Since any digital canvas corresponds to one particular number - which has the same number of digits as the number of pixels, and which is using the numerical system corresponding to the color-depth -, the amount of possible spectacles displayable on it can be easily counted. Although the number of variations is exponentially grow by increasing the resolution and/or the color-depth, using our method it is always possible to calculate the order of the image-sequence. We can also calculate the place of any existing image in the chain, so we can calculate how long does it take for a particular machine to "reach" that image running our simple counting program.

Let's recall our previous example with the 3x3 1bit digital canvas. We have two diagonal lines pictured on it: a slash and a backslash.

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0 0 1 0 1 0 1 0 0

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binary: 001010100 = decimal: 084

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1 0 0 0 1 0 0 0 1

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binary: 100010001 = decimal: 273

Our program - counting from 0 to the maximum number displayable on 9 binary digits (511), and displaying the numbers on a 1bit color depth 3x3 digital canvas - will produce much sooner the slash (which is the 84th in the sequence) than the backslash (which is the 273th). If we say that it takes one second for our program running on a particular machine to produce each image, than the distance between the two images is 273-84=189 sec.

Using this approach we can calculate the distance in time for a dull computer to get from one existing image to another one. Instead of playing with "famous images" - for example the Mona Lisa and Marilyn Monroe or cover photos of yesterday's and today's magazines - we made an attempt to calculate the distance between two frames of an animation.

Unfortunately the changes in the order of magnitude caused by increasing the image size and color depth are boundless for our terms. In 1997 we made some experiments on the image-sequence above using a very fast computer capable to calculate and render each 100x100 resolution 8 bit image in 0.003 second. (On this canvas the number of possible variations is 10000 on the power of 256.)

Searching for the runtime between the individual frames of the animation the calculated result was a more than 23000 (decimal) digits long number, using the measurment unit of seconds. According to this result it is obvious, that even by converting this time to ten milliard years, the number of digits will only decrease by some dozens. It is also obvious that by using a 10/100/1000 times faster machine the benefits in time would also be negligible.