Advanced Algorithms (II) Chihao Zhang Shanghai Jiao Tong University Mar. 4, 2019 Advanced Algorithms (II) 1/11
We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. MaxSAT C j Advanced Algorithms (II) n i y i m j z j x k k N j x i i P j y k z j Recall that we can obtain a k N j y i i P j subject to z j j m max of the two. -approximation by choosing the betuer 2/11
We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. MaxSAT C j Advanced Algorithms (II) n i y i m j z j x k k N j x i i P j y k z j k N j y i i P j subject to z j j m max of the two. 2/11 Recall that we can obtain a 3 4 -approximation by choosing the betuer
Recall that we have the following linear programming relaxation. MaxSAT C j Advanced Algorithms (II) n i y i m j z j x k k N j x i i P j y k z j k N j y i i P j subject to z j j m max of the two. 2/11 Recall that we can obtain a 3 4 -approximation by choosing the betuer We show that this ratio can also be achieved via direct rounding.
MaxSAT m Advanced Algorithms (II) x k subject to z j 2/11 max of the two. Recall that we can obtain a 3 4 -approximation by choosing the betuer We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. ∑ j =1 ∑ ∑ ∨ ∨ y i + (1 − y k ) ≥ z j , ∀ C j = ¯ x i ∨ i ∈ P j i ∈ P j k ∈ N j k ∈ N j z j ∈ [0 , 1] , ∀ j ∈ [ m ] y i ∈ [0 , 1] , ∀ i ∈ [ n ]
Instead of tossing y i -biased coins, we toss f y i -biased coins for i P j x i k N j x k , Pr C j is not satisfied Let For C j Advanced Algorithms (II) approximation. We can choose a suitable f to get f y k k N j f y i i P j . some function f j i 3/11 { } y ∗ z ∗ { } i ∈ [ n ] , j ∈ [ m ] be an optimal solution of the LP.
i P j x i k N j x k , Pr C j is not satisfied Let For C j Advanced Algorithms (II) approximation. We can choose a suitable f to get f y k k N j f y i i P j 3/11 j i { } y ∗ z ∗ { } i ∈ [ n ] , j ∈ [ m ] be an optimal solution of the LP. Instead of tossing y ∗ i -biased coins, we toss f ( y ∗ i ) -biased coins for some function f ( · ) ∈ [0 , 1] .
Let j Advanced Algorithms (II) approximation. We can choose a suitable f to get Pr x k , 3/11 i { } y ∗ z ∗ { } i ∈ [ n ] , j ∈ [ m ] be an optimal solution of the LP. Instead of tossing y ∗ i -biased coins, we toss f ( y ∗ i ) -biased coins for some function f ( · ) ∈ [0 , 1] . For C j = ∨ i ∈ P j x i ∨ ∨ k ∈ N j ¯ ∏ ∏ (1 − f ( y ∗ f ( y ∗ [ ] = i )) k ) . C j is not satisfied i ∈ P j k ∈ N j
Let j Advanced Algorithms (II) Pr x k , 3/11 i { } y ∗ z ∗ { } i ∈ [ n ] , j ∈ [ m ] be an optimal solution of the LP. Instead of tossing y ∗ i -biased coins, we toss f ( y ∗ i ) -biased coins for some function f ( · ) ∈ [0 , 1] . For C j = ∨ i ∈ P j x i ∨ ∨ k ∈ N j ¯ ∏ ∏ (1 − f ( y ∗ f ( y ∗ [ ] = i )) k ) . C j is not satisfied i ∈ P j k ∈ N j We can choose a suitable f to get 3 4 approximation.
Integrality Gap OPT Advanced Algorithms (II) is called the integrality gap of the LP relaxation. The ratio ! then we cannot have OPT LP If we alreadly know In most LP based approximation algorithms, the upper bound for OPT OPT LP OPT Then we establish OPT LP OPT the OPT is 4/11
Integrality Gap OPT Advanced Algorithms (II) is called the integrality gap of the LP relaxation. The ratio ! then we cannot have OPT LP If we alreadly know In most LP based approximation algorithms, the upper bound for OPT OPT LP OPT Then we establish the OPT is 4/11 OPT ≤ OPT ( LP ) .
Integrality Gap In most LP based approximation algorithms, the upper bound for the OPT is Then we establish If we alreadly know OPT OPT LP then we cannot have ! The ratio is called the integrality gap of the LP relaxation. Advanced Algorithms (II) 4/11 OPT ≤ OPT ( LP ) . OPT ∗ ≥ α · OPT ( LP ) ≥ α · OPT .
Integrality Gap In most LP based approximation algorithms, the upper bound for the OPT is Then we establish If we alreadly know then we cannot have ! The ratio is called the integrality gap of the LP relaxation. Advanced Algorithms (II) 4/11 OPT ≤ OPT ( LP ) . OPT ∗ ≥ α · OPT ( LP ) ≥ α · OPT . OPT ≤ β · OPT ( LP ) ,
Integrality Gap In most LP based approximation algorithms, the upper bound for the OPT is Then we establish If we alreadly know The ratio is called the integrality gap of the LP relaxation. Advanced Algorithms (II) 4/11 OPT ≤ OPT ( LP ) . OPT ∗ ≥ α · OPT ( LP ) ≥ α · OPT . OPT ≤ β · OPT ( LP ) , then we cannot have α > β !
Integrality Gap In most LP based approximation algorithms, the upper bound for the OPT is Then we establish If we alreadly know Advanced Algorithms (II) 4/11 OPT ≤ OPT ( LP ) . OPT ∗ ≥ α · OPT ( LP ) ≥ α · OPT . OPT ≤ β · OPT ( LP ) , then we cannot have α > β ! The ratio β is called the integrality gap of the LP relaxation.
The integrality gap of our LP is Integrality Gap for MaxSAT and OPT LP Advanced Algorithms (II) bound. OPT LP upper if we use OPT Corollary. We cannot beat . . OPT Consider the instance, x x x x x x x x 5/11
The integrality gap of our LP is Integrality Gap for MaxSAT Advanced Algorithms (II) bound. OPT LP upper if we use OPT Corollary. We cannot beat . . Consider the instance, and OPT LP OPT 5/11 ( x 1 ∨ x 2 ) ∧ ( x 1 ∨ ¯ x 2 ) ∧ (¯ x 1 ∨ x 2 ) ∧ (¯ x 1 ∨ ¯ x 2 )
The integrality gap of our LP is Integrality Gap for MaxSAT Consider the instance, . Corollary. We cannot beat if we use OPT OPT LP upper bound. Advanced Algorithms (II) 5/11 ( x 1 ∨ x 2 ) ∧ ( x 1 ∨ ¯ x 2 ) ∧ (¯ x 1 ∨ x 2 ) ∧ (¯ x 1 ∨ ¯ x 2 ) OPT = 3 and OPT ( LP ) = 4 .
Integrality Gap for MaxSAT Consider the instance, Corollary. We cannot beat if we use OPT OPT LP upper bound. Advanced Algorithms (II) 5/11 ( x 1 ∨ x 2 ) ∧ ( x 1 ∨ ¯ x 2 ) ∧ (¯ x 1 ∨ x 2 ) ∧ (¯ x 1 ∨ ¯ x 2 ) OPT = 3 and OPT ( LP ) = 4 . The integrality gap of our LP is 3 4 .
Integrality Gap for MaxSAT Consider the instance, bound. Advanced Algorithms (II) 5/11 ( x 1 ∨ x 2 ) ∧ ( x 1 ∨ ¯ x 2 ) ∧ (¯ x 1 ∨ x 2 ) ∧ (¯ x 1 ∨ ¯ x 2 ) OPT = 3 and OPT ( LP ) = 4 . The integrality gap of our LP is 3 4 . Corollary. We cannot beat 3 4 if we use OPT ≤ OPT ( LP ) upper
Compute a minimum set of labels L Minimum Label Cut Problem: Advanced Algorithms (II) P ). (unless NP NP -hard, and even hard to approximate with any constant ratio s and t . the removal of all edges with label in L disconnects L such that V . Minimum Label s - t Cut L ; two vertices s t e one E is labelled with L such that each e L V E ; a set of labels A graph G Input: 6/11
Minimum Label Cut Minimum Label s - t Cut Advanced Algorithms (II) P ). (unless NP NP -hard, and even hard to approximate with any constant ratio s and t . Problem: Input: A graph G 6/11 = ( V , E ) ; a set of labels [ L ] = { 1 , 2 , . . . , L } such that each e ∈ E is labelled with one ℓ ( e ) ∈ [ L ] ; two vertices s , t ∈ V . Compute a minimum set of labels L ′ ⊆ [ L ] such that the removal of all edges with label in L ′ disconnects
Minimum Label Cut Minimum Label s - t Cut Advanced Algorithms (II) NP -hard, and even hard to approximate with any constant ratio s and t . Problem: 6/11 Input: A graph G = ( V , E ) ; a set of labels [ L ] = { 1 , 2 , . . . , L } such that each e ∈ E is labelled with one ℓ ( e ) ∈ [ L ] ; two vertices s , t ∈ V . Compute a minimum set of labels L ′ ⊆ [ L ] such that the removal of all edges with label in L ′ disconnects (unless NP = P ).
We introduce a variable z j for each label j s t be the LP Relaxation P Advanced Algorithms (II) m . Q2 : What is the integrality gap of this LP? Q1 : How to solve this LP efgiciently? L j z j s t z e e P subject to z j j L min collection of paths between s and t . L . Let 7/11
LP Relaxation P Advanced Algorithms (II) m . Q2 : What is the integrality gap of this LP? Q1 : How to solve this LP efgiciently? L j z j s t e z e P subject to z j j L min collection of paths between s and t . 7/11 We introduce a variable z j for each label j ∈ [ L ] . Let P s , t be the
LP Relaxation subject to Advanced Algorithms (II) m . Q2 : What is the integrality gap of this LP? Q1 : How to solve this LP efgiciently? 7/11 z j L min collection of paths between s and t . We introduce a variable z j for each label j ∈ [ L ] . Let P s , t be the ∑ j =1 ∑ z ℓ ( e ) ≥ 1 , ∀ P ∈ P s , t e ∈ P z j ∈ [0 , 1] , ∀ j ∈ [ L ]
LP Relaxation subject to Advanced Algorithms (II) m . Q2 : What is the integrality gap of this LP? Q1 : How to solve this LP efgiciently? 7/11 z j L min collection of paths between s and t . We introduce a variable z j for each label j ∈ [ L ] . Let P s , t be the ∑ j =1 ∑ z ℓ ( e ) ≥ 1 , ∀ P ∈ P s , t e ∈ P z j ∈ [0 , 1] , ∀ j ∈ [ L ]
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