overview of the talk - mit csail · overview of the talk •hardness for frechet distance...
TRANSCRIPT
![Page 1: Overview of the talk - MIT CSAIL · Overview of the talk •Hardness for Frechet distance •Hardness for edit distance •Hardness for LCS 1 . Coordinate gadgets •If ai·bi=1 then](https://reader033.vdocuments.mx/reader033/viewer/2022053005/5f096b827e708231d426c06b/html5/thumbnails/1.jpg)
Overview of the talk
• Hardness for Frechet distance
• Hardness for edit distance
• Hardness for LCS
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Coordinate gadgets
• If ai·bi=1 then Frechet ≥ 1.01
• If ai·bi=0 then Frechet ≤ 1
• ai=1
• ai=0
• bi=0
• bi=1
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Frechet ≤ 1 Frechet ≥ 1.01
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Vector gadgets
• ai ∈ A → αi |αi|≤O(d)
• bj ∈ B → βj |βj|≤O(d)
• If ai and bj are orthogonal, then Frechet(αi,βj) ≤ 1
• If ai and bj are NOT orthogonal, then Frechet(αi,βj)≥1.01
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Vector gadgets
• a=1001 →
• b=0010 →
• Number of segments in any vector gadget: d 4
Odd coordinates Even coordinates
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Vectors are orthogonal → Frechet ≤ 1
• a=1001 →
• b=0010 →
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Vectors are not orthogonal → Frechet ≥ 1.01
• a=1001 →
• b=0011 →
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Frechet ≥ 1.01 Vectors are not orthogonal
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Construction of the final curves
• A = { 1110, 0110, 1001, 1010 }
• B = { 1011, 0010, 1111, 1100 }
• One pair of orthogonal vectors
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1110 0110 1001 1010
1011 0010 1111 1100
Construction of the final curves
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1110 0110 1001 1010
0010 1111 1100 1011
Construction of the final curves
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Reminder of our goal
• A⊆{0,1}d → curve x, |x|≤O(n·d)
• B⊆{0,1}d → curve y, |y|≤O(n·d)
• Frechet(x,y) ≤ 1, if exists a ∈ A, b ∈ B with Σiaibi =0
• Frechet(x,y) ≥ 1.01, otherwise
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
2. Traverse the second curve until the Blue vector
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
2. Traverse the second curve until the Blue vector
3. Traverse the Blue vectors in parallel
4. Traverse the second curve until end
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
2. Traverse the second curve until the Blue vector
3. Traverse the Blue vectors in parallel
4. Traverse the second curve until end
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
2. Traverse the second curve until the Blue vector
3. Traverse the Blue vectors in parallel
4. Traverse the second curve until end
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1110 0110 1001 1010
1011 0010 1111 1100
Exist orthogonal vectors → Frechet(x,y) ≤ 1
1. Traverse the first curve until the Blue vector
2. Traverse the second curve until the Blue vector
3. Traverse the Blue vectors in parallel
4. Traverse the second curve until end
5. Traverse the first curve until end
Questions?
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Reminder of our goal
• A⊆{0,1}d → curve x, |x|≤O(n·d)
• B⊆{0,1}d → curve y, |y|≤O(n·d)
• Frechet(x,y) ≤ 1, if exist a ∈ A, b ∈ B with Σiaibi =0
• Frechet(x,y) ≥ 1.01, otherwise
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No orthogonal vectors → Frechet(x,y) ≥ 1.01
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Possible positions
Frechet(x,y) < 1.01 → exist orthogonal vectors
Contraposition
Questions?
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Hardness for Frechet distance
• Theorem. If Frechet distance can be 1.001-approximated in m2-Ω(1) time (m upper-bounds the length of curves), then Orthogonal Vectors Conjecture is false
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Overview of the talk
• Hardness for Frechet distance
• Hardness for edit distance
• Hardness for LCS
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Pattern matching
• Pattern matching with respect to edit distance
• pattern(x,y), |x|≤|y|=m
• pattern(x,y)=miny’ edit(x,y’)
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x
y
y’
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High level idea
• A⊆{0,1}d → sequence x, |x|≤n·dO(1)
• B⊆{0,1}d → sequence y, |y|≤n·dO(1)
• pattern(x,y)=small, if exists a ∈ A, b ∈ B with Σiaibi=0
• pattern(x,y)=large, otherwise
• the construction time is n·dO(1)
• Theorem. If pattern matching can be computed in m2-Ω(1) time (m upper-bounds the length of sequences), then Orthogonal Vectors Conjecture is false
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Hardness for pattern matching distance
• First part: assume vector gadgets and prove hardness for pattern matching
• Second part: show vector gadget construction
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Vector gadgets
• ai ∈ A → αi |αi|≤dO(1)
• bj ∈ B → βj |βj|≤dO(1)
• If ai and bj are orthogonal, then edit(αi,βj)=S
• If ai and bj are NOT orthogonal, then edit(αi,βj)=L, S<L
• Crucial that L does not depend on vectors
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Construction of final sequences
• t – vector gadget for vector consisting of 1s
α1 $ α2 $ α3 … αn-1 $ αn
t $ t $ t … t $ t $ β1 $ β2 $ β3 … βn-1 $ βn $ t $ t $ t … t $ t
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n-1 times n-1 times
dO(1)
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Pattern matching score
• YES case: exists a ∈ A, exists b ∈ B such that Σiaibi=0
assume that a3 and b1 are orthogonal
α1 α2 α3 … αn-1 αn
t t t … t t β1 β2 β3 … βn-1 βn t t t … t t
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Score: S +(S or L) +(S or L) +(S or L) …+(S or L) +(S or L) ≤S+(n-1)·L
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Construction of final sequences
α1 $ α2 $ α3 … αn-1 $ αn
t $ t $ t … t $ t $ β1 $ β2 $ β3 … βn-1 $ βn $ t $ t $ t … t $ t
α1 $ α2 $ α3
γ1 $ γ2 $ γ3 28
n=3:
n-1 times n-1 times
Claim 1: Optimal subsequence from the lower sequence contains n vector gadgets
Claim 2: vector gadgets are aligned one by one between the upper sequence and the subsequence
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Pattern matching score
• NO case: for all a ∈ A, b ∈ B we have Σiaibi>0
α1 α2 α3 … αn-1 αn
t t t … t t β1 β2 β3 … βn-1 βn t t t … t t • Pattern matching score=n·L
• We crucially use that L does not depend on Σiaibi
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(>S+(n-1)·L)
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Hardness for edit distance
• First part: assume vector gadgets and prove hardness for pattern matching
• Second part: show vector gadget construction
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Vector gadgets
• a ∈ A → α |α|≤dO(1)
• b ∈ B → β |β|≤dO(1)
• If Σiaibi=0 then edit(α,β)=S
• If Σiaibi>0 then edit(α,β)=L, S<L
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Coordinate gadgets
• If ai·bi=1 then edit=large
• If ai·bi=0 then edit=small
• ai=1 0 0 0 1
• ai=0 0 1 1 1
• bi=0 0 0 1 1
• bi=1 1 1 1 1
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CG2(bi)
CG1(ai)
edit=1 edit=3
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Vector gadgets: first attempt
• a=a1a2…ad → CG1(a1) 22…2 CG1(a2) 22…2……22…2 CG1(ad)
• b=b1b2…bd → CG2(b1) 22…2 CG2(b2) 22…2……22…2 CG2(bd)
• edit(α’,β’)=d+2·(Σiaibi)
• edit(α’,β’) depends on the inner product
• Pattern matching for Hamming distance is easy
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β’=
α’= dO(1)
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Vector gadgets: final construction
• α’= CG1(a1) 22…2 CG1(a2) 22…2……22…2 CG1(ad)
• β’= CG2(b1) 22…2 CG2(b2) 22…2……22…2 CG2(bd)
• α’’= CG1(0) 52…2 CG1(0) 22…2……22…2 CG1(0)
• edit(α’,β’)=d+2·(Σiaibi)
• edit(α,β)≈min(d+2·(Σiaibi),d+1)
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444444444444444444 333333333333333333 444444444444444444
333333333333333333 333333333333333333
α’ α’’
β’
α=
β=
dO(1)
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Properties of vector gadgets
• a=a1a2…ad → α’ = CG1(a1) 22…2 CG1(a2) 22…2……22…2 CG1(ad)
• b=b1b2…bd → β’ = CG2(b1) 22…2 CG2(b2) 22…2……22…2 CG2(bd)
• α’’= CG1(0) 52…2 CG1(0) 22…2……22…2 CG1(0)
• edit(α’’, β’)=edit( CG1(0) 52…2 CG1(0) 22…2……22…2 CG1(0) CG2(b1) 22…2 CG2(b2) 22…2……22…2 CG2(bd) )=d+1
• edit(α’, β’)=d if Σiaibi=0
• edit(α’, β’)=d+2Σiaibi≥d+2 if Σia
ibi>0
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Vector gadgets: final construction
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444444444444444444 333333333333333333 444444444444444444
333333333333333333 333333333333333333
α’ α’’
β’
α=
β=
=min ( +delete
+replace 3333333333 with 4444444444
edit(α,β)
edit(α’,β’)
+delete 4444444444
α’’ ,
+delete
+replace 3333333333 with 4444444444
edit(α’’,β’)
+delete 4444444444
α’ ) -C
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Vector gadgets: final construction
• edit(α,β)=min(edit(α’,β’), edit(α’’,β’)) +C
• edit(α’, β’)=d if the vectors are orthogonal
• edit(α’, β’)≥d+2 if the vectors are NOT orthogonal
• edit(α’’, β’)=d+1 always
• edit(α,β)=d +C if the vectors are orthogonal
• edit(α,β)=d+1 +C if the vectors are NOT orthogonal
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Hardness for edit distance
• Theorem. If edit distance can be computed in m2-Ω(1) time (m upper-bounds the length of sequences), then Orthogonal Vectors Conjecture is false
• Gap is too small to obtain interesting approximation hardness result
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Overview of the talk
• Hardness for Frechet distance
• Hardness for edit distance
• Hardness for LCS
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Hardness for LCS
• Hardness for weighted LCS
• Reduction to unweighted LCS: write symbols in unary
• Alphabet 1,2,3,4,5,6,7
• 1=Weight(1)=Weight(2)
• 1<<Weight(3)<<Weight(4)<<Weight(5)<<Weight(6)=Weight(7)
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Coordinate gadgets
• If ai·bi=1 then LCS=small
• If ai·bi=0 then LCS=large
• ai=1 3 1 3
• ai=0 3 1 2 3
• bi=0 3 2 1 3
• bi=1 3 2 3
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CG2(bi)
CG1(ai)
LCS=large LCS=small
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Vector gadgets
• a=a1a2…ad → α=4 CG1(a1) CG1(a2) ... CG1(ad)
• b=b1b2…bd → β= CG2(b1) CG2(b2) ... CG2(bd) 4
• LCS(α,β)=large if orthogonal
• LCS(α,β)=small if NOT orthogonal
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Construction of final sequences
• t – vector gadget for vector consisting of 1s
5555555555555555555 6 α1 7 6 α2 7 … 6 αn-1 7 6 αn 7 5555555555555555555
5 6 t 7 5 6 t … t 7 5 6 β1 7 5 6 β2 … βn-1 7 5 6 βn 7 5 6 t 7 5 6 … t 7 5 6 t 7 5
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n-1 times n-1 times
Pair of orthogonal vectors → LCS is large
No pair of orthogonal vectors → LCS is small
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Hardness for LCS
• Theorem. If LCS can be computed in m2-Ω(1) time (m upper-bounds the length of sequences), then Orthogonal Vectors Conjecture is false
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Conclusion
• Quadratic hardness for Frechet, edit, LCS
• Quadratic hardness for DTWD (reduce edit to DTWD)
• Open problems: hardness of approximations for edit, LCS
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Thank you!