# Re: Putting the smith back in smith/condorcet [re-call for sponsors]

```I'm not really happy about it, but I'm not sure I can poke holes in it.

Your procedure for determining the Smith Set is dependent upon
computing the transitive closure of the pairwise victories (the
"beatpaths").  But the definition of the Smith Set doesn't depend on
beatpaths at all.

I have yet to see any attempt to show that your procedure produces the
Smith Set.  Without that, I'm not comfortable that it's equivalent.

Norman Petry did mention a procedure that would compute the Smith Set
that has two advantages I can see to yours: a) it works directly off of
the definition of a Smith Set, so it is easier to understand; and b) it
has (presumably) been discussed, tested, and critiqued by experts.

I'll also note that since you are the one proposing the method and
claiming that they are equivalent, I believe the burden of proof is on
you, not me.

Having said that...  I think this ballot set illustrates a problem.
Please verify that my work-through is correct:

4 options, 3 classes of ballots cast:

10 ABCD
10 ACDB

By inspection, the Condorcet Winner is A, and the Smith Set is also (by
inspection) {A}.

Step 2: Construct Initial Totals Table

A  B  C  D
A  - 30 30 30
B  0 -  20 10
C  0 10 -  20
D  0 20 10  -

Step 3: Construct Adjusted Totals Table

Skipped, no Supermajority Requirement

Step 4: Construct Transposed Adjusted Totals Table

A  B  C  D
A  -  0  0  0
B 30  - 10 20
C 30 20  - 10
D 30 10 20  -

Step 5: Construct "Beats Table"

A  B  C  D
A  0  1  1  1
B  0  0  1  0
C  0  0  0  1
D  0  1  0  0

Step 6: Transitive Closure

Both [B,C] and [C,D] are set, so set [B,D]
Both [C,D] and [D,B] are set, so set [C,B]
Both [D,B] and [B,C] are set, so set [D,C]
Both [B,C] and [C,B] are set, so set [B,B]
Both [C,D] and [D,C] are set, so set [C,C]
Both [D,B] and [B,D] are set, so set [D,D]
There exists no x such that both [y,x] and [x,A] is set,
so no [y,A] can be set.

Final transitive closure:

A  B  C  D
A  0  1  1  1 == 3
B  0  1  1  1 == 3
C  0  1  1  1 == 3
D  0  1  1  1 == 3

Step 6: Total rows and find largest, reduce table.

The totals are shown above, and the largest is 3.  No items are
eliminated.

It appears that the computed "Smith Set" is {A,B,C,D}, whereas the real
Smith Set is {A}.

Did I make a mistake?

--
Buddha Buck                             bmbuck@14850.com
"Just as the strength of the Internet is chaos, so the strength of our
liberty depends upon the chaos and cacophony of the unfettered speech
the First Amendment protects."  -- A.L.A. v. U.S. Dept. of Justice

```