Ballot Problem for Many Candidates

THE BALLOT PROBLEM FOR MANY CANDIDATES

Advisers :

Mr.Thammanoon Puirod

Mr. Deaw Jaibun

Members :

Mr. Norrathep Rattanavipanon

Mr. Nuttakiat Chaisettakul

Mr. Papoj Thamjaroenporn

What is the ballot of problem?

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 ?

a + ba - b

aa + b

The formula comes up to be

Why is it interesting?

The Ballot problem is one of very well-known problems. This problem had been proved for two candidates.

What about any n, which is greater than 2, candidates?

Objective

To find the formula and proof of Ballot problem for many candidates.

In case two candidates

(The Ballot problem)(The Ballot problem)

Suppose is the ballot of the 1st candidate.

is the ballot of the 2nd candidate, when .

Define “1” as the ballot given to 1st candidate. “-1” as the ballot given to 2nd candidate.

1 2a > a2a1a

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)

' '

1,1, ,1, 1, 1, , 1a s b s

=

=

Reflection Principle

1

2

Since the number of ways begin at (0,1) is the permutation of a ' 1s

and (a -1) '(-1)s and vice versa.

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:

1 2 1 2 1 21 21 2

1 2 11 2

a + a -1 a + a -1 a + aa - aB(a ,a ) = - =

a -1 a -1 aa + a

1 2B(a ,a )

In case three Candidates

st1

nd2

rd3

1 2 3

a is the ballot of the 1 candidate.

a is the ballot of the 2 candidate.

a is the ballot of the 3 candidate

when a > a > a

Suppose

How to count ?

1. Consider the path seen at the front view. Find all possible ways to “walk” within the allowed plane.

2. Consider the path seen at the side view. Find all possible ways to “walk” within the allowed plane.

3. Match each way of 1. to each of 2. (Sometimes it yields more than one complete path.)

Example

Counting 1 2 3(a ,a ,a ) = (4,3,2)

Front ViewFront View

F(1) F(2) F(3)

F(4) F(5)

Example

Counting 1 2 3(a ,a ,a ) = (4,3,2)

Side ViewSide View

S(1) S(2)

Example

Counting 1 2 3(a ,a ,a ) = (4,3,2)

MatchingMatchingF(i) * S(j) F(1) F(2) F(3) F(4) F(5)

S(1) 1 1 1 1 1

S(2) 1 1 1 2 2

F(1) F(2) F(3)

F(4) F(5)S(1) S(2)

5

7

12

Total

Dynamic ProgrammingDynamic Programming

Counting (5,4) with D.P.

0 0 0 0 0 14

0 0 0 0 5 14

0 0 0 2 5 9

0 0 1 2 3 4

1 1 1 1 1 1

Formula for three candidates

1 21 21 2

11 2

a + aa - aB(a ,a ,0) =

aa + a

Definition is the number of ways to count the ballot so that correspond to the required condition

1 2 3B(a ,a ,a )

Lemma 1.1

1 1 2B(a ,a ,a ) = 0

Lemma 1.2

1 2 31 3 2 31 21 2 3

1 2 31 2 1 3 2 3

a + a + aa - a a - aa - aB(a ,a ,a ) =

a ,a ,aa + a a + a a + a

Conjecture

1 2 3 1 2 3a ,a ,a N and (a > a > a ) Let

1 2 3 1 2 3 1 2 3 1 2 3B(a ,a ,a ) = B(a -1,a ,a ) + B(a ,a -1,a ) + B(a ,a ,a -1)Consider

Use strong induction; given is the “base”

therefore

B(3,2,1)

Proof

B(3,2,1) = B(2,2,1) + B(3,1,1) + B(3,2,0)

3+ 23- 2 1 5!= 0 + 0 + = ×

23+ 2 5 3!2!

3+ 2 +13-1 2 -1 3- 2= 2 =

3,2,13+1 2 +1 3+ 2

and hence the base case is true.

We assume that for some

all of the following are true;

1 1

2 2

3 3 1 2 3 1 2 3

0 < k a

0 < k a

0 < k a and (a ,a ,a ) (k ,k ,k ),

1 2 3k ,k andk ,

.1 2 31 3 2 31 21 2 3

1 2 31 2 1 3 2 3

k + k + kk - k k - kk - kB(k ,k ,k ) =

k ,k ,kk + k k + k k + k

We will use this assumption to prove that is true1 2 3B(a ,a ,a )

1 2 3 1 2 3 1 2 3 1 2 3

1 2 31 3 2 31 2

1 2 31 2 1 3 2 3

1 21 3 2 31 2

1 2 1 3 2 3

B(a ,a ,a ) = B(a -1,a ,a ) + B(a ,a -1,a ) + B(a ,a ,a -1)

a -1+ a + aa -1- a a - aa -1- a=

a -1,a ,aa -1+ a a -1+ a a + a

a + a -1a - a a -1- aa - a +1+

a + a -1 a + a a -1+ a

3

1 2 3

1 2 31 3 2 31 2

1 2 31 2 1 3 2 3

1 2 3 1 2 1 3 2 3 1

1 2 3 1 2 1 3 2 3 1 2 3

1 2

+ a

a ,a -1,a

a + a + a -1a - a +1 a - a +1a - a+

a ,a ,a -1a + a a + a -1 a + a -1

a + a + a (a -1- a )(a -1- a )(a - a )a= +

a ,a ,a (a -1+ a )(a -1+ a )(a + a )(a + a + a )

(a +1- a )

1 3 2 3 2 1 2 1 3 2 3 3

1 2 1 3 2 3 1 2 3 1 2 1 3 2 3 1 2 3

1 2 3 2 3 1 31 2

1 2 3 1 2 2 3 1 3

(a - a )(a -1- a )a (a - a )(a +1- a )(a - a +1)a+

(a -1+ a )(a + a )(a + a -1)(a + a + a ) (a + a )(a -1+ a )(a + a -1)(a + a + a )

a + a + a a - a a - aa - a=

a ,a ,a a + a a + a a + a

1 2 3 2 3 1 31 21 2 3

1 2 3 1 2 2 3 1 3

a + a + a a - a a - aa - aB(a ,a ,a ) =

a ,a ,a a + a a + a a + a

By strong induction, we get that.

Formula for n candidates

1 2 3 n 1 2 3 nB(a ,a ,a , ,a ,0) = B(a ,a ,a , ,a )

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

1 2 3 nB(a ,a ,a , ,a )

1 2 3 nGive P(n) is "B(a ,a ,a , ,a ) is true" n

In base case P(2) is true.

Assume P(k -1) is true. We will prove that P(k) is true

1 2 3 k i j1 2 3 k

0<i< j k1 2 3 k i j

a +a + a +…+ a a - aB(a ,a ,a ,…,a ) = .

a ,a ,a ,…,a a + a

1 2 3 k 1 1

2 2

k k

1 2 3 k 1 2 3 k

1 2 3 k i j1 2 3 k

0<i< j k1 2 3 k i j

We assume that for b ,b ,b ,…,b , 0 < b a

0 b a

0 b a

and (b ,b ,b ,…,b ) (a ,a ,a ,…,a ),

all of the following are true :

b b + b +…+ b b bB(b ,b ,b ,…,b ) = .

b ,b ,b ,…,b b + b

1 2 k 1 2 k

1 2 k

1 2 k-1

Give Q(a ,a , ,a ) is "B(a ,a , ,a ) is true" (only for k)

We need to proof that Q(a ,a , ,a ) is true.

In base case, Q(a ,a , ,a ,0) is true.

1 2 k 1 2 k 1 2 k

1 2 k

i m

i mi j

0<i< j n 0 m

(a -1) + a

n1 2 n

m=1 1 2 m n

a + a +…+ a

a ,a ,…,a -1,…,a

1 2 ni j

1 2 n0<i< j n i j

i m

1 2 ni j i m

1 2 mm i0<i< j n 0 m

(a -1)+a

n

nm=1

i j i j 1 2 n0<i< j n

ni m

i j i j m i mm=1 m i0<i< j n 0 m

We will show that is factor ofi j0<i< j n

(a -a )

i i j i j 1 2 n0<i< j n

ni m

i i j i j m i mm=1 m i0<i< j n 0 m

Case 1

Case 1Case 1

1 31 2 m m1 2 3 m m n

1 2 1 3 m m

1 2 nn 1 n

1 2 m nn 1 n

1 31 2 m m

1 2 1 3 m m

k kk k (k 1) kB(k ,k ,k , ,k 1, ,k , ,k )

k k k k (k 1) k

k k k 1k k

k ,k , ,k 1, ,kk k

k kk k k (k 1)

k k k k k (k 1)

1 2 nn 1 n

1 2 m nn 1 n

1 2 3 m m n

k k k 1k k

k ,k , ,k 1, ,kk k

B(k ,k ,k , ,k , ,k 1, ,k )

Consider

1 2 3 m m n 1 2 3 m m nB(k ,k ,k , ,k 1, ,k , ,k ) B(k ,k ,k , ,k , ,k 1, ,k )

hence;

is factor of .i j0<i< j n

(a -a )

i iF(a )andG(a )

i j i iWhen a isinstead of a ,F(a ) G(a ) 0.

Hence

We will show that is factor of .i j0<i< j n

(a +a -1)

i iF(a )andG(a )

Case 2Case 2

Consider

q j j p p q q p0< j nj p,q

p j j q q p p q0< j nj p,q

(a -a )(a -a ) (a +a )[-2a ] a

= -1

(a -a )(a -a ) (a +a )[2a ] a

1 1

i p p q

i j i j p q p i pp p q i0<i< j n 0 p,[a -(a +a )]-a

i < q,a -[a -(a +a )](a -a )[a +a -(a +a )] (a +a )

i > q,[a -(a +a

qq i0<i n

i q

= -1

a)]-a

is factor ofi j0<i< j n

(a +a -1)

i iF(a )andG(a )

i j i iWhen a instead of a ,F(a ) G(a ) 0.

Hence

We will show that is the factor of1 2 n(a +a +…+a ) i iF(a )andG(a )

iG(a )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

iaiF(a )

1 2 nm(a +a +…+a +k)

i iF(a )andG(a )1 2 n(a +a +…+a )

i iF(a )andG(a )

Case 3Case 3

From case 1,2 and 3, we now prove that

1 2 ni j

1 2 n0<i< j n i j

i m

1 2 ni j i m

1 2 mm i0<i< j n 0 m(a -1)+a

n

nm=1

By mathematical induction,

1 2 3 k i j1 2 3 k

0<i< j k1 2 3 k i j

a +a + a +…+ a a - aB(a ,a ,a ,…,a ) = .

a ,a ,a ,…,a a + a

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 BiologyLead to find the number of ways

a random graph model for angiogenesis in the renal glomerulus.

Steps to form a vascular network1. Budding2. Spliting3. Connecting

n(Budding) n(Connecting)

Application In Cryptography

Define the plaintext (code) used to send the data

Increases the security of the system

Given , when look left to right,

n(A) n(B) n(C) ...

p- sequence

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.

Ballot Problem for Many Candidates

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