The matching polytope has exponential extension complexity
Thomas Rothvoss
Department of Mathematics, UW Seattle
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The matching polytope has exponential extension complexity Thomas Rothvoss Department of Mathematics, UW Seattle Extended formulation Extended formulation Given polytope P = { x R n | Ax b } P Extended formulation Given
Thomas Rothvoss
Department of Mathematics, UW Seattle
◮ Given polytope P = {x ∈ Rn | Ax ≤ b}
P
◮ Given polytope P = {x ∈ Rn | Ax ≤ b} ◮ Write P = {x ∈ Rn | ∃y : Bx + Cy ≤ d}
P Q linear projection
◮ Given polytope P = {x ∈ Rn | Ax ≤ b}
→ many inequalities
◮ Write P = {x ∈ Rn | ∃y : Bx + Cy ≤ d}
→ few inequalities P Q linear projection
◮ Given polytope P = {x ∈ Rn | Ax ≤ b}
→ many inequalities
◮ Write P = {x ∈ Rn | ∃y : Bx + Cy ≤ d}
→ few inequalities P Q linear projection
◮ Extension complexity:
xc(P) := min #facets of Q | Q polyhedron p linear map p(Q) = P
Compact formulations:
◮ Spanning Tree Polytope [Kipp Martin ’91] ◮ Perfect Matching in planar graphs [Barahona ’93] ◮ Perfect Matching in bounded genus graphs
[Gerards ’91]
◮ O(n log n)-size for Permutahedron [Goemans ’10]
(→ tight)
◮ nO(1/ε)-size ε-apx for Knapsack Polytope [Bienstock ’08] ◮ . . .
Compact formulations:
◮ Spanning Tree Polytope [Kipp Martin ’91] ◮ Perfect Matching in planar graphs [Barahona ’93] ◮ Perfect Matching in bounded genus graphs
[Gerards ’91]
◮ O(n log n)-size for Permutahedron [Goemans ’10]
(→ tight)
◮ nO(1/ε)-size ε-apx for Knapsack Polytope [Bienstock ’08] ◮ . . .
Here: When is the extension complexity super polynomial?
◮ No symmetric compact form. for TSP [Yannakakis ’91]
Compact formulation for log n size matchings, but no symmetric one [Kaibel, Pashkovich & Theis ’10]
◮ No symmetric compact form. for TSP [Yannakakis ’91]
Compact formulation for log n size matchings, but no symmetric one [Kaibel, Pashkovich & Theis ’10]
◮ xc(random 0/1 polytope) ≥ 2Ω(n) [R. ’11]
◮ No symmetric compact form. for TSP [Yannakakis ’91]
Compact formulation for log n size matchings, but no symmetric one [Kaibel, Pashkovich & Theis ’10]
◮ xc(random 0/1 polytope) ≥ 2Ω(n) [R. ’11] ◮ Breakthrough: xc(TSP) ≥ 2Ω(√n)
[Fiorini, Massar, Pokutta, Tiwary, de Wolf ’12]
◮ No symmetric compact form. for TSP [Yannakakis ’91]
Compact formulation for log n size matchings, but no symmetric one [Kaibel, Pashkovich & Theis ’10]
◮ xc(random 0/1 polytope) ≥ 2Ω(n) [R. ’11] ◮ Breakthrough: xc(TSP) ≥ 2Ω(√n)
[Fiorini, Massar, Pokutta, Tiwary, de Wolf ’12]
◮ n1/2−ε-apx for clique polytope needs super-poly size
[Braun, Fiorini, Pokutta, Steuer ’12] Improved to n1−ε [Braverman, Moitra ’13], [Braun, P. ’13]
◮ No symmetric compact form. for TSP [Yannakakis ’91]
Compact formulation for log n size matchings, but no symmetric one [Kaibel, Pashkovich & Theis ’10]
◮ xc(random 0/1 polytope) ≥ 2Ω(n) [R. ’11] ◮ Breakthrough: xc(TSP) ≥ 2Ω(√n)
[Fiorini, Massar, Pokutta, Tiwary, de Wolf ’12]
◮ n1/2−ε-apx for clique polytope needs super-poly size
[Braun, Fiorini, Pokutta, Steuer ’12] Improved to n1−ε [Braverman, Moitra ’13], [Braun, P. ’13]
◮ (2 − ε)-apx LPs for MaxCut have size nΩ(log n/ log log n)
[Chan, Lee, Raghavendra, Steurer ’13]
◮ No symmetric compact form. for TSP [Yannakakis ’91]
Compact formulation for log n size matchings, but no symmetric one [Kaibel, Pashkovich & Theis ’10]
◮ xc(random 0/1 polytope) ≥ 2Ω(n) [R. ’11] ◮ Breakthrough: xc(TSP) ≥ 2Ω(√n)
[Fiorini, Massar, Pokutta, Tiwary, de Wolf ’12]
◮ n1/2−ε-apx for clique polytope needs super-poly size
[Braun, Fiorini, Pokutta, Steuer ’12] Improved to n1−ε [Braverman, Moitra ’13], [Braun, P. ’13]
◮ (2 − ε)-apx LPs for MaxCut have size nΩ(log n/ log log n)
[Chan, Lee, Raghavendra, Steurer ’13]
Only NP-hard polytopes!! What about poly-time problems?
G = (V, E) (complete)
G = (V, E) (complete)
x(δ(v)) = 1 ∀v ∈ V xe ≥ ∀e ∈ E G = (V, E) (complete)
x(δ(v)) = 1 ∀v ∈ V xe ≥ ∀e ∈ E
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete)
x(δ(v)) = 1 ∀v ∈ V xe ≥ ∀e ∈ E U
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete)
x(δ(v)) = 1 ∀v ∈ V x(δ(U)) ≥ 1 ∀U ⊆ V : |U| odd xe ≥ ∀e ∈ E U
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete) Quick facts:
◮ Description by [Edmonds ’65]
x(δ(v)) = 1 ∀v ∈ V x(δ(U)) ≥ 1 ∀U ⊆ V : |U| odd xe ≥ ∀e ∈ E U
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete) Quick facts:
◮ Description by [Edmonds ’65] ◮ Can optimize cT x in strongly poly-time [Edmonds ’65]
x(δ(v)) = 1 ∀v ∈ V x(δ(U)) ≥ 1 ∀U ⊆ V : |U| odd xe ≥ ∀e ∈ E U
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete) Quick facts:
◮ Description by [Edmonds ’65] ◮ Can optimize cT x in strongly poly-time [Edmonds ’65] ◮ Separation problem polytime [Padberg, Rao ’82]
x(δ(v)) = 1 ∀v ∈ V x(δ(U)) ≥ 1 ∀U ⊆ V : |U| odd xe ≥ ∀e ∈ E U
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete) Quick facts:
◮ Description by [Edmonds ’65] ◮ Can optimize cT x in strongly poly-time [Edmonds ’65] ◮ Separation problem polytime [Padberg, Rao ’82] ◮ 2Θ(n) facets
x(δ(v)) = 1 ∀v ∈ V x(δ(U)) ≥ 1 ∀U ⊆ V : |U| odd xe ≥ ∀e ∈ E U
1 2 1 2 1 2 1 2 1 2 1 2
G = (V, E) (complete) Quick facts:
◮ Description by [Edmonds ’65] ◮ Can optimize cT x in strongly poly-time [Edmonds ’65] ◮ Separation problem polytime [Padberg, Rao ’82] ◮ 2Θ(n) facets
Theorem (R.13)
xc(perfect matching polytope) ≥ 2Ω(n).
◮ Previously known: xc(P) ≥ Ω(n2)
Write: P = conv({x1, . . . , xv}) = {x ∈ Rn | Ax ≤ b}
# facets # vertices Sij Sij = bi − AT
i xj
slack-matrix P
b b b b b
Write: P = conv({x1, . . . , xv}) = {x ∈ Rn | Ax ≤ b}
# facets # vertices facet i vertex j Sij Sij = bi − AT
i xj
slack-matrix P
b b b b b Aix = bi b
xj Sij
Write: P = conv({x1, . . . , xv}) = {x ∈ Rn | Ax ≤ b}
# facets # vertices
r r Sij Sij = bi − AT
i xj
slack-matrix P
b b b b b Aix = bi b
xj Sij Non-negative rank: rk+(S) = min{r | ∃U ∈ Rf×r
≥0 , V ∈ Rr×v ≥0 : S = UV }
Theorem (Yannakakis ’91)
If S is the slack-matrix for P = {x ∈ Rn | Ax ≤ b}, then xc(P) = rk+(S). P
Theorem (Yannakakis ’91)
If S is the slack-matrix for P = {x ∈ Rn | Ax ≤ b}, then xc(P) = rk+(S). Idea: Factor S = UV with U = (conic comb. to derive constraint i)i V = (slack vector of (xj, vj))j Q Aix + 0y ≤ bi
b b b b b b b b b b b b
xj
b
(xj, yj)
b
P
rk+(S) = min
r
Ri and Ri ≥ 0 rank-1 matrix
1 1 1 1 2 2 1 2 2 3 0 3 0 0 0 3 0 3
rk+(S) = min
r
λi
Ri and 0 ≤ Ri ≤ 1 rank-1 matrix
1 2 1 2 1 2 1 2
1 1
1 2
1 1
1 0 1 0 0 0 1 0 1
rk+(S) min
r
λiRi and 0 ≤ Ri ≤ 1 rank-1 matrix
1 2 1 2 1 2 1 2
1 1
1 2
1 1
1 0 1 0 0 0 1 0 1
rk+(S) min
r
λiRi and Ri ∈ {0, 1}f×v rank-1
b b b b b
rectangles [0, 1]-rank-1 matrices
0 0 0 0 1 1 0 1 1
1 0 1 0 0 0 1 0 1
rk+(S) min
r
λi W, Ri and Ri rect.
S
b b b b b
rectangles [0, 1]-rank-1 matrices W W, R ≤ α
0 0 0 0 1 1 0 1 1
1 0 1 0 0 0 1 0 1
rk+(S) min W, S W, R : R rectangle
S
b b b b b
rectangles [0, 1]-rank-1 matrices W W, R ≤ α
0 0 0 0 1 1 0 1 1
1 0 1 0 0 0 1 0 1
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1
matchings cuts
S
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1
|δ(U) ∩ M| − 1 M U matchings cuts
S
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1 ◮ Abbreviate Qℓ := {(U, M) : |δ(U) ∩ M| = ℓ} ◮ Uniform measure: µℓ(R) := |R∩Qℓ| |Qℓ|
2 2 2 Q1 Q3 matchings cuts
S
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1 ◮ Abbreviate Qℓ := {(U, M) : |δ(U) ∩ M| = ℓ} ◮ Uniform measure: µℓ(R) := |R∩Qℓ| |Qℓ| ◮ Choose
WU,M = 0
2 2 2 Q1 Q3 matchings cuts
S
b
W
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1 ◮ Abbreviate Qℓ := {(U, M) : |δ(U) ∩ M| = ℓ} ◮ Uniform measure: µℓ(R) := |R∩Qℓ| |Qℓ| ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
2 2 2 Q1 Q3 matchings cuts
S
b
W −∞ −∞ −∞
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1 ◮ Abbreviate Qℓ := {(U, M) : |δ(U) ∩ M| = ℓ} ◮ Uniform measure: µℓ(R) := |R∩Qℓ| |Qℓ| ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
1 |Q3|
|δ(U) ∩ M| = 3
2 2 2 Q1 Q3 matchings cuts
S
b
W −∞ −∞ −∞
1 |Q3| 1 |Q3| 1 |Q3|
Goal: Find W with W,S
W,R large for each rectangle. ◮ Slack matrix SUM = |δ(U) ∩ M| − 1 ◮ Abbreviate Qℓ := {(U, M) : |δ(U) ∩ M| = ℓ} ◮ Uniform measure: µℓ(R) := |R∩Qℓ| |Qℓ| ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
1 |Q3|
|δ(U) ∩ M| = 3
2 2 2 Q1 Q3 matchings cuts
S
b
W −∞ −∞ −∞
1 |Q3| 1 |Q3| 1 |Q3|
R
Claim: There is a rectangle with W, R = Θ( 1
n4 ).
Claim: There is a rectangle with W, R = Θ( 1
n4 ).
e1 e2
◮ For e1, e2 ∈ E:
Claim: There is a rectangle with W, R = Θ( 1
n4 ).
U
e1 e2
◮ For e1, e2 ∈ E: take {U | e1, e2 ∈ δ(U)}
Claim: There is a rectangle with W, R = Θ( 1
n4 ).
U
e1 e2 M
◮ For e1, e2 ∈ E: take {U | e1, e2 ∈ δ(U)} ×{M | e1, e2 ∈ M}
Claim: There is a rectangle with W, R = Θ( 1
n4 ).
U
M e1 e2
◮ For e1, e2 ∈ E: take {U | e1, e2 ∈ δ(U)} ×{M | e1, e2 ∈ M} ◮ But µk(R) = Θ( k2 n4 )
Goal: Find W with W,S
W,R large for each rectangle. ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
1 |Q3|
|δ(U) ∩ M| = 3
2 2 2 Q1 Q3 matchings cuts
S
b
W −∞ −∞ −∞
1 |Q3| 1 |Q3| 1 |Q3|
Goal: Find W with W,S
W,R large for each rectangle. ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
1 |Q3|
|δ(U) ∩ M| = 3
2 2 2 Q1 Q3 Qk k − 1 k − 1 k − 1 matchings cuts
S
b
W −∞ −∞ −∞
1 |Q3| 1 |Q3| 1 |Q3|
Goal: Find W with W,S
W,R large for each rectangle. ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
1 |Q3|
|δ(U) ∩ M| = 3 −
1 k−1 · 1 |Qk|
|δ(U) ∩ M| = k
2 2 2 Q1 Q3 Qk k − 1 k − 1 k − 1 matchings cuts
S
b
W −∞ −∞ −∞
1 |Q3| 1 |Q3| 1 |Q3|
−
1 k−1 1 |Qk|
−
1 k−1 1 |Qk|
−
1 k−1 1 |Qk|
Goal: Find W with W,S
W,R large for each rectangle. ◮ Choose
WU,M = − ∞ |δ(U) ∩ M| = 1
1 |Q3|
|δ(U) ∩ M| = 3 −
1 k−1 · 1 |Qk|
|δ(U) ∩ M| = k
◮ Then
W, S = 0 + 2 − 1 = 1
Lemma
For k large, any rectangle R has W, R ≤ 2−Ω(n). matchings cuts W −∞ −∞ −∞
1 |Q3| 1 |Q3| 1 |Q3|
−
1 k−1 1 |Qk|
−
1 k−1 1 |Qk|
−
1 k−1 1 |Qk|
R
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
matchings cuts
S
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
R matchings cuts
S
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
R matchings cuts
S
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
R matchings cuts
S
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
R matchings cuts
S
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
R matchings cuts
S
◮ Technique: Partition scheme [Razborov ’91]
Main lemma
µ1(R) = 0 = ⇒ µ3(R) ≤ O( 1
k2 ) · µk(R) + 2−Ω(n)
R
T
matchings cuts
S
◮ Technique: Partition scheme [Razborov ’91]
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
k
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
k k
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
k − 3 nodes k k
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B)
k − 3 nodes k k 2(k − 3) nodes
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B) ◮ Edges E(T)
k − 3 nodes k k 2(k − 3) nodes
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B) ◮ Edges E(T)
k − 3 nodes k k 2(k − 3) nodes
R
T
matchings cuts
S
◮ Partition T = (A, C, D, B) ◮ Edges E(T)
k − 3 nodes k k 2(k − 3) nodes U
Imagine the following setting:
Imagine the following setting:
◮ n elements
b b b b b b
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n sets
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n sets
◮ Is it possible that ≥ 1% elements are in ≤ 49% of sets?
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n sets
◮ Is it possible that ≥ 1% elements are in ≤ 49% of sets?
NO!
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n sets
◮ Is it possible that ≥ 1% elements are in ≤ 49% of sets?
NO! Proof:
◮ Take a random set from S
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n 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
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n 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)
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n 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)
subadd
≤
n
H(xi)
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n 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)
subadd
≤
n
H(xi) ≤ n − Ω(n) 1 0.5 1.0 entropy p
Imagine the following setting:
◮ n elements ◮ set system S with 2(1−o(1))n sets
b b b b b b
Questions:
◮ Is it possible that ≥ 1% of elements are in no set at all?
NO! The 0.99n active elements form at most 20.99n sets
◮ Is it possible that ≥ 1% elements are in ≤ 49% of sets?
NO!
Lemma
If S large, for most elements i, Pr
S⊆[n][S ∈ S] ≈ Pr S⊆[n][S ∈ S | i ∈ S]
R
T
matchings cuts
S
Randomly generate (U, M) ∼ Qk: µk(R) =
R
T
matchings cuts
S
Randomly generate (U, M) ∼ Qk:
µk(R) = E
T
R
T
matchings cuts
S
Randomly generate (U, M) ∼ Qk:
µk(R) = E
T
|F|=k
R
T
matchings cuts
S
Randomly generate (U, M) ∼ Qk:
µk(R) = E
T
|F|=k
R
T
matchings cuts
S
U
Randomly generate (U, M) ∼ Qk:
µk(R) = E
T
|F|=k
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[
≤O(1/k2)
Pr[(U, M) ∈ R | T, H]
]
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H∗ ⊆ F good ⇒ |H ∩ H∗| ≥ 2
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H∗ ⊆ F good ⇒ |H ∩ H∗| ≥ 2 ◮ Suppose |H ∩ H∗| ≤ 1
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H∗ ⊆ F good ⇒ |H ∩ H∗| ≥ 2 ◮ Suppose |H ∩ H∗| ≤ 1 ◮ (T, H) good
⇒ ∃M : {u, v} ∈ M
u v
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H∗ ⊆ F good ⇒ |H ∩ H∗| ≥ 2 ◮ Suppose |H ∩ H∗| ≤ 1 ◮ (T, H) good
⇒ ∃M : {u, v} ∈ M
◮ (T, H∗) good
⇒ ∃U : u, v ∈ U
u v
◮ Suppose for a fixed (T, F):
µk(R) ≈ Pr[(U, M) ∈ R | T, F] =: p
◮ Then
µ3(R) ≈ E
H∼(F
3)
[GOOD(T, H)
· Pr[(U, M) ∈ R | T, H]
]
◮ GOOD means it doesn’t matter what condition on here ◮ Suffices to show: H, H∗ ⊆ F good ⇒ |H ∩ H∗| ≥ 2 ◮ Suppose |H ∩ H∗| ≤ 1 ◮ (T, H) good
⇒ ∃M : {u, v} ∈ M
◮ (T, H∗) good
⇒ ∃U : u, v ∈ U
◮ |δ(U) ∩ M| = 1
Contradiction!
u v
Lemma
Pr[(T, H) is M-bad] ≤ ε
Lemma
Pr[(T, H) is M-bad] ≤ ε
◮ Pick H
H
Lemma
Pr[(T, H) is M-bad] ≤ ε
◮ Pick H, A
H
Lemma
Pr[(T, H) is M-bad] ≤ ε
◮ Pick H, A, ˜
B1, . . . , ˜ Bm+1.
H
Lemma
Pr[(T, H) is M-bad] ≤ ε
◮ Pick H, A, ˜
B1, . . . , ˜ Bm+1. Split ˜ Bi = Ci ˙ ∪Di.
C2 D2 . . . . . . Cm+1 Dm+1
H
Lemma
Pr[(T, H) is M-bad] ≤ ε
◮ Pick H, A, ˜
B1, . . . , ˜ Bm+1. Split ˜ Bi = Ci ˙ ∪Di.
◮ Pick randomly i ∈ {1, . . . , m}
C2 D2 . . . . . . Cm+1 Dm+1
H
Lemma
Pr[(T, H) is M-bad] ≤ ε
◮ Pick H, A, ˜
B1, . . . , ˜ Bm+1. Split ˜ Bi = Ci ˙ ∪Di.
◮ Pick randomly i ∈ {1, . . . , m} and let C := Ci, D := Di
H
Open problem
Show that there is no small SDP representing the Correlation/TSP/matching polytope!
Open problem
Show that there is no small SDP representing the Correlation/TSP/matching polytope!