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This is using the same basic formula as exponential growth of bacteria in a petri dish.
Make robot A build 1 computer C1 and 2 robots C and D with A's own capacity. Robot's C and D make 1 computer each and 2 robots E, F and G, H each with their own capacity.
Say each computer can computer can compute Co1 additions in a second. Once robot A builds computer C1 I want C1 to start computing at Co1 additions per second. I want an equation that shows me how many additions all computers have done in, say 10 days.
Oh, the reason I ask this question.
Note: character ^ = to the power, 2 ^ 3 = 8
If you have a tic-tac-toe board. 9 boxes. Say each box can hold a 0 or a 1. The number of combinations of values on the board is 2 ^ 9 = 512.
This doesn't have enough 'resolution' to show an image of anything, except the letter H, T, I, U or L where the parts of each letter are a 1 and the remainder of the
board contains 0 at some point. It won't show any significant line art either.
1 | 0 | 1
---------
1 | 1 | 1 = H
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1 | 0 | 1
So, just increasing the resolution and color combinations:
My computer screen resolution is set to 1440 x 990. The total number of individual colors (per pixel) is 1440 x 990 = 1,425,600 pixels on my screen.
Each pixel on the screen has a possible color value between 0 and 16,777,215 at 24 bits (or 8 bits per RGB color) (2 ^ 24 = 16,777,216). Most digital cameras are 24 bits so this is a good amount of unique colors.
The number of combinations the computer must display on my screen to see ~everything~ known and to be discovered is:
result = 16,777,216 ^ 1,426,600
Here is what I came up with so far but it is not efficient.
The parallel computing grid would expand exponentially base 2. Say it takes someone one day to build the first robot. And say it takes 1 day for a robot to complete a single cycle of producing 1 computer and 2 robots. And say there were only 8 results and not (16,777,216 ^ 1,426,600), the equation 2 ^ n = 8 where n = a single start to finish cycle, a day, provides your answer. If there were only 8 results, it would take n = log(8) / log(2) = 3 days to produce at least 8 computers. If the computers only started working on the combinations to see ~ everything ~ once all computers are done, it will take 1 day for the computation to see everything once all computers where finished. Right?
So using the original numbers, 2 ^ n = result = (16,777,216 ^ 1,426,600) where n is days. n = log(16,777,216 ^ 1,426,600) / log(2). log(16,777,216 ^ 1,426,600) freezes my calculator so say A = (16,777,216 ^ 1,426,600). log(A) = 1,426,600 * log(16,777,216) =10,306,785.4035. Then n = 10,306,785.4035 / log(2) = 34238399.99986 days. 34238399.99986/365 = 93,803 years.
But of course if 93,803 initial robots were created and each created their own parallel computing grid, it would only take 1 year. Or so.
Only true if each robot stops working after producing 1 computer and 2 robots. Also, only true if the grid is run over 1 day AFTER all computers are completed.
How about a digital camera where the pixels are in a honeycomb arrangement and the
parallel printer is also hexagonal pixelly. Do you like it?
igloo myrtilles fourmis
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The number of combinations the computer must display on my screen to see ~everything~ known and to be discovered is:
Why would you assume that everything we discover must be able to be displayed on a computer screen?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Ricky: Well, ok. I'm limited by the number of computer screens, at the resolution of the computer screen. I guess I got carried away and assumed an image of anything to be discovered could be displayed on a computer screen.
The combinatorial can work at higher resolution that that of a computer monitor. If you are into photography and know what Technical Pan is, that film can resolve 200 line pairs per millimiter. If you run the combinatorial attack at that resolution at a color pixel depth of 256 (which doesn't exist yet in any camera or drums scanner) at 4 times the resolution on say a 20x30 screen. Well, more discoveries would be found?
John: LOL. You must be into the Fuji imaging sensors. Each honeycomb is still limited, with todays engineering, to 36 bit color depth. Or am I mistaken?
If you have an inexpensive Acer computer monitor like I do. And it is set to the resultion I described at 24 bit color. Well, just for fun, take your hand of the keyboard and mouse. Sit back and look at the screen. Including the icons, pictures and this text. That screen just 1 screen which will show up in result = (16,777,216 ^ 1,426,600)
And of course, if there were such a thing as a 3 dimensional computer screen. The brute force attack could be carried out on that too. Now having looked at your responses to the questions you may well be one of the few people that can imagine a computer 'screen' that is perhaps 35 dimensions.
Oh, 20 inches by 30 inches that is. Not that it matters.
The point is if what you're doing is to come up with a limit of human knowledge, without any philosophical arguments you can do no better than countably infinite.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Ricky: I just got carried away.
Ricky: The reason I now agree with you. I've heard the number of digits in pi goes on forever. No matter what resolution I set the screen to, there will be numbers in pi that will not be written across all the 'screens' even if they are written in a very small font. So, infinite knowlege.
A better example would be a function describing a random number. I've heard there is no such thing as a random number. A random number function would fill all the 'screens'. A person though could come along and add a formula that would not be written by the computers.
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