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THE BALLOT PROBLEM FOR  MANY CANDIDATES
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
What is the ballot of problem? ,[object Object],The formula comes up to be
Why is it interesting? ,[object Object],[object Object]
Objective To find the formula and proof of Ballot problem for many candidates.
In case two candidates (The Ballot problem)
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.
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) = =
 
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:
 
In case three Candidates
 
 
How to count ?
 
[object Object],[object Object],[object Object]
Example Counting Front View F(1) F(2) F(3) F(4) F(5)
Example Counting Side View S(1) S(2)
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
Dynamic Programming
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
Formula for three candidates
Definition   is the number of ways to count the ballot so that correspond to the required condition  Lemma 1.1 Lemma 1.2
Conjecture
Let Consider Use strong induction; given   is the “base” therefore Proof and hence the base case is true.
[object Object],[object Object],We will use this assumption to prove that  is true
 
 
By strong induction, we get that.
Formula for n candidates
  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
 
 
 
 
We will show that   is factor of Case 1 Case 1
Consider hence;
  is factor of . Hence We will show that  is factor of   . Case 2
Consider
  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
From case 1,2 and 3, we now prove that  By mathematical induction,
Development
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.
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.
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 .
Application
Application In Biology ,[object Object],[object Object],[object Object],[object Object],[object Object]
Application In Cryptography Define the plaintext (code) used to send the data  Increases the security of the system
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.
Thank you for  your attention

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Ballot Problem for Many Candidates

  • 1. THE BALLOT PROBLEM FOR MANY CANDIDATES
  • 2.
  • 3.
  • 4.
  • 5. Objective To find the formula and proof of Ballot problem for many candidates.
  • 6. In case two candidates (The Ballot problem)
  • 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. 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:
  • 11.  
  • 12. In case three Candidates
  • 13.  
  • 14.  
  • 16.  
  • 17.
  • 18. Example Counting Front View F(1) F(2) F(3) F(4) F(5)
  • 19. Example Counting Side View S(1) S(2)
  • 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
  • 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
  • 23. Formula for three candidates
  • 24. Definition is the number of ways to count the ballot so that correspond to the required condition Lemma 1.1 Lemma 1.2
  • 26. Let Consider Use strong induction; given is the “base” therefore Proof and hence the base case is true.
  • 27.
  • 28.  
  • 29.  
  • 30. By strong induction, we get that.
  • 31. Formula for n candidates
  • 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
  • 33.  
  • 34.  
  • 35.  
  • 36.  
  • 37. We will show that is factor of Case 1 Case 1
  • 39. is factor of . Hence We will show that is factor of . Case 2
  • 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
  • 42. From case 1,2 and 3, we now prove that By mathematical induction,
  • 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.
  • 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.
  • 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 .
  • 48.
  • 49. Application In Cryptography Define the plaintext (code) used to send the data Increases the security of the system
  • 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.
  • 51. Thank you for your attention