Simple combinatorics gone wrong...


by x2thay
Tags: combinatorics, simple
x2thay
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#1
Feb6-13, 09:04 AM
P: 14
1. So consider an 8 megapixel picture (res: 3264x2448).
Now, it seems rather simple but I just can't figure out how to calculate the entire number of possible shots/photographs one can take within that resolution, assuming each pixel can have 16777216 different values/colors.




2. Relevant equations



3. The attempt at a solution
So I realize the number has to be absolutely monstruous, so here's what I've tried so far


(3264*2448)!*2^24

Meaning, the entire possible positions all the pixels can assume times all the different values each individual pixel can have.
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jedishrfu
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#2
Feb6-13, 09:10 AM
P: 2,473
It seems your solution is correct for a JPG image but for a GIF you limited to 256 colors per image. Each color may range from 0 to 16million as a GIF uses an 8 bit value that indicates which color register from the color palette to use.

One problem I see is that you'll get shots all in one color or half one color half another...

Do you want to define what a photo is? like it must have a minimum of X colors per photo?

Next what about color changes that are imperceptibly small? Do you want to instead say that while RED can range from 0 to 255 in value we'll limit it down to a subset of {0, 16, 32, 48, 64 ... 255} in other words 64 color choices.
Dick
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#3
Feb6-13, 09:18 AM
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Quote Quote by x2thay View Post
1. So consider an 8 megapixel picture (res: 3264x2448).
Now, it seems rather simple but I just can't figure out how to calculate the entire number of possible shots/photographs one can take within that resolution, assuming each pixel can have 16777216 different values/colors.




2. Relevant equations



3. The attempt at a solution
So I realize the number has to be absolutely monstruous, so here's what I've tried so far


(3264*2448)!*2^24

Meaning, the entire possible positions all the pixels can assume times all the different values each individual pixel can have.
If you have 2^24 possible values of each pixel, then for two pixels you would have 2^24*2^24 possibilities, right? What about 3264*2448 pixels?

x2thay
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#4
Feb6-13, 09:36 AM
P: 14

Simple combinatorics gone wrong...


Quote Quote by jedishrfu View Post
It seems your solution is correct for a JPG image but for a GIF you limited to 256 colors per image. Each color may range from 0 to 16million as a GIF uses an 8 bit value that indicates which color register from the color palette to use.

One problem I see is that you'll get shots all in one color or half one color half another...

Do you want to define what a photo is? like it must have a minimum of X colors per photo?

Next what about color changes that are imperceptibly small? Do you want to instead say that while RED can range from 0 to 255 in value we'll limit it down to a subset of {0, 16, 32, 48, 64 ... 255} in other words 64 color choices.
No matter how insignificant the difference is, I meant to calculate e-v-e-r-y single possible matrix 3264x2448 arrangement, given that each entry can assume 2^24 different values. so yes, there will be an enormous amount of shots in which the only difference from the next, will be a single pixel.
x2thay
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#5
Feb6-13, 09:38 AM
P: 14
Quote Quote by Dick View Post
If you have 2^24 possible values of each pixel, then for two pixels you would have 2^24*2^24 possibilities, right? What about 3264*2448 pixels?
So... (2^24)^(3264*2448) ? Basically 17M^8M? Are you sure that's correct? It seems too simple.
Dick
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#6
Feb6-13, 09:42 AM
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Quote Quote by x2thay View Post
So... (2^24)^(3264*2448) ? Basically 17M^8M? Are you sure that's correct? It seems too simple.
Yes, I'm sure. It's actually too simple for me to be wrong. It's (number of possibilities for each choice)^(number of choices).
rollingstein
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#7
Feb6-13, 09:42 AM
P: 305
Quote Quote by x2thay View Post
So... (2^24)^(3264*2448) ? Basically 17M^8M? Are you sure that's correct? It seems too simple.
What format? You didn't answer.

Since jpeg is a lossy format a lot of those combinations ought to evaluate to the same output. I could be wrong.
x2thay
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#8
Feb6-13, 09:56 AM
P: 14
Quote Quote by rollingstein View Post
What format? You didn't answer.

Since jpeg is a lossy format a lot of those combinations ought to evaluate to the same output. I could be wrong.
I should have mentioned earlier, but I meant a bitmap format.
x2thay
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#9
Feb6-13, 10:03 AM
P: 14
Okay, got it. The solution is a number whose log is 115 805 766.
Thanks, guys.


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