[EM] Dyadic ballots (was "...encouraging truncation")

[Forest Simmons]

On Thu, 13 Mar 2003, [iso-8859-1] Kevin Venzke wrote:

> Forest,
>
> This is an interesting idea.
>
> I was trying to do some examples with it, but I'm not
> sure how to create the four matrices.  These are the
> ballots I was trying to use:
>
> 12: A at 10 (fill 8 and 2 circles)
> 11: B at 7 (fill 4, 2, 1)

The wise course for all voters is to vote at least one candidate at 15 and
at least one candidate at zero, so these two factions seem to be throwing
caution to the wind in order to express their beliefs that neither
candidate is very good.

The first faction gives binary scores of 1010 and 0000 for A and B
respectively.

Since 1>0 and 10>00 and 101>000 and 1010>0000 we see that all four of the
first faction matrices will be identical:

[0 1]
[0 0]

So from the first faction we get four matrices that look like

[0 12]
[0  0]  .

The second faction approved no candidate, so the scores for A and B,
respectively are 0000 and 0111.

Since  0=0 and 00<01 and 000<011 and 0000<0111 the three finest matrices
look like

[0 0]
[1 0] ,

while the crudest one looks like

[0 0]
[0 0] ,

reflecting the wasted opportunity of approving their approved candidate.

Multiplying by eleven and adding, we get for the respective combined
pairwise matrices (from crudest to finest)

[0 12]
[0  0],

[0 12]
[11 0],

[0 12]
[11 0], and

[0 12]
[11 0] .

The three finest matrices are indeed identical in this (degenerate)
example.

For a more helpful example consider the following preference ballot with
strength of preference indicated by the number of chevrons:

A>B>>C>D>>>E>>>>F>>G>>>H>>I>J

We translate the binary tree structure into a dyadic CR ballot:

    candidate  score
	A	1111
	B	1110
	C	1101
	D	1100
	E	1000
	F	0110
	G	0100
	H	0010
	I	0001
	J	0000

According to the most significant bit, candidates A thru E are all
preferred over F through J, so the first five rows of the crude matrix are
identical copies of [0 0 0 0 0 1 1 1 1 1]. The remaining five rows are
filled with zeros.

According to the first two significant bits, we have
A=B=C=D>E>F=G>H=I=J, so the corresponding pairwise matrix has four rows of
[0 0 0 0 1 1 1 1 1 1], one row of [0 0 0 0 0 1 1 1 1 1], two rows of
[0 0 0 0 0 0 0 1 1 1], and the last three rows filled with zeros.

According to the first three significant bits we have
A=B>C=D>E>F>G>H>I=J, so the rows of the next finest matrix are
(two of) [0 0 1 1 1 1 1 1 1 1], (two of) [0 0 0 0 1 1 1 1 1 1],
one each of [0 0 0 0 0 1 1 1 1 1], [0 0 0 0 0 0 1 1 1 1],
[0 0 0 0 0 0 0 1 1 1], and [0 0 0 0 0 0 0 0 1 1], and two rows of zeros.

When all four bits are considered we get a complete ordering so the matrix
is all ones above the main diagonal, and all zeros below the main
diagonal.

These are the contributions of one dyadic ballot to the four pairwise
matrices that would be used in the method.

I hope that clarifies things a little.

If there are fewer than five candidates there is no need to have four bit
ballots.  With three candidates two bits are sufficient, so the method
would be the same as the one Venzke described.

With five candidates four bits are necessary to distinguish the cases

A>B>>C>>>D>>>>E  and  A>>>>B>>>C>>D>E,  for example, which would have

dyadic CR ballots of

A 1111, B 1110, C 1100, D 1000, E 0000  in the first case, and

A 1000, B 0100, C 0010, D 0001, E 0000  in the second case.

These two cases yield identical matrices only at the finest level.

The most extreme difference is at the approval level:

A=B=C=D>E in the first case, and
A>B=C=D=E in the second case.

At the next finer level we have
A=B=C>D>E in the one case, and
A>B>C=D=E in the other.

At the next finer level we have
A=B>C>D>E in the first case, and
A>B>C>D=E in the other.

I'm afraid that's all I have time for now.

Forest