Re: Pitman numerology

On 8 Dic, 20:15, "R. Baldwin" <res0k...@xxxxxxxxxxxxxxxxxxxx> wrote:
Charles Brenner <cbren...@xxxxxxxxxxxx> wrote innews:cfb8c995-b1f2-4ae6-80fd-44a883fa9ed0@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

On Dec 7, 6:10 pm, "R. Baldwin" <res0k...@xxxxxxxxxxxxxxxxxxxx> wrote:
Charles Brenner <cbren...@xxxxxxxxxxxx> wrote
innews:b80fb608-8414-493c-9
768-76eea8e52...@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
distanc
es  
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.

That is correct.



But now you admit that you know that the mapping doesn't preserve
distance.

Of course it doesn't. It can't. That was the whole point from the
beginning.

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.

Actually, both the mapping and the inherent difference between the
metrics cause the different results. Hamming Distance and Euclidean
Distance are metrics on different spaces. For any given symbol set S and
length N, there is exactly one metric space S^N on which Hamming
Distance is defined. It is not an Euclidean space, it is a Hamming
space. There are many different Euclidean spaces to which each Hamming
Space can be mapped.

In other words, strings in S^N simultaneously represent points in many
different Euclidean spaces, but in only one Hamming space.

Because there is a one-to-many relationship between the Hamming space
and Euclidean spaces, the only way there could be a general relationship
between Hamming Distance and Euclidean Distance is if Euclidean Distance
were identical in all the Euclidean spaces to which S^N could be mapped
- which clearly, is not the case.

Therefore, Hamming Distance and Euclidean Distance are generally
unrelated. Only in the trivial case where |S|=2, and S^N -> E^N, can a
relationship be found. If those special constraints do not hold, then
Hamming Distance is unrelated 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 define
d
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.

You might be right about a ring. It's been too long for me to remember.



Even though strings form a metric space, they do not exist in an
Euclidea
n
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 is no particular reason to require that N=K. They are not equal in
most problems I've dealt with over my career. K=1 is far more common.



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.

That is exactly what I'm saying. But it's not an attitide - it's simply
how Hamming Distance is defined. The distinction is important in real
engineering problems, such as the best way to encode numbers in the
presence of noise.



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.

Mapping from N-dimensional Hamming space to a 1-dimensional Euclidean
space is done all the time. I really don't see why you find this
problematic.

I've given you other examples with different dimensions - is that what
you meant?

A dimention-preserving mapping would have been a better example,
IMHO.

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.

Actually, I did read it several times and it still didn't make sense to
me. Whether you call a and b 9-vectors or scalars makes no difference in
calculating the Hamming distance.

Actually it does make difference.
The Hamming distance is the number of different elements in
corresponding locations, hence if you consider a and b as vectors over
Reals^9 the Hamming distance is 2, if you consider them as real
numbers, that is, vectors over Reals, the Hamming distance is 1.

You confusion probably comes from the convenient abuse of notation
often done in Computer science of considering bit-vectors equivalent
to fixed digits integer numbers. While bijectively related, they are
different mathematical objects which are not isomorphic.

<snip>

.

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