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Geees, is this HN??? Am I the one who's too stupid to explain things properly, or is everyone under the weather because it's Monday morning? An image is a 2-di
by Rod 17y ago
Geees, is this HN??? Am I the one who's too stupid to explain things properly, or is everyone under the weather because it's Monday morning?
An image is a 2-dimensional array. Say it's a M x N array. Suppose we use RGB so that each pixel is represented by 3 bytes. Then you have an M x N x 3 array of bytes. There are infinitely many mappings from the space of all M x N x 3 arrays of bytes into the reals. Of course you can transform the array into a number. But you can do it in many, many ways. If you know what mapping was used, and if the mapping is invertible, then you can obtain the M x N x 3 array of bytes from the number. If you don't know what mapping was used, or if the mapping is non-invertible, then you can't obtain the M x N x 3 array from the number.
So, I claim that an image is NOT a number. I claim that an image is a pair (y,f), where y is a number, and f is an invertible mapping from the space of M x N x 3 arrays into the reals. You guys are looking at y, and forgetting f. That's why an image is not a number, the image is x = f^{-1} (y). In short, one must know f, and f^{-1} must exist.
- AndrewDucker 17y agoYou're merely arguing there that a number is not necessarily an image, because it cannot be converted into one without knowing how. And you'd be right. But the image is still a number, because it's stored as a series of bits, and any series of bits is interpretable as a number.
- Rod 17y agoI asked: "How can an image be a number?", when I should have asked instead: "How can a number be an image?" To use the verb to be in this scenario, I demand a one-to-one correspondence between numbers and images. If a number y can represent two distinct images x1 and x2, then we can't use "is", we can only use "can be".
- joeyo 17y agoSuch a strict definition prevents you for using "is" for anything at all! Is that your intention?
- scott_s 17y agoTo quote a former President, it depends on what your definition of "is" is. I think everyone here understands that we require information beyond the number itself to get the image. I agree that a more accurate (pedantic) verb is that a number can store or represent an image. That implies that the number alone is not sufficient to get the image. The thing is that while f can be in theory anything, it will be one of few things in practice. We can say what f is very simply. While knowing f is important, it's also easy to the point of trivial, so some people here are comfortable being less accurate and just saying "a number can be an image." This reminds me of philosophy of identity discussion me and a friend used to go through. He thought he could fit a Miata in his old-school Suburban - but maybe he'd have to take off the side mirrors. But if he's allowing the side mirrors to be taken off, then how much is too much before it's no longer a Miata?
- ErrantX 17y agoIndeed often info on which f you need is stored within y. which makes it even easier
- Rod 17y agoIndeed. It seems that for most people here is means can be, while I demand a bit more. Yes, in practice, f will be one of a few mappings, but that does not change the fact that knowledge of f is required. In practice, one merely tries all possible mappings f until one works. Such lack of precision is only tolerated because computation is fast and cheap. I would love to see people trying all f mappings with the help of an abacus! Let's be thankful that we live in this golden age.
- scott_s 17y agoWe can go further, actually. We could argue even that (y, f) is not, actually, an image and that's it's just a number and a function. Rather, the image is what's on the screen and only exists when it's displayed. I'm not trying to continue an argument, but rather just point out that the philosophy of identity can be an elusive thing.
- 17y ago
- lmkg 17y agoBut, f is fairly standardized (or rather, there's a short list of standard f's), and bitmaps aren't even hard to understand. Since the isomorphism is both natural and conventional, I would say that in practice it is not accurate to hold f to the same standards as y.