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Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Simultaneously Satisfying Linear Equations Over F 2 : Parameterized Above Average Anders Yeo anders@cs.rhul.ac.uk Department of Computer Science Royal

  1. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat The above problem was solved by Mahajan and Raman, who gave a linear kernel. It is still relatively easy due to the following: Reduce an instance by removing any two clauses of the form ( x ) and (¯ x ). Repeatadly doing this creates an instance of 2-satisfiable-SAT and does not change the problem. However ˆ φ m becomes a tight lower bound on the number of √ satisfied clausses, where ˆ φ = ( 5 − 1) / 2 ≈ 0 . 618. Therefore there is a kernel. Proof : If k < (ˆ φ − 1 2 ) m answer YES . Otherwise m ≤ k / (ˆ φ − 1 2 ). This gives rise to a new problem..... Anders Yeo Max-Lin Parameterized Above Average

  2. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat The above problem was solved by Mahajan and Raman, who gave a linear kernel. It is still relatively easy due to the following: Reduce an instance by removing any two clauses of the form ( x ) and (¯ x ). Repeatadly doing this creates an instance of 2-satisfiable-SAT and does not change the problem. However ˆ φ m becomes a tight lower bound on the number of √ satisfied clausses, where ˆ φ = ( 5 − 1) / 2 ≈ 0 . 618. Therefore there is a kernel. Proof : If k < (ˆ φ − 1 2 ) m answer YES . Otherwise m ≤ k / (ˆ φ − 1 2 ). This gives rise to a new problem..... Anders Yeo Max-Lin Parameterized Above Average

  3. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat The above problem was solved by Mahajan and Raman, who gave a linear kernel. It is still relatively easy due to the following: Reduce an instance by removing any two clauses of the form ( x ) and (¯ x ). Repeatadly doing this creates an instance of 2-satisfiable-SAT and does not change the problem. However ˆ φ m becomes a tight lower bound on the number of √ satisfied clausses, where ˆ φ = ( 5 − 1) / 2 ≈ 0 . 618. Therefore there is a kernel. Proof : If k < (ˆ φ − 1 2 ) m answer YES . Otherwise m ≤ k / (ˆ φ − 1 2 ). This gives rise to a new problem..... Anders Yeo Max-Lin Parameterized Above Average

  4. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat The above problem was solved by Mahajan and Raman, who gave a linear kernel. It is still relatively easy due to the following: Reduce an instance by removing any two clauses of the form ( x ) and (¯ x ). Repeatadly doing this creates an instance of 2-satisfiable-SAT and does not change the problem. However ˆ φ m becomes a tight lower bound on the number of √ satisfied clausses, where ˆ φ = ( 5 − 1) / 2 ≈ 0 . 618. Therefore there is a kernel. Proof : If k < (ˆ φ − 1 2 ) m answer YES . Otherwise m ≤ k / (ˆ φ − 1 2 ). This gives rise to a new problem..... Anders Yeo Max-Lin Parameterized Above Average

  5. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat Max- 2 -satisfiable-SAT parameterized above ˆ φ m Instance: A CNF formula F with n variables, m clauses and any two clauses can be simultaniously satisfied. Parameter: k . Question: Can we satisfy ≥ ˆ φ m + k clauses? The above problem was shown to have a kernel with at most O ( k ) variables, by Crowston, Gutin, Jones and AY. This approach does not seem to be eaily extendable. Anders Yeo Max-Lin Parameterized Above Average

  6. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat Max- 2 -satisfiable-SAT parameterized above ˆ φ m Instance: A CNF formula F with n variables, m clauses and any two clauses can be simultaniously satisfied. Parameter: k . Question: Can we satisfy ≥ ˆ φ m + k clauses? The above problem was shown to have a kernel with at most O ( k ) variables, by Crowston, Gutin, Jones and AY. This approach does not seem to be eaily extendable. Anders Yeo Max-Lin Parameterized Above Average

  7. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: Max-Sat Max- 2 -satisfiable-SAT parameterized above ˆ φ m Instance: A CNF formula F with n variables, m clauses and any two clauses can be simultaniously satisfied. Parameter: k . Question: Can we satisfy ≥ ˆ φ m + k clauses? The above problem was shown to have a kernel with at most O ( k ) variables, by Crowston, Gutin, Jones and AY. This approach does not seem to be eaily extendable. Anders Yeo Max-Lin Parameterized Above Average

  8. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: In general In problems parameterized above (below) tight bounds , we take a maximization (minimization) problem with a tight lower (upper) bound, and ask if we can get k above (below) this bound. Ensures the parameter is small in interesting cases. First introduced in a paper by Mahajan and Raman published in 1999. We say “above average” when the tight lower bound is the expectation of a random assignment. Anders Yeo Max-Lin Parameterized Above Average

  9. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: In general In problems parameterized above (below) tight bounds , we take a maximization (minimization) problem with a tight lower (upper) bound, and ask if we can get k above (below) this bound. Ensures the parameter is small in interesting cases. First introduced in a paper by Mahajan and Raman published in 1999. We say “above average” when the tight lower bound is the expectation of a random assignment. Anders Yeo Max-Lin Parameterized Above Average

  10. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: In general In problems parameterized above (below) tight bounds , we take a maximization (minimization) problem with a tight lower (upper) bound, and ask if we can get k above (below) this bound. Ensures the parameter is small in interesting cases. First introduced in a paper by Mahajan and Raman published in 1999. We say “above average” when the tight lower bound is the expectation of a random assignment. Anders Yeo Max-Lin Parameterized Above Average

  11. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Parameterized Above Tight Bounds: In general In problems parameterized above (below) tight bounds , we take a maximization (minimization) problem with a tight lower (upper) bound, and ask if we can get k above (below) this bound. Ensures the parameter is small in interesting cases. First introduced in a paper by Mahajan and Raman published in 1999. We say “above average” when the tight lower bound is the expectation of a random assignment. Anders Yeo Max-Lin Parameterized Above Average

  12. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Outline Parameterizing above tight bounds: Example Max-Sat 1 Max-Lin-AA 2 FPT Results 3 Related Results 4 Anders Yeo Max-Lin Parameterized Above Average

  13. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Max-Lin problem: given a system I of m linear equations in n variables over F 2 . F 2 is the Galois field with 2 elements (1 + 1 = 0). Each equation is assigned a positive integer weight. We wish to find an assignment of values to the variables in order to maximize the total weight of satisfied equations. z 1 = 1 z 1 + z 2 = 0 z 2 + z 3 = 1 z 1 + z 2 + z 3 = 1 Known bound: can satisfy at least W / 2, where W = total weight of equations. Anders Yeo Max-Lin Parameterized Above Average

  14. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Max-Lin problem: given a system I of m linear equations in n variables over F 2 . F 2 is the Galois field with 2 elements (1 + 1 = 0). Each equation is assigned a positive integer weight. We wish to find an assignment of values to the variables in order to maximize the total weight of satisfied equations. z 1 = 1 z 1 + z 2 = 0 z 2 + z 3 = 1 z 1 + z 2 + z 3 = 1 Known bound: can satisfy at least W / 2, where W = total weight of equations. Anders Yeo Max-Lin Parameterized Above Average

  15. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Max-Lin problem: given a system I of m linear equations in n variables over F 2 . F 2 is the Galois field with 2 elements (1 + 1 = 0). Each equation is assigned a positive integer weight. We wish to find an assignment of values to the variables in order to maximize the total weight of satisfied equations. z 1 = 1 z 1 + z 2 = 0 z 2 + z 3 = 1 z 1 + z 2 + z 3 = 1 Known bound: can satisfy at least W / 2, where W = total weight of equations. Anders Yeo Max-Lin Parameterized Above Average

  16. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Max-Lin problem: given a system I of m linear equations in n variables over F 2 . F 2 is the Galois field with 2 elements (1 + 1 = 0). Each equation is assigned a positive integer weight. We wish to find an assignment of values to the variables in order to maximize the total weight of satisfied equations. z 1 = 1 z 1 + z 2 = 0 z 2 + z 3 = 1 z 1 + z 2 + z 3 = 1 Known bound: can satisfy at least W / 2, where W = total weight of equations. Anders Yeo Max-Lin Parameterized Above Average

  17. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Max-Lin problem: given a system I of m linear equations in n variables over F 2 . F 2 is the Galois field with 2 elements (1 + 1 = 0). Each equation is assigned a positive integer weight. We wish to find an assignment of values to the variables in order to maximize the total weight of satisfied equations. z 1 = 1 z 1 + z 2 = 0 z 2 + z 3 = 1 z 1 + z 2 + z 3 = 1 Known bound: can satisfy at least W / 2, where W = total weight of equations. Anders Yeo Max-Lin Parameterized Above Average

  18. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Tightness of W / 2 bound W / 2 is a tight lower bound on max( I ). e.g. z 1 = 0 z 1 = 1 z 2 + z 3 = 0 z 2 + z 3 = 1 . . . Theorem (H˚ astad, 2001) For any ǫ > 0 , it is impossible to decide in polynomial time between instances of Max-Lin in which max( I ) ≤ (1 / 2 + ǫ ) m, and instances in which max( I ) ≥ (1 − ǫ ) m, unless P = NP. Anders Yeo Max-Lin Parameterized Above Average

  19. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Tightness of W / 2 bound W / 2 is a tight lower bound on max( I ). e.g. z 1 = 0 z 1 = 1 z 2 + z 3 = 0 z 2 + z 3 = 1 . . . Theorem (H˚ astad, 2001) For any ǫ > 0 , it is impossible to decide in polynomial time between instances of Max-Lin in which max( I ) ≤ (1 / 2 + ǫ ) m, and instances in which max( I ) ≥ (1 − ǫ ) m, unless P = NP. Anders Yeo Max-Lin Parameterized Above Average

  20. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Let max ( I ) denote the maximum possible weight of satisfied equations in I . Max-Lin above Average ( Max-Lin-AA ) Instance: A system I of m linear equations in n variables over F 2 , with total weight W . Parameter: k . Question: Is max( I ) ≥ W / 2 + k ? Mahajan, Raman & Sikdar (2006) asked if Max-Lin-AA is FPT. Anders Yeo Max-Lin Parameterized Above Average

  21. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Max-Lin Above Average Let max ( I ) denote the maximum possible weight of satisfied equations in I . Max-Lin above Average ( Max-Lin-AA ) Instance: A system I of m linear equations in n variables over F 2 , with total weight W . Parameter: k . Question: Is max( I ) ≥ W / 2 + k ? Mahajan, Raman & Sikdar (2006) asked if Max-Lin-AA is FPT. Anders Yeo Max-Lin Parameterized Above Average

  22. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Motivation Max- r -Lin is equivalent to Max-Lin except all equations have at most r variables. Max-Lin and Max- r -Lin are important problems, for many reasons.... H˚ astad said they were as basic as satisfiability. They are important tools for constraint satisfaction problems (such as MaxSat or Max- r -Sat ). So Max-Lin and Max- r -Lin have attracted significant interest in algorithmics. A number of papers made progress on Max- r -Lin-AA before Max-Lin-AA was shown to be FPT. Anders Yeo Max-Lin Parameterized Above Average

  23. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Motivation Max- r -Lin is equivalent to Max-Lin except all equations have at most r variables. Max-Lin and Max- r -Lin are important problems, for many reasons.... H˚ astad said they were as basic as satisfiability. They are important tools for constraint satisfaction problems (such as MaxSat or Max- r -Sat ). So Max-Lin and Max- r -Lin have attracted significant interest in algorithmics. A number of papers made progress on Max- r -Lin-AA before Max-Lin-AA was shown to be FPT. Anders Yeo Max-Lin Parameterized Above Average

  24. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Motivation Max- r -Lin is equivalent to Max-Lin except all equations have at most r variables. Max-Lin and Max- r -Lin are important problems, for many reasons.... H˚ astad said they were as basic as satisfiability. They are important tools for constraint satisfaction problems (such as MaxSat or Max- r -Sat ). So Max-Lin and Max- r -Lin have attracted significant interest in algorithmics. A number of papers made progress on Max- r -Lin-AA before Max-Lin-AA was shown to be FPT. Anders Yeo Max-Lin Parameterized Above Average

  25. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Motivation Max- r -Lin is equivalent to Max-Lin except all equations have at most r variables. Max-Lin and Max- r -Lin are important problems, for many reasons.... H˚ astad said they were as basic as satisfiability. They are important tools for constraint satisfaction problems (such as MaxSat or Max- r -Sat ). So Max-Lin and Max- r -Lin have attracted significant interest in algorithmics. A number of papers made progress on Max- r -Lin-AA before Max-Lin-AA was shown to be FPT. Anders Yeo Max-Lin Parameterized Above Average

  26. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Motivation Max- r -Lin is equivalent to Max-Lin except all equations have at most r variables. Max-Lin and Max- r -Lin are important problems, for many reasons.... H˚ astad said they were as basic as satisfiability. They are important tools for constraint satisfaction problems (such as MaxSat or Max- r -Sat ). So Max-Lin and Max- r -Lin have attracted significant interest in algorithmics. A number of papers made progress on Max- r -Lin-AA before Max-Lin-AA was shown to be FPT. Anders Yeo Max-Lin Parameterized Above Average

  27. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Outline Parameterizing above tight bounds: Example Max-Sat 1 Max-Lin-AA 2 FPT Results 3 Related Results 4 Anders Yeo Max-Lin Parameterized Above Average

  28. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Overview Notation Reduction Rules Main Results Proof of the Main Results Anders Yeo Max-Lin Parameterized Above Average

  29. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Notation For a given assignment, the excess = total weight of satisfied equations − total weight of falsified equations. Max-Lin-AA is equivalent to asking if the max excess is at least 2 k . Example: z 1 = 1 z 2 = 1 z 1 + z 2 = 1 Anders Yeo Max-Lin Parameterized Above Average

  30. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Notation For a given assignment, the excess = total weight of satisfied equations − total weight of falsified equations. Max-Lin-AA is equivalent to asking if the max excess is at least 2 k . Example: z 1 = 1 z 2 = 1 z 1 + z 2 = 1 Anders Yeo Max-Lin Parameterized Above Average

  31. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (LHS rule) Suppose we have two equations, � i ∈ S z i = b 1 (weight w 1 ) and � i ∈ S z i = b 2 (weight w 2 ), where w 1 ≥ w 2 . If b 1 = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 + w 2 ). If b 1 � = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 − w 2 ). z 1 + z 2 = 1 ( w = 1) ⇒ z 1 + z 2 = 1 ( w = 3) z 1 + z 2 = 1 ( w = 2) z 2 + z 3 + z 4 = 0 ( w = 3) ⇒ z 2 + z 3 + z 4 = 0 ( w = 1) z 2 + z 3 + z 4 = 1 ( w = 2) Allows us to assume no two equations have the same left-hand side. Anders Yeo Max-Lin Parameterized Above Average

  32. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (LHS rule) Suppose we have two equations, � i ∈ S z i = b 1 (weight w 1 ) and � i ∈ S z i = b 2 (weight w 2 ), where w 1 ≥ w 2 . If b 1 = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 + w 2 ). If b 1 � = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 − w 2 ). z 1 + z 2 = 1 ( w = 1) ⇒ z 1 + z 2 = 1 ( w = 3) z 1 + z 2 = 1 ( w = 2) z 2 + z 3 + z 4 = 0 ( w = 3) ⇒ z 2 + z 3 + z 4 = 0 ( w = 1) z 2 + z 3 + z 4 = 1 ( w = 2) Allows us to assume no two equations have the same left-hand side. Anders Yeo Max-Lin Parameterized Above Average

  33. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (LHS rule) Suppose we have two equations, � i ∈ S z i = b 1 (weight w 1 ) and � i ∈ S z i = b 2 (weight w 2 ), where w 1 ≥ w 2 . If b 1 = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 + w 2 ). If b 1 � = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 − w 2 ). z 1 + z 2 = 1 ( w = 1) ⇒ z 1 + z 2 = 1 ( w = 3) z 1 + z 2 = 1 ( w = 2) z 2 + z 3 + z 4 = 0 ( w = 3) ⇒ z 2 + z 3 + z 4 = 0 ( w = 1) z 2 + z 3 + z 4 = 1 ( w = 2) Allows us to assume no two equations have the same left-hand side. Anders Yeo Max-Lin Parameterized Above Average

  34. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (LHS rule) Suppose we have two equations, � i ∈ S z i = b 1 (weight w 1 ) and � i ∈ S z i = b 2 (weight w 2 ), where w 1 ≥ w 2 . If b 1 = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 + w 2 ). If b 1 � = b 2 , replace with one equation � i ∈ S z i = b 1 (weight w 1 − w 2 ). z 1 + z 2 = 1 ( w = 1) ⇒ z 1 + z 2 = 1 ( w = 3) z 1 + z 2 = 1 ( w = 2) z 2 + z 3 + z 4 = 0 ( w = 3) ⇒ z 2 + z 3 + z 4 = 0 ( w = 1) z 2 + z 3 + z 4 = 1 ( w = 2) Allows us to assume no two equations have the same left-hand side. Anders Yeo Max-Lin Parameterized Above Average

  35. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (Rank rule) Let A be the matrix over F 2 corresponding to the set of equations in I , such that a ji = 1 if z i appears in equation j, and 0 otherwise. Let t = rank A and suppose columns a i 1 , . . . , a i t of A are linearly independent. Then delete all variables not in { z i 1 , . . . , z i t } from the equations of S.   z 1 + z 3 + z 4 = 1 1 0 1 1 z 1 + z 4 = 1 z 2 + z 3 + z 4 = 0 ⇒ 0 1 1 1 ⇒ z 2 + z 4 = 0     z 2 + z 3 = 0 0 1 1 0 z 2 = 0   z 1 + z 2 = 1 1 1 0 0 z 1 + z 2 = 1 Anders Yeo Max-Lin Parameterized Above Average

  36. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (Rank rule) Let A be the matrix over F 2 corresponding to the set of equations in I , such that a ji = 1 if z i appears in equation j, and 0 otherwise. Let t = rank A and suppose columns a i 1 , . . . , a i t of A are linearly independent. Then delete all variables not in { z i 1 , . . . , z i t } from the equations of S.   z 1 + z 3 + z 4 = 1 1 0 1 1 z 1 + z 4 = 1 z 2 + z 3 + z 4 = 0 ⇒ 0 1 1 1 ⇒ z 2 + z 4 = 0     z 2 + z 3 = 0 0 1 1 0 z 2 = 0   z 1 + z 2 = 1 1 1 0 0 z 1 + z 2 = 1 Anders Yeo Max-Lin Parameterized Above Average

  37. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Reduction Rule (Rank rule) Let A be the matrix over F 2 corresponding to the set of equations in I , such that a ji = 1 if z i appears in equation j, and 0 otherwise. Let t = rank A and suppose columns a i 1 , . . . , a i t of A are linearly independent. Then delete all variables not in { z i 1 , . . . , z i t } from the equations of S.   z 1 + z 3 + z 4 = 1 1 0 1 1 z 1 + z 4 = 1 z 2 + z 3 + z 4 = 0 ⇒ 0 1 1 1 ⇒ z 2 + z 4 = 0     z 2 + z 3 = 0 0 1 1 0 z 2 = 0   z 1 + z 2 = 1 1 1 0 0 z 1 + z 2 = 1 Anders Yeo Max-Lin Parameterized Above Average

  38. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Why does the Rank Rule work? I ′ I z 1 + z 3 + z 4 = 1   z 1 + z 4 = 1 1 0 1 1 ⇒ ⇒ z 2 + z 3 + z 4 = 0 0 1 1 1 z 2 + z 4 = 0     z 2 + z 3 = 0 z 2 = 0 0 1 1 0   z 1 + z 2 = 1 z 1 + z 2 = 1 1 1 0 0 Set z 3 = 0 and add a solution for I ′ to get a solution of equal weight for I . Consider a solution for I . If z 3 = 1, then change the values of z 1 , z 2 , z 3 to get an equivalent solution with z 3 = 0. Why does this work? So z 3 = 0, and we have a solution for I ′ of equal weight. Anders Yeo Max-Lin Parameterized Above Average

  39. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Why does the Rank Rule work? I ′ I z 1 + z 3 + z 4 = 1   z 1 + z 4 = 1 1 0 1 1 ⇒ ⇒ z 2 + z 3 + z 4 = 0 0 1 1 1 z 2 + z 4 = 0     z 2 + z 3 = 0 z 2 = 0 0 1 1 0   z 1 + z 2 = 1 z 1 + z 2 = 1 1 1 0 0 Set z 3 = 0 and add a solution for I ′ to get a solution of equal weight for I . Consider a solution for I . If z 3 = 1, then change the values of z 1 , z 2 , z 3 to get an equivalent solution with z 3 = 0. Why does this work? So z 3 = 0, and we have a solution for I ′ of equal weight. Anders Yeo Max-Lin Parameterized Above Average

  40. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Why does the Rank Rule work? I ′ I z 1 + z 3 + z 4 = 1   z 1 + z 4 = 1 1 0 1 1 ⇒ ⇒ z 2 + z 3 + z 4 = 0 0 1 1 1 z 2 + z 4 = 0     z 2 + z 3 = 0 z 2 = 0 0 1 1 0   z 1 + z 2 = 1 z 1 + z 2 = 1 1 1 0 0 Set z 3 = 0 and add a solution for I ′ to get a solution of equal weight for I . Consider a solution for I . If z 3 = 1, then change the values of z 1 , z 2 , z 3 to get an equivalent solution with z 3 = 0. Why does this work? So z 3 = 0, and we have a solution for I ′ of equal weight. Anders Yeo Max-Lin Parameterized Above Average

  41. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Why does the Rank Rule work? I ′ I z 1 + z 3 + z 4 = 1   z 1 + z 4 = 1 1 0 1 1 ⇒ ⇒ z 2 + z 3 + z 4 = 0 0 1 1 1 z 2 + z 4 = 0     z 2 + z 3 = 0 z 2 = 0 0 1 1 0   z 1 + z 2 = 1 z 1 + z 2 = 1 1 1 0 0 Set z 3 = 0 and add a solution for I ′ to get a solution of equal weight for I . Consider a solution for I . If z 3 = 1, then change the values of z 1 , z 2 , z 3 to get an equivalent solution with z 3 = 0. Why does this work? So z 3 = 0, and we have a solution for I ′ of equal weight. Anders Yeo Max-Lin Parameterized Above Average

  42. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction Rules Why does the Rank Rule work? I ′ I z 1 + z 3 + z 4 = 1   z 1 + z 4 = 1 1 0 1 1 ⇒ ⇒ z 2 + z 3 + z 4 = 0 0 1 1 1 z 2 + z 4 = 0     z 2 + z 3 = 0 z 2 = 0 0 1 1 0   z 1 + z 2 = 1 z 1 + z 2 = 1 1 1 0 0 Set z 3 = 0 and add a solution for I ′ to get a solution of equal weight for I . Consider a solution for I . If z 3 = 1, then change the values of z 1 , z 2 , z 3 to get an equivalent solution with z 3 = 0. Why does this work? So z 3 = 0, and we have a solution for I ′ of equal weight. Anders Yeo Max-Lin Parameterized Above Average

  43. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction rule What we would like to show: For reduced instances, if m is large enough the answer is Yes . Sadly this is not true... Consider a ’complete’ system on n variables with all RHS = 1. x 1 = 1 x 2 = 1 x 1 + x 2 = 1 x 3 = 1 x 1 + x 3 = 1 x 2 + x 3 = 1 x 1 + x 2 + x 3 = 1 Anders Yeo Max-Lin Parameterized Above Average

  44. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction rule What we would like to show: For reduced instances, if m is large enough the answer is Yes . Sadly this is not true... Consider a ’complete’ system on n variables with all RHS = 1. x 1 = 1 x 2 = 1 x 1 + x 2 = 1 x 3 = 1 x 1 + x 3 = 1 x 2 + x 3 = 1 x 1 + x 2 + x 3 = 1 Anders Yeo Max-Lin Parameterized Above Average

  45. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction rule What we would like to show: For reduced instances, if m is large enough the answer is Yes . Sadly this is not true... Consider a ’complete’ system on n variables with all RHS = 1. x 1 = 1 x 2 = 1 x 1 + x 2 = 1 x 3 = 1 x 1 + x 3 = 1 x 2 + x 3 = 1 x 1 + x 2 + x 3 = 1 Anders Yeo Max-Lin Parameterized Above Average

  46. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction rule What we would like to show: For reduced instances, if m is large enough the answer is Yes . Sadly this is not true... Consider a ’complete’ system on n variables with all RHS = 1. x 1 = 1 x 2 = 1 x 2 = 0 x 3 = 1 x 3 = 0 x 2 + x 3 = 1 x 2 + x 3 = 0 Anders Yeo Max-Lin Parameterized Above Average

  47. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Reduction rule What we would like to show: For reduced instances, if m is large enough the answer is Yes . Sadly this is not true... Consider a ’complete’ system on n variables with all RHS = 1. x 1 = 1 x 2 = 1 x 2 = 0 x 3 = 1 x 3 = 0 x 2 + x 3 = 1 x 2 + x 3 = 0 The maximum excess is 1 but m = 2 n − 1. Anders Yeo Max-Lin Parameterized Above Average

  48. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Main Results Theorem A: [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] Max-Lin-AA can be solved in time O ∗ ( n 2 k ). Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. The above results can be combined to show the following Theorem (Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011) Max-Lin-AA is fixed-parameter tractable, and has a kernel with O ( k 2 log k ) variables. Anders Yeo Max-Lin Parameterized Above Average

  49. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem A (Algorithm H ) Algorithm H (More detail) Choose an equation e, which can be written as � i ∈ S z i = b, 1 with weight w ( e ) . Choose some j ∈ S. 2 Simplify the system under the assumption that e is true: 3 Remove equation e. 1 Perform the substitution z j = � ( i ∈ S \ j ) z i + b for all 2 equations containing z j . Reduce the system by LHS Rule. 3 Reduce k by w ( e ) / 2 . 4 Anders Yeo Max-Lin Parameterized Above Average

  50. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example z 1 + z 3 + z 5 = 1 ⇒ z 1 = z 3 + z 5 + 1 z 2 + z 3 = 1 ⇒ z 2 + z 3 = 1 ⇒ ⇒ z 1 + z 2 = 0 z 3 + z 5 + 1 + z 2 = 0 z 2 + z 3 + z 5 = 1 ⇒ z 3 + z 4 + z 5 = 1 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 4 = 0 z 3 + z 5 + 1 + z 4 = 0 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 2 + z 5 = 1 z 3 + z 5 + 1 + z 2 + z 5 = 1 z 2 + z 3 = 0 Now we simplify...... Anders Yeo Max-Lin Parameterized Above Average

  51. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example z 1 + z 3 + z 5 = 1 ⇒ z 1 = z 3 + z 5 + 1 z 2 + z 3 = 1 ⇒ z 2 + z 3 = 1 ⇒ ⇒ z 1 + z 2 = 0 z 3 + z 5 + 1 + z 2 = 0 z 2 + z 3 + z 5 = 1 ⇒ z 3 + z 4 + z 5 = 1 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 4 = 0 z 3 + z 5 + 1 + z 4 = 0 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 2 + z 5 = 1 z 3 + z 5 + 1 + z 2 + z 5 = 1 z 2 + z 3 = 0 Now we simplify...... Anders Yeo Max-Lin Parameterized Above Average

  52. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example z 1 + z 3 + z 5 = 1 ⇒ z 1 = z 3 + z 5 + 1 z 2 + z 3 = 1 ⇒ z 2 + z 3 = 1 ⇒ ⇒ z 1 + z 2 = 0 z 3 + z 5 + 1 + z 2 = 0 z 2 + z 3 + z 5 = 1 ⇒ z 3 + z 4 + z 5 = 1 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 4 = 0 z 3 + z 5 + 1 + z 4 = 0 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 2 + z 5 = 1 z 3 + z 5 + 1 + z 2 + z 5 = 1 z 2 + z 3 = 0 Now we simplify...... Anders Yeo Max-Lin Parameterized Above Average

  53. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example z 1 + z 3 + z 5 = 1 ⇒ z 1 = z 3 + z 5 + 1 z 2 + z 3 = 1 ⇒ z 2 + z 3 = 1 ⇒ ⇒ z 1 + z 2 = 0 z 3 + z 5 + 1 + z 2 = 0 z 2 + z 3 + z 5 = 1 ⇒ z 3 + z 4 + z 5 = 1 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 4 = 0 z 3 + z 5 + 1 + z 4 = 0 z 3 + z 4 + z 5 = 1 ⇒ ⇒ z 1 + z 2 + z 5 = 1 z 3 + z 5 + 1 + z 2 + z 5 = 1 z 2 + z 3 = 0 Now we simplify...... Anders Yeo Max-Lin Parameterized Above Average

  54. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example ( z 1 + z 3 + z 5 = 1) z 2 + z 3 + z 5 = 1 z 3 + z 4 + z 5 = 1 ( w = 2) So under the assumption that e = ” z 1 + z 3 + z 5 = 1” is true we have reduced I to a smaller problem I ′ such that we can do w ( e ) / 2 more above average in I than in I ′ . Why? Answer: For any solution of I ′ , set z 1 = z 3 + z 5 + 1..... Anders Yeo Max-Lin Parameterized Above Average

  55. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example ( z 1 + z 3 + z 5 = 1) z 2 + z 3 + z 5 = 1 z 3 + z 4 + z 5 = 1 ( w = 2) So under the assumption that e = ” z 1 + z 3 + z 5 = 1” is true we have reduced I to a smaller problem I ′ such that we can do w ( e ) / 2 more above average in I than in I ′ . Why? Answer: For any solution of I ′ , set z 1 = z 3 + z 5 + 1..... Anders Yeo Max-Lin Parameterized Above Average

  56. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example ( z 1 + z 3 + z 5 = 1) z 2 + z 3 + z 5 = 1 z 3 + z 4 + z 5 = 1 ( w = 2) So under the assumption that e = ” z 1 + z 3 + z 5 = 1” is true we have reduced I to a smaller problem I ′ such that we can do w ( e ) / 2 more above average in I than in I ′ . Why? Answer: For any solution of I ′ , set z 1 = z 3 + z 5 + 1..... Anders Yeo Max-Lin Parameterized Above Average

  57. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Example ( z 1 + z 3 + z 5 = 1) z 2 + z 3 + z 5 = 1 z 3 + z 4 + z 5 = 1 ( w = 2) So under the assumption that e = ” z 1 + z 3 + z 5 = 1” is true we have reduced I to a smaller problem I ′ such that we can do w ( e ) / 2 more above average in I than in I ′ . Why? Answer: For any solution of I ′ , set z 1 = z 3 + z 5 + 1..... Anders Yeo Max-Lin Parameterized Above Average

  58. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results So what does Algorithm H give us Assume our instance is reduced. If we can mark equations of total weight R then the maximum excess is at least R (we can get at least R / 2 above the average). If the maximum excess is R then if we keep choosing equations which are true in a given optimal solution, we will mark equations of total weight R . How can this be used to prove Theorem A...... Anders Yeo Max-Lin Parameterized Above Average

  59. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results So what does Algorithm H give us Assume our instance is reduced. If we can mark equations of total weight R then the maximum excess is at least R (we can get at least R / 2 above the average). If the maximum excess is R then if we keep choosing equations which are true in a given optimal solution, we will mark equations of total weight R . How can this be used to prove Theorem A...... Anders Yeo Max-Lin Parameterized Above Average

  60. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results So what does Algorithm H give us Assume our instance is reduced. If we can mark equations of total weight R then the maximum excess is at least R (we can get at least R / 2 above the average). If the maximum excess is R then if we keep choosing equations which are true in a given optimal solution, we will mark equations of total weight R . How can this be used to prove Theorem A...... Anders Yeo Max-Lin Parameterized Above Average

  61. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results So what does Algorithm H give us Assume our instance is reduced. If we can mark equations of total weight R then the maximum excess is at least R (we can get at least R / 2 above the average). If the maximum excess is R then if we keep choosing equations which are true in a given optimal solution, we will mark equations of total weight R . How can this be used to prove Theorem A...... Anders Yeo Max-Lin Parameterized Above Average

  62. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem A Theorem A [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] There exists an O ∗ ( n 2 k )-time algorithm for Max-Lin-AA . Proof (sketch): Let e 1 , . . . e n be a set of equations in I which are ’independent’. (LHSs correspond to independent rows in matrix A .) Check unique assignment in which e 1 , . . . e n all false. If this assignment achieves excess 2 k , return Yes . Otherwise, one of e 1 , . . . e k must be true. Branch n ways. In branch i mark equation e i in Algorithm H and solve resulting system. Since we can stop after 2 k iterations of H , search tree has n 2 k leaves. Anders Yeo Max-Lin Parameterized Above Average

  63. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem A Theorem A [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] There exists an O ∗ ( n 2 k )-time algorithm for Max-Lin-AA . Proof (sketch): Let e 1 , . . . e n be a set of equations in I which are ’independent’. (LHSs correspond to independent rows in matrix A .) Check unique assignment in which e 1 , . . . e n all false. If this assignment achieves excess 2 k , return Yes . Otherwise, one of e 1 , . . . e k must be true. Branch n ways. In branch i mark equation e i in Algorithm H and solve resulting system. Since we can stop after 2 k iterations of H , search tree has n 2 k leaves. Anders Yeo Max-Lin Parameterized Above Average

  64. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem A Theorem A [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] There exists an O ∗ ( n 2 k )-time algorithm for Max-Lin-AA . Proof (sketch): Let e 1 , . . . e n be a set of equations in I which are ’independent’. (LHSs correspond to independent rows in matrix A .) Check unique assignment in which e 1 , . . . e n all false. If this assignment achieves excess 2 k , return Yes . Otherwise, one of e 1 , . . . e k must be true. Branch n ways. In branch i mark equation e i in Algorithm H and solve resulting system. Since we can stop after 2 k iterations of H , search tree has n 2 k leaves. Anders Yeo Max-Lin Parameterized Above Average

  65. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem A Theorem A [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] There exists an O ∗ ( n 2 k )-time algorithm for Max-Lin-AA . Proof (sketch): Let e 1 , . . . e n be a set of equations in I which are ’independent’. (LHSs correspond to independent rows in matrix A .) Check unique assignment in which e 1 , . . . e n all false. If this assignment achieves excess 2 k , return Yes . Otherwise, one of e 1 , . . . e k must be true. Branch n ways. In branch i mark equation e i in Algorithm H and solve resulting system. Since we can stop after 2 k iterations of H , search tree has n 2 k leaves. Anders Yeo Max-Lin Parameterized Above Average

  66. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem A Theorem A [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] There exists an O ∗ ( n 2 k )-time algorithm for Max-Lin-AA . Proof (sketch): Let e 1 , . . . e n be a set of equations in I which are ’independent’. (LHSs correspond to independent rows in matrix A .) Check unique assignment in which e 1 , . . . e n all false. If this assignment achieves excess 2 k , return Yes . Otherwise, one of e 1 , . . . e k must be true. Branch n ways. In branch i mark equation e i in Algorithm H and solve resulting system. Since we can stop after 2 k iterations of H , search tree has n 2 k leaves. Anders Yeo Max-Lin Parameterized Above Average

  67. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. If we can run algorithm H for 2 k iterations, we can get an excess of at least 2 k . Problem: After running H a few times all the remaining equations may ’cancel out’ under LHS Rule. One solution: M -sum-free vectors. Let K and M be sets of vectors in F n 2 such that K ⊆ M . K is M -sum-free if no sum of two or more vectors in K is equal to a vector in M . Anders Yeo Max-Lin Parameterized Above Average

  68. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. If we can run algorithm H for 2 k iterations, we can get an excess of at least 2 k . Problem: After running H a few times all the remaining equations may ’cancel out’ under LHS Rule. One solution: M -sum-free vectors. Let K and M be sets of vectors in F n 2 such that K ⊆ M . K is M -sum-free if no sum of two or more vectors in K is equal to a vector in M . Anders Yeo Max-Lin Parameterized Above Average

  69. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. If we can run algorithm H for 2 k iterations, we can get an excess of at least 2 k . Problem: After running H a few times all the remaining equations may ’cancel out’ under LHS Rule. One solution: M -sum-free vectors. Let K and M be sets of vectors in F n 2 such that K ⊆ M . K is M -sum-free if no sum of two or more vectors in K is equal to a vector in M . Anders Yeo Max-Lin Parameterized Above Average

  70. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. If we can run algorithm H for 2 k iterations, we can get an excess of at least 2 k . Problem: After running H a few times all the remaining equations may ’cancel out’ under LHS Rule. One solution: M -sum-free vectors. Let K and M be sets of vectors in F n 2 such that K ⊆ M . K is M -sum-free if no sum of two or more vectors in K is equal to a vector in M . Anders Yeo Max-Lin Parameterized Above Average

  71. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. If we can run algorithm H for 2 k iterations, we can get an excess of at least 2 k . Problem: After running H a few times all the remaining equations may ’cancel out’ under LHS Rule. One solution: M -sum-free vectors. Let K and M be sets of vectors in F n 2 such that K ⊆ M . K is M -sum-free if no sum of two or more vectors in K is equal to a vector in M . Anders Yeo Max-Lin Parameterized Above Average

  72. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma View the LHSs of equations in I as a set M of vectors in F n 2 . Let e 1 , . . . e t be a set of equations in I that correspond to an M-sum-free set of vectors. Then we can run algorithm H for t iterations, choosing equations e 1 , . . . e t in turn, and get an excess of at least t. Why? Assume for the sake of contradiction e i gets cancelled out. Then by picking e 1 , . . . , e i − 1 in Algorithm H we have created a different equation, say f i , with the same LHS as e i . So considering LHSs we get: e i = f i = e j 1 + e j 2 + · · · + e j a + e ′ for some { j 1 , . . . , j a } ⊆ { 1 , . . . , i − 1 } and e ′ is any equation. However this implies that e ′ = e j 1 + e j 2 + · · · + e j a + e i , a contradiction. Anders Yeo Max-Lin Parameterized Above Average

  73. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma View the LHSs of equations in I as a set M of vectors in F n 2 . Let e 1 , . . . e t be a set of equations in I that correspond to an M-sum-free set of vectors. Then we can run algorithm H for t iterations, choosing equations e 1 , . . . e t in turn, and get an excess of at least t. Why? Assume for the sake of contradiction e i gets cancelled out. Then by picking e 1 , . . . , e i − 1 in Algorithm H we have created a different equation, say f i , with the same LHS as e i . So considering LHSs we get: e i = f i = e j 1 + e j 2 + · · · + e j a + e ′ for some { j 1 , . . . , j a } ⊆ { 1 , . . . , i − 1 } and e ′ is any equation. However this implies that e ′ = e j 1 + e j 2 + · · · + e j a + e i , a contradiction. Anders Yeo Max-Lin Parameterized Above Average

  74. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma View the LHSs of equations in I as a set M of vectors in F n 2 . Let e 1 , . . . e t be a set of equations in I that correspond to an M-sum-free set of vectors. Then we can run algorithm H for t iterations, choosing equations e 1 , . . . e t in turn, and get an excess of at least t. Why? Assume for the sake of contradiction e i gets cancelled out. Then by picking e 1 , . . . , e i − 1 in Algorithm H we have created a different equation, say f i , with the same LHS as e i . So considering LHSs we get: e i = f i = e j 1 + e j 2 + · · · + e j a + e ′ for some { j 1 , . . . , j a } ⊆ { 1 , . . . , i − 1 } and e ′ is any equation. However this implies that e ′ = e j 1 + e j 2 + · · · + e j a + e i , a contradiction. Anders Yeo Max-Lin Parameterized Above Average

  75. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma View the LHSs of equations in I as a set M of vectors in F n 2 . Let e 1 , . . . e t be a set of equations in I that correspond to an M-sum-free set of vectors. Then we can run algorithm H for t iterations, choosing equations e 1 , . . . e t in turn, and get an excess of at least t. Why? Assume for the sake of contradiction e i gets cancelled out. Then by picking e 1 , . . . , e i − 1 in Algorithm H we have created a different equation, say f i , with the same LHS as e i . So considering LHSs we get: e i = f i = e j 1 + e j 2 + · · · + e j a + e ′ for some { j 1 , . . . , j a } ⊆ { 1 , . . . , i − 1 } and e ′ is any equation. However this implies that e ′ = e j 1 + e j 2 + · · · + e j a + e i , a contradiction. Anders Yeo Max-Lin Parameterized Above Average

  76. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma View the LHSs of equations in I as a set M of vectors in F n 2 . Let e 1 , . . . e t be a set of equations in I that correspond to an M-sum-free set of vectors. Then we can run algorithm H for t iterations, choosing equations e 1 , . . . e t in turn, and get an excess of at least t. Why? Assume for the sake of contradiction e i gets cancelled out. Then by picking e 1 , . . . , e i − 1 in Algorithm H we have created a different equation, say f i , with the same LHS as e i . So considering LHSs we get: e i = f i = e j 1 + e j 2 + · · · + e j a + e ′ for some { j 1 , . . . , j a } ⊆ { 1 , . . . , i − 1 } and e ′ is any equation. However this implies that e ′ = e j 1 + e j 2 + · · · + e j a + e i , a contradiction. Anders Yeo Max-Lin Parameterized Above Average

  77. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma C [Crowston, Gutin, Jones, Kim and Ruzsa (2010)] Let M be a proper subset in F n 2 such that span ( M ) = F n 2 . If k is a positive integer and t ≤ | M | ≤ 2 n / t then, in time | M | O (1) , we can find an M -sum-free subset K of M s.t. | K | = t . Theorem B: If 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Suppose I is reduced and 2 k ≤ m ≤ 2 n / 2 k . Let M be the set of vectors in F n 2 corresponding to LHSs of equations in I Find an M -sum-free subset K of M s.t. | K | = 2 k . Let e 1 , . . . e 2 k be the equations corresponding to K , and run algorithm H marking e 1 , . . . e 2 k in turn. Then we get excess 2 k , so the answer is Yes . Anders Yeo Max-Lin Parameterized Above Average

  78. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma C [Crowston, Gutin, Jones, Kim and Ruzsa (2010)] Let M be a proper subset in F n 2 such that span ( M ) = F n 2 . If k is a positive integer and t ≤ | M | ≤ 2 n / t then, in time | M | O (1) , we can find an M -sum-free subset K of M s.t. | K | = t . Theorem B: If 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Suppose I is reduced and 2 k ≤ m ≤ 2 n / 2 k . Let M be the set of vectors in F n 2 corresponding to LHSs of equations in I Find an M -sum-free subset K of M s.t. | K | = 2 k . Let e 1 , . . . e 2 k be the equations corresponding to K , and run algorithm H marking e 1 , . . . e 2 k in turn. Then we get excess 2 k , so the answer is Yes . Anders Yeo Max-Lin Parameterized Above Average

  79. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma C [Crowston, Gutin, Jones, Kim and Ruzsa (2010)] Let M be a proper subset in F n 2 such that span ( M ) = F n 2 . If k is a positive integer and t ≤ | M | ≤ 2 n / t then, in time | M | O (1) , we can find an M -sum-free subset K of M s.t. | K | = t . Theorem B: If 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Suppose I is reduced and 2 k ≤ m ≤ 2 n / 2 k . Let M be the set of vectors in F n 2 corresponding to LHSs of equations in I Find an M -sum-free subset K of M s.t. | K | = 2 k . Let e 1 , . . . e 2 k be the equations corresponding to K , and run algorithm H marking e 1 , . . . e 2 k in turn. Then we get excess 2 k , so the answer is Yes . Anders Yeo Max-Lin Parameterized Above Average

  80. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma C [Crowston, Gutin, Jones, Kim and Ruzsa (2010)] Let M be a proper subset in F n 2 such that span ( M ) = F n 2 . If k is a positive integer and t ≤ | M | ≤ 2 n / t then, in time | M | O (1) , we can find an M -sum-free subset K of M s.t. | K | = t . Theorem B: If 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Suppose I is reduced and 2 k ≤ m ≤ 2 n / 2 k . Let M be the set of vectors in F n 2 corresponding to LHSs of equations in I Find an M -sum-free subset K of M s.t. | K | = 2 k . Let e 1 , . . . e 2 k be the equations corresponding to K , and run algorithm H marking e 1 , . . . e 2 k in turn. Then we get excess 2 k , so the answer is Yes . Anders Yeo Max-Lin Parameterized Above Average

  81. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma C [Crowston, Gutin, Jones, Kim and Ruzsa (2010)] Let M be a proper subset in F n 2 such that span ( M ) = F n 2 . If k is a positive integer and t ≤ | M | ≤ 2 n / t then, in time | M | O (1) , we can find an M -sum-free subset K of M s.t. | K | = t . Theorem B: If 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Suppose I is reduced and 2 k ≤ m ≤ 2 n / 2 k . Let M be the set of vectors in F n 2 corresponding to LHSs of equations in I Find an M -sum-free subset K of M s.t. | K | = 2 k . Let e 1 , . . . e 2 k be the equations corresponding to K , and run algorithm H marking e 1 , . . . e 2 k in turn. Then we get excess 2 k , so the answer is Yes . Anders Yeo Max-Lin Parameterized Above Average

  82. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of Theorem B Lemma C [Crowston, Gutin, Jones, Kim and Ruzsa (2010)] Let M be a proper subset in F n 2 such that span ( M ) = F n 2 . If k is a positive integer and t ≤ | M | ≤ 2 n / t then, in time | M | O (1) , we can find an M -sum-free subset K of M s.t. | K | = t . Theorem B: If 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Suppose I is reduced and 2 k ≤ m ≤ 2 n / 2 k . Let M be the set of vectors in F n 2 corresponding to LHSs of equations in I Find an M -sum-free subset K of M s.t. | K | = 2 k . Let e 1 , . . . e 2 k be the equations corresponding to K , and run algorithm H marking e 1 , . . . e 2 k in turn. Then we get excess 2 k , so the answer is Yes . Anders Yeo Max-Lin Parameterized Above Average

  83. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Recall Theorem A and Theorem B Theorem A: [Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011] Max-Lin-AA can be solved in time O ∗ ( n 2 k ). Theorem B: [Crowston, Gutin, Jones, Kim, Ruzsa, 2010] If I is reduced and 2 k ≤ m < 2 n / 2 k , then I is a Yes -instance. Anders Yeo Max-Lin Parameterized Above Average

  84. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of our main result Theorem (Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011) Max-Lin-AA has a kernel with at most O ( k 2 log k ) variables. Proof: Let I be a reduced system. Case 1: m ≥ n 2 k . Then using O ∗ ( n 2 k ) algorithm, can solve in polynomial time. Case 2: 2 k ≤ m ≤ 2 n / 2 k . By earlier Theorem return yes . Case 3: m < 2 k . Since I reduced by Rank Rule, n ≤ m so n = O ( k 2 log k ) . Only remaining case is 2 n / 2 k < m < n 2 k . Anders Yeo Max-Lin Parameterized Above Average

  85. Parameterizing above tight bounds: Example Max-Sat Max-Lin-AA FPT Results Related Results Proof of our main result Theorem (Crowston, Fellows, Gutin, Jones, Rosamond, Thomasse, Yeo, 2011) Max-Lin-AA has a kernel with at most O ( k 2 log k ) variables. Proof: Let I be a reduced system. Case 1: m ≥ n 2 k . Then using O ∗ ( n 2 k ) algorithm, can solve in polynomial time. Case 2: 2 k ≤ m ≤ 2 n / 2 k . By earlier Theorem return yes . Case 3: m < 2 k . Since I reduced by Rank Rule, n ≤ m so n = O ( k 2 log k ) . Only remaining case is 2 n / 2 k < m < n 2 k . Anders Yeo Max-Lin Parameterized Above Average

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