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“Catalan Numbers and Pascal’s Triangle” Jim Olsen and Allison McGann Western Illinois University IMACC Robert Allerton.

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Presentation on theme: "“Catalan Numbers and Pascal’s Triangle” Jim Olsen and Allison McGann Western Illinois University IMACC Robert Allerton."— Presentation transcript:

1 “Catalan Numbers and Pascal’s Triangle” Jim Olsen and Allison McGann Western Illinois University http://www.wiu.edu/users/mfjro1/wiu/ IMACC Robert Allerton Park, IL March 27, 2010

2 Catalan Numbers are: 1, 2, 5, 14, 42, 132, 429, 1430, … Notation: C 1 = 1 C 2 = 2 C 3 = 5 C 4 = 14 Also, C 0 = 1 1,

3 Pascal’s Triangle 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 The n th entry in row r is (n and r start with 0) Example:

4 Catalan Numbers in Pascal’s Triangle 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 by subtraction 1 2 4 6 15 20 Characterization #18 of Pascal’s Triangle The difference of the middle number in an even row and its neighbor is a Catalan Number.

5 Catalan Numbers in Pascal’s Triangle 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 by division 2 6 20 Characterization #19 of Pascal’s Triangle The middle number in an even row, divided by half the row number plus 1, is a Catalan Number.

6 ProblemsFormulas Difference Quotient Product Recursive Summation

7 n +1’s and n –1’s Problems Find the number of ways we can order 2n numbers from a list made up of n +1’s and n -1’s such that the sum (from the beginning) at any point is  0. Example: n = 3 1, 1, 1, -1, -1, -1 1, 1, -1, -1, 1, -1 1, 1, -1, 1, -1, -1 1, -1, 1, 1, -1, -1 1, -1, 1, -1, 1, -1 5 ways  C 3 = 5

8 n +1’s and n –1’s Problems Find the number of ways a rook in the lower left corner of an (n+1) by (n+1) chess board can get to the upper right corner, without ever crossing the diagonal. Example: n = 3 5 ways  C 3 = 5 Rook moves n+1 by n+1 board

9 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon Problems Find the number of ways to triangulate an (n+2)-gon Example: n = 3 5 ways  C 3 = 5

10 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses Problems Find the number of ways to put n sets of parentheses in a list of (n+1) letters (to indicate the n multiplications). Example: n = 3 (((ab)c)d) ((ab)(cd)) (((a(bc))d) (a((bc)d)) (a(b(cd))) 5 ways  C 3 = 5

11 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses Problems The number of trivalent rooted trees with n trivalent vertices (it has n+1 branches). Example: n = 3 5 ways  C 3 = 5 ABCD ABCD ABCD ABCD ABCD Trivalent trees n+1 branches

12 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation …But WHY? Now we will show the connections Why are the formulas equivalent? Why are the problems equivalent? Why do the formulas solve the problems? First, a few easy ones. Trivalent trees n+1 branches

13 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation equivalent Left as an exercise (in simplifying factorials). Trivalent trees n+1 branches

14 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation equivalent “obvious” Trivalent trees n+1 branches

15 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation equivalent +1 corresponds to a rook move to the right. – 1 corresponds to a rook move up. Trivalent trees n+1 branches

16 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses Trivalent trees n+1 branches ProblemsFormulas Difference Quotient Product Recursive Summation solves

17 Prove the Summation formula solves the Triangulations of an n+2-gon Problem Prove: is a formula for the number of triangulations of an n+2-gon Let Proof by induction on n. n=1: This is a formula for the number of triangulations of a 3-gon, because there is 1 triangulation of a triangle.

18 Induction Step Inductive hypothesis: Assume is a formula for the number of triangulations of an h+2-gon, for all h { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/3592108/12/slides/slide_17.jpg", "name": "Induction Step Inductive hypothesis: Assume is a formula for the number of triangulations of an h+2-gon, for all h

19 Consider a k+2-gon. Here are a couple of triangulations.

20 Consider a k+2-gon. XY Label two consecutive vertices X and Y. This leaves k additional points. We label these points P 0, P 1, P 2,… P k-1. Every triangulation of the polygon will include a triangle with the points X, Y, and P i. P0P0 P1P1 P2P2 PiPi …

21 XY P0P0 P1P1 P2P2 PiPi To form an entire triangulation of the polygon containing triangle XYP i, we need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle. …

22 XY P0P0 P1P1 P2P2 PiPi We need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle. The number of points to the right and to the left is less than k+2, so we know how many triangulations there are to the right and to the left. We will multiply these, by the fundamental counting principle. In fact, the number of triangulations to right is C i and the number of triangulations to left is C k-1-i …

23 XY P0P0 P1P1 P2P2 P i =P 3 We need to triangulate the 5 points to the right of triangle XYP 3, and triangulate the 4 points to the left of the triangle. The number of triangulations to right is C 3 =5. The number of triangulations to left is C 2 =2. So, there are 2  5=10 triangulations which contain this particular triangle (XYP 3 ). For example, k=6 Octagon, i=3 (Only one triangulation on each side is shown.)

24 We let the point P i go from P 0 to P k-1. We see that The number of triangulations of a k+2-gon is Therefore, by the principle of strong induction, is a formula for the number of triangulations of an n+2-gon. XY P0P0 P1P1 P2P2 PiPi … (The formula is valid for all natural numbers.)

25 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation equivalent Trivalent trees n+1 branches

26

27 A B C D E ABCDE

28 A B C D E ABCDE

29 A B C D E ABCDE (CD)

30 A B C D E ABCDE (B(CD))

31 A B C D E ABCDE (CD) (B(CD)) (A(B(CD))) (A(B(CD)))E)

32 A B C D E ABCDE (CD) (B(CD)) (A(B(CD))) (A(B(CD)))E)

33 >> Second Example <<

34

35 A B C D E ABCDE

36 A B C D E ABCDE

37 A B C D E ABCDE (AB)

38 A B C D E ABCDE (DE)

39 A B C D E ABCDE (AB) (DE) ((AB)C) (((AB)C)(DE))

40 A B C D E ABCDE (AB) (DE) ((AB)C) (((AB)C)(DE))

41 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation equivalent See Handout (Math Induction) Trivalent trees n+1 branches

42 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses Planted trivalent Trees n branches ProblemsFormulas Difference Quotient Product Recursive Summation solves See Rook Moves Proofs Handout

43 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses Planted trivalent Trees n branches ProblemsFormulas Difference Quotient Product Recursive Summation solves See Handout on the n +1’s and n -1’s problem

44 Trivalent trees n+1 branches n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation Solves For further Investigation: Interesting proof (can be found online). Involves building new triangulations from old.

45 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses ProblemsFormulas Difference Quotient Product Recursive Summation Equivalent For further Investigation: Pretty easy – See Martin Gardner article. Trivalent trees n+1 branches

46 References Mathematical Games: Catalan numbers: an integer sequence that materializes in unexpected places. By Martin Gardner. Scientific American. June 1976, Vol 234, No. 6. pp. 120-125. Wikipedia.com – Catalan Numbers Catalan Numbers with Applications. Book by Thomas Koshy. Oxford Press. 2009. 422 pages. Enumerative Combinatorics (http://math.mit.edu/~rstan/ec/ ) two-volume book and website by Richard Stanley. Includes 66 combinatorial interpretations of Catalan numbers!http://math.mit.edu/~rstan/ec/ The On-Line Encyclopedia of Integer Sequences (by AT&T) - http://www.research.att.com/~njas/sequences/index.html Type in 1, 2, 5, 14, 42 http://www.research.att.com/~njas/sequences/index.html

47 Thank You Jim Olsen and Allison McGann Western Illinois University http://www.wiu.edu/users/mfjro1/wiu/ JR-Olsen@wiu.edu


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