log n), being against health care
  conservative is good enough

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Easy to check
n/2

• Count fraction of sample outside blue
• poly(log n) samples checks this accurately

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Getting a conservative splitter
n/2

• We can sample (log n) different candidates and
  check which of them disbelieves evolution is
  conservative
• What if no conservative splitter exists?

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A liberal paradise
Choose any line
No. of points outside at least µn
n/2

• So we know that LIS lt (1-µ) n
• Leads to the next idea. Boosting approximations!
• Given d-approx to LIS, can we get improve to d?

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Boosting approximations
Run dn-approx on points in box
No. of points outside at least µn
n2
Run dn-approx on points in box
Real splitter
n1
n/2

• Take sum of outputs as total LIS estimate
• LIS LIS1 LIS2, Est Est1 Est2
• Est1 LIS1 lt dn1 Est2 LIS2 lt dn2
• So Est LIS lt d(n1 n2)
• n1n2 lt (1-µ)n, so Est LIS lt d(1-µ)n !

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Putting it together
Conservative splitter?
n/2

• Check if each is conservative splitter
• If it is, were found right subproblems
• Otherwise

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Putting it together
Run dn-approx on points in box
S
Run dn-approx on points in box
n/2

• One of these is close enough to real splitter
• Est(S) Left-Est(S) Right-Est(S)

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Putting it together
Run dn-approx on points in box
S
Run dn-approx on points in box
n/2

• One of these is close enough to real splitter
• Est(S) Left-Est(S) Right-Est(S)
• Final Estimate maxS Est(S)
• Looks like a great idea!
• We go from dn to d(1- µ)n. Recur to keep
  improving approximation

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It fails, miserably
Alg
d0 d1(1-µ)
Alg
Alg
d1
1/µ
Alg
Alg
Alg
Alg
d2
½
Alg
Alg
Alg
Alg

• As we go up each level, approx gets better by
  (1-µ).
• So to get d0 ¼, how many levels needed?
• ¼ ½ (1-µ)t So t 1/µ
• We have running time at least 21/µ.
• So, µ needs to be gt 1/log log n.

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Find a splitter
Total no. of points outside bluelt µn
Conservative splitter
n/2

• If µ lt 1/(100 log n), being against health care
  conservative is good enough

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The basic dichotomy
Continue IP
P
We find splitter
Cannot find splitter
The Interactive Protocol phase
The Dynamic Programming phase

• For IP, we need µ lt 1/log n
• µn is error in each level of IP
• For DP, we need µ gt 1/log log n
• (1-µ) is decrease in approximation

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The basic dichotomy
Strengthen
Continue IP
Weaken
P
We find splitter
Cannot find splitter
The Interactive Protocol phase
The Dynamic Programming phase

• For IP, we need µ lt 1/log n
• µn is error in each level of IP
• For DP, we need µ gt 1/log log n
• (1-µ) is decrease in approximation

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Reducing to smaller DP!
n/(log n)
Run d-approx to get LIS estimate inside box
n/(log n)

• Run d-approx on all poly(log n) such boxes

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Reducing to smaller DP!
n/(log n)

• Run d-approx on all poly(log n) such boxes
• Use Dynamic Program to find chain with largest
  sum of estimates
• Longest path in DAG
• Can solve in poly(log n) time

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Dichotomy theorem
OR
One can go from d-approx to (d-d2)-approx by a
(log n) sized DP
Either it is easy to find the right subproblems
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The algorithm, in one slide
Continue IP
P
We find splitter
Cannot find splitter
Make poly(log n) calls to d-approx. Solve DP of
poly(log n) size.

• Overall running time becomes (log n)1/d
• miracle that the math works out

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The even better version

• Dont exactly solve this dynamic program!
• Use our sublinear algo to approximately solve in
  (loglog n) time. Then do it recursively
• Its painful
• Its all Greek a ß ? d e ? ? µ ?
• We had ?, but got rid of it

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What next?

• Sublinear dynamic programming!
• We get (1/d)1/d (log n)c time. Can we get
• (log n)/d time?
• Would be extremely cool. Completely optimal
• Applications for other dynamic programs?
• How does one find the right subproblems in
  sublinear time?
• Generalize the dichotomy
• Longest common subsequence/edit distance?

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Ask and you shall know