1 / 47

“Catalan Numbers and Pascal’s Triangle”

“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. Catalan Numbers are:. 1,. 1, 2, 5, 14, 42, 132, 429, 1430, … Notation: C 1 = 1 C 2 = 2 C 3 = 5

ozzy
Télécharger la présentation

“Catalan Numbers and Pascal’s Triangle”

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Catalan Numbers are: 1, 1, 2, 5, 14, 42, 132, 429, 1430, … Notation: C1 = 1 C2 = 2 C3 = 5 C4 = 14 Also, C0 = 1

  3. Pascal’s Triangle 1 1 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 nth entry in row r is (n and r start with 0) Example:

  4. Catalan Numbers in Pascal’s Triangle by subtraction 1 1 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 1 2 4 6 15 20 Characterization #18 of Pascal’s Triangle The difference of the middle number in an even rowand its neighbor is a Catalan Number.

  5. Catalan Numbers in Pascal’s Triangle by division 1 1 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 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. Formulas Problems Difference Quotient Product Recursive Summation

  7. Problems n +1’s and n –1’s 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  C3 = 5

  8. 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 Problems n +1’s and n –1’s Rook moves n+1 by n+1 board 5 ways  C3 = 5

  9. Problems n +1’s and n –1’s Find the number of ways to triangulate an (n+2)-gon Example: n = 3 5 ways  C3 = 5 Rook moves n+1 by n+1 board Triangulations n+2 - gon

  10. 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  C3 = 5 n +1’s and n –1’s Rook moves n+1 by n+1 board Triangulations n+2 - gon n pair of parentheses

  11. Problems The number of trivalent rooted trees with n trivalent vertices (it has n+1 branches). Example: n = 3 5 ways  C3 = 5 n +1’s and n –1’s Rook moves n+1 by n+1 board A B C D A B C D A B C D Triangulations n+2 - gon n pair of parentheses A B C D A B C D Trivalent trees n+1 branches

  12. Formulas Problems Difference n +1’s and n –1’s • …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. Rook moves n+1 by n+1 board Quotient Triangulations n+2 - gon Product Recursive n pair of parentheses Summation Trivalent trees n+1 branches

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

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

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

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

  17. Prove the Summation formula solves the Triangulations of an n+2-gon Problem is a formula for the number of triangulations of an n+2-gon Prove: 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 is a formula for the number of triangulations of an h+2-gon, for all h<k. Inductive hypothesis: Assume is a formula for the number of triangulations of a k+2-gon Prove

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

  20. … P2 Pi P1 Label two consecutive vertices X and Y. This leaves k additional points. We label these points P0, P1, P2,… Pk-1. Consider a k+2-gon. P0 X Y Every triangulation of the polygon will include a triangle with the points X, Y, and Pi.

  21. … P2 Every triangulation of the polygon will include a triangle with the points X, Y, and Pi. Pi P1 P0 X Y To form an entire triangulation of the polygon containing triangle XYPi, we need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle.

  22. … P2 We need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle. Pi P1 P0 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. X Y In fact, the number of triangulations to right is Ci and the number of triangulations to left is Ck-1-i

  23. For example, k=6 Octagon, i=3 Pi =P3 P2 We need to triangulate the 5 points to the right of triangle XYP3, and triangulate the 4 points to the left of the triangle. P1 P0 X Y (Only one triangulation on each side is shown.) The number of triangulations to right is C3=5. The number of triangulations to left is C2=2. So, there are 25=10 triangulations which contain this particular triangle (XYP3).

  24. … P2 We let the point Pi go from P0 to Pk-1. We see that The number of triangulations of a k+2-gon is Pi P1 P0 X Y Therefore, by the principle of strong induction, is a formula for the number of triangulations of an n+2-gon. (The formula is valid for all natural numbers.)

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

  26. C A B C D E B D E A

  27. C A B C D E B D E A

  28. C A B C D E B D (CD) E A

  29. C A B C D E B D (CD) (B(CD)) E A

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

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

  32. >> Second Example <<

  33. C A B C D E B D E A

  34. C A B C D E B D E A

  35. C A B C D E B D (AB) E A

  36. C A B C D E B D (DE) (AB) E A

  37. C A B C D E B D (DE) (AB) ((AB)C) E A (((AB)C)(DE))

  38. C A B C D E B D (DE) (AB) ((AB)C) E A (((AB)C)(DE))

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

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

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

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

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

  44. 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! • 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

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

More Related