# 3TA (correction)

P.Hovenkamp Hovenkamp at NHN.LEIDENUNIV.NL
Fri Apr 11 11:09:54 CDT 2003

```At 10:48 AM 4/10/03 +0200, I wrote:
>At 08:55 AM 4/10/03 +0100, David Williams wrote:
>
>(...)
>>My interest in three-item data has nothing to do with any of the above and
>>is almost painfully simple. What might be the answer to the simple two
>>character problem AB(CD) + AC(BD)? As far as I am aware most (all?) methods
>>yield no result, or return the same two trees, as if the solution to 2 + 2
>>= 2 + 2. I see the answer as A(BCD) [2 + 2 = 4]. I need no method, no
>>matrix, no program (all my MSc students suggest the same answer as well,
>>before they learn that computer programs know better). Now, to me,
>>something is wrong.
>
>Yes - but where?
>When I run this data through Nona using all the usual trappings (heuristic
>search settings etc.) I get the following results: two most parsimonious
>trees (corresponding with the two trees alluded to by David), consensus of
>which is A(BCD). What most people would take to be the result of the
>analysis is exactly what David would accept as the result of the analysis.
>So why is he not satisfied with the result being produced by a strict
>consensus?

Off-list David informed me that I had, indeed, made a beginner's mistake.
Oops. Sorry.
I had mistaken the resolution of the basal node for a real supported
resolution. In fact, it is only supported by the decision to use an
outgroup - which defines the in-group.

When analyzed properly with an outgroup, Davids example indeed holds:
parsimony does not reconstruct a group (BCD). Using parsimony, there is no
character support for this group. That raises an interesting point: what
makes the presence of this group so glaringly obvious to most people
looking at this matrix:

Data:
O 00
A 00
B 01
C 10
D 11

Intuitively, we appear to argue that B and C are both closer to D than A is
to D, so they must be in the same cluster. Is our intuition wrong, or is