1. Half-Integrality, LP-Branching, and FPT Algorithms “ FPT Algorithms via Discrete Relaxation” Yuichi Yoshida NII Joint work with Yoichi Iwata and Magnus Wahlström

2. 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, …

3. This talk Introduce a unified technique to get FPT algorithms: Discrete relaxation Generalizes the branching method based on half-integrality Many applications Bisubmodularity and k-submodularity play a crucial role. Odd Cycle Transversal Almost 2-SAT Vertex Cover (above LP) Multiway Cut (above LP) Unique Label Cover (Group/Subset) Feedback Vertex Set

4. 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)? Current best: O * (2.6181 p ) [Narayanaswamy et al’13] ½ ½ ½ ½ ½ ½ 1 1 1 1 0 0 0 0 LP = 3.5 0 0 1 1 1 1 1 1 1 1 0 0 0 0 OPT = 4

5. Branching guided by LP Let’s look at a simple O * (4 p )-time algorithm. [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. min Σ v ∈ V x v s.t. x u + x v ≥ 1 (uv ∈ E) 0 ≤ x v ≤ 1 (v ∈ V)

6. Branching guided by LP ½ ½½ 1 1 00 Solve LPFix integral values Fixed to 1 Fixed to 0 LP value = 3.5 Check other variables can be fixed without increasing the LP value. 1 1 ½½ 1 1 1 1 0 0 0 0 LP value = 4 ⇒ cannot fix

7. Branching guided by LP For each branching, the LP value increases by at least ½ ⇒ The size of the search tree is 2 2p = 4 p ⇒ O * (4 p )-time algorithm. Pick a free variable v. Fixed to 1 Fixed to 0 Branching v v

8. 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 half- integrality and persistency of LP relaxations. Let’s design relaxations that are inherently half-integral!

9. Discrete relaxation Original function f: D V → 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 ∈ D V. Persistency: For any solution x’ to f’, one can fix integral variables without changing the optimum of f. Tractability: f’ can be efficiently minimized

10. FPT Algorithms via discrete relaxation f 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 * (2 2p ) time [Iwata-Wahlström-Y.’14] f: D V → 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.

11. 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.

12. 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]

13. 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. Can be minimized in polynomial time [Fujishige-Iwata’05] Persistent when D = {0, 1} [Kolmogorov’12] 1 ½ 0 ∨ 01/ 2 1 000 0 1 1 11 ∧ 0 1 00 1 1

14. Vertex cover Original problem (D = {0, 1}) min f(x) := Σ v ∈ E x v + M × Σ uv ∈ E max(0, 1 – x u – x v ) g(a, b) := max(0, 1 – a – b) is not submodular… g(1, 0) + g(0, 1) < g(0, 0) + g(1, 1)

15. Vertex cover Relaxed problem (D’ = {0, 1/2, 1}) min f’(x) := Σ v ∈ E x v + M × Σ uv ∈ E max(0, 1 – x u – x v ) 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 * (4 min f-min f’ ) time. ⇒ As min f’ = LP value, VC above LP is solvable in O * (4 p ) time.

16. 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 * (4 min f ) time. – Odd Cycle Transversal – Almost 2-SAT Same time complexity as previous works, but the proof is much simpler.

17. Other applications: Multiway Cut above LP Multiway cut: edge set separating all the terminals T = {t 1,…,t k }. MC above LP: Is there a multiway cut of size ≤ (LP value + p)? O * (4 p )-time FPT based on LP-branching [Cygan et al. 2012] How can we prove this via discrete relaxation? t1t1 t1t1 t2t2 t2t2 t3t3 t3t3

18. 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. 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] 2 0 ∨ 0ij 00ij iii0 jj0j 1 ∧ 0ij 0000 i0i0 j00j k …

19. Multiway Cut and k-submodularity Original problem (D = {1, …, k}) min f(x) := Σ uv ∈ E [x u ≠ x v ] s.t. x ti = i. 1 1 4 4 t1t1 t1t1 t2t2 t2t2 t3t3 t3t3 t4t4 t4t4 1

20. Multiway Cut and k-submodularity Relaxed problem (D’ = {0, 1, …, k}) min f’(x) := Σ uv ∈ E [x u ≠ x v ] s.t. x ti = i. 1 1 4 4 0 0 t1t1 t1t1 t2t2 t2t2 t3t3 t3t3 t4t4 t4t4 1 ½

21. Multiway Cut and k-submodularity Relaxed problem (D’ = {0, 1, …, k}) min f’(x) := Σ uv ∈ E [x u ≠ x v ] s.t. x ti = i. f’ is k-submodular! ⇒ MC can be solved in O * (k 2(min f-min f’) ) time. ⇒ As min f’ = LP value, MC above LP is solvable in O * (k 2p ) time. ⇒ A slightly careful analysis gives O * (2 2p ) = O * (4 p ) time.

22. Node Multiway Cut Original problem (D = {1, …, k, C}, C = remove the vertex) min f(x) := Σ v ∈ V g(x v ) + M × Σ uv ∈ E h(x u, x v ) s.t. x ti = i No k-submodular relaxation… hijC i010 j100 C000 g i0 j0 C1

23. Relaxation of Node Multiway Cut Relaxed problem (D’ = {1, …, k, C, 0, 1’, …, k’}) min f’(x) := Σ v ∈ V g’(x v ) + M × Σ uv ∈ E h’(x u, x v ) s.t. x ti = i h'ijCi'j’0 i0100½½ j100½0½ C000000 i’0½0000 j’½00000 0½½0000 g' i0 j0 C1 i’½ j’½ 00

24. Intuition 1 1 4 4 C C C C C C C C 2’ 4’ 3’ 1’ t1t1 t1t1 t4t4 t4t4 t3t3 t3t3 t2t2 t2t2

25. Relaxation of Node Multiway Cut f’(x) + f’(y) ≥ f’(x ∨ y) + f’(x ∧ y) holds for any x, y ∈ D’ V where Tractable because ( ∨, ∧ ) is a binary symmetric fractional polymorphism [Thapper-Živný’13] Persistency easily follows ∨ ijCi'j’0 ii0i’i i j0jj’ jj Ci’j’CCCC i’ij’Ci’C j’i’jCCj’ 0ijCi’j’0 ∧ ijCi'j’0 ii0i’ 00 j0jj’0 0 Ci’j’Ci’j’0 i’ 0 00 j’0 0 0 0000000

26. 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 * (k 2p ) time (k = domain size) – Previous best: O * (k O(p^2log p) ) [Chitnis et al.’12] Group FVS: O * (4 p ) time – Works even when the group is exponentially large. – Previous best: O * (2 O(plog p) ) [Cygan et al.’12]) Linear-time (= O(c p (n+m))) FPT algorithms using max flow for VC above LP, Almost 2SAT, and Unique Label Cover.

27. 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 k- submodular function f’: {0,1,…,k} V → ℚ [Hirai-Iwamasa’15] – But f’ can take a negative value even when f is non-negative… 27

28. Multiway Cut and k-submodularity Relaxed problem (D’ = {0, 1, …, |T|}) min f(x) := Σ uv ∈ E [x u ≠ x v ] s.t. x ti = i. [a ≠ b] is |T|-submodular! ⇒ The relaxed function f’ is also |T|-submodular. ⇒ MC can be solved in O * (4 min f-min f’ )-time. ≠12…|T|0 10111/2 2101 … |T|1101/2 0 0