Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � 0 ≤ 1 − 2 v 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 − 2 v = 1 − 4 v + 4 v + 4 v v × v = v + v 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 = 1 � � 1 − 2 v 1 − 4 v + 4 v + 4 n n v v v 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 = 1 � � 1 − 2 v 1 − 4 v + 4 v + 4 n n v v v + 4 = 1 − 4 + 4 3 1 1 1 = � = � v ∈ V ( G ) v ∈ V ( G ) 3 | V ( G ) | | V ( G ) | v v 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 = 1 � � 1 − 2 v 1 − 4 v + 4 v + 4 n n v v v + 4 = 1 − 4 + 4 3 = 2 + 1 + 3 3 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 = 1 � � 1 − 2 v 1 − 4 v + 4 v + 4 n n v v v + 4 = 1 − 4 3 = 2 + 1 3 3 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 = 1 � � 1 − 2 v 1 − 4 v + 4 v + 4 n n v v v + 4 = 1 − 4 3 − 2 = 1 − 2 3 = 4 + 2 2 3 3 13
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 1st try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) � 2 � � � 0 ≤ 1 = 1 � � 1 − 2 v 1 − 4 v + 4 v + 4 n n v v v + 4 = 1 − 4 3 − 2 = 1 − 2 3 = 4 + 2 2 3 3 ≤ 1 − 2 13
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Example - stability for Mantel = 1 Assume = 0 and 2 . Goal is G = . − 2 0 ≤ 1 − 2 3 0 ≤ − 2 3 0 ≥ Only and appear in G . 14
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) 15
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) • p n ( F ) := probability that random | F | vertices of G n induces F 15
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) • p n ( F ) := probability that random | F | vertices of G n induces F • sequence ( G n ) is convergent if p n ( F ) converge for all F 15
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) • p n ( F ) := probability that random | F | vertices of G n induces F • sequence ( G n ) is convergent if p n ( F ) converge for all F • limit object – function q : all finite 2-edge-colored graphs → [0 , 1] 15
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) • p n ( F ) := probability that random | F | vertices of G n induces F • sequence ( G n ) is convergent if p n ( F ) converge for all F • limit object – function q : all finite 2-edge-colored graphs → [0 , 1] • q yields homomorphism from linear combinations of graphs to R 15
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) • p n ( F ) := probability that random | F | vertices of G n induces F • sequence ( G n ) is convergent if p n ( F ) converge for all F • limit object – function q : all finite 2-edge-colored graphs → [0 , 1] • q yields homomorphism from linear combinations of graphs to R • the set of limit objects LIM = homomorphisms q : q ( F ) ≥ 0 15
Flag Algebras First try for Mantel More automatic approach Applications Flag Algebras - formal approach • consider 2-edge-colored complete graphs G 1 , G 2 , . . . ( | G n | → ∞ ) • p n ( F ) := probability that random | F | vertices of G n induces F • sequence ( G n ) is convergent if p n ( F ) converge for all F • limit object – function q : all finite 2-edge-colored graphs → [0 , 1] • q yields homomorphism from linear combinations of graphs to R • the set of limit objects LIM = homomorphisms q : q ( F ) ≥ 0 � � � � • we optimize on LIM T = q ∈ LIM : q = 0 1 � � 2 ≥ q ∈ LIM T q max 15
Flag Algebras First try for Mantel More automatic approach Applications More automatic approach • How to use computer to guess the right equation for you. � 2 � 0 ≤ 1 − 2 v 16
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 � � ≤ 2 + + 3 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 � � ≤ 2 + + 3 1 = + + + 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 � � ≤ 2 + + 3 1 = + + 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try Theorem (Mantel 1907) 4 n 2 edges. A triangle-free graph contains at most 1 Assume edges are red and non-edges are blue. ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 � � ≤ 2 + + 3 1 = + + ≤ 2 3 17
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 18
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 Idea: find c 1 , c 2 , c 3 ∈ R such that for every graph G 0 ≤ c 1 + c 2 + c 3 . 18
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 Idea: find c 1 , c 2 , c 3 ∈ R such that for every graph G 0 ≤ c 1 + c 2 + c 3 . After summing together � 1 � � 2 � ≤ c 1 + 3 + c 2 + 3 + c 3 and � � (0 + c 1 ) , 1 3 + c 2 , 2 ≤ max 3 + c 3 . 18
Flag Algebras First try for Mantel More automatic approach Applications Example - Mantel’s theorem, 2nd try ≤ 1 Assume = 0. (We want to conclude 2 .) + 1 + 2 = 0 3 3 Idea: find c 1 , c 2 , c 3 ∈ R such that for every graph G 0 ≤ c 1 + c 2 + c 3 . After summing together � 1 � � 2 � ≤ c 1 + 3 + c 2 + 3 + c 3 and � � (0 + c 1 ) , 1 3 + c 2 , 2 ≤ max 3 + c 3 . c 3 < 0 18
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � c 0 ≤ , , c b v v v v � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � c 0 ≤ , , c b v v v v ? ? ? ? + 1 + 1 = a + b 2 c 2 c v v v v ? ? v = 1 v × v = v × 2 v v � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � c 0 ≤ , , c b v v v v ? ? ? = a + b + c v v v ? ? v = 1 v × v = v × 2 v v � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � c 0 ≤ , , c b v v v v ? ? ? = a + b + c v v v � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � 0 ≤ 1 c � , , n c b v v v v v ? ? ? = 1 � a + b + c n v v v v � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � 0 ≤ 1 c � , , n c b v v v v v ? ? ? = 1 � a + b + c n v v v v + a + 2 c + b + 2 c = a + b 3 3 1 1 1 = � = � 3 v ∈ V ( G ) v ∈ V ( G ) | V ( G ) | | V ( G ) | v v � a � c 2 1 = � � 0 (matrix is positive semidefinite) v ∈ V ( G ) 3 | V ( G ) | v c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � 0 ≤ 1 c � , , n c b v v v v v ? ? ? = 1 � a + b + c n v v v v + a + 2 c + b + 2 c = a 3 3 1 1 1 = � = � 3 v ∈ V ( G ) v ∈ V ( G ) | V ( G ) | | V ( G ) | v v � a � c 2 1 = � � 0 (matrix is positive semidefinite) v ∈ V ( G ) 3 | V ( G ) | v c b 19
Flag Algebras First try for Mantel More automatic approach Applications Candidates for c 1 , c 2 , c 3 � � a � T � � � 0 ≤ 1 c � , , n c b v v v v v ? ? ? = 1 � a + b + c n v v v v + a + 2 c + b + 2 c = a 3 3 c 1 = a , c 2 = a + 2 c , c 3 = b + 2 c 3 3 � a � c � 0 (matrix is positive semidefinite) c b 19
Flag Algebras First try for Mantel More automatic approach Applications Using c 1 , c 2 , c 3 + 1 + 2 = 3 3 + a + 2 c + b + 2 c 0 ≤ a 3 3 � a � c � 0 (matrix is positive semidefinite) c b 20
Flag Algebras First try for Mantel More automatic approach Applications Using c 1 , c 2 , c 3 + 1 + 2 = 3 3 + a + 2 c + b + 2 c 0 ≤ a 3 3 � a � c � 0 (matrix is positive semidefinite) c b 20
Flag Algebras First try for Mantel More automatic approach Applications Using c 1 , c 2 , c 3 + 1 + 2 = 3 3 + a + 2 c + b + 2 c 0 ≤ a 3 3 � a , 1 + a + 2 c , 2 + b + 2 c � ≤ max . 3 3 � a � c � 0 (matrix is positive semidefinite) c b 20
Flag Algebras First try for Mantel More automatic approach Applications Using c 1 , c 2 , c 3 + 1 + 2 = 3 3 + a + 2 c + b + 2 c 0 ≤ a 3 3 � a , 1 + a + 2 c , 2 + b + 2 c � ≤ max . 3 3 Try � a � � � c 1 / 2 − 1 / 2 = . c b − 1 / 2 1 / 2 20
Flag Algebras First try for Mantel More automatic approach Applications Using c 1 , c 2 , c 3 + 1 + 2 = 3 3 + a + 2 c + b + 2 c 0 ≤ a 3 3 � a , 1 + a + 2 c , 2 + b + 2 c � ≤ max . 3 3 Try � a � � � c 1 / 2 − 1 / 2 = . c b − 1 / 2 1 / 2 It gives � 1 2 , 1 6 , 1 � = 1 ≤ max 2 . 2 20
Flag Algebras First try for Mantel More automatic approach Applications Optimizing a , b , c � a , 1 + a + 2 c , 2 + b + 2 c � ≤ max 3 3 Minimize d subject to a ≤ d 1+ a +2 c ≤ d 3 ( SDP ) 2+ b +2 c ≤ d 3 � � a c � 0 c b ( SDP ) can be solved on computers using CSDP or SDPA. Rounding may be needed for exact results. 21
Flag Algebras First try for Mantel More automatic approach Applications Applications Recall is the probability that 3 randomly chosen vertices form a red triangle. 22
Flag Algebras First try for Mantel More automatic approach Applications J. Balogh P. Hu Hypercubes and posets H. Liu B. L. Application to sparse structure. 23
Flag Algebras First try for Mantel More automatic approach Applications Hypercube Q n is n -dimensional hypercube ( n -cube) Q 1 Q 2 Q 3 24
Flag Algebras First try for Mantel More automatic approach Applications Hypercube Q n is n -dimensional hypercube ( n -cube) Q 1 Q 2 Q 3 Problem (Erd˝ os 1984) What is the maximum number of edges in a subgraph of Q n with no Q 2 ? 24
Flag Algebras First try for Mantel More automatic approach Applications Hypercube Q n is n -dimensional hypercube ( n -cube) Q 1 Q 2 Q 3 Problem (Erd˝ os 1984) What is the maximum number of edges in a subgraph of Q n with no Q 2 ? maximize subject to = 0 24
Flag Algebras First try for Mantel More automatic approach Applications Lower bound Conjecture (Erd˝ os 1984) In Q n where n → ∞ : ≤ 1 = 0 ⇒ 2 . 25
Flag Algebras First try for Mantel More automatic approach Applications Lower bound Conjecture (Erd˝ os 1984) In Q n where n → ∞ : ≤ 1 = 0 ⇒ 2 . As posets Q 7 Q 7 By removing every second layer, ≥ 1 / 2. 25
Flag Algebras First try for Mantel More automatic approach Applications Results about hypercubes If = 0 then Theorem (Chung 1992) ≤ 0 . 62284 . 26
Flag Algebras First try for Mantel More automatic approach Applications Results about hypercubes If = 0 then Theorem (Chung 1992) ≤ 0 . 62284 . Theorem (Thomason and Wagner 2009) ≤ 0 . 62256 . 26
Flag Algebras First try for Mantel More automatic approach Applications Results about hypercubes If = 0 then Theorem (Chung 1992) ≤ 0 . 62284 . Theorem (Thomason and Wagner 2009) ≤ 0 . 62256 . ≤ 0 . 62083 . 26
Flag Algebras First try for Mantel More automatic approach Applications Results about hypercubes If = 0 then Theorem (Chung 1992) ≤ 0 . 62284 . Theorem (Thomason and Wagner 2009) ≤ 0 . 62256 . ≤ 0 . 62083 . Theorem (Balogh, Hu, L. , Liu 2014; Baber 2014+) ≤ 0 . 6068 . (Uses Q 3 instead of Q 2 .) 26
Flag Algebras First try for Mantel More automatic approach Applications Related results - boolean lattice Let B n denote n -dimensional boolean lattice. Let F be a subposet of B n not containing ♦ . B 7 27
Flag Algebras First try for Mantel More automatic approach Applications Related results - boolean lattice Let B n denote n -dimensional boolean lattice. Let F be a subposet of B n not containing ♦ . Theorem n � � | F | ≤ ( c + o (1)) , where ⌊ n / 2 ⌋ c ≤ 2 . 3 [Griggs, Lu 2009] c ≤ 2 . 284 [Axenovich, Manske, Martin 2012] c ≤ 2 . 273 [Griggs, Li, Lu 2011] c ≤ 2 . 25 [Kramer, Martin, Young 2013] B 7 27
Flag Algebras First try for Mantel More automatic approach Applications Related results - boolean lattice Let B n denote n -dimensional boolean lattice. Let F be a subposet of B n not containing ♦ . Theorem n � � | F | ≤ ( c + o (1)) , where ⌊ n / 2 ⌋ c ≤ 2 . 3 [Griggs, Lu 2009] c ≤ 2 . 284 [Axenovich, Manske, Martin 2012] c ≤ 2 . 273 [Griggs, Li, Lu 2011] c ≤ 2 . 25 [Kramer, Martin, Young 2013] B 7 If F is a subposet of only the middle three layers of B n , then c ≤ 2 . 1547 [Manske, Shen 2013] c ≤ 2 . 15121 [Balogh, Hu, L., Liu 2014] 27
Flag Algebras First try for Mantel More automatic approach Applications Related results - boolean lattice Let B n denote n -dimensional boolean lattice. Let F be a subposet of B n not containing ♦ . Theorem n � � | F | ≤ ( c + o (1)) , where ⌊ n / 2 ⌋ c ≤ 2 . 3 [Griggs, Lu 2009] c ≤ 2 . 284 [Axenovich, Manske, Martin 2012] c ≤ 2 . 273 [Griggs, Li, Lu 2011] c ≤ 2 . 25 [Kramer, Martin, Young 2013] B 7 If F is a subposet of only the middle three layers of B n , then c ≤ 2 . 1547 [Manske, Shen 2013] c ≤ 2 . 15121 [Balogh, Hu, L., Liu 2014] c = 2 [Kramer, Martin 2015, announced] 27
Flag Algebras First try for Mantel More automatic approach Applications J. Balogh P. Hu B. L. Permutations O. Pikhurko B. Udvari J. Volec Application with exact result. 28
Flag Algebras First try for Mantel More automatic approach Applications Permutations and extremal problems Problem What is the minimum number of monotone subsequences of size k in a permutation of [ n ] ? 29
Flag Algebras First try for Mantel More automatic approach Applications Permutations and extremal problems Problem What is the minimum number of monotone subsequences of size k in a permutation of [ n ] ? (5,4,1),(5,4,2),(5,4,3) k = 3 (1,2,3) n = 5 (5,4,1,2,3) 29
Flag Algebras First try for Mantel More automatic approach Applications Permutations and extremal problems Problem What is the minimum number of monotone subsequences of size k in a permutation of [ n ] ? (5,4,1),(5,4,2),(5,4,3) k = 3 (1,2,3) n = 5 (5,4,1,2,3) (1,2,3) (4,5,1,2,3) 29
Flag Algebras First try for Mantel More automatic approach Applications Conjecture Conjecture (Myers 2002) The number of monotone subsequences of length k is minimized by a permutation on [ n ] with k − 1 increasing runs of as equal lengths as possible. k = 4 , n = 15 30
Flag Algebras First try for Mantel More automatic approach Applications Extremal case is not unique 31
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