Room change! The Thursday recitation at 11:15 now meets in EE/CS 3-115.

Solution to HW #62, p. 39 and #78 p. 40 (Condorcet loser question)

62. Try this

15 9  4  3
A  B  D  B
C  D  C  C
B  C  B  D
D  A  A  A

C is a Condorcet candidate: C beats A: 16-15. C beats B: 19-12 C beats D:18-13.

B is the Borda count winner: points are A:76, B: 86, C:84, and D:64.

78. (a), (b) Here is a preference schedule with distinct numbers:

8  5  3  2
A  B  C  D
B  C  B  C
C  D  D  B
D  A  A  A

For question (b) use the same preference schedule. In plurality with elimination, D is the first candidate eliminated, so D's 2 first place votes go to C. Now C and D have 5 votes, vs. A's 8, so C and D are eliminated, and A wins.

(c) Let's modify the point system, say the last place candidate gets 0 points, the next to last candidate 1 points, then 2 points, etc. So if there are 5 candidates, the top choice gets 4 points. With this new point system, the winner does not change, because each candidate's point total is decreased by the same amount, namely the total number of voters.

So if there are 5 candidates, the top choice gets 4 points, which is the number of other candidates he beats! So we see that the each candidate's point total is the total number of wins he has in the 1-1 battles.

A Concorcet loser, call him Mr. X, has a losing record in all of these 1-1 battles, so if we find Mr. X's overall record it would be a losing one, something like 63-99, for 63 points. Each of the other candidates, who have the same total number of 1-1 comparisons, would have to have a worse record, say for example 59-103, 59 points. If Mr. X has more points than they do. But it is impossible for all candidates to have losing records, so Mr. X could not have the best record and be the Borda count winner.