# There are only so many photos you can take!!!

Discussion in 'Digital Photography' started by chibitul, May 27, 2004.

1. ### chibitulGuest

Hi, what are you guys gonna do when you run out of photos??? "What the
hell are you talking about you crazy stupid?" yells one annoyed r.p.d.
reader. Well, let me explain my friend.

let's say 3000x2000 (6 MegaPixels) is plenty enough of information. Of
course, there are cameras with 8, 11 or even 14 mega pixels. but that is
besides the point. the final image is not that much different.

so we're stuck with this 6 megapixels camera. each pixel has 3 colors,
and each color 256 (8 bit) levels. So each pixels has 24 bits, or 2^24
=16.7 million colors.

Good. Now each pixel can have any color, so the total number of
different shots yu can take is (6 million)^(16.7 million). Now, most of
you don't realize, but this *IS* a finite number. Large, yes. Very
large, of course. Super mega large, sure. But it is *finite*. At some
point you will run out of pictures. No matter where you go, no matter
what lens you use or how you compose the photo, that photo will be
identical to another photo someone else took before you. whre is the fun
then???

sure, I have too much time on my hands, but did you ever think about
it??? now back to my wine

chibitul, May 27, 2004

2. ### Lisa HortonGuest

You are correct, you have too much time on your hands

Lisa

Lisa Horton, May 27, 2004

3. ### RalphGuest

Shouldn't that be 16.7 million factorial or something similar? I vaguely
remember 3! = 3x2x1 = 6 but think that there was a way to factor (sorry) in
the 6 million also.

Ralph, May 27, 2004
4. ### Ken WeitzelGuest

hehehe.... 16.7! is a kinda large number

Nary a one of us will last long enough to take that
many pictures.... doesn't matter though, 'cause the
sun will have gone out by the time we've barely started

Ken

Ken Weitzel, May 27, 2004
5. ### BryceGuest

It's not finite. It could only be finite if things never moved and things
didn't die.

Bryce, May 27, 2004
6. ### Al DenelsbeckGuest

Here's the really, really frightening part:

Fully half of those photos will be of someone's cat!

I'm gonna have problems sleeping tonight...

- Al.

Al Denelsbeck, May 27, 2004
7. ### gsumGuest

Sorry mate, your maths is incorrect. A sensor with 6mpixels in which
each pixel could only record black or white has 6million factorial (!)
combinations (100! is 100*99*98 .... *3*2*1*1).
Some years ago it was calculated that there were 110! unique games
of chess. 110! is thought to be greater than the number of photons in
the universe. 6 million! is a bit bigger!!!
And that doesn't take colour into account so keep snapping.

Graham

gsum, May 27, 2004
8. ### Dave MartindaleGuest

No, factorials come into it when you have a finite set of blocks or
balls (or whatever) to start with, so when a particular one is used up
in a particular position it isn't available for later choices. For
example, say you have 256 block labelled 0 to 255. The total number of
arrangements of these blocks in a row is (256 choices for the first) *
(255 choices for the second) * (254 choices for the third) *...
which is 256!

But for the image, each pixel can take any value. So you have
2^24 choices for each pixel, and 6 million of these all multiplied
together, for (2^24)^6e6 possible different 6 megapixel images.

Note that the original formula of (6 million)^(16.7 million) is backwards.
It has a value of about 10^(114 million), while the correct formula of
(2^24)^6e6 has a value of about 10^(43 million). So his number is a
factor of 10^(70 million) too large. That's a 1 followed by 70 million
zeros, which is a *really big* error.

Dave

Dave Martindale, May 27, 2004
9. ### Dave MartindaleGuest

No it doesn't. 100 factorial is the number of ways you can arrange 100
blocks, all identifiably different.

For a 6 megapixel 1-bit image, the total possibilities are 2^(6e6), or
about 10^(1.8 million). For an 8-bit monochrome image, it's
These are all ridiculously large numbers. No one can ever take even a
tiny fraction of the total number of images. A computer program
cranking them out as fast as possible (it's not hard, after all) would
still take an enormous time.

The depressing thing is that virtually all of these possible images are
extremely boring. I don't think I'll give up using a camera to create
carefully *selected* images.

Dave

Dave Martindale, May 27, 2004
10. ### Lisa HortonGuest

Hey Al, I have new cat pictures...

Lisa

Lisa Horton, May 27, 2004
11. ### gsumGuest

You had me going there. I was about to and crawl under my desk in
shame but I satnd by my assertion that the number of possible 1 bit
images is 6,000,000!

The number of combinations in a 4 pixel (2x2), 1 bit square is
4! i.e. 24 unique 'pictures'. .
The 'pixels' are all identifiably different as they are designated
0 or 1.
This can be proved by hand for 4 pixels but I wouldn't like
to do it for 16.

Graham

gsum, May 27, 2004
12. ### LionelGuest

Nope, because our planet will be destroyed when our sun dies, which will
be well before you generate your (assuming an EOS 10D DSLR photo format)
2^10871635968 images[0]. You will also be be unable to display or store
these images, as there will be more of them than there are atoms in the
universe.
So, I don't think we have anything to worry about /just/ yet.

[0] 2^(12x12x12x3072x2048) = 12 bits of red, 12 bits of green, 12 bits
of blue, by 3072 x 2048 pixels.

Lionel, May 27, 2004
13. ### LionelGuest

Kibo informs me that (Dave Martindale) stated that:
Indeed.
Assuming a 10D, the numbers are 2^10871635968 in RAW format, or
2^3221225472 if you prefer JPEGs.

Lionel, May 27, 2004
14. ### LionelGuest

Nope. I'm afraid that you're confusing Combinations with Permutations.
The correct number is 2^6000000. This is because 6 million 1 bit pixels
is exactly equivalent to a 6M bit binary variable, which can store any
one of 2^6000000 values, each of which would be a distinct image.

Lionel, May 27, 2004
15. ### gsumGuest

But I'm currently under my desk wondering why I can't do simple
Maffs any more. Good job I'm on holiday next week.
You are of course correct Dave.

Graham

gsum, May 27, 2004
16. ### mch42Guest

If you figure out how to desribe 24 different combinations using 4 bits,
you'll get rich over night. Most people only figure out 16 different ways

1: 0000
2: 0001
3: 0010
4: 0011
5: 0100
6: 0101
7: 0110
8: 0111
9: 1000
10: 1001
11: 1010
12: 1011
13: 1100
14: 1101
15: 1110
16: 1111

mch42, May 27, 2004
17. ### GorfGuest

It's not finite. It could only be finite if things never moved and things
I think you missed the point.

Given that there's a finite number of pixels on the CCD, and there's a
finite number of values each pixel can contain, there's a finite number of
combinations of those colours and pixels.

Admittedly it's a huge number - Excel blew up trying to calculate the number
of combinations at somewhere between 30 and 50 pixels (that's right -
pixels, not megapixels) - but it's still finite.

Gorf, May 27, 2004
18. ### Dan PidcockGuest

It's not factorial.

If you had a 4 pixel black and white image you get the following
possible images:
00 00 00 00
00 01 10 11

01 01 01 01
00 01 10 11

10 10 10 10
00 01 10 11

11 11 11 11
00 01 10 11

16 images, which is 2^4, where 4 is the number of pixels, and 2 is the
bit depth of each pixel.

If we had a 6MP mono sensor we would have 2^6000000 possible images.
If each pixel was an 8-bit grey sensor that could record 256 levels we
could have
256^6 million images.
Assume shooting in 24-bit jpeg mode that means we have
16.7 million ^ 6 million which is the opposite of the original poster,
and nothing to do with factorial.
This is less than originally suggested but still pretty huge.

Dan

Dan Pidcock, May 27, 2004
19. ### DJGuest

And most of the rest will be of someone's pussy.

DJ, May 27, 2004
20. ### DJGuest

<snip>

Well George, aren't you going to inform us that a Sigma can take 3 times that
many pictures?

DJ, May 27, 2004