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# Ballot Problem for Many Candidates

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### Ballot Problem for Many Candidates

1. 1. THE BALLOT PROBLEM FOR MANY CANDIDATES
2. 2. <ul><li>Advisers : </li></ul><ul><li>Mr.Thammanoon Puirod </li></ul><ul><li>Mr. Deaw Jaibun </li></ul><ul><li>Members : </li></ul><ul><li>Mr. Norrathep Rattanavipanon </li></ul><ul><li>Mr. Nuttakiat Chaisettakul </li></ul><ul><li>Mr. Papoj Thamjaroenporn </li></ul>
3. 3. What is the ballot of problem? <ul><li>Suppose A and B are candidates and there are a+b voters, a voting for A and b for B. In how many ways can the ballots be counted so that A is always ahead of B ? </li></ul>The formula comes up to be
4. 4. Why is it interesting? <ul><li>The Ballot problem is one of very well-known problems. This problem had been proved for two candidates. </li></ul><ul><li>What about any n, which is greater than 2, candidates? </li></ul>
5. 5. Objective To find the formula and proof of Ballot problem for many candidates.
6. 6. In case two candidates (The Ballot problem)
7. 7. Suppose is the ballot of the 1 st candidate. is the ballot of the 2 nd candidate, when . Define “1” as the ballot given to 1 st candidate. “ -1” as the ballot given to 2 nd candidate.
8. 8. The number of ways to count the ballots for required condition. Permutation of the sequence: such that the partial sum is always positive. The number of ways to walking on the lattice plane with start at (0,0) and finish at (a,b), and can’t pass line y=x except (0,0) = =
9. 10. Reflection Principle The way to count the number of path is using “reflection principle”, one can show that the number of bad ways which begin at (1,0) is equal to the number of ways begin at (0,1). It implies that, if we denote as the number of ways as required:
10. 12. In case three Candidates
11. 15. How to count ?
12. 17. <ul><li>1. Consider the path seen at the front view. Find all possible ways to “walk” within the allowed plane. </li></ul><ul><li>2. Consider the path seen at the side view. Find all possible ways to “walk” within the allowed plane. </li></ul><ul><li>3. Match each way of 1. to each of 2. (Sometimes it yields more than one complete path.) </li></ul>
13. 18. Example Counting Front View F(1) F(2) F(3) F(4) F(5)
14. 19. Example Counting Side View S(1) S(2)
15. 20. Example Counting Matching 2 2 1 1 1 S(2) 1 1 1 1 1 S(1) F(5) F(4) F(3) F(2) F(1) F(i) * S(j) F(1) F(2) F(3) F(4) F(5) S(1) S(2) 5 7 12 Total
16. 21. Dynamic Programming
17. 22. Counting (5,4) with D.P. 1 1 1 1 1 1 4 3 2 1 0 0 9 5 2 0 0 0 14 5 0 0 0 0 14 0 0 0 0 0
18. 23. Formula for three candidates
19. 24. Definition is the number of ways to count the ballot so that correspond to the required condition Lemma 1.1 Lemma 1.2
20. 25. Conjecture
21. 26. Let Consider Use strong induction; given is the “base” therefore Proof and hence the base case is true.
22. 27. <ul><li>We assume that for some </li></ul><ul><li>all of the following are true; </li></ul>We will use this assumption to prove that is true
23. 30. By strong induction, we get that.
24. 31. Formula for n candidates
25. 32. is the number of ways to count the ballots of the n candidates such that, while the ballots were counting, the ballots of higher-complete-balloted candidate are always greater than that of smaller-complete-balloted. Definition Lemma 3
26. 37. We will show that is factor of Case 1 Case 1
27. 38. Consider hence;
28. 39. is factor of . Hence We will show that is factor of . Case 2
29. 40. Consider
30. 41. is factor of Hence We will show that is the factor of Consider the degree of each of each term of is one less than that of , so we can conclude that there must be the factor when m,k is constant. By comparing the coefficient of , it yields that k,m=1 . Therefore, is the factor of Case 3
31. 42. From case 1,2 and 3, we now prove that By mathematical induction,
32. 43. Development
33. 44. 1. The number of ways to count the ballots of the n candidates such that, while the ballots were counting, the ballots of higher-complete-balloted candidate are never less than that of smaller-complete-balloted candidate.
34. 45. 2. The number of ways to count the ballots of the n candidates such that, while the ballots were counting, in m candidates (m<n) the ballots of higher-complete-balloted candidate are always greater than that of smaller-complete-balloted.
35. 46. 3. The number of ways to count the ballots of the n candidates such that, while the ballots were counting, the K candidate are always greater than that of the M candidate and the P candidate are always greater than that of the Q candidate .
36. 47. Application
37. 48. Application In Biology <ul><li>Lead to find the number of ways a random graph model for angiogenesis in the renal glomerulus. </li></ul><ul><li>Steps to form a vascular network </li></ul><ul><li>1. Budding </li></ul><ul><li>2. Spliting </li></ul><ul><li>3. Connecting </li></ul>
38. 49. Application In Cryptography Define the plaintext (code) used to send the data Increases the security of the system
39. 50. Reference Miklos Bona, Unimodality, Introduction to Enumerative Combinatorics, McGrawHill, 2007. Chen Chuan-Chong and Koh Khee-Meng, Principles and Techniques in Combinatorics , World Scientific, 3rd ed., 1999. Michael L. GARGANO, Lorraine L. LURIE Louis V. QUINTAS, and Eric M. WAHL, The Ballot Problem, U.S.A.,2005. Sriram V. Pemmaraju, Steven S. Skienay, A System for Exploring Combinatorics and Graph Theory in Mathematica, U.S.A., 2004. Marc Renault, Four Proofs of the Ballot Theorem, U.S.A., 2007.
40. 51. Thank you for your attention