Lower Bounds for Subexponential Parameterized Complexity of
Lower bounds for the parameterized complexity of Minimum Fill-in and other completion problems Michal Pilipczuk Institute of Informatics, University of Warsaw Joint work with Ivan Bliznets, Marek Cygan, Pawel Komosa and Luk´ aˇs Mach
Simons Institute, November 4th , 2015
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
1/16
Motivation GKKMPRR: O ? (k O(k
1/2 )
)
Split
⊂ ⊂
⊂
Threshold
Trivially perfect
DFPV: ?
O (k
O(k 1/2 )
⊂
DFPV: )
?
O (k
O(k 1/2 )
) ⊂
Proper interval
⊂
Interval BFPP: O ? (k O(k
1/2 )
Chordal FV:
)
O ? (k O(k
1/2 )
)
BFPP: O ? (k O(k
Bliznets, Cygan, Komosa, Mach, Pilipczuk
2/3 )
)
Completion problems: lower bounds
2/16
Motivation GKKMPRR: O ? (k O(k
1/2 )
)
Split
⊂ ⊂
⊂
Threshold
Trivially perfect
DFPV: ?
O (k
O(k 1/2 )
⊂
DFPV: )
?
O (k
O(k 1/2 )
) ⊂
Proper interval
⊂
Interval BFPP: O ? (k O(k
1/2 )
Chordal FV:
)
O ? (k O(k
1/2 )
)
BFPP: O ? (k O(k
˜
2/3 )
Is O? (2O(k Bliznets, Cygan, Komosa, Mach, Pilipczuk
)
1/2 )
) the correct answer?
Completion problems: lower bounds
2/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
k)
) lower
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
k)
) lower
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: Known reductions give O? (2o(n
Bliznets, Cygan, Komosa, Mach, Pilipczuk
1/6
)
) and O? (2o(k
Completion problems: lower bounds
1/9
)
k)
) lower
) lower bounds, or worse.
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 :
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : An O? (2o(k
1/2
)
) algorithm implies also 2o(n) .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum. Note: Ignore polylog factors.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum. Note: Ignore polylog factors. Personal opinion: log k in the exponent can be shaved off.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum. Note: Ignore polylog factors. Personal opinion: log k in the exponent can be shaved off.
Goal: Prove a 2o(n) lower bound for Minimum Fill-in.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
Results
Under ETH:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: no 2O(n
1/2
/ logc n)
algorithm for Minimum Fill-in;
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions: no 2o(n) algorithm for Minimum Fill-in;
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions: no 2o(n) algorithm for Minimum Fill-in; 1/2 consequently, no O? (2o(k ) ) FPT algorithm.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions: no 2o(n) algorithm for Minimum Fill-in; 1/2 consequently, no O? (2o(k ) ) FPT algorithm.
Same lower bounds for all the other completion problems.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Known reductions
3SAT
MaxCut
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA
Completion problems: lower bounds
Fill-in
5/16
Known reductions
3SAT
MaxCut
OLA
Fill-in
n0 = O(n + m) m0 = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
5/16
Known reductions
3SAT
MaxCut n0 = O(n + m) 0
m = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA
Fill-in
n00 = O((n0 )3 ) m00 = O((n0 )6 )
Completion problems: lower bounds
5/16
Known reductions
3SAT
MaxCut n0 = O(n + m) 0
m = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA n00 = O((n0 )3 ) m
00
Fill-in n000 = O(∆ · n00 )
0 6
= O((n ) )
Completion problems: lower bounds
5/16
Known reductions
3SAT
MaxCut n0 = O(n + m) 0
m = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA n00 = O((n0 )3 ) m
00
Fill-in n000 = O(∆ · n00 )
0 6
= O((n ) )
Completion problems: lower bounds
5/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )|
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )| P The cost of π is e∈E (G ) c(e).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )| P The cost of π is e∈E (G ) c(e). Optimum Linear Arrangement: Find π with minimum cost.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )| P The cost of π is e∈E (G ) c(e). Optimum Linear Arrangement: Find π with minimum cost. Note: It can be as large as cubic.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Reduction MaxCut
OLA
Complement the graph.
G K
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G K
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges. We want to maximize the number of non-edges flying over K .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges. We want to maximize the number of non-edges flying over K . Every edge flying over K has to gain more than the total noise on the sides.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges. We want to maximize the number of non-edges flying over K . Every edge flying over K has to gain more than the total noise on the sides. Hence K must be large to make this work.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Idea
Start with an instance that has a gap.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1. 2 Then set |K | = d β−α e · n.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1. 2 Then set |K | = d β−α e · n.
Gap in MaxCut
Gap of ≥ 2nm on edges flying over K .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1. 2 Then set |K | = d β−α e · n.
Gap in MaxCut
Gap of ≥ 2nm on edges flying over K .
Maximum noise is smaller than nm, so the gap amortizes the noise.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
c
m)
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
c
m)
Already observed and used by Marx in 2007 for proving lower bounds on the running times of geometric PTASes.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
c
m)
Already observed and used by Marx in 2007 for proving lower bounds on the running times of geometric PTASes. Standard reductions to MaxCut preserve the gap ⇒ c 2O(m/ log m) hardness of Gap MaxCut[α,β] .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
c
m)
Already observed and used by Marx in 2007 for proving lower bounds on the running times of geometric PTASes. Standard reductions to MaxCut preserve the gap ⇒ c 2O(m/ log m) hardness of Gap MaxCut[α,β] . Cor: Under ETH, there is no 2O(n/ log
Bliznets, Cygan, Komosa, Mach, Pilipczuk
c
n)
algorithm for OLA, for some c.
Completion problems: lower bounds
9/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
m
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
m
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
We obtained a 2O(n/ log
m
c
n)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
This proves 2
1/2
c
n)
c
/ log n)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs. lower bound for Minimum Fill-in.
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation: Plan 1: Show hardness of OLA on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation: Plan 1: Show hardness of OLA on bounded degree graphs. Plan 2: Find a better reduction from OLA to Minimum Fill-in.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation: Plan 1: Show hardness of OLA on bounded degree graphs. Plan 2: Find a better reduction from OLA to Minimum Fill-in.
Let’s look at the reduction OLA
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Minimum Fill-in first.
Completion problems: lower bounds
10/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). A
B
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph. Chain Completion Chordal/Interval/Proper Interval Completion: Make both A and B into cliques.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph. Chain Completion Chordal/Interval/Proper Interval Completion: Make both A and B into cliques. Chain Completion Threshold/Trivially Perfect Completion: Make A into a clique.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph. Chain Completion Chordal/Interval/Proper Interval Completion: Make both A and B into cliques. Chain Completion Threshold/Trivially Perfect Completion: Make A into a clique.
Cor: Suffices to get reduction OLA Bliznets, Cygan, Komosa, Mach, Pilipczuk
Chain Completion
Completion problems: lower bounds
11/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )|
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )| The sum of π(u) + π(v ) summands is constant if the input graph is regular.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )| The sum of π(u) + π(v ) summands is constant if the input graph is regular. Can be easily achieved by adding loops.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )| The sum of π(u) + π(v ) summands is constant if the input graph is regular. Can be easily achieved by adding loops. Ergo: Minimization of the number of fill edges is equivalent to minimization of the OLA cost.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs. Our approach:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs. Our approach: Introduce a new hypothesis about approximability of Minimum Bisection.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs. Our approach: Introduce a new hypothesis about approximability of Minimum Bisection. Prove that starting with this hypothesis we can make this plan work.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT. No 2o(n) lower bound on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT. No 2o(n) lower bound on bounded degree graphs.
Hypothesis There exist 0 ≤ α < β ≤ 1 and d ∈ N such that there is no 2o(n) -time algorithm for Gap MinBisection[α,β] on d-regular graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT. No 2o(n) lower bound on bounded degree graphs.
Hypothesis There exist 0 ≤ α < β ≤ 1 and d ∈ N such that there is no 2o(n) -time algorithm for Gap MinBisection[α,β] on d-regular graphs. Intuition: MinBisection on bounded degree graphs does not admit a subexponential-time approximation scheme.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Reduction MinBisection
OLA
First attempt:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees.
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner.
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner. Replace the full join with ‘balanced’ connections between expanders and G .
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner. Replace the full join with ‘balanced’ connections between expanders and G . The neighborhoods of G are ‘uniformly distributed’ in K .
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner. Replace the full join with ‘balanced’ connections between expanders and G . The neighborhoods of G are ‘uniformly distributed’ in K . Do the maths to make sure that the gap swallows the possible noise.
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
1/4
/ logc k)
) lower
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
1/4
/ logc k)
) lower
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis? Different route to hardness of Minimum Fill-in?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis? Different route to hardness of Minimum Fill-in? Tight lower bounds for Feedback Arc Set in Tournaments?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis? Different route to hardness of Minimum Fill-in? Tight lower bounds for Feedback Arc Set in Tournaments?
Thanks for your attention!
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Simons Institute, November 4th , 2015
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
1/16
Motivation GKKMPRR: O ? (k O(k
1/2 )
)
Split
⊂ ⊂
⊂
Threshold
Trivially perfect
DFPV: ?
O (k
O(k 1/2 )
⊂
DFPV: )
?
O (k
O(k 1/2 )
) ⊂
Proper interval
⊂
Interval BFPP: O ? (k O(k
1/2 )
Chordal FV:
)
O ? (k O(k
1/2 )
)
BFPP: O ? (k O(k
Bliznets, Cygan, Komosa, Mach, Pilipczuk
2/3 )
)
Completion problems: lower bounds
2/16
Motivation GKKMPRR: O ? (k O(k
1/2 )
)
Split
⊂ ⊂
⊂
Threshold
Trivially perfect
DFPV: ?
O (k
O(k 1/2 )
⊂
DFPV: )
?
O (k
O(k 1/2 )
) ⊂
Proper interval
⊂
Interval BFPP: O ? (k O(k
1/2 )
Chordal FV:
)
O ? (k O(k
1/2 )
)
BFPP: O ? (k O(k
˜
2/3 )
Is O? (2O(k Bliznets, Cygan, Komosa, Mach, Pilipczuk
)
1/2 )
) the correct answer?
Completion problems: lower bounds
2/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
k)
) lower
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
k)
) lower
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: Known reductions give O? (2o(n
Bliznets, Cygan, Komosa, Mach, Pilipczuk
1/6
)
) and O? (2o(k
Completion problems: lower bounds
1/9
)
k)
) lower
) lower bounds, or worse.
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 :
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : An O? (2o(k
1/2
)
) algorithm implies also 2o(n) .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum. Note: Ignore polylog factors.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum. Note: Ignore polylog factors. Personal opinion: log k in the exponent can be shaved off.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
The square root phenomenon √
Planar graphs: Known NP-hardness reduction usually give O? (2o( bounds under ETH. Completion problems: 1/6
k)
) lower
1/9
Known reductions give O? (2o(n ) ) and O? (2o(k ) ) lower bounds, or worse. Fomin & Villanger: Here the big gap between what we suspect and what we know is frustrating.
Arguments for the optimality of k 1/2 : 1/2
An O? (2o(k ) ) algorithm implies also 2o(n) . Fundamental trade-off between cheap and expensive vertices.
Our answer: We corroborate the suspicion that k 1/2 is optimum. Note: Ignore polylog factors. Personal opinion: log k in the exponent can be shaved off.
Goal: Prove a 2o(n) lower bound for Minimum Fill-in.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
3/16
Results
Under ETH:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: no 2O(n
1/2
/ logc n)
algorithm for Minimum Fill-in;
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions: no 2o(n) algorithm for Minimum Fill-in;
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions: no 2o(n) algorithm for Minimum Fill-in; 1/2 consequently, no O? (2o(k ) ) FPT algorithm.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Results
Under ETH: 1/2
c
no 2O(n / log n) algorithm for Minimum Fill-in; 1/4 c consequently, no O? (2O(k / log k) ) FPT algorithm.
Under stronger assumptions: no 2o(n) algorithm for Minimum Fill-in; 1/2 consequently, no O? (2o(k ) ) FPT algorithm.
Same lower bounds for all the other completion problems.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
4/16
Known reductions
3SAT
MaxCut
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA
Completion problems: lower bounds
Fill-in
5/16
Known reductions
3SAT
MaxCut
OLA
Fill-in
n0 = O(n + m) m0 = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
5/16
Known reductions
3SAT
MaxCut n0 = O(n + m) 0
m = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA
Fill-in
n00 = O((n0 )3 ) m00 = O((n0 )6 )
Completion problems: lower bounds
5/16
Known reductions
3SAT
MaxCut n0 = O(n + m) 0
m = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA n00 = O((n0 )3 ) m
00
Fill-in n000 = O(∆ · n00 )
0 6
= O((n ) )
Completion problems: lower bounds
5/16
Known reductions
3SAT
MaxCut n0 = O(n + m) 0
m = O(n + m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
OLA n00 = O((n0 )3 ) m
00
Fill-in n000 = O(∆ · n00 )
0 6
= O((n ) )
Completion problems: lower bounds
5/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )|
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )| P The cost of π is e∈E (G ) c(e).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )| P The cost of π is e∈E (G ) c(e). Optimum Linear Arrangement: Find π with minimum cost.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Optimum Linear Arrangement
Let π : V (G ) → {1, 2, . . . , n} be an ordering of V (G ). The cost of an edge uv is c(uv ) = |π(u) − π(v )| P The cost of π is e∈E (G ) c(e). Optimum Linear Arrangement: Find π with minimum cost. Note: It can be as large as cubic.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
6/16
Reduction MaxCut
OLA
Complement the graph.
G K
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G K
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges. We want to maximize the number of non-edges flying over K .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges. We want to maximize the number of non-edges flying over K . Every edge flying over K has to gain more than the total noise on the sides.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Reduction MaxCut
OLA
Complement the graph. Add a clique K of size N = nc for a large c, and make it fully adjacent to the rest of the graph. G A
K K
B
Easy: K can be assumed to be consecutive. Instead of minimizing the cost, maximize the cost of the non-edges. We want to maximize the number of non-edges flying over K . Every edge flying over K has to gain more than the total noise on the sides. Hence K must be large to make this work.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
7/16
Idea
Start with an instance that has a gap.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1. 2 Then set |K | = d β−α e · n.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1. 2 Then set |K | = d β−α e · n.
Gap in MaxCut
Gap of ≥ 2nm on edges flying over K .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Idea
Start with an instance that has a gap. Gap MaxCut[α,β] : distinguish between OPT ≤ αm and OPT ≥ βm. Suppose Gap MaxCut[α,β] is hard for some 0 ≤ α < β ≤ 1. 2 Then set |K | = d β−α e · n.
Gap in MaxCut
Gap of ≥ 2nm on edges flying over K .
Maximum noise is smaller than nm, so the gap amortizes the noise.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
8/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
c
m)
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
c
m)
Already observed and used by Marx in 2007 for proving lower bounds on the running times of geometric PTASes.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
c
m)
Already observed and used by Marx in 2007 for proving lower bounds on the running times of geometric PTASes. Standard reductions to MaxCut preserve the gap ⇒ c 2O(m/ log m) hardness of Gap MaxCut[α,β] .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
9/16
Hardness of Gap MaxCut Sparsification lemma ⇒ 2o(m) hardness of 3SAT Almost linear PCPs: Reduction from 3SAT to Gap 3SAT[r ,1] that increases the size by logc m. Corollary Under ETH, there exists constants r < 1 and c such that there is no 2O(m/ log algorithm for Gap 3SAT[r ,1] .
c
m)
Already observed and used by Marx in 2007 for proving lower bounds on the running times of geometric PTASes. Standard reductions to MaxCut preserve the gap ⇒ c 2O(m/ log m) hardness of Gap MaxCut[α,β] . Cor: Under ETH, there is no 2O(n/ log
Bliznets, Cygan, Komosa, Mach, Pilipczuk
c
n)
algorithm for OLA, for some c.
Completion problems: lower bounds
9/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
m
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
m
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
We obtained a 2O(n/ log
m
c
n)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
This proves 2
1/2
c
n)
c
/ log n)
Bliznets, Cygan, Komosa, Mach, Pilipczuk
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs. lower bound for Minimum Fill-in.
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation: Plan 1: Show hardness of OLA on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation: Plan 1: Show hardness of OLA on bounded degree graphs. Plan 2: Find a better reduction from OLA to Minimum Fill-in.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
10/16
Reductions
3SAT
Gap MaxCut
OLA
0
c
n
0
c
00
n = O(m log m) m = O(m log m)
m
We obtained a 2O(n/ log O(n
1/2
c
n)
00
0
= O(n )
Fill-in n
000
00
= O(∆ · n )
0 2
= O((n ) )
lower bound for OLA, but not on sparse graphs.
c
/ log n)
This proves 2 lower bound for Minimum Fill-in. Two routes to rescue the situation: Plan 1: Show hardness of OLA on bounded degree graphs. Plan 2: Find a better reduction from OLA to Minimum Fill-in.
Let’s look at the reduction OLA
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Minimum Fill-in first.
Completion problems: lower bounds
10/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B .
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). A
B
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph. Chain Completion Chordal/Interval/Proper Interval Completion: Make both A and B into cliques.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph. Chain Completion Chordal/Interval/Proper Interval Completion: Make both A and B into cliques. Chain Completion Threshold/Trivially Perfect Completion: Make A into a clique.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
11/16
Reduction of Yannakakis Def. A bipartite graph (A, B, E ) is a chain graph if {N(u)}u∈A form a chain in the inclusion order. Eq., this holds for {N(v )}v ∈B . Eq., there is an ordering π of A such that N(u) ⊇ N(v ) for π(u) ≤ π(v ). Eq., neighborhoods of vertices of B are downward closed w.r.t. π. A
B
Chain Completion: Add at most k edges to a given bipartite graph with a fixed bipartition to obtain a chain graph. Chain Completion Chordal/Interval/Proper Interval Completion: Make both A and B into cliques. Chain Completion Threshold/Trivially Perfect Completion: Make A into a clique.
Cor: Suffices to get reduction OLA Bliznets, Cygan, Komosa, Mach, Pilipczuk
Chain Completion
Completion problems: lower bounds
11/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )|
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )| The sum of π(u) + π(v ) summands is constant if the input graph is regular.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )| The sum of π(u) + π(v ) summands is constant if the input graph is regular. Can be easily achieved by adding loops.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Reduction of Yannakakis Set A = V (G ) and B = E (G ). Connect u ∈ A with e ∈ B iff u is an endpoint of e. Fix some ordering π : V (G ) → {1, 2, . . . , n}, and suppose this is the target ordering of neighborhoods. For every uv ∈ E (G ) = B, we need to add max(π(u), π(v )) − 2 edges. Crucial observation: 2 · max(π(u), π(v )) = (π(u) + π(v )) + |π(u) − π(v )| The sum of π(u) + π(v ) summands is constant if the input graph is regular. Can be easily achieved by adding loops. Ergo: Minimization of the number of fill edges is equivalent to minimization of the OLA cost.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
12/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs. Our approach:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs. Our approach: Introduce a new hypothesis about approximability of Minimum Bisection.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Issues
Wanted: 2o(n) hardness for OLA on bounded degree graphs. Route via MaxCut: We would need hardness of MaxCut on co-bounded degree graphs. Our approach: Introduce a new hypothesis about approximability of Minimum Bisection. Prove that starting with this hypothesis we can make this plan work.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
13/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ).
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT. No 2o(n) lower bound on bounded degree graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT. No 2o(n) lower bound on bounded degree graphs.
Hypothesis There exist 0 ≤ α < β ≤ 1 and d ∈ N such that there is no 2o(n) -time algorithm for Gap MinBisection[α,β] on d-regular graphs.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Hypothesis MinBisection: Partition the vertex set into equal halves so that the number of edges crossing the partition is minimized. What is known: NP-hard and APX-hard on bounded degree graphs. The best known approximation has apx factor O(log OPT ). The problem is FPT. No 2o(n) lower bound on bounded degree graphs.
Hypothesis There exist 0 ≤ α < β ≤ 1 and d ∈ N such that there is no 2o(n) -time algorithm for Gap MinBisection[α,β] on d-regular graphs. Intuition: MinBisection on bounded degree graphs does not admit a subexponential-time approximation scheme.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
14/16
Reduction MinBisection
OLA
First attempt:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees.
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner.
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner. Replace the full join with ‘balanced’ connections between expanders and G .
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner. Replace the full join with ‘balanced’ connections between expanders and G . The neighborhoods of G are ‘uniformly distributed’ in K .
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Reduction MinBisection
OLA
First attempt: Start with a hard instance G of Gap MinBisection. Replace K with a large constant-degree expander. What to do with the full join between K and the rest of the graph?
Second attempt: Replace K with a careful construction consisting of a constant-length chain of expanders of increasing degrees. The chain behaves in a rigid manner. Replace the full join with ‘balanced’ connections between expanders and G . The neighborhoods of G are ‘uniformly distributed’ in K . Do the maths to make sure that the gap swallows the possible noise.
G Bliznets, Cygan, Komosa, Mach, Pilipczuk
H1
H2
Completion problems: lower bounds
H3
HZ 15/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
1/4
/ logc k)
) lower
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
1/4
/ logc k)
) lower
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis.
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis? Different route to hardness of Minimum Fill-in?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis? Different route to hardness of Minimum Fill-in? Tight lower bounds for Feedback Arc Set in Tournaments?
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
Conclusions
We proved suboptimal lower bounds under ETH and almost tight lower bounds under the Hypothesis. Message: Look at approximation hardness for MinBisection first. 1/2
c
Approach can be also used to get 2O(n / log n) and O? (2O(k bounds for Feedback Arc Set in Tournaments. Open problems:
1/4
/ logc k)
) lower
Investigate the Hypothesis. Equivalence of OLA hardness and the Hypothesis? Different route to hardness of Minimum Fill-in? Tight lower bounds for Feedback Arc Set in Tournaments?
Thanks for your attention!
Bliznets, Cygan, Komosa, Mach, Pilipczuk
Completion problems: lower bounds
16/16
