The matching polytope has exponential extension complexity Thomas - PowerPoint PPT Presentation

Applying the Hyperplane bound Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Slack matrix S UM = | δ ( U ) ∩ M | − 1 M S | δ ( U ) ∩ M | − 1 cuts U matchings

Applying the Hyperplane bound Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Slack matrix S UM = | δ ( U ) ∩ M | − 1 ◮ Abbreviate Q ℓ := { ( U, M ) : | δ ( U ) ∩ M | = ℓ } ◮ Uniform measure : µ ℓ ( R ) := | R ∩ Q ℓ | | Q ℓ | Q 1 Q 3 S 0 2 cuts 2 0 0 2 matchings

b Applying the Hyperplane bound Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Slack matrix S UM = | δ ( U ) ∩ M | − 1 ◮ Abbreviate Q ℓ := { ( U, M ) : | δ ( U ) ∩ M | = ℓ } ◮ Uniform measure : µ ℓ ( R ) := | R ∩ Q ℓ | | Q ℓ | ◮ Choose    W U,M =   0 otherwise . Q 1 Q 3 S 0 2 W cuts 2 0 0 2 matchings

b Applying the Hyperplane bound Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Slack matrix S UM = | δ ( U ) ∩ M | − 1 ◮ Abbreviate Q ℓ := { ( U, M ) : | δ ( U ) ∩ M | = ℓ } ◮ Uniform measure : µ ℓ ( R ) := | R ∩ Q ℓ | | Q ℓ | ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1   W U,M =   0 otherwise . Q 1 Q 3 −∞ S 0 2 W cuts 2 −∞ 0 −∞ 0 2 matchings

b Applying the Hyperplane bound Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Slack matrix S UM = | δ ( U ) ∩ M | − 1 ◮ Abbreviate Q ℓ := { ( U, M ) : | δ ( U ) ∩ M | = ℓ } ◮ Uniform measure : µ ℓ ( R ) := | R ∩ Q ℓ | | Q ℓ | ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1   1 W U,M = | δ ( U ) ∩ M | = 3 | Q 3 |   0 otherwise . Q 1 Q 3 1 −∞ S 0 2 W | Q 3 | cuts 1 2 | Q 3 | −∞ 0 1 −∞ 0 2 | Q 3 | matchings

b Applying the Hyperplane bound Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Slack matrix S UM = | δ ( U ) ∩ M | − 1 ◮ Abbreviate Q ℓ := { ( U, M ) : | δ ( U ) ∩ M | = ℓ } ◮ Uniform measure : µ ℓ ( R ) := | R ∩ Q ℓ | | Q ℓ | ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1   1 W U,M = | δ ( U ) ∩ M | = 3 | Q 3 |   0 otherwise . Q 1 Q 3 1 −∞ S 0 2 W | Q 3 | cuts 1 2 | Q 3 | −∞ 0 R 1 −∞ 0 2 | Q 3 | matchings

Rectangle covering for matching Claim: There is a rectangle with � W, R � = Θ( 1 n 4 ).

Rectangle covering for matching Claim: There is a rectangle with � W, R � = Θ( 1 n 4 ). e 1 e 2 ◮ For e 1 , e 2 ∈ E :

Rectangle covering for matching Claim: There is a rectangle with � W, R � = Θ( 1 n 4 ). e 1 U e 2 ◮ For e 1 , e 2 ∈ E : take { U | e 1 , e 2 ∈ δ ( U ) }

Rectangle covering for matching Claim: There is a rectangle with � W, R � = Θ( 1 n 4 ). e 1 U M e 2 ◮ For e 1 , e 2 ∈ E : take { U | e 1 , e 2 ∈ δ ( U ) } ×{ M | e 1 , e 2 ∈ M }

Rectangle covering for matching Claim: There is a rectangle with � W, R � = Θ( 1 n 4 ). e 1 U M e 2 ◮ For e 1 , e 2 ∈ E : take { U | e 1 , e 2 ∈ δ ( U ) } ×{ M | e 1 , e 2 ∈ M } ◮ But µ k ( R ) = Θ( k 2 n 4 )

b Applying the Hyperplane bound (II) Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1     1 | δ ( U ) ∩ M | = 3  | Q 3 | W U,M =      0 otherwise . Q 1 Q 3 1 −∞ S 0 2 W | Q 3 | cuts 1 2 | Q 3 | 0 −∞ 1 −∞ 0 2 | Q 3 | matchings

b Applying the Hyperplane bound (II) Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1     1 | δ ( U ) ∩ M | = 3  | Q 3 | W U,M =      0 otherwise . Q 1 Q 3 1 −∞ S 0 2 W | Q 3 | Q k k − 1 k − 1 cuts 1 2 | Q 3 | 0 −∞ 1 −∞ 0 2 k − 1 | Q 3 | matchings

b Applying the Hyperplane bound (II) Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1     1 | δ ( U ) ∩ M | = 3  | Q 3 | W U,M = 1 1 − k − 1 · | δ ( U ) ∩ M | = k   | Q k |    0 otherwise . Q 1 Q 3 1 −∞ S 0 2 W | Q 3 | 1 1 1 1 Q k − − k − 1 k − 1 k − 1 | Q k | k − 1 | Q k | cuts 1 2 | Q 3 | 0 −∞ 1 1 1 −∞ − 0 2 k − 1 | Q 3 | k − 1 | Q k | matchings

Applying the Hyperplane bound (II) Goal: Find W with � W,S � � W,R � large for each rectangle. ◮ Choose  − ∞ | δ ( U ) ∩ M | = 1     1 | δ ( U ) ∩ M | = 3  | Q 3 | W U,M = 1 1 − k − 1 · | δ ( U ) ∩ M | = k   | Q k |    0 otherwise . 1 −∞ W | Q 3 | ◮ Then 1 1 1 1 − − � W, S � = 0 + 2 − 1 = 1 k − 1 | Q k | k − 1 | Q k | cuts 1 | Q 3 | Lemma −∞ R For k large, any rectangle R 1 1 1 −∞ − has � W, R � ≤ 2 − Ω( n ) . | Q 3 | k − 1 | Q k | matchings

Applying the Hyperplane bound (III)

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts matchings

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts R matchings

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts R matchings

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts R matchings

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts R matchings

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts R matchings ◮ Technique: Partition scheme [Razborov ’91]

Applying the Hyperplane bound (III) Main lemma ⇒ µ 3 ( R ) ≤ O ( 1 k 2 ) · µ k ( R ) + 2 − Ω( n ) µ 1 ( R ) = 0 = S cuts T R matchings ◮ Technique: Partition scheme [Razborov ’91]

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R matchings

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R A matchings

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R A B matchings

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R A C B matchings k

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R A C D B matchings k k

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R A C D B matchings A 1 . . . k − 3 A m nodes k k

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R A C D B matchings . . . A 1 B 1 B m . . . k − 3 A m nodes 2( k − 3) k k nodes

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R ◮ Edges E ( T ) A C D B matchings . . . A 1 B 1 B m . . . k − 3 A m nodes 2( k − 3) k k nodes

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R ◮ Edges E ( T ) A C D B matchings . . . A 1 B 1 B m . . . k − 3 A m nodes 2( k − 3) k k nodes

Partitions S cuts ◮ Partition T = ( A, C, D, B ) T R ◮ Edges E ( T ) A C D B matchings U . . . A 1 B 1 B m . . . k − 3 A m nodes 2( k − 3) k k nodes

Pseudo-random behaviour of large set systems Imagine the following setting:

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all?

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets?

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO!

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO! Proof: ◮ Take a random set from S

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO! Proof: ◮ Take a random set from S ◮ Denote char. vector as x ∈ { 0 , 1 } n

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO! Proof: ◮ Take a random set from S ◮ Denote char. vector as x ∈ { 0 , 1 } n log |S| = H ( x )

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO! Proof: ◮ Take a random set from S ◮ Denote char. vector as x ∈ { 0 , 1 } n n subadd � log |S| = H ( x ) ≤ H ( x i ) i =1

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO! Proof: entropy 1 ◮ Take a random set from S ◮ Denote char. vector as x ∈ { 0 , 1 } n n subadd � log |S| = H ( x ) ≤ H ( x i ) ≤ n − Ω( n ) p 0 i =1 0 0 . 5 1 . 0

b b b b b b Pseudo-random behaviour of large set systems Imagine the following setting: ◮ n elements ◮ set system S with 2 (1 − o (1)) n sets Questions: ◮ Is it possible that ≥ 1 % of elements are in no set at all? NO! The 0 . 99 n active elements form at most 2 0 . 99 n sets ◮ Is it possible that ≥ 1 % elements are in ≤ 49 % of sets? NO! Lemma If S large, for most elements i , S ⊆ [ n ] [ S ∈ S ] ≈ Pr Pr S ⊆ [ n ] [ S ∈ S | i ∈ S ]

S Rewriting µ k ( R ) cuts T R matchings Randomly generate ( U, M ) ∼ Q k : µ k ( R ) =

S Rewriting µ k ( R ) cuts A C D B T R . . . A 1 B 1 B m matchings . . . A m Randomly generate ( U, M ) ∼ Q k : 1. Choose T � � µ k ( R ) = E T

S Rewriting µ k ( R ) cuts A C D B T R F . . . A 1 B 1 B m matchings . . . A m Randomly generate ( U, M ) ∼ Q k : 1. Choose T 2. Choose k edges F ⊆ C × D � � �� µ k ( R ) = E E T | F | = k

S Rewriting µ k ( R ) cuts A C D B T R F . . . A 1 B 1 B m matchings . . . A m Randomly generate ( U, M ) ∼ Q k : 1. Choose T 2. Choose k edges F ⊆ C × D 3. Choose M ⊇ F � � �� µ k ( R ) = E Pr[ M ∈ R | T, H ] E T | F | = k

S Rewriting µ k ( R ) cuts A C D B T R U F . . . A 1 B 1 B m matchings . . . A m Randomly generate ( U, M ) ∼ Q k : 1. Choose T 2. Choose k edges F ⊆ C × D 3. Choose M ⊇ F 4. Choose U ⊇ C (not cutting any A i ) � � �� µ k ( R ) = E Pr[ M ∈ R | T, H ] · Pr[ U ∈ R | T, H ] E T | F | = k

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) H . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) H . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) ◮ GOOD means it doesn’t matter what condition on here H . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) ◮ GOOD means it doesn’t matter what condition on here H . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) ◮ GOOD means it doesn’t matter what condition on here H . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) ◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H ∗ ⊆ F good ⇒ | H ∩ H ∗ | ≥ 2 H . . . A 1 B 1 B m . . . F A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) ◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H ∗ ⊆ F good ⇒ | H ∩ H ∗ | ≥ 2 ◮ Suppose | H ∩ H ∗ | ≤ 1 H . . . A 1 B 1 B m . . . H ∗ A m

How does an average partition look like ◮ Suppose for a fixed ( T, F ): µ k ( R ) ≈ Pr[( U, M ) ∈ R | T, F ] =: p ◮ Then µ 3 ( R ) ≈ [ GOOD ( T, H ) · Pr[( U, M ) ∈ R | T, H ] ] E H ∼ ( F 3 ) � �� � ≈ p � �� � ≤ O (1 /k 2 ) ◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H ∗ ⊆ F good ⇒ | H ∩ H ∗ | ≥ 2 ◮ Suppose | H ∩ H ∗ | ≤ 1 H . . . A 1 B 1 B m ◮ ( T, H ) good ⇒ ∃ M : { u, v } ∈ M u . . . H ∗ v A m