Half-Integrality, LP-Branching,
and FPT Algorithms
“FPT Algorithms via Discrete Relaxation”
Yuichi Yoshida
NII
Joint work with Yoichi Iwata and Magnus Wahlström
Fixed Parameter Tractability
FPT (fixed parameter tractability)
• Introduce a parameter p (e.g., solution size)
• A problem is FPT if can be solved in f(p)poly(n) time.
Many NP-hard problems become FPT for appropriate params:
• Vertex cover, multiway cut, …
Many techniques for proving FPT:
• Branching, important separator, color coding, iterative
compression, half-integrality, …
This talk
Introduce a unified technique to get FPT algorithms:
Discrete relaxation
• Generalizes the branching method based on half-integrality
• Many applications
Vertex Cover
(above LP)
Odd Cycle
Transversal
Unique Label
Cover
Multiway Cut
(above LP)
Almost 2-SAT
(Group/Subset)
Feedback Vertex Set
• Bisubmodularity and k-submodularity play a crucial role.
Motivating example: Vertex Cover above LP
Vertex cover: vertex set incident to every edge.
VC above LP: Is there a vertex cover of size ≤ (LP value + p)?
0
½
1
1
½
½
1
1
1
1
0
0
0
0
OPT = 4
LP = 3.5
• Current best: O*(2.6181p) [Narayanaswamy et al’13]
Branching guided by LP
Let’s look at a simple O*(4p)-time algorithm.
min Σv∈V xv
s.t. xu + xv ≥ 1 (uv ∈ E)
0 ≤ xv ≤ 1 (v ∈ V)
[Nemhauser-Trotter’75]
The LP above satisfy
• Half-integrality: a {0, ½, 1}-valued LP solution exists.
• Persistency: For any LP solution x, one can fix integral
variables without changing the integer optimum.
Branching guided by LP
Solve LP
Fix integral values
Check other variables
can be fixed without
increasing the LP value.
½
1
½
½
½
½
1
1
1
1
0
0
0
0
LP value = 3.5
LP value = 4
⇒ cannot fix
Fixed to 1
Fixed to 0
Branching guided by LP
Fixed to 1
Pick a free variable v.
v
Branching
v
Fixed to 0
For each branching, the LP value increases by at least ½
⇒ The size of the search tree is 22p = 4p
⇒ O*(4p)-time algorithm.
Analysis
Can we similarly show FPT of other problems?
• Yes for (node) Multiway cut above LP. [Cygan et al.’13]
• Not sure for other problems because it is hard to prove halfintegrality and persistency of LP relaxations.
Let’s design relaxations that are inherently half-integral!
Discrete relaxation
Original function f: DV →ℤ+
Relaxed function f’: D’V→ℤ+/2 for D ⊆ D’.
D = integral part, D’ \ D = fractional part
We ask for f’ to satisfy:
• Consistency: f’(x) = f(x) for any x ∈ DV.
• Persistency: For any solution x’ to f’, one can fix integral
variables without changing the optimum of f.
• Tractability: f’ can be efficiently minimized
FPT Algorithms via discrete relaxation
• f
[Iwata-Wahlström-Y.’14]
f: DV→ℤ+ has a consistent, persistent, and tractable relaxation
f’:D’V→ℤ+/2
⇒ f can be minimized in O*(|D|2(min f – min f’)) time.
The algorithm and proof are the same as the LP-branching.
E.g.: VC above LP
• Set D’ = {0, ½, 1} and f’ = LP (restricted to half-integral
assignments).
• f’ is consistent, persistent, and tractable because so is the LP.
⇒ O*(22p) time
Discrete relaxation and submodularity
Is this useful? Isn’t it just a rephrasing of LP relaxation?
Many problems admit consistent discrete relaxations that are
Bisubmodular or k-submodular
⇒ Persistency and tractability automatically follow.
Submodularity
A function f: {0, 1}V → ℝ is submodular if, for any x, y ∈ {0, 1} V,
f(x) + f(y) ≥ f(x ∨ y) + f(x ∧ y),
where ∨ and ∧ = coordinate-wise max and min.
• Can be minimized in polynomial time. [Iwata et al.’01] [Schrijver’00]
Bisubmodularity
A function f: {0, ½, 1}V → ℝ is bisubmodular if
f(x) + f(y) ≥ f(x ∨ y) + f(x ∧ y)
for any x, y ∈ {0, ½, 1}V.
∨
0
1/
2
0
1
∧
0
1/ 1
0
2
0 0
1/
0 0 1/ 1/
2
2 2
½
1/ 0 1/ 1
1/ 1/ 1/ 1/
2
2
2 2 2 2
• Can 1be minimized
time1[Fujishige-Iwata’05]
1/ 1 1 in polynomial
1 1/ 1/
2
2 2
• Persistent when D = {0, 1} [Kolmogorov’12]
1
Vertex cover
Original problem (D = {0, 1})
min f(x) := Σv∈E xv + M × Σuv∈E max(0, 1 – xu – xv)
g(a, b) := max(0, 1 – a – b) is not submodular…
g(1, 0) + g(0, 1) < g(0, 0) + g(1, 1)
Vertex cover
Relaxed problem (D’ = {0, 1/2, 1})
min f’(x) := Σv∈E xv + M × Σuv∈E max(0, 1 – xu – xv)
g’(a, b) := max(0, 1 – a – b) is now bisubmodular!
g(1 ∨ 0, 0 ∨ 1) = g(1 ∧ 0, 0 ∧ 1) = g(1/2, 1/2) = 0
⇒ The relaxed function f’ is also bisubmodular.
⇒ VC is solvable in O*(4min f-min f’) time.
⇒ As min f’ = LP value, VC above LP is solvable in O*(4p) time.
Other applications: binary Boolean functions
Any binary function: {0, 1}2 → ℤ+ can be relaxed to a half-integral
bisubmodular function.
⇒ Any function f that is a sum of binary functions can be
minimized in O*(4min f) time.
– Odd Cycle Transversal
– Almost 2-SAT
Same time complexity as previous works, but the proof is
much simpler.
Other applications: Multiway Cut above LP
Multiway cut: edge set separating all the terminals T = {t1,…,tk}.
MC above LP: Is there a multiway cut of size ≤ (LP value + p)?
t2
t1
t3
O*(4p)-time FPT based on LP-branching [Cygan et al. 2012]
How can we prove this via discrete relaxation?
k-Submodularity
A function f: {0, 1, 2, …, k}V → ℝ is k-submodular if
f(x) + f(y) ≥ f(x ∨ y) + f(x ∧ y)
for any x, y ∈ {0, 1, 2, …, k}V.
∨
0
i
j
∧
0
i
j
0
0
i
j
0
0
0
0
i
i
i
0
i
0
i
0
j
j
0
j
j
0
0
j
1
…
2
k
0
• No poly-time algorithm known in the value oracle model.
– The sum of constant-arity k-submodular functions can be
minimized by LP. [Thapper-Živný‘13]
• Persistent when D = {1, 2, …, k} [Kolmogorov’12]
Multiway Cut and k-submodularity
Original problem (D = {1, …, k})
min f(x) := Σuv∈E [xu ≠ xv] s.t. xti = i.
t4
4
1
t1
t3
1
3
2
t2
Multiway Cut and k-submodularity
Relaxed problem (D’ = {0, 1, …, k})
min f’(x) := Σuv∈E [xu ≠ xv] s.t. xti = i.
1
t1
t4
4
½
t3
0
1
2
3
t2
Multiway Cut and k-submodularity
Relaxed problem (D’ = {0, 1, …, k})
min f’(x) := Σuv∈E [xu ≠ xv] s.t. xti = i.
f’ is k-submodular!
⇒ MC can be solved in O*(k2(min f-min f’)) time.
⇒ As min f’ = LP value, MC above LP is solvable in O*(k2p) time.
⇒ A slightly careful analysis gives O*(22p) = O*(4p) time.
Node Multiway Cut
Original problem (D = {1, …, k, C}, C = remove the vertex)
min f(x) := Σv∈V g(xv) + M × Σuv∈E h(xu, xv) s.t. xti = i
g
h
i
j
C
i
0
i
0
1
0
j
0
j
1
0
0
C
1
C
0
0
0
• No k-submodular relaxation…
Relaxation of Node Multiway Cut
Relaxed problem (D’ = {1, …, k, C, 0, 1’, …, k’})
min f’(x) := Σv∈V g’(xv) + M × Σuv∈E h’(xu, xv) s.t. xti = i
g'
h'
i
j
C
i'
j’
0
i
0
i
0
1
0
0
½
½
j
0
j
1
0
0
½
0
½
C
1
C
0
0
0
0
0
0
i’
½
i’
0
½
0
0
0
0
j’
½
j’
½
0
0
0
0
0
0
0
0
½ ½
0
0
0
0
Intuition
4
C
t4
C
4’
t3
t1
1’
1
0
3
3’
2’
C
C
2
t2
Relaxation of Node Multiway Cut
f’(x) + f’(y) ≥ f’(x ∨ y) + f’(x ∧ y) holds for any x, y ∈ D’V where
∨
i
j
C
i'
j’
0
∧
i
j
C
i'
j’
0
i
i
0
i’
i
i’
i
i
i
0
i’
i’
0
0
j
0
j
j’
j’
j
j
j
0
j
j’
0
j’
0
C
i’
j’
C C C C
C
i’
j’
C
i’
j’
0
i’
i
j’
C
i’
C
i’
i’
i’
0
i’
i’
0
0
j’
i’
j
C C
j’
j’
j’
0
j’
j’
0
j’
0
0
i
j
C
j’
0
0
0
0
0
0
0
0
i’
• Tractable because (∨, ∧) is a binary symmetric fractional
polymorphism [Thapper-Živný’13]
• Persistency easily follows
Conclusion
Discrete relaxation: a unified technique to get FPT algorithms.
• If the relaxation admits some nice fractional polymorphism,
then we automatically achieve FPT.
Other results
• Unique Label Cover: O*(k2p) time (k = domain size)
– Previous best: O*(kO(p^2log p)) [Chitnis et al.’12]
• Group FVS: O*(4p) time
– Works even when the group is exponentially large.
– Previous best: O*(2O(plog p)) [Cygan et al.’12])
• Linear-time (= O(cp(n+m))) FPT algorithms using max flow for
VC above LP, Almost 2SAT, and Unique Label Cover.
Future work
• Directed graph cut problems?
– Directed FVS, Directed Multiway Cut, …
⇒ Magnus’ talk
• Which VCSPs can be minimized on subdomain D⊆D’ in FPT time?
– Any function f: {1,…,k}V → ℤ can be extended to a ksubmodular function f’: {0,1,…,k}V → ℚ [Hirai-Iwamasa’15]
– But f’ can take a negative value even when f is non-negative…
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