how many possible combonations?

[SonicIce]

how many possible screens are there on a monitor at 640x480 with 256 colors? or how about 1024x768 with 24bit color?

 

[bobsmith1492]

Oogles. 🙂

 

[bobsmith1492]

Well, for 640x480, there are 307200 pixels, with 256 possibilities each, so there are 78643200 options there; you would need 78643200!, which I don't know of anything that could calculate that... I'm still sticking with oogles as the best answer. I don't even want to think about the other one (mainly because I don't know how many colors "24-bit" can do - 2^24 probably(?) which would make for 13194139533312! possibilities....)

 

[SonicIce]

is it really that simple? 13 trillion possible images :Q. can someone confirm this?

 

Originally posted by: SonicIce
is it really that simple? 13 trillion possible images :Q. can someone confirm this?

its not 13 trillion...its 13194139533312 x 13194139533311 x 13194139533310...x 1

 

[nsadhal]

Goodness, I hope I'm not missing something utterly ridiculous, but I don't think I understand bobsmith's solution. While this is something I should totally know... I'm coming up with a different answer (?)

The OP asked "how many combinations" but then asked "how many possible screens are there?" so to get every possible screen you have to consider every pixel being every possible value.

With 307200 pixels in 640x480, we have 256^307200 possible screens.
Isn't this exactly the same way we figure out that 8-bit color has 2^8, or 256 possible colors?

With 1024x768, we have 786432 pixels with 2^24 possible colors for each pixel. The result I get is (2^24)^786432.

I don't understand how bobsmith says that with N values and M possibilities for each value there are (M*N)! possible outcomes.
That would mean that an 8 bit field could represent (2*8)! possible values. That comes out to a lot more than 256 by my trusty TI-89.

Am I misintepreting the question?

 

[Calin]

Originally posted by: bobsmith1492
Well, for 640x480, there are 307200 pixels, with 256 possibilities each, so there are 78643200 options there; you would need 78643200!, which I don't know of anything that could calculate that... I'm still sticking with oogles as the best answer. I don't even want to think about the other one (mainly because I don't know how many colors "24-bit" can do - 2^24 probably(?) which would make for 13194139533312! possibilities....)

For the first case, there are 640*480 independent pixels, and each can have 256 colors (states). As a result, the number of possibilities is 640*480*256.

For the second case, there are 1024*768*16.7 millions

Things get much more interesting when you use a screen like the Z80 based computers had - they had a resolution of some 176*224 pixels (aprox), but there are 8x8 cells (characters) that can have only two colors.
I mean you can address every pixel individually, but on each of the 8x8 characters on the screen there are two colors top.

 

[Mark R]

1) 2.08 x 10 ^ 739811
2) 8.26 x 10 ^ 5681750

Oogles is probably about as descriptive as is necessary though.

Those are some quite large numbers.

 

[nsadhal]

Originally posted by: Calin
For the first case, there are 640*480 independent pixels, and each can have 256 colors (states). As a result, the number of possibilities is 640*480*256.

For the second case, there are 1024*768*16.7 millions

I don't understand how you are just multiplying the number of state holding units (pixels) by the number of states each of those acn hold.
Note that if each pixel has an 8 bit color value, then we have 8*640*480 bits. A bit can be either on or off, so we have 2^(640*480*8) possible screens. This is the same as what I posted earlier... 256^(640*480). This is far greater than 640*480*256.

Actually... I just realized where you may be getting this multiplication from. Rows*cols*colors represents the number of points in the 3d space where 2 dimensions represent the space and 1 dimension represents the color. This is the size of the space we are working in, but what you're only covering is the number of screens that have 1 pixel different from some base screen. That is, if my base is an all black screen, you've calculated how many screens have 1 colored pixel on them in a sea of black, including all the screens that have 1 black pixel in a sea of black.

 

[sdifox]

Because the question specified 256 colours. Each of the pixels is capable of representing 1 of the 256 colours, from black to white. Each of the pixels is an independent event.

1 pixel with 256 colour yields 256 possibilities, 2 pixel yields 256x2 possibilities, 3 pixels yields 256x3. From that, we figure out 640x480x256.

Originally posted by: nsadhal

Originally posted by: Calin
For the first case, there are 640*480 independent pixels, and each can have 256 colors (states). As a result, the number of possibilities is 640*480*256.

For the second case, there are 1024*768*16.7 millions

I don't understand how you are just multiplying the number of state holding units (pixels) by the number of states each of those acn hold.
Note that if each pixel has an 8 bit color value, then we have 8*640*480 bits. A bit can be either on or off, so we have 2^(640*480*8) possible screens. This is the same as what I posted earlier... 256^(640*480). This is far greater than 640*480*256.

Actually... I just realized where you may be getting this multiplication from. Rows*cols*colors represents the number of points in the 3d space where 2 dimensions represent the space and 1 dimension represents the color. This is the size of the space we are working in, but what you're only covering is the number of screens that have 1 pixel different from some base screen. That is, if my base is an all black screen, you've calculated how many screens have 1 colored pixel on them in a sea of black, including all the screens that have 1 black pixel in a sea of black.

 

[Born2bwire]

Originally posted by: sdifox
Because the question specified 256 colours. Each of the pixels is capable of representing 1 of the 256 colours, from black to white. Each of the pixels is an independent event.

1 pixel with 256 colour yields 256 possibilities, 2 pixel yields 256x2 possibilities, 3 pixels yields 256x3. From that, we figure out 640x480x256.

If we had one pixel of two possiblities, and the second of 3 possibilities, then for each of possibility of the first pixel, we have three for the second. So the actual possibilities for the two together is 2*3. So for a 2 pixel 256 color screen, we have 256^2 possibilities.

Nsadhal's second solution is a little more interesting to think about. The value of each pixel is held as a number in memory. Multiple pixels are just contiguous numbers next to each other (since we have a unique color value for every single possible number for that pixel). So for each screen, we can think of the screen being uniquely represented by a single binary number of X bits. So the number of possible screens would be 2^X.

Either way of thinking about it, you arrive at the same result, that is for 640x480 256 colors, we have 256^(640*480) possible screens.

EDIT: A better way of thinking about Nsadhal's second solution is with the combination lock on a briefcase. You have three numbers, 0-9, how many combinations? The answer is easy, 1000, because you can represent all numbers between 000-999. Same thing here, you have a single screen that is represented by a 640*480 numbers with values from 0-255. So it's the same problem and solution, just in base 256 (or we can easily change it to base two) instead of base 10.

 

[sdifox]

I guess I am not thinking straight... it is indeed 256^(640*480)...good thing I am not teaching math.

Originally posted by: Born2bwire

Originally posted by: sdifox
Because the question specified 256 colours. Each of the pixels is capable of representing 1 of the 256 colours, from black to white. Each of the pixels is an independent event.

1 pixel with 256 colour yields 256 possibilities, 2 pixel yields 256x2 possibilities, 3 pixels yields 256x3. From that, we figure out 640x480x256.

If we had one pixel of two possiblities, and the second of 3 possibilities, then for each of possibility of the first pixel, we have three for the second. So the actual possibilities for the two together is 2*3. So for a 2 pixel 256 color screen, we have 256^2 possibilities.

Nsadhal's second solution is a little more interesting to think about. The value of each pixel is held as a number in memory. Multiple pixels are just contiguous numbers next to each other (since we have a unique color value for every single possible number for that pixel). So for each screen, we can think of the screen being uniquely represented by a single binary number of X bits. So the number of possible screens would be 2^X.

Either way of thinking about it, you arrive at the same result, that is for 640x480 256 colors, we have 256^(640*480) possible screens.

EDIT: A better way of thinking about Nsadhal's second solution is with the combination lock on a briefcase. You have three numbers, 0-9, how many combinations? The answer is easy, 1000, because you can represent all numbers between 000-999. Same thing here, you have a single screen that is represented by a 640*480 numbers with values from 0-255. So it's the same problem and solution, just in base 256 (or we can easily change it to base two) instead of base 10.

 

[linuxconvert]

To me it sounds like it is being inferred that every pixel is being used because if only one pixel was on a screen it has 640*480 possible places for that one pixel. if there were two pixels then...

the formula for combinations is:
n!
nCr = -------
(n-r)! r!

with n being the amount of pixels to choose from (in this case 640*480*256, for each color choice) and r being the amount of pixels you wish to find combinations for.

so you would need to go through the formula n amount of times.

 

[Diademed]

The above equation:

n!
-------
(n-r)! r!

 

[Rhombuss]

256^(640x480) and 16E6^(1024x768)

Although...for LCDs, don't discount the possibility of having a dead pixel!

In which case it would be...

257^(640x480) and (16E6+1)^(1024x768)

😛

 

[kevinthenerd]

Originally posted by: bobsmith1492
Well, for 640x480, there are 307200 pixels, with 256 possibilities each, so there are 78643200 options there; you would need 78643200!, which I don't know of anything that could calculate that... I'm still sticking with oogles as the best answer. I don't even want to think about the other one (mainly because I don't know how many colors "24-bit" can do - 2^24 probably(?) which would make for 13194139533312! possibilities....)

In that calculation, you're assuming two pixels can't be the same value, as if you picked them out of a deck of cards.

The answer for 640x480x256c is

256 ^ (640*480)

and for 1024x768x16777216 is

16777216 ^ (1024 * 768)

 

[kevinthenerd]

Originally posted by: nsadhal

Originally posted by: Calin
For the first case, there are 640*480 independent pixels, and each can have 256 colors (states). As a result, the number of possibilities is 640*480*256.

For the second case, there are 1024*768*16.7 millions

I don't understand how you are just multiplying the number of state holding units (pixels) by the number of states each of those acn hold.
Note that if each pixel has an 8 bit color value, then we have 8*640*480 bits. A bit can be either on or off, so we have 2^(640*480*8) possible screens. This is the same as what I posted earlier... 256^(640*480). This is far greater than 640*480*256.

Actually... I just realized where you may be getting this multiplication from. Rows*cols*colors represents the number of points in the 3d space where 2 dimensions represent the space and 1 dimension represents the color. This is the size of the space we are working in, but what you're only covering is the number of screens that have 1 pixel different from some base screen. That is, if my base is an all black screen, you've calculated how many screens have 1 colored pixel on them in a sea of black, including all the screens that have 1 black pixel in a sea of black.

Originally posted by: sdifox
Because the question specified 256 colours. Each of the pixels is capable of representing 1 of the 256 colours, from black to white. Each of the pixels is an independent event.

1 pixel with 256 colour yields 256 possibilities, 2 pixel yields 256x2 possibilities, 3 pixels yields 256x3. From that, we figure out 640x480x256.

No. 2 pixels will hold 256 * 256 possibilities.

Flip a coin. (This is my favorite example recently, so bear with me.) If you flip it once, you have to possibilites. If you flip it twice, you have four possibilties:

TT
TH
HT
HH

and three times...

TTT
TTH
THT
THH
HTT
HTH
HHT
HHH

which gives 8 possibiities.

3 - 8
4 - 1
...
n - 2^n


Similarly, n pixels over c colors gives c^n possibilities.

 

[DrPizza]

Originally posted by: Mark R
1) 2.08 x 10 ^ 739811
2) 8.26 x 10 ^ 5681750

Oogles is probably about as descriptive as is necessary though.

Those are some quite large numbers.

oh crud. I just spent a minute with my calculator figuring that out; I missed your post.


For what it's worth, I don't think any human can even come close to comprehending how large those numbers are.

FWIW, if we said the Universe was a sphere with a 15 billion light year radius,
the volume of the universe would be about
2.07 * 10^74 cubic meters.

As the size of a proton is roughly 10^-45 meters, the entire universe, (ignoring compression into singularities) could only hold about 10^119 protons!
Compare this to 10^739811...
The number of protons that would fit in the known universe is absolutely insignificant compared to the number of possible images.

 

[kevinthenerd]

Originally posted by: DrPizza

Originally posted by: Mark R
1) 2.08 x 10 ^ 739811
2) 8.26 x 10 ^ 5681750

Oogles is probably about as descriptive as is necessary though.

Those are some quite large numbers.

oh crud. I just spent a minute with my calculator figuring that out; I missed your post.


For what it's worth, I don't think any human can even come close to comprehending how large those numbers are.

FWIW, if we said the Universe was a sphere with a 15 billion light year radius,
the volume of the universe would be about
2.07 * 10^74 cubic meters.

As the size of a proton is roughly 10^-45 meters, the entire universe, (ignoring compression into singularities) could only hold about 10^119 protons!
Compare this to 10^739811...
The number of protons that would fit in the known universe is absolutely insignificant compared to the number of possible images.

There are approximately 10^88 particles in the known universe. I won't pretend to be able to truly comprehend anything over 10^6, and even that's quite a stretch. (A million inches is about 15.8 miles. A million seconds is about 11.6 days, which would take about four weeks to count. A million ounces equals about 31 tons, or the weight of about 24 economy cars at 2600lb each.)

 

[ones3k]

Originally posted by: SonicIce
how many possible screens are there on a monitor at 640x480 with 256 colors? or how about 1024x768 with 24bit color?

256^(307200)

A more interesting question might be: what percentage of these images are actually recognizable and wouldn't be considered "noise"

 

[DanTMWTMP]

I LOL'ed @ the title, sorry 😛. "combonations" roflmao!! 😛


(i really apologize for giving an offtopic answer to a highly technical discussion, but... )

 

[BrownTown]

its kinda funny how many people got this wrong becasue they replied too fast without thinking. It really is quite simple, you have N pixels (screen width x screen height), and you have X different color possibilites for each pixel. The number of combinations is just X^N.

 

[mobobuff]

If you wanna feel REALLY tripped out*, you could...

1. Set your digital camera to 640x480.
2. Get naked, smear grape jelly all over yourself, put a bright orange road cone on your head, wrap yourself in christmas lights, and stand in front of a Twister mat, and have the camera take your picture with a 10 second delay.
3. Downsample the color palette to 256 colors.

4. Realize that the resulting image of you would HAVE to be one of the possible random images.





* and very sticky

 

[kevinthenerd]

Originally posted by: mobobuff
If you wanna feel REALLY tripped out*, you could...

1. Set your digital camera to 640x480.
2. Get naked, smear grape jelly all over yourself, put a bright orange road cone on your head, wrap yourself in christmas lights, and stand in front of a Twister mat, and have the camera take your picture with a 10 second delay.
3. Downsample the color palette to 256 colors.

4. Realize that the resulting image of you would HAVE to be one of the possible random images.





* and very sticky

Whoa.... what if you wrote a program that created random images, and it accidently produced one such picture?

Ok, OFF TOPIC. For the first time ever, in over three years of my time in Anandtech, I vote a lock. This is, after all HT.