Squares and Sticks
James
LYONSW at UCONNVM.UCONN.EDU
Thu Jun 16 13:05:50 EST 1994
RAS = RAP. I said that if two squares UNIQUELY share a stick, then
there score would be N-2. If three squares share a stick, RAP (=RAS)
if N-3. N=25.
All sticks shared by any number of squares ARE counted (including
those sticks shared by 3,4,5...24 squares.
I slipped when I wrote RAS; I originated this problem in a phylogenetic
(systematics) context, where similarities unique to groups of
organsims are called "synapomorphies" ( = uniquely derived character
states).
I constructed this analogy to clarify the problem. Here, synapomorphies
are uniquely shared sticks. 2+ squares can share a stick uniquely.
The reason why I call the measure relative apparent proximity is that
1. Counting each stick as a "synapomorphy" to a pair each time
makes the SIGMA RAP score RELATIVE, and
2. Sometimes as stick will be shared by two squares which do not
share a contimerous border, making an individual RAP score APPARENT
(not absolute).
The problem really is that I am the colleague, and evolution is
the sender. All we can tell from overall similarity is that
organism A and B are more similar overall. Parsimony measures
are relative, but we constrained by assumptions of economy of
evolution, and because each hypothetical ancestor is untestable,
the realtive distance measurements are only subject to the original
parsimony criterion/assumption.
If you've no idea about any of the assertions I am making about
parsimony or phylogenetics, please ignore them.
What the sender knows when compiling the matrix is whether any given
stick shared by two squares is shared doe to immediate proximity, or
proximity once removed.
The receiver does not know if any of the SIGMA RAP scores include
scores which do not denote contact between squares.
A B C
______ ______ ______
| | | |
| +++|++++++|+++ | 1
______ ______ ______
Here, if I tell you that A1 and C1 share a stick, you know that
A1 and B1 share a stick because you know the relative positions of
A1, B1 and C1. You also know that B1 and C1 share the stick.
But if I send the matrix
Square Pair Stick SIGMA RAP
1 2 3 4...100
A1, B1 23 Score
A1, C1 23 Score
B1, C1 23 Score
You don't know which pair of squares (A1,B1,C1 choose 2) is contimerous.
So the question is, can my colleague proceed with trying the reconstruct
the arrangement of sqaures based on RAP scores _knowing that some of the
scores are not independent but not knowing which_?
As before, HELP!
James
I must apologize for the scrambled nature of my thoughts. It is probably
genetic or something.
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