I'm no scholar nor historian in the age-old war between Microsoft and the Liberalists (or should I say - Librelists). But having studied and worked in decently technologically oriented college campuses, I do have an inkling about this rift. Call it adding fuel to the fire, or what you will, here I add a few observations of my own.
Recently, a friend of mine bought a laptop. Apparently, these days you do not get laptops with WinXP. More like, you cannot get laptops with WinXP. You can only get laptops (by which I mean branded ones, like HP Pavillion series or IBM Lenovo series, etc..) pre-installed with WinVista. Microsoft's first step, in pushing users towards Vista. This is not a business strategy. This is bonded labour. When you outsource Business Processing to oriental or indian trained labour markets, it is not bonded labour. That is business strategy, since on the one hand, IT companies get to reduce their costs, and Indian/Chinese/EastAsian employees are salaried in dollars, and everybody's happy. What Microsoft's pushing, though, is not a business strategy. Because the users take a hit. They have to pay extra, for something they don't need, indeed something they don't want. (Apparently, it'll cost more to even get the Vista removed from the laptop before buying it.)
Agreed, that Vista has a lot more capabilities and functional capacities than XP. But the customer bracket which includes my parents, like people who just started operating on computers just two to three years back, and would like to use computers solely for the purpose of chatting with their sons/daughters living abroad, clearly are not very adjustable to the change in operating systems. Especially when they can't really figure out the differences between a computer and an operating system, and verily, don't need to. And the customer bracket which includes those sons/daughters, who have studied in decently technologically oriented campuses, would clearly know the distinction between pests such as Microsoft Softwares and Free (both gratis and libre) Softwares. Sometimes I really am confused whether people buy Microsoft products because they like them, or because they're forced to. At other times, I'm confused whether Microsoft thinks its buyers really like its products, or perhaps it just sits smugly in the knowledge that people are forced to buy its products.
Coming to the motivation for this post, I noticed that once you install gtalk on Vista, all the browser links from gtalk open in IE, eventhough you've selected Firefox as your default browser. Investigating this issue a little more closely, I found that Vista has a different concept of a default browser. Infact, it has a different view of default programs. What used to be the "File Types" tab in the Tools->Folder Options dialog box, no longer exists in Vista. What used to be the "Set Program and Access defaults" has also been slightly changed. All this is now included in the Default Programs tool in the Control Panel, in Vista. With a twist. This twist being, even though Firefox is the default browser, it is not allowed to open files of type .mht, .mhtml, or .url. Why? Firstly, Firefox 2.0+ does not support mhtml. I really don't understand/know why. I'm illiterate in this area. So when i google it up, i come across an article on Wikipedia (where else?) saying that MHTML has something to do with internet mail extensions, and what i learnt in my fleetingly small attention span is that none of the current browsers are any good at supporting this thing. But the issue still remains. IE has the sole rights to .mht and .mhtml files, and you can't even remove it. You cannot even change the program associated with the .url files! I mean c'mon! they're just text files!! What is the sense in preventing, infact disabling the user from doing whatever he means to do? Isn't there the Vista's system restore and HP's backup files or restore schema to overcome any damage that the user might inadvertently do? I dunno. Call me uneducated and cavemanish, but all I see is Microsoft's monopoly. Coz when I tried to change the default program associated with each filetype, the Windows Explorer all of a sudden started acting up, and had to restart itself (thereby exiting Default Programs tool of the Control Panel). It might have been a bug, but does it really have to be activated only when one uses the Default Programs tool? That too to ensure IE is maintained the default program for three filetypes? As I've said before, sometimes I keep wondering...
The Wailing Wall is where Jews lament the destruction of the Temple, atone for their sins, and pray for forgiveness. This, in not so religious terms, is where i pour out all my anguish, concern, and curiosity about the world, and about me.
August 18, 2007
August 17, 2007
Of n-dimensions and Euclidean distances
This is more of an excercise in writing formulae on my blog than actually doing some math. But then again, who needs a reason to write equations?
Well anyway, there was this doubt that a friend asked me, which was put to the students by a prof in his class. We know that the Euclidean distance, between two points and in n-dimensions, is given by:
We also know that for any three points in 2-dimensional space, , , and , the Triangle Inequality holds, namely,
This is a well known result in Euclidean Geometry, and I'm loathe to go into the proof of this here. Now the question, was to find three points , and in 3-dimensional space, which violate the Triangle Inequality!
At first glance, one will easily see that even in 3-dimensional space, the three points will still form a triangle, and so the Triangle Inequality should hold. But how do we prove it? This is where geometric visualization aided me.
You see, in 3-d space, one has a lot more freedom to twist and turn shapes. Maybe it has to do with the degrees of freedom, which I never understood clearly. In 2-d, all that one can do with a shape, say a square, is to translate, rotate, or scale it. Put it in more layman terms, you can put the square at any place on the paper, turn the paper whichever way, and even shrink or expand the paper. Translation is possible in two directions, and rotation in only one ('coz , which basically says that rotating anticlockwise is the same as rotating clockwise through a larger angle). All this, while still retaining the properties of the square. In 3-d, on the other hand, all that one can do is still essentially the same, translate, rotate, or scale it. But here, you can translate or rotate in 3 different directions (unlike 2-d, rotation is possible in 3 directions, since rotations in different directions are not commutative).
I'm pretty sure all this previous paragraph would give a sore headache, to any layman. What I've put forward, in not-so-informal terms, is the justification for the following transformations:
Translate the origin to , thus making , and .
Rotate the axes, such that lies on the x-axis, thus making , and .
Rotate the axes, such that lies on the xy-plane, thus making .
All these transformations are perfectly legitimate, in the sense that the relative positions of the three points do not change. But by doing this, we notice that we've transformed our axes such that all three points now lie on the same plane (the xy-plane here). The truth is that the three points always lay on the same plane, ('coz any three points in 3-d lie on a plane) but we've shifted our axes such that this plane became our xy-plane. In any case, I claim that these three points are the same as three points in 2-d space, since I've reduced the component of all of them to zero. And we know that the Triangle Inequality will hold for any three points in 2-d space.
Thus we see, that the Triangle Inequality should indeed hold for any set of three points in 3-d space as well. And so the original question now takes a further leap: Will the Triangle Inequality hold for any set of three points in n-dimensional space? With the help of reasoning similar to the one above, I claim yes!
Key concept used: Any set of n points in n-dimensional space lie on the same (n-1)-dimensional hyperplane. Is this really true? you tell me :P
Well anyway, there was this doubt that a friend asked me, which was put to the students by a prof in his class. We know that the Euclidean distance, between two points and in n-dimensions, is given by:
We also know that for any three points in 2-dimensional space, , , and , the Triangle Inequality holds, namely,
This is a well known result in Euclidean Geometry, and I'm loathe to go into the proof of this here. Now the question, was to find three points , and in 3-dimensional space, which violate the Triangle Inequality!
At first glance, one will easily see that even in 3-dimensional space, the three points will still form a triangle, and so the Triangle Inequality should hold. But how do we prove it? This is where geometric visualization aided me.
You see, in 3-d space, one has a lot more freedom to twist and turn shapes. Maybe it has to do with the degrees of freedom, which I never understood clearly. In 2-d, all that one can do with a shape, say a square, is to translate, rotate, or scale it. Put it in more layman terms, you can put the square at any place on the paper, turn the paper whichever way, and even shrink or expand the paper. Translation is possible in two directions, and rotation in only one ('coz , which basically says that rotating anticlockwise is the same as rotating clockwise through a larger angle). All this, while still retaining the properties of the square. In 3-d, on the other hand, all that one can do is still essentially the same, translate, rotate, or scale it. But here, you can translate or rotate in 3 different directions (unlike 2-d, rotation is possible in 3 directions, since rotations in different directions are not commutative).
I'm pretty sure all this previous paragraph would give a sore headache, to any layman. What I've put forward, in not-so-informal terms, is the justification for the following transformations:
Translate the origin to , thus making , and .
Rotate the axes, such that lies on the x-axis, thus making , and .
Rotate the axes, such that lies on the xy-plane, thus making .
All these transformations are perfectly legitimate, in the sense that the relative positions of the three points do not change. But by doing this, we notice that we've transformed our axes such that all three points now lie on the same plane (the xy-plane here). The truth is that the three points always lay on the same plane, ('coz any three points in 3-d lie on a plane) but we've shifted our axes such that this plane became our xy-plane. In any case, I claim that these three points are the same as three points in 2-d space, since I've reduced the component of all of them to zero. And we know that the Triangle Inequality will hold for any three points in 2-d space.
Thus we see, that the Triangle Inequality should indeed hold for any set of three points in 3-d space as well. And so the original question now takes a further leap: Will the Triangle Inequality hold for any set of three points in n-dimensional space? With the help of reasoning similar to the one above, I claim yes!
Key concept used: Any set of n points in n-dimensional space lie on the same (n-1)-dimensional hyperplane. Is this really true? you tell me :P
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