Article: 863 of cornell.class.cs100m
From: tyan@twinkie.cs.cornell.edu (Thomas Yan)
Newsgroups: cornell.class.cs100m
Subject: some sample data for P5.2
Date: 31 Mar 2001 01:23:17 GMT
Message-ID: <9a3bi5$8i4$1@news01.cit.cornell.edu>

election1 = [
    'A>D>E>C>B'
    'C=B=E>D=A'
    'A>E>C=D=B'
    'A>B=E=D=C'
    'D>C>E>B>A'
    'D=A=C>B=E'
    'A>C>E>B=D'
    'A>D=C>E>B'
    'A=C>E>B=D'
    'B=C>E=D>A'];

% $tallies(election1)$ should be equal to $pwm1$

pwm1 = [
     0     7     5     6     7
     3     0     0     2     1
     3     6     0     4     6
     2     4     2     0     4
     3     6     2     4     0
];

% Plain Condorcet winner of $election1$ is $'A'$


election2 = [
    'B>C=A>E>D'
    'A>E>C=B=D'
    'D>E>B=C=A'
    'C=E=A>B>D'
    'A>B=D>C=E'
    'D=C=B=A>E'
    'A=D=C>B>E'
    'A>B=C=D=E'
    'B=A=D=E>C'
    'C>A>E=D=B'
    'E>C>B=A>D'
    'D=E=A=C>B'
    'C=A>E=D=B'
    'C>B>D=E=A'
    'B=A>E>C>D'
    'D>E=B>A>C'
    'D>B>A=C>E'
    'A=D>C=B>E'
    'D=C>E>B=A'
    'C>D>A=B>E'];

% $tallies(election2)$ should be equal to $pwm2$

pwm2 = [
     0     9     7     9    12
     4     0     6     5     9
     5     9     0     8    10
     5     8     6     0     9
     4     6     6     5     0
];

% Plain Condorcet winner for $election2$ is again $'A'$


election3 = [
 0 47 52 58 60
37  0 77 37 33
57 39  0 30 60
42 67 65  0 57
75 52 47 35  0];

% $tallies(election3)$ should just return $election3$ -- $ischar$ is helpful!

% Plain Condorcet winner for $election3$ is $'D'$


% more details of examples above and additional examples to follow...


Article: 864 of cornell.class.cs100m
From: tyan@twinkie.cs.cornell.edu (Thomas Yan)
Newsgroups: cornell.class.cs100m
Subject: Re: some sample data for P5.2
Date: 31 Mar 2001 01:48:34 GMT
Message-ID: <9a3d1i$97m$1@news01.cit.cornell.edu>

[below, letters are shown since they are easier to think about and
because that is what $condorcet$ should return.]

pairwise matrix for election1:

      A     B     C     D     E
  A   0     7     5     6     7
  B   3     0     0     2     1
  C   3     6     0     4     6
  D   2     4     2     0     4
  E   3     6     2     4     0

$alldefeats$ should return:

   D  C  A  C  E  A  C  A  A 
i: 4  3  1  3  5  1  3  1  1 

   B  D  C  B  B  D  E  B  E 
j: 2  4  3  2  2  4  5  2  5 

v: 4  4  5  6  6  6  6  7  7 

since A is never defeated, it is the winner.



pairwise matrix for election2:

      A     B     C     D     E
  A   0     9     7     9    12
  B   4     0     6     5     9
  C   5     9     0     8    10
  D   5     8     6     0     9
  E   4     6     6     5     0

$alldefeats$ should return:

   A  D  C  A  C  A  B  D  C  A 
i: 1  4  3  1  3  1  2  4  3  1 

   C  B  D  B  B  D  E  E  E  E 
j: 3  2  4  2  2  4  5  5  5  5 

v: 7  8  8  9  9  9  9  9 10 12 

again, since A is never defeated, it is the winner.


pairwise matrix for election3:

    A  B  C  D  E
 A  0 47 52 58 60
 B 37  0 77 37 33
 C 57 39  0 30 60
 D 42 67 65  0 57
 E 75 52 47 35  0

$alldefeats$ should return:

    A  E  C  D  A  C  D  D  E  B 
i:  1  5  3  4  1  3  4  4  5  2 

    B  B  A  E  D  E  C  B  A  C 
j:  2  2  1  5  4  5  3  2  1  3 

v: 47 52 57 57 58 60 65 67 75 77 

since every candidate is defeated, Plain Condorcet must drop defeats
in ascending order of magnitude until there is an unbeaten candidate.
this happens after A/B (47), E/B (52), C/A (57), D/E (57), A/D (58)
are dropped: $D$ is no longer defeated by anyone, so $D$ is the winner.