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Transversals in hypergraphs Anders Yeo AndersYeo@gmail.com Engineering Systems and Design Singapore University of Technology and Design 8 Somapah Road 487372, Singapore Based on several papers co-authored by people such as M. A. Henning and

  1. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  2. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  3. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  4. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  5. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  6. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  7. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  8. Applications 3-SAT Colinear points in the plane. Total domination Phylogenetic trees in Biology. Feedback vertex sets (in tournaments). Coloring in hypergraphs. 2-coloring is equivalent to vertex-disjoint transversals. And many more..... Why? Anders Yeo Transversals in hypergraphs

  9. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  10. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  11. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  12. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  13. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  14. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  15. 3 -SAT Consider an instance, I , of 3-SAT with n vriables and m clauses. For example, I = ( v 1 ∨ v 2 ∨ v 3 ) ∧ ( v 2 ∨ v 3 ∨ v 4 ) ∧ · · · . v 1 v 2 v 3 v 4 v 5 v n · · · v 1 v 2 v 3 v 4 v 5 v n We build a hypergraph, H , with 2 n vertices as above. Each variable gives rise to a 2-edge. Each clause gives rise to a 3-edge. H has a transversal of size n if and only if I is satisfiable. We believe this reduction may be used to improve currently best algorithms for 3-SAT. Anders Yeo Transversals in hypergraphs

  16. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  17. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  18. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  19. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  20. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  21. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  22. Colinear points in the plane We are given a number of points in the plane. Can we find the largest set of points where no 3 are co-linear (lie on a line)? Add 3-edges between all sets of 3 points that are colinear. Remove the points of a minimum transversal. If no four points are co-linear then the resulting hypergraph is linear. Anders Yeo Transversals in hypergraphs

  23. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  24. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  25. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  26. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  27. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  28. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  29. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  30. Total Domination S is a total dominating set in G , if and only if every vertex in G has a neighbour in S . v 1 v 2 v 3 u 4 u 5 v 6 v 5 v 4 u 1 u 2 u 3 G ′ G γ t ( G ) denotes the size of a smallest total dominating set in G . Anders Yeo Transversals in hypergraphs

  31. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  32. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  33. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  34. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  35. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  36. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  37. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  38. Total Domination, Open Neighbourhood hypergraph Given G , the Open Neighbourhood Hypergraph, ONH ( G ), of G has vertex-set V ( G ) and edge-set { N ( v ) | v ∈ V ( G ) } . u 4 u 5 u 4 u 5 u 1 u 2 u 3 u 1 u 2 u 3 ONH ( G ) G A transversal in ONH ( G ) is equivalent to a total dominating set in G . Most recent results on Total Domination are found using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  39. More.... In Phylogenetic trees we want to remove as few species as possible such that the known Phylogenetic trees dont contradict each other. This can be modelled as a transversal problem in 3-uniform hypergraphs. Feedback vertex sets (in tournaments). Given a graph or digraph put a hyperdege around the vertex set of all chordless cycles. A transversal in the resulting hypergraph is a minimum feedback vertex set. For tournaments all chordless cycles are 3-cycles so our hypergraph is 3-uniform. We have best known algorithms for both the above using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  40. More.... In Phylogenetic trees we want to remove as few species as possible such that the known Phylogenetic trees dont contradict each other. This can be modelled as a transversal problem in 3-uniform hypergraphs. Feedback vertex sets (in tournaments). Given a graph or digraph put a hyperdege around the vertex set of all chordless cycles. A transversal in the resulting hypergraph is a minimum feedback vertex set. For tournaments all chordless cycles are 3-cycles so our hypergraph is 3-uniform. We have best known algorithms for both the above using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  41. More.... In Phylogenetic trees we want to remove as few species as possible such that the known Phylogenetic trees dont contradict each other. This can be modelled as a transversal problem in 3-uniform hypergraphs. Feedback vertex sets (in tournaments). Given a graph or digraph put a hyperdege around the vertex set of all chordless cycles. A transversal in the resulting hypergraph is a minimum feedback vertex set. For tournaments all chordless cycles are 3-cycles so our hypergraph is 3-uniform. We have best known algorithms for both the above using transversals in hypergraphs. Anders Yeo Transversals in hypergraphs

  42. Known results and conjectures, order and size For connected k -uniform hypergraphs there are bounds just based on the order, n , and size, m . os & Tuza) If k = 2 then τ ( H ) ≤ 2( n + m +1) (Erd¨ . 7 atal & McDiarmid) If k ≥ 2 then τ ( H ) ≤ n + ⌊ k / 2 ⌋ m (Chv´ ⌊ 3 k / 2 ⌋ . The above bounds are tight for hy- k 2 n = 3 k / 2 pergraphs of average degree 2. m = 3 For larger average degrees not τ = 2 k k much is known. 2 2 The following is tight for average degrees 3-4. e & Yeo) If k = 4 then τ ( H ) ≤ 5 n +4 m (Thomass´ . 21 Anders Yeo Transversals in hypergraphs

  43. Known results and conjectures, order and size For connected k -uniform hypergraphs there are bounds just based on the order, n , and size, m . os & Tuza) If k = 2 then τ ( H ) ≤ 2( n + m +1) (Erd¨ . 7 atal & McDiarmid) If k ≥ 2 then τ ( H ) ≤ n + ⌊ k / 2 ⌋ m (Chv´ ⌊ 3 k / 2 ⌋ . The above bounds are tight for hy- k 2 n = 3 k / 2 pergraphs of average degree 2. m = 3 For larger average degrees not τ = 2 k k much is known. 2 2 The following is tight for average degrees 3-4. e & Yeo) If k = 4 then τ ( H ) ≤ 5 n +4 m (Thomass´ . 21 Anders Yeo Transversals in hypergraphs

  44. Known results and conjectures, order and size For connected k -uniform hypergraphs there are bounds just based on the order, n , and size, m . os & Tuza) If k = 2 then τ ( H ) ≤ 2( n + m +1) (Erd¨ . 7 atal & McDiarmid) If k ≥ 2 then τ ( H ) ≤ n + ⌊ k / 2 ⌋ m (Chv´ ⌊ 3 k / 2 ⌋ . The above bounds are tight for hy- k 2 n = 3 k / 2 pergraphs of average degree 2. m = 3 For larger average degrees not τ = 2 k k much is known. 2 2 The following is tight for average degrees 3-4. e & Yeo) If k = 4 then τ ( H ) ≤ 5 n +4 m (Thomass´ . 21 Anders Yeo Transversals in hypergraphs

  45. Known results and conjectures, order and size For connected k -uniform hypergraphs there are bounds just based on the order, n , and size, m . os & Tuza) If k = 2 then τ ( H ) ≤ 2( n + m +1) (Erd¨ . 7 atal & McDiarmid) If k ≥ 2 then τ ( H ) ≤ n + ⌊ k / 2 ⌋ m (Chv´ ⌊ 3 k / 2 ⌋ . The above bounds are tight for hy- k 2 n = 3 k / 2 pergraphs of average degree 2. m = 3 For larger average degrees not τ = 2 k k much is known. 2 2 The following is tight for average degrees 3-4. e & Yeo) If k = 4 then τ ( H ) ≤ 5 n +4 m (Thomass´ . 21 Anders Yeo Transversals in hypergraphs

  46. Known results and conjectures, corollaries Given the results on the previous slide, we automatically get results for total domination in graphs with given minimum degree. If k = 3 and n = m then τ ( H ) ≤ n / 2. So, γ t ( G ) ≤ n / 4 when δ ( G ) ≥ 3. If k = 4 and n = m then τ ( H ) ≤ 3 n / 7. So, γ t ( G ) ≤ 3 n / 7 when δ ( G ) ≥ 4. Conjecture: If k = 5 and n = m then τ ( H ) ≤ 4 n / 11. The above conjecture would be best possible. Open Problem: For k ≥ 4 prove better bounds for linear hypergraphs. Anders Yeo Transversals in hypergraphs

  47. Known results and conjectures, corollaries Given the results on the previous slide, we automatically get results for total domination in graphs with given minimum degree. If k = 3 and n = m then τ ( H ) ≤ n / 2. So, γ t ( G ) ≤ n / 4 when δ ( G ) ≥ 3. If k = 4 and n = m then τ ( H ) ≤ 3 n / 7. So, γ t ( G ) ≤ 3 n / 7 when δ ( G ) ≥ 4. Conjecture: If k = 5 and n = m then τ ( H ) ≤ 4 n / 11. The above conjecture would be best possible. Open Problem: For k ≥ 4 prove better bounds for linear hypergraphs. Anders Yeo Transversals in hypergraphs

  48. Known results and conjectures, corollaries Given the results on the previous slide, we automatically get results for total domination in graphs with given minimum degree. If k = 3 and n = m then τ ( H ) ≤ n / 2. So, γ t ( G ) ≤ n / 4 when δ ( G ) ≥ 3. If k = 4 and n = m then τ ( H ) ≤ 3 n / 7. So, γ t ( G ) ≤ 3 n / 7 when δ ( G ) ≥ 4. Conjecture: If k = 5 and n = m then τ ( H ) ≤ 4 n / 11. The above conjecture would be best possible. Open Problem: For k ≥ 4 prove better bounds for linear hypergraphs. Anders Yeo Transversals in hypergraphs

  49. Known results and conjectures, corollaries Given the results on the previous slide, we automatically get results for total domination in graphs with given minimum degree. If k = 3 and n = m then τ ( H ) ≤ n / 2. So, γ t ( G ) ≤ n / 4 when δ ( G ) ≥ 3. If k = 4 and n = m then τ ( H ) ≤ 3 n / 7. So, γ t ( G ) ≤ 3 n / 7 when δ ( G ) ≥ 4. Conjecture: If k = 5 and n = m then τ ( H ) ≤ 4 n / 11. The above conjecture would be best possible. Open Problem: For k ≥ 4 prove better bounds for linear hypergraphs. Anders Yeo Transversals in hypergraphs

  50. Known results and conjectures, corollaries Given the results on the previous slide, we automatically get results for total domination in graphs with given minimum degree. If k = 3 and n = m then τ ( H ) ≤ n / 2. So, γ t ( G ) ≤ n / 4 when δ ( G ) ≥ 3. If k = 4 and n = m then τ ( H ) ≤ 3 n / 7. So, γ t ( G ) ≤ 3 n / 7 when δ ( G ) ≥ 4. Conjecture: If k = 5 and n = m then τ ( H ) ≤ 4 n / 11. The above conjecture would be best possible. Open Problem: For k ≥ 4 prove better bounds for linear hypergraphs. Anders Yeo Transversals in hypergraphs

  51. Known results and conjectures, corollaries Given the results on the previous slide, we automatically get results for total domination in graphs with given minimum degree. If k = 3 and n = m then τ ( H ) ≤ n / 2. So, γ t ( G ) ≤ n / 4 when δ ( G ) ≥ 3. If k = 4 and n = m then τ ( H ) ≤ 3 n / 7. So, γ t ( G ) ≤ 3 n / 7 when δ ( G ) ≥ 4. Conjecture: If k = 5 and n = m then τ ( H ) ≤ 4 n / 11. The above conjecture would be best possible. Open Problem: For k ≥ 4 prove better bounds for linear hypergraphs. Anders Yeo Transversals in hypergraphs

  52. Known results and conjectures, regularity Very little is known regarding the regularity of a hypergraphs and the following has been conjectured. Conjecture For all k ≥ 2, if H is a k -uniform hypergraph, then the following holds. (a) If H is 3-regular then τ ( H ) ≤ 3 n 2 k (holds for k ≤ 4). (b) If H is 4-regular then τ ( H ) ≤ 12 n 7 k (holds for k ≤ 4). (c) If H is 5-regular then τ ( H ) ≤ 20 n 11 k (holds for k ≤ 3). (d) If H is 6-regular then τ ( H ) ≤ 24 n 13 k (holds for k ≤ 2). We now believe we have a proof of (a) for when k = 6. That is we can prove, Tuza-Vestergaard Conjecture (Now proved?): Every 3-regular 6-uniform hypergraph, H , satisfies τ ( H ) ≤ | V ( H ) | / 4. Anders Yeo Transversals in hypergraphs

  53. Known results and conjectures, regularity Very little is known regarding the regularity of a hypergraphs and the following has been conjectured. Conjecture For all k ≥ 2, if H is a k -uniform hypergraph, then the following holds. (a) If H is 3-regular then τ ( H ) ≤ 3 n 2 k (holds for k ≤ 4). (b) If H is 4-regular then τ ( H ) ≤ 12 n 7 k (holds for k ≤ 4). (c) If H is 5-regular then τ ( H ) ≤ 20 n 11 k (holds for k ≤ 3). (d) If H is 6-regular then τ ( H ) ≤ 24 n 13 k (holds for k ≤ 2). We now believe we have a proof of (a) for when k = 6. That is we can prove, Tuza-Vestergaard Conjecture (Now proved?): Every 3-regular 6-uniform hypergraph, H , satisfies τ ( H ) ≤ | V ( H ) | / 4. Anders Yeo Transversals in hypergraphs

  54. Known results and conjectures, regularity Very little is known regarding the regularity of a hypergraphs and the following has been conjectured. Conjecture For all k ≥ 2, if H is a k -uniform hypergraph, then the following holds. (a) If H is 3-regular then τ ( H ) ≤ 3 n 2 k (holds for k ≤ 4). (b) If H is 4-regular then τ ( H ) ≤ 12 n 7 k (holds for k ≤ 4). (c) If H is 5-regular then τ ( H ) ≤ 20 n 11 k (holds for k ≤ 3). (d) If H is 6-regular then τ ( H ) ≤ 24 n 13 k (holds for k ≤ 2). We now believe we have a proof of (a) for when k = 6. That is we can prove, Tuza-Vestergaard Conjecture (Now proved?): Every 3-regular 6-uniform hypergraph, H , satisfies τ ( H ) ≤ | V ( H ) | / 4. Anders Yeo Transversals in hypergraphs

  55. Known results and conjectures, regularity Very little is known regarding the regularity of a hypergraphs and the following has been conjectured. Conjecture For all k ≥ 2, if H is a k -uniform hypergraph, then the following holds. (a) If H is 3-regular then τ ( H ) ≤ 3 n 2 k (holds for k ≤ 4). (b) If H is 4-regular then τ ( H ) ≤ 12 n 7 k (holds for k ≤ 4). (c) If H is 5-regular then τ ( H ) ≤ 20 n 11 k (holds for k ≤ 3). (d) If H is 6-regular then τ ( H ) ≤ 24 n 13 k (holds for k ≤ 2). We now believe we have a proof of (a) for when k = 6. That is we can prove, Tuza-Vestergaard Conjecture (Now proved?): Every 3-regular 6-uniform hypergraph, H , satisfies τ ( H ) ≤ | V ( H ) | / 4. Anders Yeo Transversals in hypergraphs

  56. Progress on the Tuza-Vestergaard Conjecture The well-known Tuza-Vestergaard conjecture was published in 2002. Bounds of 0 . 27778 n , 0 . 27037 n and 0 . 268116 n have been showed. There are infinitely many hypergraphs achieving equality in the bound n / 4. (Take any 3-regular 3-uniform hyper- graph, H , with τ ( H ) = n / 2 and du- plicate every vertex). Our proof of the Tuza-Vestergaard conjecture also implies a number of other results. For example, the following for k = 6. Conjecture: For all k ≥ 2, if H is a k -uniform hypergraph with maximum degree ∆( H ) ≤ 3 then τ ( H ) ≤ | V ( H ) | + | E ( H ) | . 6 k Anders Yeo Transversals in hypergraphs

  57. Progress on the Tuza-Vestergaard Conjecture The well-known Tuza-Vestergaard conjecture was published in 2002. Bounds of 0 . 27778 n , 0 . 27037 n and 0 . 268116 n have been showed. There are infinitely many hypergraphs achieving equality in the bound n / 4. (Take any 3-regular 3-uniform hyper- graph, H , with τ ( H ) = n / 2 and du- plicate every vertex). Our proof of the Tuza-Vestergaard conjecture also implies a number of other results. For example, the following for k = 6. Conjecture: For all k ≥ 2, if H is a k -uniform hypergraph with maximum degree ∆( H ) ≤ 3 then τ ( H ) ≤ | V ( H ) | + | E ( H ) | . 6 k Anders Yeo Transversals in hypergraphs

  58. Progress on the Tuza-Vestergaard Conjecture The well-known Tuza-Vestergaard conjecture was published in 2002. Bounds of 0 . 27778 n , 0 . 27037 n and 0 . 268116 n have been showed. There are infinitely many hypergraphs achieving equality in the bound n / 4. (Take any 3-regular 3-uniform hyper- graph, H , with τ ( H ) = n / 2 and du- plicate every vertex). Our proof of the Tuza-Vestergaard conjecture also implies a number of other results. For example, the following for k = 6. Conjecture: For all k ≥ 2, if H is a k -uniform hypergraph with maximum degree ∆( H ) ≤ 3 then τ ( H ) ≤ | V ( H ) | + | E ( H ) | . 6 k Anders Yeo Transversals in hypergraphs

  59. Progress on the Tuza-Vestergaard Conjecture The well-known Tuza-Vestergaard conjecture was published in 2002. Bounds of 0 . 27778 n , 0 . 27037 n and 0 . 268116 n have been showed. There are infinitely many hypergraphs achieving equality in the bound n / 4. (Take any 3-regular 3-uniform hyper- graph, H , with τ ( H ) = n / 2 and du- plicate every vertex). Our proof of the Tuza-Vestergaard conjecture also implies a number of other results. For example, the following for k = 6. Conjecture: For all k ≥ 2, if H is a k -uniform hypergraph with maximum degree ∆( H ) ≤ 3 then τ ( H ) ≤ | V ( H ) | + | E ( H ) | . 6 k Anders Yeo Transversals in hypergraphs

  60. Progress on the Tuza-Vestergaard Conjecture The well-known Tuza-Vestergaard conjecture was published in 2002. Bounds of 0 . 27778 n , 0 . 27037 n and 0 . 268116 n have been showed. There are infinitely many hypergraphs achieving equality in the bound n / 4. (Take any 3-regular 3-uniform hyper- graph, H , with τ ( H ) = n / 2 and du- plicate every vertex). Our proof of the Tuza-Vestergaard conjecture also implies a number of other results. For example, the following for k = 6. Conjecture: For all k ≥ 2, if H is a k -uniform hypergraph with maximum degree ∆( H ) ≤ 3 then τ ( H ) ≤ | V ( H ) | + | E ( H ) | . 6 k Anders Yeo Transversals in hypergraphs

  61. Progress on the Tuza-Vestergaard Conjecture In order to prove the Tuza-Vestergaard Conjecture we prove a much stronger result. If H is a 6-uniform hypergraph on n vertices and m edges with ∆( H ) ≤ 3, then 20 τ ( H ) ≤ 2 n + 6 m + ρ , where ρ is a number that depends on the existence of certain hypergraphs, which will be described in more detail below. SB 1 B 1 B 2 B 3 B 4 Anders Yeo Transversals in hypergraphs

  62. Progress on the Tuza-Vestergaard Conjecture In order to prove the Tuza-Vestergaard Conjecture we prove a much stronger result. If H is a 6-uniform hypergraph on n vertices and m edges with ∆( H ) ≤ 3, then 20 τ ( H ) ≤ 2 n + 6 m + ρ , where ρ is a number that depends on the existence of certain hypergraphs, which will be described in more detail below. SB 1 B 1 B 2 B 3 B 4 Anders Yeo Transversals in hypergraphs

  63. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  64. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  65. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  66. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  67. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  68. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  69. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  70. Progress on the Tuza-Vestergaard Conjecture We recusively define an infinite family B of bad hypergraphs as follows. Let B = { B 1 , B 2 , B 3 , B 4 } and recursively do the following.... No minimum transversal contains a red and green In B → vertex. Any hypergraph in B is 2 short in the formula 20 τ ≤ 2 n + 6 m . What about SB 1 ? Anders Yeo Transversals in hypergraphs

  71. Progress on the Tuza-Vestergaard Conjecture SB 1 is 4 short in the formula 20 τ ≤ 2 n + 6 m . If C is a SB 1 subgraph in H , then define f ( C ) as follows: # degree 2’s in C 0 1 2 3 4 5 6 7 8 9 f ( C ) 0 0 0 1 1 1 2 2 3 4  2 R is a component in H  If R ∈ B then f ( R ) = 1 R is intersected by 1 edge . 0 R is intersected by ≥ 2 edges  Let ρ = max � Q ∈Q f ( Q ) over all families of vertex-disjoint subgraphs Q in H ( F ( Q ) = 0 if Q �∈ B ∪ SB 1 ). Anders Yeo Transversals in hypergraphs

  72. Progress on the Tuza-Vestergaard Conjecture SB 1 is 4 short in the formula 20 τ ≤ 2 n + 6 m . If C is a SB 1 subgraph in H , then define f ( C ) as follows: # degree 2’s in C 0 1 2 3 4 5 6 7 8 9 f ( C ) 0 0 0 1 1 1 2 2 3 4  2 R is a component in H  If R ∈ B then f ( R ) = 1 R is intersected by 1 edge . 0 R is intersected by ≥ 2 edges  Let ρ = max � Q ∈Q f ( Q ) over all families of vertex-disjoint subgraphs Q in H ( F ( Q ) = 0 if Q �∈ B ∪ SB 1 ). Anders Yeo Transversals in hypergraphs

  73. Progress on the Tuza-Vestergaard Conjecture SB 1 is 4 short in the formula 20 τ ≤ 2 n + 6 m . If C is a SB 1 subgraph in H , then define f ( C ) as follows: # degree 2’s in C 0 1 2 3 4 5 6 7 8 9 f ( C ) 0 0 0 1 1 1 2 2 3 4  2 R is a component in H  If R ∈ B then f ( R ) = 1 R is intersected by 1 edge . 0 R is intersected by ≥ 2 edges  Let ρ = max � Q ∈Q f ( Q ) over all families of vertex-disjoint subgraphs Q in H ( F ( Q ) = 0 if Q �∈ B ∪ SB 1 ). Anders Yeo Transversals in hypergraphs

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