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**The Catalan Numbers and their Applications**

1, 2, 5, 14, 42, 132, … Christen Mirando Western New England University

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**In this presentation… History of the Catalan Numbers**

Examples/Applications Euler’s triangulation of convex polygons Catalan’s parenthesization problem Mountain ranges on a graph Hankel matrix Relationship to computer science Pascal’s Triangle Formulas Recursive Explicit Generating Function

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**The Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13…**

… is a set of numbers with which most people are familiar. These numbers appear in several areas of mathematics, in nature and in many other places. One set of numbers people may not be as familiar with is the set of Catalan numbers. This set of numbers, like the Fibonacci numbers, has many unexpected occurrences and beautiful relationships in mathematics.

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**The first few Catalan Numbers**

Cn 1 2 3 5 4 14 42 6 132 7 429 8 1,430 9 4,862 10 16,796 11 58,786 12 208,012

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**History of the Catalan Numbers**

Eugene Charles Catalan ( ) was a Belgian mathematician who “discovered” the Catalan Numbers in 1838 while studying well- formed sequences of parentheses. Although this set of numbers is named after him, he was not the first to discover it. Leonhard Euler ( ) discovered this set of numbers while studying triangulations of convex polygons. Published a recursive formula in 1761. Worked with Hungarian mathematician, Johann von Segner ( ) , to develop a second order recurrence relation. Gabriel Lame ( ), was a French mathematician who used Euler’s recursive formula to find a explicit formula in 1838. Although Euler and Catalan are given credit for this sequence, there is speculation that a Chinese mathematician Antu Ming ( ) discovered them in the 1730’s through geometric models.

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**Euler’s triangulation of a convex n-gon**

Let Tn be the number of ways a convex n-gon can be divided into n-2 triangles by drawing n-3 nonintersecting diagonals where n ≥ 3. When n = 4, Tn = 2 When n = 5, Tn = 5 When n = 6, Tn = 14

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**Euler’s Recursive Formula**

Euler let Tn be the number of triangulations of a convex n-gon Tn = 2610…(4n – 10) n ≥ 3 (n – 1)! T3 = 2 = 1, T4 = 26 = 2, and T5 = 2610 = 5 2! ! ! These look like the Catalan numbers, except the subscript of Tn is shifted two places to the right, so Cn = Tn+2. Substituting n + 2 into Tn yields: Cn = 2610…(4n – 2) (n + 1)! Cn = 4n – 2 2610…(4n – 6) n n! Cn = 4n – 2 Tn+1 n + 1 = 4n – 2 Cn – 1

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**Catalan’s Parenthesization Problem**

Suppose we are given n pairs of left and right parentheses. Find the number of correctly parenthesized sequences, Pn, that can be formed. Correct: () () Incorrect: ())( () When n = 1, Pn = 1 ()(), (()) When n = 2, Pn = 2 ()()(), (())(), ()(()), (()()), ((())) When n = 3, Pn = 5 (((()))), ((()())), ((())()), (()(())), When n = 4, Pn = 14 (()()()), (())(()), ()()()(), ()(())(), ()()(()), ()(()()), ()((())), (())()(), (()())(), ((()))()

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**Since |Pn| = |Tn| = Cn, a bijection must exist between Tn and Pn**

In 1961, H.G. Forder found a solution to this problem. He let each diagonal spanning adjacent sides be labeled with the parenthesized concatenation of the labels of the sides. c b d a e c b d (ab) a e c b d (ab) (de) a e c b ((ab)c) d (ab) (de) a e c b ((ab)c) d (ab) (de) a e ((ab)c)(de)) Since n = 6, there are 14 different ways to triangulate a hexagon

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**Classic Problem – Mountain Ranges**

Find the number of mountain ranges that can be drawn with n upstrokes and n downstrokes. That is, find the number of different paths, Mn, we can choose from the origin to the lattice point (2n, 0). Can touch the x-axis but cannot cross it. From the point (x, y) we can climb up to the point (x+1, y +1) or climb down to the point (x+1, y −1). n = 0, Mn = 1 n = 1, Mn = 1 n = 2, Mn = 2 n = 3, Mn = 5 Do you want me to make these shapes in powerpoint instead of using the scanned copy, I can do it, just takes more time

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**For n = 4, there are 14 possible mountain ranges.**

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Hankel Matrix is a nxn matrix whose (i, j) entry is the Catalan number Ci+j-2. These matrices always have determinant 1. (1,1) (1,2) (2,1) (2,2) C(1+1-2) C(1+2-2) C(2+1-2) C(2+2-2) C0 C1 C1 C2 1 1 1 2 (1)(2) – (1)(1) = 2 – 1 = 1 Finish this

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**Det = [(2)(14) – (5)(5)] – [(1)(14) – (2)(5)] + 2[(1)(5) – (2)(2)]**

Now for a 3x3 matrix: (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3) C(1+1-2) C(1+2-2) C(1+3-2) C(2+1-2) C(2+2-2) C(2+3-2) C(3+1-2) C(3+2-2) C(3+3-2) C0 C1 C2 C1 C2 C3 C2 C3 C4 Det = [(2)(14) – (5)(5)] – [(1)(14) – (2)(5)] + 2[(1)(5) – (2)(2)] = [28 – 25 – ] = 1

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**Application to Computer Science**

Stacks, Pushing, Popping Using a stack, determine the number of different ways of the ordered list of integers, 1, 2, 3, ..., n can be pushed onto an empty stack and then popped resulting in an empty stack. For n = 1, there is only one way to push/pop data. For n = 2, there are two ways to push/pop data. Output: 12 empty stack push 1 pop 1 push 2 pop 2 Output: 21 2 empty stack push 1 push 2 pop 2 pop 1

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**When n = 3, there are 6 permutations, but only 5 satisfy the stack structure. **

1 2 3 empty stack push 1 pop 1 push 2 pop 2 push 3 pop 3 output: 123 output: 132 output: 213 output: 231

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3 2 1 empty stack push 1 push 2 push 3 pop 3 pop 2 pop 1 output: 321 Notice that the permutation, 312, cannot be made. It would have to start with the steps shown below, but there is no way to pop 1 before popping 2. 3 2 1 empty stack push 1 push 2 push 3 pop 3

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**Explicit Formula from Euler’s Recursive Formula**

Cn = 4n – 2 Cn – 1 n + 1 = (4n – 2)(4n – 6) Cn – 2 (n + 1)n = (4n – 2)(4n – 6)(4n-10) Cn – 3 (n + 1)n(n – 1) This continues on until the last term is C0 = (4n – 2)(4n – 6)(4n-10)…62 C0 (n + 1)n(n – 1)… 32

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**(2n)(2n -2)(2n - 4) … (2) (n+1)! 2n n! (n+1)! n! (n+1)! (n+1)!**

2 is factored out from each term in the numerator , and the denominator becomes (n + 1)! Also, remember C0 = 1 = (2n – 1)(2n – 3)(2n – 5)… 31 2n (n+1)! In order to make the numerator (2n)!, terms are inserted in red = (2n)(2n – 1)(2n - 2)(2n – 3)(2n - 4)(2n – 5)… 321 2n (2n)(2n -2)(2n - 4) … (2) (n+1)! Note, what is inserted in red, (2n)(2n – 2)(2n-4)…2 = 2n n! by factoring out a 2 from all n terms. We can substitute this in the denominator. = (2n)! 2n 2n n! (n+1)! = (2n)! n! (n+1)!

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**This is the explicit formula from Euler’s recursive formula.**

= (2n)! (n + 1)! n! Note: = (2n)! (n + 1) n! n! 2n = (2n)! n n! (2n – n)! = 1 2n (n + 1) n = (2n)! n! n! This is the explicit formula from Euler’s recursive formula.

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**Pascal's Triangle and the Catalan Numbers**

Explicit formula for the Catalan numbers; Cn = 2n n n Example C4: = (70) = (4+1) row

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**Another Example, Cn = 1 2n n n -1**

2n is the term directly to the right/left of the CBC . n -1 Now, we divide this term by row number n. 1 row 0 Example C4: = 1 (56) = 14

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**Generating Function for the Catalan Numbers**

We have given a recursive and an explicit formula for finding the nth Catalan number, Cn. We will now find the generating function for the Catalan numbers. Segner’s Recursive Formula: Cn = (C0, C1, C2,…, Cn-1) (Cn-1 ,Cn-1,…,C0 ) = C0Cn-1 + C1Cn-2 +…+ Cn-1C0 Let C(x) = C0 + C1x + C2x2 + C2x2 + … + Cnxn [C(x)]2 = (C0 + C1x + C2x2 + C2x2 + … + Cnxn) (C0 + C1 + C2x2 + C2x2 + … + Cnxn) = (C0C0) + (C0C1 + C1C0) x + (C0C2 + C1C1 + C2C0) x2 +….. = (11) + (11 + 11) x + (12 + 11 + 21) x2 +….. = 1 + 2x + 5 x2 +….. = C1x0 + C2x1 + C3x2 +…. + Cn+1xn

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**C(x) 2 = C1 + C2x + C3x2 +…. + Cn+1xn C(x) 2 = C1x + C2x2 + C3x3 …**

C(x) 2 = C1 + C2x + C3x2 +…. + Cn+1xn C(x) 2 = C1x + C2x2 + C3x3 …. + Cnxn x C(x) 2 = (C0 + C1x + C2x2 + C3x3 …. + Cnxn) – C0 C(x) 2 = C(x) – C0 xC(x) 2 = C(x) – C0 xC(x) 2 - C(x) + 1 = 0 Using the quadratic formula, a = x, b = -1, c = 1 C(x) = 1 +/- (-1)2 – 4(x)(1) 2x C(x) = – 4(x)

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**… the solution is the Catalan Numbers!**

If any of these problems appear: Number of binary trees with n vertices Number of full binary trees with n vertices Number of n multisets {a1, a2, …, an} of elements ai such that a1 + a2 + … + an = 0 Number of ways n semicircles can be arranged such that no two intersect Number of permutations for the set {1, 2, …, 2n} where f is an involution that has no fixed points and satisfies the noncrossing condition Number of ways to arrange 2n soldiers in two rows in ascending height order Number of tilings of the staircase shape that can be made with n rectangles Number of paths a rook can take from the upper left-hand corner to the lower right-hand corner on an nxn chessboard without crossing the main diagonal Number of ideals in a ring is Cn+1 Number of nilpotent ideals in a ring ….. Many more! … the solution is the Catalan Numbers!

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Any Questions?

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References Guy, R.K. Letters to the Editor. Mathematics Magazine 61 (Oct. 1988): 269. Larcombe, P. “The 18th Century Chinese Discovery of the Catalan Numbers.” Mathematical Spectrum 32 ( ): 5-6. Jarvis, F. “Catalan Numbers.” Mathematical Spectrum 36 ( ): 9-12. Koshy, Thomas. “The Ubiquitous Catalan Numbers.” Mathematics Teacher (2006): Gardner, M. “Mathematical Games.” Scientific American 234 (June 1976): Grimaldi, Ralph P. Fibonacci and Catalan Numbers: An Introduction. Hoboken, NJ: John Wiley & Sons, Print. Koshy, Thomas. Catalan Numbers with Applications. Oxford: Oxford UP, Print. Brualdi, Richard A. Introductory Combinatorics. New York: North-Holland, Print.

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