Re: Pitman numerology

On Dec 7, 6:10 pm, "R. Baldwin" <res0k...@xxxxxxxxxxxxxxxxxxxx> wrote:
Charles Brenner <cbren...@xxxxxxxxxxxx> wrote innews:b80fb608-8414-493c-9768-76eea8e527e7@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:



On Dec 7, 12:39 pm, "R. Baldwin" <res0k...@xxxxxxxxxxxxxxxxxxxx>
wrote:
Charles Brenner <cbren...@xxxxxxxxxxxx> wrote
innews:db5b8b07-bf81-482c-a
dcc-6b0443e58...@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

[snip]

You made one interpretation instead of another you could have made.
You interpreted the two strings as elements of D^9 where D=digits.
Yo
u
could instead have interpreted them as elements of R^1 as you had
advertised and as you interpreted them for your Euclidean
calculation.

It simply requires that two strings
have the same number of symbols, and counts how many are
different.

But you have to know what constitutes a symbol (a token).

However, if you prefer, the fallacy of Sean's declaration can be
shown in other ways.

No dispute that Sean's claim was wrong. I picked on him too.

Your original post in this thread was excellent and insightful. For
that reason I was astonished at the carelessness of the later one.

I'm still having trouble understanding your objection. The D^9 digits
as numbers with positional notation map to the Euclidean R^1 space
just fine
.

Mapping doesn't preserve the distances in question, so it's not an
innocuous operation.

Mapping isn't supposed to preserve the distances in question. The distances  
are different metrics on different spaces.

You wanted to show the lack of relationship between Hamming and
Euclidean distance. In an effort to demonstrate you took an example in
some 9-dimensional space, evaluated the Hamming distance, then
*mapped* (by reinterpretation of the symbols) the example to some 1-
dimensional space and evaluated the Euclidean distance. You pointed
out that the two distances were unrelated.

But now you admit that you know that the mapping doesn't preserve
distance. Therefore your demonstration proves nothing because it is
logically possible, from your procedure, that the mapping, rather than
any difference between H & E, caused the different results.

Indeed, one could use your procedure to prove that Euclidean distance
bears no relationship to Euclidean distance!

Hamming Distance is a metric d: S^N X S^N -> R on the a metric space for
the set S^N of ordered lists (strings), where S is a symbol set and N is an
integer length. The metric space S^N is not an Euclidean space, because S
is not a field. It is only a set of ordered lists. Addition is not defined
for S, neither is multiplication, nor the additive and multiplicative
identities.

Euclidean Distance is a metric for a pair of points in an Euclidean space..
To have an Euclidean space it is necessary to first have a field.

Very well, have a field (though I think a ring is sufficient). A
vector space over a field is in particular a module on a set, on which
you have Hamming distance.

Even though strings form a metric space, they do not exist in an Euclidean
space until they are mapped to one (M: S^N -> E^K).

If you would take N=K, then I would say this is pointless abstraction
but ok. Unfortunately you seem though to like N=9 and K=1.

There are many
different ways to do this for the same set of strings S^N. The two most
obvious mappings are S^N -> R^N

That would be fine.

and S^N -> R. Hamming Distance is
unaffected by the mapping, because it is a metric on the original metric
space, not the Euclidean space onto which the original metric space is
mapped.

Apparently you have decided that you won't consider Hamming distance
on the same space on which you consider Euclidean distance. If that is
your attitude obviously you can't make any meaningful comparison
between them. But the rest of us can.

The interpretation therefore permits definition for both Euclidean
Distance and Hamming Distance.
What is incorrect about the interpretation, in your opinion?.

I'm not 100% sure what you mean by "the" interpretation since you
switch back and forth, but if you stick to one interpretation at a
time then it's possible to compute both distances.

I've had one all along. Perhaps what I've written above will clarify.

You've had two all along. Sometimes 9 dimensions, sometimes 1
dimension.


Returning to your original example:

Let's take an example from the Euclidean space R^1, radix 10.

Ok, we are by your account dealing with 1-dimensional space. The radix
10 part is redundant since no one is going to assume base 8. However
it seems to have caused some confusion in you own mind and led you to
think more than you should (i.e. more than not at all) about the
individual digits in the notation. Forget about the digits. These are
just points on a ray from zero to infinity.

So for now "the interpretation" is 1 dimension.

a = 100000000
b = 000000070

Hamming distance = 2

No. The Hamming distance is 1 -- the numbers are different. They're
not 9-vectors, they're scalars (or 1-vectors if you prefer). They are
different in the only position there is.

How on Earth do you get 1? The strings a and b differ in two positions,
therefore the Hamming Distance is 2.

I've used the mathematical terms for sets and dimensions which you
introduced, and I've been very explicit. I read your message carefully
and in full. If you don't understand how I got 1, then I can only
suppose you didn't return the favor. I cannot explain it again better.

Again, the Hamming Distance is a metric on the original metric space - in
this case, 9-symbol ordered lists over {0..9}. The Euclidean distance is a
metric on the Euclidean space to which the original metric space is mapped.



Euclidean distance = sqrt( (a-b)^2 ) = |a-b| = 99999930

I agree that's the Euclidean distance.

-----------------------------

Alternatively, one could regard the above notation for a and b as
denoting points in 9 dimensions -- now the interpretation is D^9 --
although that's not what you said you were going to do. So regarded,

Hamming distance =2  because two digit positions differ, and
Euclidean distance = sqrt( 1^2 + 7^2 ) = 7.07...

Yes, that could be done if you decide to map D^9 -> R^9.

And this multi-dimensional example would be more to the point in
debating with Sean; all the palaver with a mere 1-dimensional space
over a large set like R is really a pedantic digression. His situation
involves 1000-dimensional spaces essentially over a binary set (point
mutations). That's a high-dimensional unit cube. Hamming distance is
number of mutations. Nobody (but you?) would assume that when Sean
mentions Euclidean distance he means that the coordinates of a point
are to be considered as bits in a large binary number. No, Euclidean
distance in this context obviously means diagonal distances through
the cube -- in fact the square root of Hamming distance.

.

Relevant Pages

  • Re: Neighbors in Space
    ... generalization of what we commonly call the distance between two ... There are lots of different metrics that we commonly use ... apart in Euclidean space might be neighbors in the Wormhole Nexus. ... Another strange one is the TARDIS metric, ...
    (rec.arts.sf.written)
  • Re: Pitman numerology
    ... digits as numbers with positional notation map to the Euclidean ... Euclidean distance. ... Hamming Distance and Euclidean ... strings in S^N simultaneously represent points in many ...
    (talk.origins)
  • Re: Pitman numerology
    ... look like a complete idiot by continuing to use the word "Euclidean" ... Hamming distance is intended as synonymous with "Manhatten distance" ...
    (talk.origins)
  • Re: Fudge Factors and a new physics.
    ... for the length L, in the x direction, of a rod in its own rest frame: ... A lightyear is the distance that light travels ... you do a rotation you'd like the distance to be unchanged. ... But for relativity you don't care about rotating and leaving euclidean ...
    (sci.physics.relativity)
  • Re: Help in Classification with k Nearest Neighbor
    ... >> What formulas do you use for Euclidean and L2? ... This is SQUARED EUCLIDEAN DISTANCE ... I want to take the first K minimum distance ... % find the second norm for each row ...
    (comp.soft-sys.matlab)