Avoidance in (2+2)-Free Posets

Transcription

1 Avoidance in (2+2)-Free Posets Nihal Gowravaram Acton Boxborough Regional High School Mentor: Wuttisak Trongsiriwat PRIMES Annual Conference Massachusetts Institute of Technology May 18-19, / 13

2 What is a Poset? Poset Hasse Diagrams Avoidance Partially Ordered Set (P, ) 2 / 13

3 What is a Poset? Poset Hasse Diagrams Avoidance Partially Ordered Set (P, ) Reflexivity i i for all i P Antisymmetry If i j and j i, then i = j for i,j P Transitivity If i j and j k, then i k for i,j,k P 2 / 13

4 What is a Poset? Poset Hasse Diagrams Avoidance Partially Ordered Set (P, ) Reflexivity i i for all i P Antisymmetry If i j and j i, then i = j for i,j P Transitivity If i j and j k, then i k for i,j,k P Call i,j P comparable if i j or j i 2 / 13

5 Hasse Diagrams Poset Hasse Diagrams Avoidance (P({x,y,z}), ) {x,y,z} {x,y} {x,z} {y,z} {x} {y} {z} {} 3 / 13

6 Hasse Diagrams Poset Hasse Diagrams Avoidance (P({x,y,z}), ) {x,y,z} {x,y} {x,z} {y,z} {x} {y} {z} {} {} and {x,z} are comparable {y} and {x, z} are incomparable 3 / 13

7 Hasse Diagrams Poset Hasse Diagrams Avoidance (P({x,y,z}), ) {x,y,z} {x,y} {x,z} {y,z} {x} {y} {z} {} {} and {x,z} are comparable {y} and {x, z} are incomparable {}, {x}, {x,z}, and {x,y,z} form a chain of length 4. 3 / 13

8 Avoidance in Posets Poset Hasse Diagrams Avoidance A poset P is said to contain a poset S if there exists some subposet W of P that is isomorphic to S. P is said to avoid S if P does not contain S. 4 / 13

9 Avoidance in Posets Poset Hasse Diagrams Avoidance A poset P is said to contain a poset S if there exists some subposet W of P that is isomorphic to S. P is said to avoid S if P does not contain S. 4 / 13

10 Avoidance in Posets Poset Hasse Diagrams Avoidance A poset P is said to contain a poset S if there exists some subposet W of P that is isomorphic to S. P is said to avoid S if P does not contain S. Contains (3+1) Avoids (2+2) 4 / 13

11 Why (2+2)-Free Posets? (2+2)-Free Posets Previous Interval Orders A poset is an interval order if it is isomorphic to some set of intervals on the real line ordered by left-to-right precedence. Interval orders are important in mathematics, computer science, and engineering (Ex. task distributions in complex manufacturing processes). (Fishburn 1970) (2+2)-Free Posets are precisely interval orders. 5 / 13

12 Why (2+2)-Free Posets? (2+2)-Free Posets Previous Interval Orders A poset is an interval order if it is isomorphic to some set of intervals on the real line ordered by left-to-right precedence. Interval orders are important in mathematics, computer science, and engineering (Ex. task distributions in complex manufacturing processes). (Fishburn 1970) (2+2)-Free Posets are precisely interval orders. Ascent Sequences An ascent sequence is a sequence x 1 x 2 x n satisfies x 1 = 0 and, for all i with 1 < i n, x i asc(x 1 x 2 x i 1 )+1. (Bousquet-Mélou et al 2009) The ascent sequences are in bijection with the (2+2)-free posets. 5 / 13

13 Previous with (2+2)-Free posets (2+2)-Free Posets Previous Let P n (x) refer to the set of posets of size n that avoid the poset x. Define a function a(x) to return the ascent sequence associated with a poset x. Let A n (y) refer to the set of posets of size n whose ascent sequences avoid the ascent sequence y. 6 / 13

14 Previous with (2+2)-Free posets (2+2)-Free Posets Previous Let P n (x) refer to the set of posets of size n that avoid the poset x. Define a function a(x) to return the ascent sequence associated with a poset x. Let A n (y) refer to the set of posets of size n whose ascent sequences avoid the ascent sequence y. (Stanley 1997) P n (2+2,3+1) = C n. (Enum. Comb. 1) P n (2+2,N) = C n. 6 / 13

15 Previous with (2+2)-Free posets (2+2)-Free Posets Previous Let P n (x) refer to the set of posets of size n that avoid the poset x. Define a function a(x) to return the ascent sequence associated with a poset x. Let A n (y) refer to the set of posets of size n whose ascent sequences avoid the ascent sequence y. (Stanley 1997) P n (2+2,3+1) = C n. (Enum. Comb. 1) P n (2+2,N) = C n. (Trongsiriwat) P n (2+2,N,p 1,,p k ) = A n (0101,a(p 1 ),,a(p k )). 6 / 13

16 Previous with (2+2)-Free posets (2+2)-Free Posets Previous Let P n (x) refer to the set of posets of size n that avoid the poset x. Define a function a(x) to return the ascent sequence associated with a poset x. Let A n (y) refer to the set of posets of size n whose ascent sequences avoid the ascent sequence y. (Stanley 1997) P n (2+2,3+1) = C n. (Enum. Comb. 1) P n (2+2,N) = C n. (Trongsiriwat) P n (2+2,N,p 1,,p k ) = A n (0101,a(p 1 ),,a(p k )). Question 1: Can we explicitly compute P n (2+2,p) for other posets p? 6 / 13

17 Previous with (2+2)-Free posets (2+2)-Free Posets Previous Let P n (x) refer to the set of posets of size n that avoid the poset x. Define a function a(x) to return the ascent sequence associated with a poset x. Let A n (y) refer to the set of posets of size n whose ascent sequences avoid the ascent sequence y. (Stanley 1997) P n (2+2,3+1) = C n. (Enum. Comb. 1) P n (2+2,N) = C n. (Trongsiriwat) P n (2+2,N,p 1,,p k ) = A n (0101,a(p 1 ),,a(p k )). Question 1: Can we explicitly compute P n (2+2,p) for other posets p? Question 2: For what posets p is it true that P n (2+2,p) = A n (a(p))? 6 / 13

18 (2+2)-Free Posets and Ascent Sequences Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free (2+2)-Free Poset p Ascent Sequence a(p) P n (2+2,p) n n n n 1 7 / 13

19 (2+2, )-Free and (2+2, )-Free Posets P n (2+2, ) = P n (2+2, ) (Inverting Procedure) Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free 8 / 13

20 (2+2, )-Free and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2, ) = P n (2+2, ) (Inverting Procedure) P n (2+2, ) 8 / 13

21 (2+2, )-Free and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2, ) = P n (2+2, ) (Inverting Procedure) P n (2+2, ) Add a free node. Add a maximal node. or 8 / 13

22 (2+2, )-Free and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2, ) = P n (2+2, ) (Inverting Procedure) P n (2+2, ) Add a free node. Add a maximal node. or P n (2+2, ) = P n (2+2, ) = 2 n 1 8 / 13

23 (2+2, 3)-Free Posets P n (2+2,3) are posets with level at most 2. Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free 9 / 13

24 (2+2, 3)-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2,3) are posets with level at most 2. Level 2: a nodes: x 1, x 2,, x a. Level 1: b nodes: y 1, y 2,, y b. Define S i = {y j y j x i }. Assign x 1, x 2,, x a to the a nodes such that S 1 S 2 S a. 9 / 13

25 (2+2, 3)-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2,3) are posets with level at most 2. Level 2: a nodes: x 1, x 2,, x a. Level 1: b nodes: y 1, y 2,, y b. Define S i = {y j y j x i }. Assign x 1, x 2,, x a to the a nodes such that S 1 S 2 S a. S 1 S 2 S a. { S 1, S 2,, S a } (( )) b a = ( a+b 1 a ). 9 / 13

26 (2+2, 3)-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2,3) are posets with level at most 2. Level 2: a nodes: x 1, x 2,, x a. Level 1: b nodes: y 1, y 2,, y b. Define S i = {y j y j x i }. Assign x 1, x 2,, x a to the a nodes such that S 1 S 2 S a. S 1 S 2 S a. (( )) ( ) b a+b 1 { S 1, S 2,, S a } =. a a P n (2+2,3) = ( ) a+b 1 = n 1 ( ) n 1 a a a+b=n a=0 = 2 n 1 9 / 13

27 Bijection between (2+2, 3) and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free Poset P (A,B). A = {Maximal Nodes in P}. B = P \A. 10 / 13

28 Bijection between (2+2, 3) and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free Poset P (A,B). A = {Maximal Nodes in P}. B = P \A. In (2+2, )-Free, B forms a chain. 10 / 13

29 Bijection between (2+2, 3) and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free Poset P (A,B). A = {Maximal Nodes in P}. B = P \A. In (2+2, )-Free, B forms a chain. In (2+2, 3)-Free, B forms the lower level. 10 / 13

30 Bijection between (2+2, 3) and (2+2, )-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free Poset P (A,B). A = {Maximal Nodes in P}. B = P \A. In (2+2, )-Free, B forms a chain. In (2+2, 3)-Free, B forms the lower level. Maintains all order relations between A and B. 10 / 13

31 (2+2, 4) and (2+2, Y)-Free Posets Posets and Ascent Sequences (2+2, )-Free and (2+2, )-Free (2+2, 3)-Free Bijection (2+2, 4)-Free and (2+2, Y)-Free P n (2+2,Y) P n (2+2,4). Theorem. P n (2+2,Y) = P n (2+2,4) = 1 + r+m<n ( ) ( ) ( ) ( ) n + mr + 1 n + m(r 1) + 1 n + r(m 1) n + (r 1)(m 1) + n m r n m r n m r n m r, where r 0 and m > / 13

32 Future Directions of Research P n (2+2, ) P n (2+2,3). P n (2+2,Y) P n (2+2,4). Future Directions Acknowledgements 12 / 13

33 Future Directions of Research Future Directions Acknowledgements P n (2+2, ) P n (2+2,3). P n (2+2,Y) P n (2+2,4). Conjecture. Define a function Y(n), n 3 as follows. Y(3) =. Y(n) is the result of adding a minimal node to Y(n 1). Then, P n (2+2,Y(k)) P n (2+2,k). 12 / 13

34 Future Directions of Research Future Directions Acknowledgements P n (2+2, ) P n (2+2,3). P n (2+2,Y) P n (2+2,4). Conjecture. Define a function Y(n), n 3 as follows. Y(3) =. Y(n) is the result of adding a minimal node to Y(n 1). Then, P n (2+2,Y(k)) P n (2+2,k). P n (2+2,3+1) = P n (2+2,N). P n (2+2,Y) = P n (2+2,4) 12 / 13

35 Future Directions of Research Future Directions Acknowledgements P n (2+2, ) P n (2+2,3). P n (2+2,Y) P n (2+2,4). Conjecture. Define a function Y(n), n 3 as follows. Y(3) =. Y(n) is the result of adding a minimal node to Y(n 1). Then, P n (2+2,Y(k)) P n (2+2,k). P n (2+2,3+1) = P n (2+2,N). P n (2+2,Y) = P n (2+2,4) Query. Do there exist other nontrivial Wilf-Equivalences in (2+2)-Free Posets? What other posets p, q exist such that P n (2+2,p) = P n (2+2,q) for all n N? 12 / 13

36 Acknowledgements Future Directions Acknowledgements Thanks to My mentor Wuttisak Trongsiriwat for his valuable insight and guidance. The PRIMES program for making this experience possible. My parents for their support. Thanks to all of you for listening. 13 / 13