EKR for Matchings

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EKR for Matchings

  1. 1. EKR for Matchings.........
  2. 2. Mrn: the family of all matchings of size r in K2n.(Note that r n.)
  3. 3. Mrn: the family of all matchings of size r in K2n.(Note that r n.)..1.2.3.4.5.6.7.8
  4. 4. Mrn: the family of all matchings of size r in K2n.(Note that r n.)..1.2.3.4.5.6.7.8
  5. 5. Mrn: the family of all matchings of size r in K2n.(Note that r n.)..1.2.3.4.5.6.7.8
  6. 6. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)
  7. 7. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)..1.2.3.4.5.6.7.8
  8. 8. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)..1.2.3.4.5.6.7.8
  9. 9. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)..1.2.3.4.5.6.7.8
  10. 10. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)..1.2.3.4.5.6.7.8
  11. 11. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)..1.2.3.4.5.6.7.8
  12. 12. Mrn(e): all matchings of K2n, on r edges, containing e.(This is the star centered at e.)..1.2.3.4.5.6.7.8
  13. 13. The number of matchings of size r in K2n is:
  14. 14. The number of matchings of size r in K2n is:(2n2)
  15. 15. The number of matchings of size r in K2n is:(2n2)(2n − 22)
  16. 16. The number of matchings of size r in K2n is:(2n2)(2n − 22)(2n − 42)
  17. 17. The number of matchings of size r in K2n is:(2n2)(2n − 22)(2n − 42)· · ·
  18. 18. The number of matchings of size r in K2n is:(2n2)(2n − 22)(2n − 42)· · ·(2n − 2(r − 1)2)r choices
  19. 19. The number of matchings of size r in K2n is:(2n2)(2n − 22)(2n − 42)· · ·(2n − 2(r − 1)2)r choicesr!
  20. 20. The number of matchings of size r in K2n is:|Mrn| =(2n2)(2n − 22)(2n − 42)· · ·(2n − 2(r − 1)2)r choicesr!
  21. 21. The number of matchings of size r in a star centered at e is:
  22. 22. The number of matchings of size r in a star centered at e is:(2n − 22)
  23. 23. The number of matchings of size r in a star centered at e is:(2n − 22)(2n − 42)
  24. 24. The number of matchings of size r in a star centered at e is:(2n − 22)(2n − 42)· · ·
  25. 25. The number of matchings of size r in a star centered at e is:(2n − 22)(2n − 42)· · ·(2n − 2(r − 1)2)(r−1) choices
  26. 26. The number of matchings of size r in a star centered at e is:(2n − 22)(2n − 42)· · ·(2n − 2(r − 1)2)(r−1) choices(r − 1)!
  27. 27. The number of matchings of size r in a star centered at e is:|Mrn(e)| =(2n − 22)(2n − 42)· · ·(2n − 2(r − 1)2)(r−1) choices(r − 1)!
  28. 28. The EKR Statement
  29. 29. The EKR StatementIf A is an intersecting family of r-matchings in K2n, then:|A| |Mrn(e)|,
  30. 30. The EKR StatementIf A is an intersecting family of r-matchings in K2n, then:|A| |Mrn(e)|,with equality holding if and only if A is a star.
  31. 31. EKR for Set Systems
  32. 32. The EKR Statement for Families of Sets
  33. 33. The EKR Statement for Families of SetsIf A is an intersecting family of r-subsets of [n], then:|A|(n − 1r − 1),
  34. 34. The EKR Statement for Families of SetsIf A is an intersecting family of r-subsets of [n], then:|A|(n − 1r − 1),with equality holding if and only if A is a star.
  35. 35. Proof by PictureKatona (1972)
  36. 36. Proof by PictureKatona (1972)Arrange the elements of the universe on a circle.Call this arrangement σ.
  37. 37. Proof by PictureKatona (1972)Arrange the elements of the universe on a circle.Call this arrangement σ.Aσ: those sets of A that happen to be intervals on this circulararrangement.
  38. 38. Proof by PictureKatona (1972)Arrange the elements of the universe on a circle.Call this arrangement σ.Aσ: those sets of A that happen to be intervals on this circulararrangement.How big can Aσ be, given that A is intersecting?
  39. 39. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15
  40. 40. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  41. 41. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  42. 42. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  43. 43. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  44. 44. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  45. 45. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  46. 46. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  47. 47. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  48. 48. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  49. 49. Proof by PictureKatona (1972)Arrange the elements of the universe on a circle.Call this arrangement σ.Aσ: those sets of A that happen to be intervals on this circulararrangement.How big can Aσ be, given that A is intersecting?
  50. 50. Proof by PictureKatona (1972)Arrange the elements of the universe on a circle.Call this arrangement σ.Aσ: those sets of A that happen to be intervals on this circulararrangement.|Aσ| r
  51. 51. The CountKatona (1972)How many pairs (S, σ) are there,where S ∈ A, and σ is a cyclic permutation of [n]?
  52. 52. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  53. 53. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  54. 54. ..1.15.2.4.5.6.7.8.9.10.11.12.13.14.3.r!.∈ A
  55. 55. ..3.14.3.4.5.6.7.8.9.10.11.12.13.2.15.r!.∈ A
  56. 56. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.r!.∈ A
  57. 57. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.r! · (n − r)!.∈ A
  58. 58. The CountKatona (1972)How many pairs (S, σ) are there,where S ∈ A, σ is a cyclic permutation of [n], and S occurs as aninterval in σ?
  59. 59. The CountKatona (1972)How many pairs (S, σ) are there,where S ∈ A, σ is a cyclic permutation of [n], and S occurs as aninterval in σ?|A| · r! · (n − r)!
  60. 60. The CountKatona (1972)How many pairs (S, σ) are there,where S ∈ A, σ is a cyclic permutation of [n], and S occurs as aninterval in σ?|A| · r! · (n − r)! (n − 1)!r
  61. 61. The CountKatona (1972)How many pairs (S, σ) are there,where S ∈ A, σ is a cyclic permutation of [n], and S occurs as aninterval in σ?|A| · r! · (n − r)! (n − 1)!r|A| (n−1)!rr!(n−r)!
  62. 62. The CountKatona (1972)How many pairs (S, σ) are there,where S ∈ A, σ is a cyclic permutation of [n], and S occurs as aninterval in σ?|A| · r! · (n − r)! (n − 1)!r|A| (n−1)!rr!(n−r)! =(n−1r−1).
  63. 63. Setting Up Katona-Like Local EnvironmentsOur universe is now the set of all edges in K2n.
  64. 64. Setting Up Katona-Like Local EnvironmentsOur universe is now the set of all edges in K2n.We might begin by considering cyclic permutations of these edges.
  65. 65. Setting Up Katona-Like Local EnvironmentsOur universe is now the set of all edges in K2n.We might begin by considering cyclic permutations of these edges.A direct approach only leads to a weak bound...
  66. 66. ...and informally, the bounds are loose becausean interval of an arbitrary permutation of E(K2n)is not automatically a matching.
  67. 67. ...and informally, the bounds are loose becausean interval of an arbitrary permutation of E(K2n)is not automatically a matching.Our first goal, therefore, is to come up with a more suitableselection of cyclic permutations.
  68. 68. Baranyai PartitionsA Decomposition of the Edges of K2ninto (2n − 1) perfect matchings.
  69. 69. Baranyai Partitions..1.2.3.4.5.6.7.8
  70. 70. Baranyai Partitions..1.2.3.4.5.6.7.8
  71. 71. Baranyai Partitions..1.2.3.4.5.6.7.8
  72. 72. Baranyai Partitions..1.2.3.4.5.6.7.8
  73. 73. Baranyai Partitions..1.2.3.4.5.6.7.8
  74. 74. Baranyai Partitions..1.2.3.4.5.6.7.8
  75. 75. Baranyai Partitions..1.2.3.4.5.6.7.8
  76. 76. Baranyai Partitions..1.2.3.4.5.6.7.8
  77. 77. ..1.2.3.4.5.6.7.8.(4, 5) .(3, 6) .(2, 7) .(1, 8)
  78. 78. ..1.2.3.4.5.6.7.8.(6, 5) .(7, 4) .(1, 3) .(2, 8)
  79. 79. ..1.2.3.4.5.6.7.8.(7, 6) .(1, 5) .(2, 4) .(3, 8)
  80. 80. ..1.2.3.4.5.6.7.8.(7, 1) .(2, 6) .(3, 5) .(4, 8)
  81. 81. ..1.2.3.4.5.6.7.8.(1, 2) .(3, 7) .(6, 4) .(5, 8)
  82. 82. ..1.2.3.4.5.6.7.8.(2, 3) .(4, 1) .(7, 5) .(6, 8)
  83. 83. ..1.2.3.4.5.6.7.8.(3, 4) .(5, 2) .(1, 6) .(7, 8)
  84. 84. ..(4, 5) .(3, 6) .(2, 7) .(1, 8)
  85. 85. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8)
  86. 86. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8).(7, 6) .(1, 5) .(2, 4) .(3, 8)
  87. 87. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8).(7, 6) .(1, 5) .(2, 4) .(3, 8).(7, 1) .(2, 6) .(3, 5) .(4, 8)
  88. 88. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8).(7, 6) .(1, 5) .(2, 4) .(3, 8).(7, 1) .(2, 6) .(3, 5) .(4, 8).(1, 2) .(3, 7) .(6, 4) .(5, 8)
  89. 89. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8).(7, 6) .(1, 5) .(2, 4) .(3, 8).(7, 1) .(2, 6) .(3, 5) .(4, 8).(1, 2) .(3, 7) .(6, 4) .(5, 8).(2, 3) .(4, 1) .(7, 5) .(6, 8)
  90. 90. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8).(7, 6) .(1, 5) .(2, 4) .(3, 8).(7, 1) .(2, 6) .(3, 5) .(4, 8).(1, 2) .(3, 7) .(6, 4) .(5, 8).(2, 3) .(4, 1) .(7, 5) .(6, 8).(3, 4) .(5, 2) .(1, 6) .(7, 8)
  91. 91. ..(4, 5) .(3, 6) .(2, 7) .(1, 8).(6, 5) .(7, 4) .(1, 3) .(2, 8).(7, 6) .(1, 5) .(2, 4) .(3, 8).(7, 1) .(2, 6) .(3, 5) .(4, 8).(1, 2) .(3, 7) .(6, 4) .(5, 8).(2, 3) .(4, 1) .(7, 5) .(6, 8).(3, 4) .(5, 2) .(1, 6) .(7, 8)
  92. 92. We have just described one cyclic permutation of E(K2n).
  93. 93. We have just described one cyclic permutation of E(K2n).We can generate other cyclic permutations of E(K2n) using thismethod.
  94. 94. We have just described one cyclic permutation of E(K2n).We can generate other cyclic permutations of E(K2n) using thismethod.Let σ be a permutation of [2n].Start with the following Baranyai Partition...(illustration for n = 4):
  95. 95. ..σ(1).σ(2).σ(3).σ(4).σ(5).σ(6).σ(7).σ(8)
  96. 96. ...and generate the following cyclic order, just as before:
  97. 97. ..(σ(4), σ(5)) .(σ(3), σ(6)) .(σ(2), σ(7)) .(σ(1), σ(8)).(σ(6), σ(5)) .(σ(7), σ(4)) .(σ(1), σ(3)) .(σ(2), σ(8)).(σ(7), σ(6)) .(σ(1), σ(5)) .(σ(2), σ(4)) .(σ(3), σ(8)).(σ(7), σ(1)) .(σ(2), σ(6)) .(σ(3), σ(5)) .(σ(4), σ(8)).(σ(1), σ(2)) .(σ(3), σ(7)) .(σ(6), σ(4)) .(σ(5), σ(8)).(σ(2), σ(3)) .(σ(4), σ(1)) .(σ(7), σ(5)) .(σ(6), σ(8)).(σ(3), σ(4)) .(σ(5), σ(2)) .(σ(1), σ(6)) .(σ(7), σ(8))
  98. 98. The cyclic orders that we have just generated will serve aswireframes on which we can project elements of A as intervals, a laKatona’s proof for the classic version of the theorem.
  99. 99. The cyclic orders that we have just generated will serve aswireframes on which we can project elements of A as intervals, a laKatona’s proof for the classic version of the theorem.Preliminary Observation.Every interval of length r is a matching, as long as r < n.
  100. 100. The cyclic orders that we have just generated will serve aswireframes on which we can project elements of A as intervals, a laKatona’s proof for the classic version of the theorem.Preliminary Observation.Every interval of length r is a matching, as long as r < n.
  101. 101. .................
  102. 102. .................
  103. 103. .................
  104. 104. .................
  105. 105. .................
  106. 106. Proving the EKR boundLet A be an intersecting family of r-matchings of K2n, where r < n.
  107. 107. Proving the EKR boundLet A be an intersecting family of r-matchings of K2n, where r < n.Let σ be a permutation of [2n] - consider the cyclic permutationsof E(K2n) that we generated based on σ - let’s call this χσ.
  108. 108. Proving the EKR boundLet A be an intersecting family of r-matchings of K2n, where r < n.Let σ be a permutation of [2n] - consider the cyclic permutationsof E(K2n) that we generated based on σ - let’s call this χσ.Aσ: those sets of A that happen to be intervals on this circulararrangement.
  109. 109. Proving the EKR boundLet A be an intersecting family of r-matchings of K2n, where r < n.Let σ be a permutation of [2n] - consider the cyclic permutationsof E(K2n) that we generated based on σ - let’s call this χσ.Aσ: those sets of A that happen to be intervals on this circulararrangement.How big can Aσ be, given that A is intersecting?
  110. 110. Proving the EKR boundLet A be an intersecting family of r-matchings of K2n, where r < n.Let σ be a permutation of [2n] - consider the cyclic permutationsof E(K2n) that we generated based on σ - let’s call this χσ.Aσ: those sets of A that happen to be intervals on this circulararrangement.|Aσ| r, for the same reasons as before.
  111. 111. Proving the EKR bound (contd.)As before, consider the set of pairs (M, σ), where:♣ M is a r-matching of K2n,
  112. 112. Proving the EKR bound (contd.)As before, consider the set of pairs (M, σ), where:♣ M is a r-matching of K2n,♣ M belongs to A,
  113. 113. Proving the EKR bound (contd.)As before, consider the set of pairs (M, σ), where:♣ M is a r-matching of K2n,♣ M belongs to A,♣ σ is a permutation of [2n],
  114. 114. Proving the EKR bound (contd.)As before, consider the set of pairs (M, σ), where:♣ M is a r-matching of K2n,♣ M belongs to A,♣ σ is a permutation of [2n],♣ and M occurs as an interval in χσ.
  115. 115. Proving the EKR bound (contd.)As before, consider the set of pairs (M, σ), where:♣ M is a r-matching of K2n,♣ M belongs to A,♣ σ is a permutation of [2n],♣ and M occurs as an interval in χσ.Clearly,#(M, σ) r · (2n)!
  116. 116. Now, let us investigate the following question:In how many cyclic orders χσ can a given matching M occur as aninterval?
  117. 117. To address the question, let us recall the cyclic orders χσ.
  118. 118. To address the question, let us recall the cyclic orders χσ.It is useful to think of χσ as being composed of (2n − 1) chunks, asthey arise from the (2n − 1) parts of the Baranyai partitions.
  119. 119. To address the question, let us recall the cyclic orders χσ.It is useful to think of χσ as being composed of (2n − 1) chunks, asthey arise from the (2n − 1) parts of the Baranyai partitions.In the picture that follows, each row is a “chunk”.
  120. 120. ..(σ(4), σ(5)) .(σ(3), σ(6)) .(σ(2), σ(7)) .(σ(1), σ(8)).(σ(6), σ(5)) .(σ(7), σ(4)) .(σ(1), σ(3)) .(σ(2), σ(8)).(σ(7), σ(6)) .(σ(1), σ(5)) .(σ(2), σ(4)) .(σ(3), σ(8)).(σ(7), σ(1)) .(σ(2), σ(6)) .(σ(3), σ(5)) .(σ(4), σ(8)).(σ(1), σ(2)) .(σ(3), σ(7)) .(σ(6), σ(4)) .(σ(5), σ(8)).(σ(2), σ(3)) .(σ(4), σ(1)) .(σ(7), σ(5)) .(σ(6), σ(8)).(σ(3), σ(4)) .(σ(5), σ(2)) .(σ(1), σ(6)) .(σ(7), σ(8))
  121. 121. Task 1.Enumerate all σ whereM belongs to χσ as an interval,and M lies entirely inside one of the chunks of χσ.
  122. 122. If M has r edges, then we have r! ways to order the edges of M.For a fixed ordering p of the edges of M, let us devise a σ such thatχσ will contain M as an interval in its ith chunk, with the edges ofM appearing in the order prescribed by p.
  123. 123. ..i..............
  124. 124. ..i..............
  125. 125. ..i..............
  126. 126. ..i..............
  127. 127. ..i..............
  128. 128. We have (n − r) choices to begin the placement of the edges of M.
  129. 129. We have (n − r) choices to begin the placement of the edges of M.The ordering of the vertices that are not incident to M areimmaterial, and there are (2n − 2r)! such orderings.
  130. 130. We have (n − r) choices to begin the placement of the edges of M.The ordering of the vertices that are not incident to M areimmaterial, and there are (2n − 2r)! such orderings.Finally, for a fixed realization of M respecting the order p, we maystill swap the endpoints of M to get a different permutation withthe same realization.
  131. 131. ..i..............
  132. 132. ..i..............
  133. 133. ..i..............
  134. 134. ..i..............
  135. 135. Note that there are 2r choices for swapping the endpoints of theedges of M.
  136. 136. Note that there are 2r choices for swapping the endpoints of theedges of M.Putting everything together, we have....
  137. 137. These many ways in whichthe chunk of σ in which we realize Mcan be chosen:(2n − 1)
  138. 138. These many ways in whichthe ordering of the edges of Mcan be chosen:(2n − 1) · r!
  139. 139. These many ways in whichthe starting point of Mcan be chosen:(2n − 1) · r! · (n − r)
  140. 140. These many ways in whichthe ordering the vertices not incident to Mcan be chosen:(2n − 1) · r! · (n − r) · (2n − 2r)!
  141. 141. These many ways in which“swapped” vertices within edges of Mcan be chosen:(2n − 1) · r! · (n − r) · (2n − 2r)! · 2r
  142. 142. These many ways in whichthe permutation σcan be chosen:(2n − 1) · r! · (n − r) · (2n − 2r)! · 2r
  143. 143. Task 2.Enumerate all σ whereM belongs to χσ as an interval,and M “splits across” two chunks of χσ.
  144. 144. If M has r edges, then we have r! ways to order the edges of M.For a fixed ordering p of the edges of M, let us devise a σ such thatχσ will contain M as an interval starting in its ith chunk, with theedges of M appearing in the order prescribed by p,and spilling over to the (i + 1)th chunk.
  145. 145. ..i..............(Crossover edge not depicted for clarity.)
  146. 146. ..i..............(Crossover edge not depicted for clarity.)
  147. 147. ..i..............(Crossover edge not depicted for clarity.)
  148. 148. ..i..............(Crossover edge not depicted for clarity.)
  149. 149. This time, we have r choices to begin the placement of the edges ofM.
  150. 150. This time, we have r choices to begin the placement of the edges ofM.As before, the ordering of the vertices that are not incident to Mare immaterial, and there are (2n − 2r)! such orderings.
  151. 151. This time, we have r choices to begin the placement of the edges ofM.As before, the ordering of the vertices that are not incident to Mare immaterial, and there are (2n − 2r)! such orderings.And again, for a fixed realization of M respecting the order p, wemay still swap (in 2r ways) the endpoints of M to get a differentpermutation with the same realization.
  152. 152. These many ways in whichthe chunk of σ in which we realize Mcan be chosen:(2n − 1)
  153. 153. These many ways in whichthe ordering of the edges of Mcan be chosen:(2n − 1) · r!
  154. 154. These many ways in whichthe starting point of Mcan be chosen:(2n − 1) · r! · r
  155. 155. These many ways in whichthe ordering the vertices not incident to Mcan be chosen:(2n − 1) · r! · r · (2n − 2r)!
  156. 156. These many ways in which“swapped” vertices within edges of Mcan be chosen:(2n − 1) · r! · r · (2n − 2r)! · 2r
  157. 157. These many ways in whichthe permutation σcan be chosen:(2n − 1) · r! · r · (2n − 2r)! · 2r
  158. 158. We are now in a position to tackle this:In how many cyclic orders χσ can a given matching M occur as aninterval?
  159. 159. We are now in a position to tackle this:In how many cyclic orders χσ can a given matching M occur as aninterval?(2n − 1) · r! · r · (2n − 2r)! · 2r+(2n − 1) · r! · (n − r) · (2n − 2r)! · 2r
  160. 160. We are now in a position to tackle this:In how many cyclic orders χσ can a given matching M occur as aninterval?(2n − 1) · r! · r · (2n − 2r)! · 2r+(2n − 1) · r! · (n − r) · (2n − 2r)! · 2r= (2n − 1) · r! · n · (2n − 2r)! · 2r
  161. 161. Finally, we have:#(M, σ) (2n)!r
  162. 162. Finally, we have:|A| · (2n − 1)r!n(2n − 2r)!2r= #(M, σ) (2n)!r
  163. 163. Finally, we have:|A| · (2n − 1)r!n(2n − 2r)!2r= #(M, σ) (2n)!r|A|(2n)!r(2n − 1)r!n(2n − 2r)!2r
  164. 164. Finally, we have:|A| · (2n − 1)r!n(2n − 2r)!2r= #(M, σ) (2n)!r|A|(2n)!r(2n − 1)r!n(2n − 2r)!2r= |Mrn(e)|
  165. 165. |A|(2n)!r(2n − 1)r!n(2n − 2r)!2r
  166. 166. |A|(2n)!r(2n − 1)r!n(2n − 2r)!2r|A|1(r − 1)!·(2n)!(2n − 1)n(2n − 2r)!2r
  167. 167. |A|(2n)!r(2n − 1)r!n(2n − 2r)!2r|A|1(r − 1)!·(2n)!(2n − 1)n(2n − 2r)!2r|A|1(r − 1)!·(2n)(2n − 1) · · · (2n − 2r + 1)(2n − 1)n2r
  168. 168. |A|(2n)!r(2n − 1)r!n(2n − 2r)!2r|A|1(r − 1)!·(2n)!(2n − 1)n(2n − 2r)!2r|A|1(r − 1)!·(2n)(2n − 1) · · · (2n − 2r + 1)(2n − 1)n2r|A|1(r − 1)!·(2n − 2)(2n − 3) · · · (2n − 2r + 1)2r−1
  169. 169. |A|(2n)!r(2n − 1)r!n(2n − 2r)!2r|A|1(r − 1)!·(2n)!(2n − 1)n(2n − 2r)!2r|A|1(r − 1)!·(2n)(2n − 1) · · · (2n − 2r + 1)(2n − 1)n2r|A|1(r − 1)!·(2n − 2)(2n − 3) · · · (2n − 2r + 1)2r−1|A|1(r − 1)!·(2n − 22)· · ·(2n − 2(r − 1)2)= |Mrn(e)|
  170. 170. The structural claimLet A be an extremal intersecting family of r-matchings of K2n.
  171. 171. The structural claimLet A be an extremal intersecting family of r-matchings of K2n.Then there must be an edge that is common to all matchings in A.
  172. 172. High Level Proof StrategyWe know that if A is an extremal intersecting family, then:|Aσ| = r, for all σ ∈ S2n.Therefore, the matchings of the subfamily Aσ necessarily have acommon edge...
  173. 173. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15
  174. 174. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  175. 175. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  176. 176. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  177. 177. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  178. 178. ..1.2.3.4.5.6.7.8.9.10.11.12.13.14.15 .∈ A
  179. 179. High Level Proof Strategy (Contd.)Now that we know that every Aσ has a common edge,it remains to show that the common edge is the same for every σ.
  180. 180. High Level Proof Strategy (Contd.)Now that we know that every Aσ has a common edge,it remains to show that the common edge is the same for every σ.Begin by assuming that Aσ is centered at the edge e, where σ is theidentity permutation.
  181. 181. High Level Proof Strategy (Contd.)Now that we know that every Aσ has a common edge,it remains to show that the common edge is the same for every σ.Begin by assuming that Aσ is centered at the edge e, where σ is theidentity permutation.
  182. 182. High Level Proof Strategy (Contd.)Let σi be obtained by σ by a transposition of the element at i.σi(j) =i + 1 if j = i,i if j = i + 1,j otherwise .
  183. 183. High Level Proof Strategy (Contd.)Let σi be obtained by σ by a transposition of the element at i.σi(j) =i + 1 if j = i,i if j = i + 1,j otherwise .It suffices to show that, for every 1 i 2n,Aσiand Aσ have the same common edge.
  184. 184. High Level Proof Strategy (Contd.)Let σi be obtained by σ by a transposition of the element at i.σi(j) =i + 1 if j = i,i if j = i + 1,j otherwise .It suffices to show that, for every 1 i 2n,Aσiand Aσ have the same common edge.
  185. 185. High Level Proof Strategy (Contd.)Let σi be obtained by σ by a transposition of the element at i.σi(j) =i + 1 if j = i,i if j = i + 1,j otherwise .It suffices to show that, for every 1 i 2n,Aσiand Aσ have the same common edge.The proof follows by an analysis on the structure of χσi,based on position of i.
  186. 186. Thank You!.........

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