explain main methods to solve Condorcet paradox and find the winner

1. Aleppo University
Faculty of Electrical and Electronic Engineering
Computer Engineering Department
Supervised By:
PHD. Dima Mufti Alshawafa
Prepared by:
Abdulrahman Haidar Mohammed Haj Hilal
Second Semester
2017/2018

2. 2
Condorcet method
Condorcet consistent
Young’s Solution The kemeny Rule
The Mallow’s Model
The Plackett-Luce
Model1.1-Pairwise comparison
1.2-Plurality
2.1-Majority Graph
2.2-Copland’s Rule
2.3-Black’s Rule

3. 3
‫طريقتين‬ ‫عن‬ ‫سنتكلم‬ ‫هنا‬:
1-PAIRWISE COMPARISION
2-PLURALITY
1-:PAIRWISE COMPARISION
‫اآلخرين‬ ‫المرشحين‬ ‫بقية‬ ‫مع‬ ‫مرشح‬ ‫كل‬ ‫مقارنة‬ ‫يتم‬(ONE ON ONE)
‫الفائز‬ ‫هو‬ ‫يكون‬ ‫األكثر‬ ‫النقاط‬ ‫ذو‬ ‫والمرشح‬ ‫فوز‬ ‫لكل‬ ‫نقطة‬ ‫وإعطاء‬ ‫جدول‬ ‫وانشاء‬
‫أجل‬ ‫من‬ ‫المقارنات‬ ‫عدد‬ ‫ويكون‬N‫هو‬ ‫مرشح‬N(N-1)/2.
2-PLURALITY:
‫ذو‬ ‫للمرشح‬ ‫التصويت‬ ‫يتم‬‫األعلى‬ ‫المعدل‬‫األكثر‬ ‫األصوات‬ ‫على‬ ‫يحصل‬ ‫الذي‬ ‫والمرشح‬
‫الفائز‬ ‫هو‬ ‫يكون‬.

4. 4
71191210# OF
VOTERS
ADBCA
DACEB
BEADC
CCEAD
EBDBE
‫مثال‬:
‫أجل‬ ‫من‬ ‫المقارنات‬ ‫عدد‬5‫هو‬ ‫مرشحين‬:
5(5-1)/2=10
EDCBA
WWWWA
WLWLB
WWLLC
WLWLD
LLLLE
4-0=4
2-2=0
2-2=0
2-2=0
0-4=-4
‫الفائ‬‫ز‬
‫أن‬ ‫وبما‬A‫ا‬ ‫فهو‬ ‫المرشحين‬ ‫جميع‬ ‫على‬ ‫فاز‬ ‫قد‬‫لـ‬
Condorcet winner

5. 5
‫مجموع‬ ‫لدينا‬ ‫حيث‬
‫هو‬ ‫الكلي‬ ‫األصوات‬300
‫يفضلون‬ ‫المصوتين‬ ‫أغلب‬ ‫وبالتالي‬
‫المرشح‬C
254312519475#of
voters
WWOOCC1st place
OCCWWO2nd place
COWCOW3rd place
C: 75+94 = 169.
O: 51+12 = 63.
W: 43+25 = 68.
‫مثال‬:

6. 6
‫الـ‬ ‫في‬ ‫بينما‬ ‫رابح‬ َ‫ا‬‫دائم‬ ‫هناك‬ ‫يكون‬ ‫أن‬ ‫يجب‬Condorcet method
‫على‬ ‫يتغلب‬ ‫من‬ ‫فقط‬ ‫الرابح‬ ‫يكون‬‫المرشحين‬ ‫كل‬‫اآلخرين‬
‫حالة‬ ‫في‬ ‫ندخل‬ ‫أن‬ ‫وممكن‬Condorcet Paradox‫رابح‬ ‫وجود‬ ‫عدم‬ ‫عند‬
‫الحلقات‬ ‫وجود‬ ‫بسبب‬.

7. 7
‫الـ‬ ‫أساس‬ ‫على‬ ‫المرشحين‬ ‫بين‬ ‫العالقات‬ ‫يبين‬ ‫غراف‬ ‫مخطط‬ ‫رسم‬ ‫يتم‬Majority
1251221# OF VOTERS
BDCAbest
DCDB
AABC
CBADworst
A
C D
B
‫األولى‬ ‫حلقتين‬ ‫لدينا‬ ‫هنا‬A B D‫والثانية‬A C D‫الـ‬ ‫دور‬ ‫يأتي‬ ‫وهنا‬CONDORCET CONSISTENT
‫فائز‬ ‫وجود‬ ‫لعدم‬.

8. 8
The win-lose record for candidate X is
WL(X)=|{Y|X>mY}|-|{Y|Y>mX}|
The Copeland winner is the candidate that maximizes WL
1251221# OF VOTERS
BDCAbest
DCDB
AABC
CBADworst
A
C D
B
A=2(W)-1(L)=1.
B=2(W)-1(L)=1.
C=1(W)-2(L)=-1.
D=1(W)-2(L)=-1.
‫فائز‬ ‫يوجد‬ ‫ال‬ ‫هنا‬

9. 9
‫وجد‬ ‫اذا‬CONDORCET WINNER‫فنستخدم‬ ‫يوجد‬ ‫لم‬ ‫واذا‬ ‫الفائز‬ ‫فهو‬
BORDA COUNT‫الفائز‬ ‫هو‬ ‫األعلى‬ ‫الرقم‬ ‫صاحب‬ ‫والمرشح‬.
1251221# OF VOTERS
BDCAbest
DCDB
AABC
CBADworst
A
C D
B
B(A) =3*21+0*12+1*5+1*12=80.
B(B) =2*21+1*12+0*5+3*12=90.
B(C) =1*21+3*12+2*5+0*12=67.
B(D) =0*21+2*12+3*5+2*12=63.
‫فائز‬
3
2
1
0

10. 10
M= matrix of votes.
Suppose true ranking is a ≻ b ≻ c:
Pr[M| ≻]: 𝟏𝟑
𝟖
𝒑 𝟖(𝟏 − 𝒑) 𝟓. 𝟏𝟑
𝟔
𝒑 𝟔(𝟏 − 𝒑) 𝟕. 𝟏𝟑
𝟏𝟏
𝒑 𝟏𝟏(𝟏 − 𝒑) 𝟐
for a ≻ c ≻ b Pr[M| ≻] is:
𝟏𝟑
𝟖
𝒑 𝟖
(𝟏 − 𝒑) 𝟓
. 𝟏𝟑
𝟔
𝒑 𝟔
(𝟏 − 𝒑) 𝟕
. 𝟏𝟑
𝟏𝟏
𝒑 𝟏𝟏
(𝟏 − 𝒑) 𝟐
coefficients are identical so :
Pr[𝑀|≻]∝ 𝒑#𝒂𝒈𝒓𝒆𝒆.(𝟏 − 𝒑)#𝒅𝒊𝒔𝒂𝒈𝒓𝒆𝒆
cba
68a
115b
27c

11. 11
The Kendall tau rank distance between two rank orders is the total number of rank
disagreements over all unordered pairs.
For example,
the Kendall tau distance between rank orders
b ≻ e ≻ d ≻ a ≻ c and b ≻ a ≻ e ≻ d ≻ c is two,
since the orders disagree on pairs {a, e} and {a, d} but no other pairs.
The Kemeny rule chooses a consensus ranking that minimizes the sum total Kendall tau
rank distance to the preference order of each voter.
Kemeny : Given preference profile ≻ = (≻1...... ≻n)the Kemeny rule selects as the social
ranking a preference order ≻ R that minimizes the Kendall tau rank.
distance between ≻ R and ≻ i, summed over all agents i  N.
When used as a social choice rule, the alternative selected is the top alternative in the
social rank order ≻ R.

12. 12
Example
Suppose there are three alternatives
{a, b, c} and sixty agents, with the
following votes:
81021723
ccbba
baacb
abcac
the majority graph
A
C B
6
24
10
a ≻ b ≻ c = +6+24-10 = 20
a ≻ c ≻ b = -10-24+6 = -28
b ≻ a ≻ c = -6-10+24 = 8
b ≻ c ≻ a = +24+10-6 = 28
c ≻ a ≻ b = +10+6-24 = -8
c ≻ b ≻ a =-24-6+10 = -20
the
kemeny
winner

13. 13
The Mallows model for the rank order on a set of alternatives A has two
parameters:
(i) a rank order ≻ 0 on alternatives, and
(ii) a probability p > 0.5.
The model defines a distribution on rank orders.
The probability of rank order ≻  P≻! given parameters
 = (≻ 0, p), is
Pr (≻) =
𝟏
𝒁𝟏
𝒑 𝒏 𝒂𝒈𝒈(≻ ,, ≻ 0).(𝟏 − 𝒑) 𝒏 𝒅𝒊𝒔(≻ ,, ≻ 0)
where Z1 is a normalization constant that ensures that the probabilities sum to one

14. 14
Example
Suppose there are three alternatives {a, b, c} and the parameters are ≻ 0:
b, a, c and p = 0.7.
The Mallows model defines the following probability distribution on rank orders:
a ≻ b ≻ c = 1/Z1(1-0.7)(0.7)(0.7) = 0.186
a ≻ c ≻ b = 1/Z1(0.7)(1-0.7)(1-0.7) = 0.079
b ≻ a ≻ c = 1/Z1(0.7)(0.7)(0.7) = 0.434
b ≻ c ≻ a = 1/Z1(0.7)(1-0.7)(0.7) = 0.186
c ≻ a ≻ b = 1/Z1(1-0.7)(1-0.7)(1-0.7) = 0.034
c ≻ b ≻ a = 1/Z1(1-0.7)(0.7)(1-0.7) = 0.079
Straightforward calculations
reveal that the
normalization constant is
Z1 =0.79.
Based on this,
example probabilities are
Pr(a ≻ b ≻ c) =0.186 and
Pr(b ≻ a ≻ c) = 0.434.
the Mallow's
winner

15. 15
The Plackett-Luce model for the rank order on a set of alternatives
A = {1, . . . ,m}, has m parameters:
• a scoreᵞR+ for each alternative j.
The parameters are normalized so that ∑. ᵞ = 1.
The model defines a distribution on rank orders.
Let ≻[k] denote the alternative ranked in kth place by rank order ≻  P≻ .
The probability of rank order ≻  P! given parameters
ᵞ = (ᵞ 1, . . . , ᵞ m), is
Pr(≻)=
ᵞ≻[𝟏]
ᵞ≻ 𝟏 +⋯+ᵞ≻[𝒎]
.
ᵞ≻[𝟐]
ᵞ≻ 𝟐 +⋯+ᵞ≻[𝒎]
. ........ .
ᵞ≻[𝒎−𝟏]
ᵞ≻ 𝒎−𝟏 +ᵞ≻[𝒎]
.
ᵞ≻[𝒎]
ᵞ≻[𝒎]

16. 16
Example
Suppose the alternatives are A = {a, b, c}, and consider Plackett-Luce
with scores ᵞa = 0.6, ᵞb = 0.3, ᵞc = 0.1 .
The model assigns the following probabilities to rank orders:
Pr(a ≻ b ≻c)=
𝟎.𝟔
𝟎.𝟔+𝟎.𝟑+𝟎.𝟏
.
𝟎.𝟑
𝟎.𝟑+𝟎.𝟏
.
𝟎.𝟏
𝟎.𝟏
= 𝟎.𝟒𝟓
Pr(b ≻ a ≻c)=
𝟎.𝟑
𝟎.𝟔+𝟎.𝟑+𝟎.𝟏
.
𝟎.𝟔
𝟎.𝟔+𝟎.𝟏
.
𝟎.𝟏
𝟎.𝟏
= 𝟎.𝟐𝟔
Pr(c ≻ a ≻b)=
𝟎.𝟏
𝟎.𝟔+𝟎.𝟑+𝟎.𝟏
.
𝟎.𝟔
𝟎.𝟔+𝟎.𝟑
.
𝟎.𝟑
𝟎.𝟑
= 0.07
the
Plackett-luce
winner

17. 17