CO330A4solns

# CO330A4solns - C&O 330 - SOLUTIONS #4 PROFESSOR D.M....

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Unformatted text preview: C&O 330 - SOLUTIONS #4 PROFESSOR D.M. JACKSON (1) (15 points) A rise in a sequence σ1 · · · σp is a pair (σj , σj +1 ) such that σj < σj +1 . Prove that the number of permutations on n symbols with exactly k rises is u−1 xn . uk n! u − e(u−1)x [Hint: Use the Maximal Decomposition Theorem and the Permutation Lemma.] Solution: Let fk mark the occurrence of a maximal <-substring of length k. Such a substring contains k − 1 rises so uk−1 xk = 1 + f (x) = 1 + k ≥1 1 + (1 − u) x x = . 1 − ux 1−x Then, by the Maximal Decomposition Theorem, the generating series for sequences with respect to rises is F = f −1 ◦ γ < = = = = −1 1−x 1 + (1 − u) x −1 ◦ γ< −1 i i≥0 (u − 1) xi − xi+1 ◦ γ < 1 − u− 1 u−1 i i≥1 u−1 i≥0 −1 (u − 1) xi ◦ γ < i < (u − 1) γi . Thus the generating series for permutations with respect to rises is, by the Permutation Lemma, ∆ u− u−1 i< i≥0 (u − 1) γi = u− u−1 i≥0 (u − 1)i xi i! and the result follows. (2) (15 points) Prove that the number of permutations of length n, with precisely k maximal <-substrings having length greater than or equal to 2, 1 2 PROFESSOR D.M. JACKSON is uk xn n! cosh(zx) − z −1 sinh (zx) where z= √ 1 − u. −1 2k 2k+1 y y [Comment: Recall that cosh (y ) = k≥0 (2k)! and sinh (y ) = k≥0 (2k+1)! .] Solution: Let u mark the occurrence of a maximal <-substring of length greater than or equal to 2. Then, following the argument in the previous problem, f (x) = 1 + x + u x2 + x3 + · · · ux2 1−x 1 + (u − 1) x2 = 1−x so, by the Maximal Decomposition Theorem and the Permutation Lemma, the desired generating series is = 1+x+ −1 F = = 1−x ◦ γ< 1 + (u − 1) x2 k (u − 1) x2k ◦ γ < − ∆ ∆ k ≥0 = cosh (zx) − z −1 sinh (zx) k ≥0 −1 −1 k (u − 1) x2k+1 ◦ γ < (3) (15 points) Let D (x1 , x2 , . . .) be the ordinary generating series for the number d (k1 , . . . , kn ) of sequences with ki occurrences of i for i = 1, . . . , n such that adjacent symbols in the sequence are not equal. (a) (6 points) Find the generating series F for the set of all sequences over Nn = {1, . . . , n} with respect to the number of symbols of each type, where xi marks the occurrence of the symbol i, for i = 1, . . . , n. Solution: F is the generating series for {1. . . . , n} , which is 1 . n 1 − i=1 xi (b) (9 points) Find D by ﬁrst determining how to construct each string over Nn from a unique sequence counted by d (k1 , . . . , kn ) . Solution: The set {1. . . . , n} can be constructed by selecting a sequence σ in it, and replacing each symbol i by i {i} for each i, and for each σ. Thus, at the level of generating series, 1 D (x1 , . . . , xn ) |xi →xi (1−xi )−1 ,i=1,2,... = F = n 1 − i=1 xi from the previous part, whence D xn x1 ,..., 1 − x1 1 − xn = 1 1− n i=1 xi . SOLUTIONS #4 Let yi = xi . 1 − xi xi = 3 yi 1 + yi Then so D (y1 , . . . , yn ) = 1 1− n yi i=1 1+yi (4) (15 points) Let D (x1 , x2 , . . .) be the ordinary generating series for the number d (k1 , . . . , kn ) of sequences with ki occurrences of i for i = 1, . . . , n such that adjacent symbols in the sequence are not equal. (a) (7 points) Use the Maximal Decomposition Theorem to prove that −1 d (k1 , . . . , kn ) = xk1 · · · xkn 1 − n 1 −1 xi (1 + xi ) i≥1 . [Comment: This question is the same as the previous one. However. this time you are asked to use the maximal Decomposition Theorem.] Solution: Let π1 = ” = ”. Then f (x) = 1 + x. Then by the Maximal Decomposition Theorem, the required generating series is F = (1 + x) = 1 + −1 i≥1 i≥1 (−1) xi ◦ γ = = (−1)i γi i i≥1 k≥1 −1 (−1) xi k −1 = 1 − −1 i = 1 + −1 −1 = 1 + ◦ γ= k ≥1 xk (1 + xk )−1 . (b) (8 points) Let u ↔< and d ↔≥ . By using part (a), or otherwise, state a combinatorial interpretation of φ D d, ud, u2 d, u3 d, . . . . Solution: This is the generating series for the number of sequences such that no pairof adjacent maximal <-substrings have the same length. (5) (25 points) Let u ↔< and d ↔≥, and let x1 , x2 , . . . be commuting indeterminates, and let φ be the partial homomorphism associated with the Pattern Algebra. 4 PROFESSOR D.M. JACKSON (a) (5 points) Prove that 3 2 φ (ud) u = φ (ud) u γ2 − φ ((ud) u) γ4 + φ (u) γ6 − γ8 . Solution: (ud)3 u = (ud)2 (ud) u 2 (ud) u (w − u) u = 2 2 (ud) uwu − (ud) u3 = 2 (ud) uwu − (ud) u (w − u) u3 = 2 (ud) uwu − (ud) uwu3 + (ud) u5 = (ud)2 uwu − (ud) uwu3 + uwu5 − u7 , = so, applying the partial homomorphism φ, we obtain the result. 2 (b) (5 points)By deducing similar expressions for φ (ud) u and φ ((ud) u), prove that these expressions satisfy the system of simultaneous equations φ (ud)3 u 1 −γ2 γ4 −γ6 −γ8 0 1 −γ2 γ4 φ (ud)2 u γ6 = −γ4 . 0 0 1 −γ2 φ ((ud) u) 0 0 0 1 γ2 φ (u) Solution: In similar fashion, 2 (ud) u = = (ud) uwu − (ud) u3 (ud) uwu − uwu3 + u5 , so 2 φ (ud) u = φ ((ud) u) γ2 − φ (u) γ4 + γ6 , and (ud) u = uwu − u3 , so φ ((ud) u) = φ (u) γ2 − γ4 . Finally, φ (u) = γ2 . The result now follows by presenting these four equations matricially. 3 (c) (10 points) By solving this equation for φ (ud) u , or otherwise, prove that 8 2 x8 sec (x) = 8! 8 4 8 6 8 8 1 6 2 6 4 6 6 0 1 4 2 4 4 0 0 1 2 2 . SOLUTIONS #4 5 Solution: By Cr´mer’s Rule a 3 φ (ud) u = −γ8 γ6 −γ4 γ2 −γ2 1 0 0 −γ6 γ4 −γ2 1 γ4 −γ2 1 0 since the coeﬃcient matrix is unidiagonal. Interchange the order of the columns and use elementary row and column operations to obtain 3 φ (ud) u = γ2 1 0 0 γ4 γ2 1 0 γ6 γ4 γ2 1 γ8 γ6 γ4 γ2 . Then, applying the permutation homomorphism we have x4 4! x2 2! x2 2! 3 ∆φ (ud) u = 1 0 0 x6 6! x4 4! x2 2! 1 0 x8 8! x6 6! x4 4! x2 2! 4 1 . Now multiply columns 1,2 and 3 by x6 , x and x2 , respectively, and divide rows 2, 3 and 4 by x6 , x4 and x2 , respectively, to obtain, 1 2! 1 0 0 ∆φ (ud)3 u = x8 1 4! 1 2! 1 0 1 6! 1 4! 1 2! 1 1 8! 1 6! 1 4! 1 2! . Now divide columns 1,2 and 3 by 6!, 4! and 2!, respectively, and multiply rows 2, 3 and 4 by 6!, 4! and 2!, respectively, to obtain, 8 2 x8 8! 8 6 8 8 1 6 2 6 4 6 6 0 1 4 2 4 4 0 3 ∆φ (ud) u = 8 4 0 1 2 2 . But the left hand side is the generating series for the number of <alternating permutations of length 8. The result follows. (d) (5 points) Try to see how an expression of this sort may be deduced for x2n sec (x) , (2n)! or conjecture what this might be. ...
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## This note was uploaded on 11/18/2010 for the course CO 330 taught by Professor R.metzger during the Spring '05 term at Waterloo.

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