Compression - Binary Image Reduction Theorem

Author

Binary Image Reduction Theorem

Plato777

2004-08-29, 3:55 pm

When using a specific, most-efficient,
image reducing technique on all binary
images of a series class, all images of
that class have a degree of redundancy
and can can be reduced to a specific
image of a lower-order series class
save for one image of the series class
in question.

Erik T. Evenson

Willem

2004-08-29, 3:55 pm

Plato777 wrote:
) When using a specific, most-efficient,
) image reducing technique on all binary
) images of a series class, all images of
) that class have a degree of redundancy
) and can can be reduced to a specific
) image of a lower-order series class
) save for one image of the series class
) in question.

For this to be a real theorem, you need to define all the terminology
you're using, otherwise it's meaningless. Also, you'll need a proof.

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

Plato777

2004-09-05, 3:55 pm

Willem <willem@stack.nl> wrote in message news:<slrncj49n8.1gcv.willem@toad.stack.nl>...
> Plato777 wrote:
> ) When using a specific, most-efficient,
> ) image reducing technique on all binary
> ) images of a series class, all images of
> ) that class have a degree of redundancy
> ) and can can be reduced to a specific
> ) image of a lower-order series class
> ) save for one image of the series class
> ) in question.
>
> For this to be a real theorem, you need to define all the terminology
> you're using, otherwise it's meaningless. Also, you'll need a proof.
>
>
> SaSW, Willem

Sure thing.
This theorem is absolutely consistent with the Counting Theorem.

A buffer containing 10 bits has an order-class of 10.
In this class there are 2^10, or 1024 images in the
set together representing all the possible images in
the series.

The simplest and most efficient image-reducing process
is a simple one-to-one mapping of all images in the
base series to all the other possible images (including
the empty-set) of all the lower-class series.

For example, if we take 2^10 as the base series then,

2^10, 1024 elements
2^9, 512 elements; a 1-bit reduction in image
2^8, 256 elements; a 2-bit reduction in image
2^7, 128 elements; a 3-bit reduction in image
2^6, 64 elements; a 4-bit reduction in image
2^5, 32 elements; a 5-bit reduction in image
2^4, 16 elements; a 6-bit reduction in image
2^3, 8 elements; a 7-bit reduction in image
2^2, 4 elements; an 8-bit reduction in image
2^1, 2 elements; a 9-bit reduction in image
2^0, 1 element (the empty set); a 10-bit reduction in image
There are no other possible elements in the lower-order series.

Add the total number of lower-order elements to give 1023.
Since there are 1024 in the base series it follows that
one of the base series elements cannot be reduced, therefore
this one image has no redundacy giving the current mapping
scheme. However that one image is more than probable of
being reduced in another wholly different mapping scheme.

The conclusion is that all binary images of every size has
redundancy. Any talk of a perfect random sample, having as
a requirement that it has no redundancy is hogwash. The
one image that has no redundancy as proved is totally
dependent on the acting reduction method.

QED.

Erik T. Evenson.

Willem

2004-09-05, 3:55 pm

Plato777 wrote:
) <snip>
) Add the total number of lower-order elements to give 1023.
) Since there are 1024 in the base series it follows that
) one of the base series elements cannot be reduced, therefore
) this one image has no redundacy giving the current mapping
) scheme. However that one image is more than probable of
) being reduced in another wholly different mapping scheme.

You're assuming that every 'image' has a certain length that can
somehow be divined. If I were to concatenate two of your 'lower-order'
elements, how can you tell where one ends and the next one starts ?

) The conclusion is that all binary images of every size has
) redundancy. Any talk of a perfect random sample, having as
) a requirement that it has no redundancy is hogwash. The
) one image that has no redundancy as proved is totally
) dependent on the acting reduction method.

Nonsense. If you restrict the input set to have a certain given length,
but you assume that the output set can not only have different lengths,
but that the length is *known*, then that does not mean there is
redundancy, just that the output set, being less restrictive, is smaller.

Duh.

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

Plato777

2004-09-06, 3:55 am

Willem <willem@stack.nl> wrote in message news:<slrncjmp8c.gf1.willem@toad.stack.nl>...
> Plato777 wrote:
> ) <snip>
> ) Add the total number of lower-order elements to give 1023.
> ) Since there are 1024 in the base series it follows that
> ) one of the base series elements cannot be reduced, therefore
> ) this one image has no redundacy giving the current mapping
> ) scheme. However that one image is more than probable of
> ) being reduced in another wholly different mapping scheme.
>
> You're assuming that every 'image' has a certain length that can
> somehow be divined. If I were to concatenate two of your 'lower-order'
> elements, how can you tell where one ends and the next one starts ?

This theorem does not apply to concatenated outputs. This theorem
deals with the overall image in particular. You are assuming that
the overall image can be divided into manageable groups to be
mapped with various outputs and concatenating them together as
an output and then from the output recreating the input.
Can't be done. This theorem garantees one output for each input.
Of course for large binary images the inter-image mapping is
unmanagable.

>
> ) The conclusion is that all binary images of every size has
> ) redundancy. Any talk of a perfect random sample, having as
> ) a requirement that it has no redundancy is hogwash. The
> ) one image that has no redundancy as proved is totally
> ) dependent on the acting reduction method.
>
> Nonsense. If you restrict the input set to have a certain given length,
> but you assume that the output set can not only have different lengths,
> but that the length is *known*, then that does not mean there is
> redundancy, just that the output set, being less restrictive, is smaller.

With only one output image the length is self-evident.
I'm talking about one input image not a division of the
input image into managable sections for reduction and
then concatenation of the reduced parts into a final
output form for the reduced image.

How can a smaller output set not indicate a
reduction of redundancy? I'm sure for any
image mapping scheme used in a reduction
scheme there are any number of usable Logic
Gate maps that can be implemented.
The counting of "consecutive" bits or bytes
in a binary image to determine redundancy
is wholly arbitrary and is only
one of numerous methods that can be used.

What's the current definition of redundancy anyway?

Erik.

Willem

2004-09-06, 3:55 am

Plato777 wrote:
) This theorem does not apply to concatenated outputs. This theorem
) deals with the overall image in particular. You are assuming that
) the overall image can be divided into manageable groups to be
) mapped with various outputs and concatenating them together as
) an output and then from the output recreating the input.
) Can't be done. This theorem garantees one output for each input.
) Of course for large binary images the inter-image mapping is
) unmanagable.

Your theorem assumes that the input be one fixed length.
It also assumes that the output can be any length.
In other words: you're making use of a restriction you put upon the
input images to find 'redundancy' when there is none.

) With only one output image the length is self-evident.

I refer you to the discussion between David Scott and Dale King.

) How can a smaller output set not indicate a
) reduction of redundancy?

If the input set is arbitrarily restricted to a required length, but the
output set isn't, then this 'redundancy' lies wholly in this restriction
you put upon the input set. Normally, it is assumed that the input and
output 'images' are in the same domain.

SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

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