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18.212 Course Notes 18.212 Course Notes Wanlin Li Spring 2019 1 Catalan Numbers 1.1 February 6 • Stanley-style (enumerative/algebraic/geometric) vs Erdos-style combinatorics (ex- tremal/probabilistic) • Enumerative results tend to be of the form A = B; extremal results tend to be asymptotic • Catalan numbers Cn: number of sequences "1;"2;:::;"2n with "i = ±1 such that any partial sum "1 + ··· + "i ≥ 0 and "1 + ··· + "2n = 0 • Definition. Dyck path: path on Cartesian plane starting at (0; 0); ending at (2n; 0); and traveling (1; 1) or (1; −1) at each step always staying in the upper half plane • Starting at index 0: 1; 1; 2; 5; 14; 42;::: • Drunk man problem: random walk on number line, cliff to one side and infinite in other direction; find probability of survival • 1 P Cn Probability that man falls is 2 22n n≥0 • Simple solution: probability that person starting at i reaches position N before i falling off cliff at 0 is N ; in original problem i = 1 and N ! 1 so drunk man falls off cliff Pn • Catalan recurrence: Cn = i=1 Ci−1Cn−i with C0 = 1 • 2 Generatingp function for Catalan numbers satisfies f(x) = xf(x) + 1 so f(x) = 1− 1−4x 2x 1 18.212 Wanlin Li 1.2 February 8 p 1− 1−4x • Generating function f(x) = 2x for Catalan numbers • Biased random walk with probability p of moving one step toward safety and 1 − p of moving toward cliff: probability of falling is (1 − p)f(p(1 − p)) which is 1 1 1−p 1 if p ≤ 2 ; p if p > 2 n • Proof of generalized binomial expansion (1 + x) by Taylor series expansion 2n • 1 Generating function gives Cn = n+1 n • Number of total lattice paths from A to B with n up steps and n down steps is 2n n • Path crosses x-axis at first point C; reflect path from C to B = (2n; 0) about y = −1 to get path from (0; 0) to (2n; −2) with n−1 up steps and n+1 down steps 2n • Procedure is invertible so number of non-Dyck paths from A to B is n−1 2n 2n 2n • 1 Thus number of Dyck paths is n − n−1 = n+1 n 2n+1 • Necklace and cyclic shift proof: n is number of sequences with n positive 1s and n + 1 −1s • 2n + 1 cyclic shifts of the sequence 2n+1 • 1 Claim 1: all 2n + 1 shifts are unique, so the number of necklaces is 2n+1 n • Claim 2: for each necklace there is a unique sequence among all 2n + 1 arrange- ments such that "1 + ··· + "i ≥ 0 for i ≤ 2n and "2n+1 = −1 2n+1 2n • 1 1 Therefore Cn is the number of such necklaces 2n+1 n = n+1 n • Other interpretations of Cn: number of triangulations of (n + 2)-gon, number of valid parenthesizations of n + 1 letters, number of plane binary trees with n non-leaf vertices and n + 1 leaves • Bijection between triangulation and binary tree: start with side of (n + 2)-gon, exactly one triangle is adjacent to that side; then children of triangle are the triangles/sides adjacent to it • Letters correspond to n + 1 leaves of binary tree, binary tree gives order of mul- tiplication 2 18.212 Wanlin Li 1.3 February 11 • Definition. Queue: first in first out • Definition. Stack: last in first out • Claim: number of queue-sortable permutations of 1; 2; : : : ; n is Cn • Queue-sortable: each number is read and either placed in queue or directly sent to output • Example. 2; 4; 1; 3 is queue-sortable: 2 goes into queue, 4 goes into queue, 1 goes directly to output, 2 is removed from queue and appended, 3 goes directly to output, 4 comes out of queue and is appended • Example. 3; 2; 1 is not queue-sortable • Claim: number of stack-sortable permutations of size n is also Cn • Example. 4; 1; 3; 2 is stack-sortable: 4 goes into stack, 1 goes directly to output, 3 goes into stack, 2 goes into stack; 2; 3; 4 taken out of stack and appended to output • Example. 2; 3; 1 is not stack-sortable • w = (w1; w2; : : : ; wn) is a permutation of size n and π = (π1; π2; : : : ; πk) is a per- mutation of size k ≤ n • Permutation w contains pattern π if there exists some (not necessarily consecu- tive) subsequence of w whose entries appear in the same relative order as entries of π • Call w π-avoiding if w does not contain pattern π • Exercise 1. The queue-sortable permutations are exactly the (321)-avoiding per- mutations. • Exercise 2. The stack-sortable permutations are exactly the (231)-avoiding per- mutations. • Exercise 3. For all patterns π of size 3; the number of π-avoiding permutations of size n is Cn: 3 18.212 Wanlin Li 2 Young Tableaux 2.1 February 11 • Definition. Partition: λ = (λ1 ≥ λ2 ≥ · · · ≥ λe > 0) such that n = λ1 + λ2 + ··· + λe • Young diagram/Ferrers shape: row i contains λi boxes where λi are a non- increasing partition • Definition. Standard Young tableau: way of filling boxes of Young diagram with numbers 1; 2; : : : ; n without repetition such that the numbers increase within the rows and columns of the Young diagram λ • f is the number of standard Young tableaux of shape λ λ Lemma. Given λ = (n; n); f = Cn: • Proof: construct a sequence "1;:::;"2n of ±1 such that "i = 1 if i is in the first row of the tableaux and −1 if i is in the second row • Hook-Length Formula due to Frame-Robinson-Thrall 1954 2.2 February 13 λ • f is number of standard Young tableaux of shape λ Frobenuius-Young Identity. X (f λ)2 = n! λ:jλj=n • RHS is number of permutations of Sn; LHS is number of pairs (P; Q) where P and Q are standard Young tableaux of the same shape • (1961) Schensted correspondence LHS $ RHS • Schensted insertion algorithm: T some intermediate standard Young tableau filled with numbers in S ⊂ Z≥1; x 2 Z≥1nS • T x is the SYT with one extra box given by following procedure: 1. x1 := x; i := 1 2. If xi is greater than all the entries in the ith row of T; add a new box at the end of the ith row filled with xi and stop 4 18.212 Wanlin Li 3. Otherwise find the smallest entry y in the ith row such that y > xi; replace y by xi and set xi+1 = y; go back to step 1 • P is the insertion tableau, Q is the recording tableau • w1; w2; : : : ; wn 2 Sn 7! (P; Q) where P = ((; w1) w2) · · · wn and Q : i 2 [n] goes to the box that was added at the ith step in construction of P • Example. w = 3; 5; 2; 4; 7; 1; 6; P goes in sequence ; 3 3 5 2 5 2 4 2 4 7 1 4 7 1 4 6 3 3 5 3 5 2 5 2 5 7 3 3 and Q is 1 2 5 3 4 7 6 Theorem. This map w 7! (P; Q) gives a bijection between Sn and pairs (P; Q): • Proof: process is reversible by example • Definition. Schensted shape: given w 7! (P; Q); the shape λ of P and Q is called the Schensted shape of w Schensted’s Theorem. The length λ1 of the first row is size of the longest in- creasing subsequence in w: The length of the first column of λ is the size of the longest decreasing subsequence in w: • Corollary: the number of (123)-avoiding permutations in Sn is the nth Catalan number. • Proof: w is (123)-avoiding iff the length of the longest increasing subsequence is at most 2: w matches with a pair (P; Q) of tableaux with the same shape; P and Q can be glued together (Q is flipped by 180◦ and its entry i is replaced by 2n + 1 − i) to form a SYT of shape (n; n); of which there are exactly Cn Green’s Theorem. λ1 + ··· + λi is the length of the longest subsequence that can be split into i increasing subsequences. 5 18.212 Wanlin Li 2.3 February 15 • Schensted correspondence w 2 Sn 7! (P; Q) • Definition. jth basic subsequence: for j 2 1; : : : ; λ1;Bj is the set of all entries of permutation w that were originally inserted in the jth position of the first row Lemma. Each Bj is decreasing. Lemma. For every x 2 Bj with j ≥ 2; there exists y 2 Bj−1 such that y < x and y is located to the left of x in w: • Proof: at the moment of insertion of x; y is the entry located to the left of x: • Proof of Schensted’s Theorem: λ1 is the number of basic sequences, and if x1 < x2 ··· < xr is an increasing subsequence in w; r < λ1 because each xi must come from a different Bj: By the second lemma there does in fact exist an increasing subsequence of size λ1: • Hook-length formula for number of SYT with shape λ • Hook-walk proof by induction on n (Greene, Nijenhuis, Wilf 1979) • H(λ) is product of hook-lengths of λ λ P λ−v • f = v f where v is a bottom right corner of λ λ (4;4;4;1) (5;4;3;1) (5;4;4) • Example. λ = (5; 4; 4; 1) so f = f + f + f • Easy to see hook-length holds for n = 1; want to show n! X (n − 1)! = H(λ) H(λ − v) v X H(λ) 1 = nH(λ − v) v • Idea to construct random process with probabilities represented by RHS • Definition.
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