2. OMv Lower Bounds · Chapter 2. OMv Lower Bounds Danupon Nanongkai KTH, Sweden ADFOCS 2018 1 Last...
Transcript of 2. OMv Lower Bounds · Chapter 2. OMv Lower Bounds Danupon Nanongkai KTH, Sweden ADFOCS 2018 1 Last...
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Chapter 2.
OMv Lower Bounds
Danupon NanongkaiKTH, Sweden
1ADFOCS 2018Last edited: Aug. 15, 2018
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THE CONJECTURESPart 1
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Input: Boolean matrix MThen: Boolean vectors
Output:
Conjecture: No algorithms with total time
…
𝒗𝟏 𝒗𝟐 𝒗𝒏
𝑴𝒗𝟏 𝑴𝒗𝟐
…
𝑴𝒗𝒏
OMv Conjecture(Online Matrix‐Vector Multiplication) [Henzinger, Krinninger, N, Saranurak, STOC’15]
Current Best: √ [Larsen‐Williams SODA’17]
(OR,AND)‐mult.not , ‐mult.
Answer 𝑀𝑣 beforegetting 𝑣
Example on board?
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OuMv Conjecture (Matrix Form)
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Input: Boolean matrix MThen: pairs of Boolean vectors 𝒊 𝒊
Output: Answer u 𝑀𝑣 beforegetting u , 𝑣
Conjecture: No algorithms with total time even with polynomial time to process M!
Example on board?
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Input:
Output:
Preprocess:
(𝐿 , 𝑅…(𝐿 , 𝑅
Any edge linking 𝐿 and 𝑅 ?… Any edge linking 𝐿 and 𝑅 ?
𝑝𝑜𝑙𝑦 𝑛 time
e.g.
yes NoOuMv ConjNo n time
Write on boardWrite on board
OuMv as Independent set
𝐿 ∪ 𝑅 is independent set?
Output before next input arrives
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‐OuMv Conjecture (or just a “free‐form” of OuMv)
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Input: Boolean matrix M, 𝟐𝜸, .
Then: pairs of Boolean vectors 𝒊 𝒊
Output:
Conjecture: No algorithms with total time even with polynomial time to process M!
Example on board?
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Formal Statements
OMv Conjecture: For any constant , there is no ‐time algorithm that solves OMv with an error probability of at most 1/3.
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‐OuMv Conjecture: For any constant , there is no algorithm for ‐OuMv with parameters
using preprocessing time and computational time that has error probability of at most 1/3.
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Plan• Some lower bounds from OuMv
• Prove above Theorem
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Theorem: OMv implies ‐OuMv
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Part 2
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Some Update Time Bounds
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Example set 1: Fully‐Dynamic Graphs
After each edge insertions/deletion check:1. st‐reachability2. undirected st‐shortest paths
– Unweighted/weighted3. strong edge‐connectivity
These bounds hold against amortization & randomization!Main reason: ‐OuMv allows arbitrary (polynomial) preprocessing time and number of updates.
13𝑛 = # of nodes, 𝑚=# of edges What lower bound can you prove under OMv?
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Example 1.1
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st‐Reachability
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Dynamic st‐Reachability Problem
Input:Update in G
insert(1,3) delete(3,t) insert(2,t)
Picture
Output:s reach t?
No Yes No Yes
1
t2
3
s
1
t2
3
s
1
t2
3
s
1
t2
3
s
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Thanks Thatchaphol Saranurak for slides
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Known Results for st‐Reach
• Incremental: O(1) amortized update time•𝛀 𝒏 lower bound assuming OuMv
• Hold against randomized and amortized algorithms• … even with oblivious‐adversary & empty‐start assumptions• Higher lower bound for a related problem called #SSR• Ω 𝑛 lower bound for “combinatorial” algorithms
• Fully‐dynamic: 𝚯 𝒏𝟏.𝟒𝟎𝟕 worst‐case update time• Lower bound assumes a variant of OuMv
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𝑛 = # of nodes, 𝑚=# of edges
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st‐Reach‐ Preprocess: ‐ Update: (amortized)
Independent Set‐ Preprocess: ‐ Time (for 𝑛 queries):
Will show…
𝑛 = # of nodes, 𝑚=# of edgesThanks Thatchaphol Saranurak for slides
Impossible!assuming OMv
So this cannot exist
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Independent Set st‐Reach
s t
Same graph but directed
Preprocess
Thanks Thatchaphol Saranurak for slides
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s t
After O(n) updates…an edge linking and s can reach t
st‐Reach
From 𝑹𝟏To 𝑳𝟏
Edge( )?
Thanks Thatchaphol Saranurak for slides
Independent Set
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s t
st‐Reach
From 𝑹𝟏To 𝑳𝟏
Edge( )?
After knowing “Edge( )?”, UNDO.
Thanks Thatchaphol Saranurak for slides
Independent Set
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s t
Use O(n) updates.
st‐Reach
From 𝑹𝟏To 𝑳𝟏
Edge( )?
Thanks Thatchaphol Saranurak for slides
Independent Set
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ts
Edge( )? (another example)
After O(n) updates…Not an edge linking
and s can not reach t
From 𝑹𝟐
To 𝑳𝟐
st‐Reach
Thanks Thatchaphol Saranurak for slides
Independent Set
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Check: The lower bound hold for amortized update time?
• Suppose that an algorithm 𝑨 for st‐reach takes 𝑶 𝒏𝟎.𝟗𝒕 time after 𝒕updates, when start from an empty graph.
• Setting up the original bipartite graph: Take 𝑶 𝒏𝟐.𝟗 time to insert 𝒏𝟐edges.
• Handling one pair of 𝑢 , 𝑣 : Take 𝑶 𝒏𝟏.𝟗 time to insert 𝒏 edges.
Take 𝑶 𝒏𝟐.𝟗 time to handle all pairs of vectors
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Check: The lower bound hold against randomized algorithms?
• The conjecture was also for randomized algorithms. • The reduction is between decision problems. There is no difference between oblivious and non‐oblivious adversary. • Must be more careful for, e.g. approximation algorithms.
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Example 1.2
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st‐Distance(Undirected)
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Input:Update in G insert(1,3) delete(3,t) insert(1,t)
Picture
Output:st‐distance 3 2
1
t2
3
s
1
t2
3
s
1
t2
3
s
1
t2
3
s
• Easy: lower bound for exact version• How about approximate version?
Dynamic st‐Distance Problem
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for unweighted (5/3‐ )‐approximation
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𝑛 = # of nodes, 𝑚=# of edges
uMv = 1 dist(s,t) 3
uMv = 0 dist(s,t) 5
Same reduction as st‐ReachabilityOutput number 𝑥 s.t.
𝑑𝑖𝑠𝑡 𝑠, 𝑡 𝑥53 𝜖 𝑑𝑖𝑠𝑡 𝑠, 𝑡
Algorithm’s output 𝜖 3 5
Algorithm’s output 5
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for weighted (3‐ )‐approximation
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𝑛 = # of nodes, 𝑚=# of edges
uMv = 1 dist(s,t) 1
uMv = 0 dist(s,t) 3
Same reduction as st‐Reachability
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Known Results
Fully‐dynamic•𝛀 𝒏 lower bound assuming OuMv for (5/3‐𝜖)‐approx
• Hold against randomized and amortized algorithms• … even with oblivious‐adversary & empty‐start assumptions• Hold against (small‐)approximation algorithms
•𝐎 𝒏𝟏.𝟕𝟐𝟒 worst‐case update time for 𝟏 𝝐 ‐approx
Incremental/decremental: • Exact: Θ(n) amortized update time, Θ 𝑚 worst‐case• 𝟏 𝝐 ‐approx: 𝑂 𝑛 amortized
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𝑛 = # of nodes, 𝑚=# of edges
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Example 1.3
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Strong Edge‐Connectivity
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Dynamic Strong Edge‐Connectivity Problem
Input Update Output
A directed graph Edge insertions/deletions Is the graph strongly connected?(Every s can reach every t)
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for strong edge‐connectivity
• Reduce from st‐Reachability by adding• edges 𝐸 from t to every node, and• edges 𝐸 from every node to s.
• Observe: Adding edges pointing to s and from t does not change st‐reachability.
• If t is not reachable from s, this remains the case. • If t is reachable from s, then
• s can reach all nodes via 𝐸 , and• all nodes can reach s via 𝐸
• Easy: Extend to Ω 𝑚 lower bound
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Example set 2: Non‐Graph Problems
1. Erickson’s Problem2. Pagh’s Problem
These bounds hold against amortization & randomization!
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𝑛 = # of nodes, 𝑚=# of edges What lower bound can you prove under OMv?
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Example 2.1
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Erickson’s problem
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Erickson’s problem
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Reduction
1 1
1
1 1
1
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Example 2.2
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Pagh’s problem
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• Input: 𝑘 subsets 𝑋 , 𝑋 ,… , 𝑋 over a universe 𝑈 1,… , 𝑘• Update: Given a pointer to two subsets 𝑋 and 𝑋 , create a new subset 𝑋 ∩ 𝑋
• Output: After each update outputs whether the new subset is empty or not.
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Pagh’s problem (a variant)
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Pagh’s problem ‐‐ Reduction
1 1
1
0 0
0
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Pagh’s problem ‐‐ Reduction
1 1
1
0 0
0
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Pagh’s problem ‐‐ Reduction
1 1
1
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1 1
1
0 0
0
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Questions?
This project has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme under grant agreement No 715672
Thanks to co‐authors: Sayan Bhattacharya, Jan van den Brand, Deeparnab Chakraborty, Sebastian Forster, Monika Henzinger, Christian Wulff‐Nilsen, Thatchaphol Saranurak
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