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Can anyone recommend some good starting points for understanding Transformation Matrices for dummies like me with poor math skills.
I'm willing to learn the math, and I'm not a complete idiot (I hope) but the examples I'm finding seem to require a huge leap from what I know, to what I need to know.
I wrote a web program that can be used to play around with transformation matrices. It allows preset types and custom ones.
Play with it online or grab the source.
It should be easy to play with the numbers and instantly see how it affects the house drawing. Look at the code available online to determine what it's doing, and you should be able to understand what's going on.
If you're having trouble, realise that the 3×3 matrix is simply being multiplied by each vertex (X & Y coordinate) in the house shape. Matrix multiplication with the vertex (we will now refer to it as a vector) and a transformation matrix looks like so...
1 0 0 1
0 1 0 * 2
0 0 1 0
On the left is an identity matrix (an idempotent matrix that doesn't affect the vector) and a vector of 1, 2, 0 (assume this maps to position X1 and Y2 in the program mentioned above's graph and ignore the final 0).
Matrix multiplication can be visualised like so...
a b c x a * x + b * y + c * z
d e f + y = d * x + e * y + f * z
g h i z g * x + h * y + i * z
So, in our example, that would be...
1 0 0 1 1 * 1 + 0 * 2 + 0 * 0
0 1 0 * 2 = 0 * 1 + 1 * 2 + 0 * 0
0 0 1 0 0 * 1 + 0 * 2 + 1 * 0
Do that math and we get the final vector...
1
2
0
Since we said our identity matrix shouldn't modify the values, we can see above that that is the case as the resulting vector matched the original.
To explain further, consider when you need to translate the vector. Let's say we want to translate the house by 5 pixels along the X axis. We want to start with the identity matrix, but change the top right number to 5 and make the extra dimension in the vector 1 (you will see why briefly).
1 0 5 1 1 * 1 + 0 * 2 + 5 * 1
0 1 0 * 2 = 0 * 1 + 1 * 2 + 0 * 1
0 0 1 1 0 * 1 + 0 * 2 + 1 * 1
We do the math again...
6
2
1
We can see that the first number (X in our coordinates) has been translated along the X axis by 5. Try it in the program linked above.
The reason we made the third value 1 is so when the math was performed, the translation was considered. Had it been 0, it will be ignored, as any number multiplied by 0 results in 0.
If you're still having trouble, check out videos online (this one, for example) which can help explain it in a more visual fashion.
Remember: pretty much anyone can drive a car, and pretty much anyone can learn this, despite any self-evaluated poor understanding of math. Just keep at it: persistence is key. Good luck.
Like duffymo has pointed out, Matrix Transformations is nothing more but (pre)multiplying a vector (like a 3d point) by a matrix. However, that is pure mathematics, and hard for some people to visualise.
The best way to understand transformation matrices (at least for me) is to get an example code, get it running, and play around with the numbers. Try to see if you can place an object farther away, or rotated by 45 degrees. Try putting the transformations in different order and see what the results are.
All working? Good.
Once you get a feel of that, and if you're brave enough to tackle the maths, you could take these steps:
First, understand how matrix multiplications work. Some links:
http://en.wikipedia.org/wiki/Matrix_multiplication
http://www.gamedev.net/reference/list.asp?categoryid=28#259
Also borrow highschool math textbooks from any of your friends.
Google is your friend.
Once you are comfortable with multiplying a matrix by hand, you will get a feel of why transformations are written that way. As you use them, the understanding of matrices will eventually come to you.
Secondly, I always recommend spending an afternoon trying to implement your own Matrix class and define a few common operations like mul(Vector v), transpose() or even createTranslationMatrix(float x, float y, float z). Put in a few tests and see if the results are the same with what you did by hand.
If you've come that far, try implementing your own Perspective Transformation! It's the most amazing thing we never come to appreciate. A useful explanation here:
Deriving Projection Matrices
You will be very proud of yourself once you have accomplished one of the most labourous, yet under-appreciated tasks of implementing a matrix object. Good luck!
A transformation is nothing more than a matrix multiplying a vector to produce the transformed vector, so if you don't understand matrix multiplication and addition you can't get very far.
Start with matricies and linear algebra. There are lots of books out there, but realize that based on the statement that I made above you don't need to read that whole book. You won't need eigenvalues or Gaussian elimination or vector spaces or any of the other stuff that will be advanced and difficult.
You just need to know how to extend what you know about multiplying and adding numbers to matricies.
Getting the entries in that transformation matrix is another matter altogether. You'll need a good book on mathematics and computer graphics. You won't find that in a linear algebra textbook.
Related
Disclaimer: This question is not the same question as other projection matrix questions.
So Projection Matrices are 4x4 Matrices that are multiplied with 4D vectors to flatten them onto a 2D plane. Like this one:
1 0 0 0
0 1 0 0
0 0 0 0
0 0 1 0
But in the explanation, it says that the x and y coordinates of the vector are divided by Z. But I don't understand how this works because each part of the matrix that is multiplied by Z is 0. A comment in another question on this subject said, "The hardware does this for you." And I didn't quite get what it meant by that. Thank you in advance!
I was confounded by this nomenclature issue, too. Here is a bit better explanation in regards to Vulkan: https://matthewwellings.com/blog/the-new-vulkan-coordinate-system/
After the programmable vertex stage a set of fixed function vertex operations are run. During this process your homogeneous coordinates in clip space are divided by wc
Clearly, calling those matrices projection matrices is very misleading if the actual perspective correction isn't actually done by them. :)
I'm trying to implement movement through four points, while recording which points I visit. Think of it as a square. I can move from corner to corner or diagonally.
If you 'unwrap' the square you get a straight line with four points, which can be thought of as 1-2-3-4- where after 4 it goes back to 1. So if I'm at point 2 I can move to 1 and 3 directly or 4 diagonally. I'd implement that as 2-1 / 2+1 for corner-to-corner or 2+/-2 for diagonally. The problem occurs when I'm at 2 and will try to subtract 2 where I'll end up outside of the list.
The thought I've had is that if I could somehow translate my "out of bounds" numbers to in bounds this would be solved. One solution is hard coding that:
0=4
-1=3
5=1
6=2
but I'm pretty sure there is a better way to do this, however I can't seem to find it.
It seems to me all you want is modular arithmetic (bless the lord for math)
magicFun <- function (x) x %% 4
Here is a simple test run
> magicFun(0:6)
[1] 0 1 2 3 0 1 2
Addendum
It's more about math but the reason it works for negatives is that in Z/nZ ("the world where n is equal to 0") n is "identified" to 0.
This means you can add n as many times as you wish to a given number without changing it's "value".
Also, by convention the numbers in Z/nZ are listed as {0, 1, ..., n-1}.
So suppose n = 4 and x = -6, by the above x = x + 2*4 = 2.
i came across the following fuzzy logic example about fuzzy logic.
Representing Age
Problem 2-1. Fuzzy sets can be used to represent fuzzy concepts. Let U be a reasonable age interval of human beings.
U = {0, 1, 2, 3, ... , 100}
Solution 2-1. This interval can be interpreted with fuzzy sets by setting the universal space for age to range from 0 to 100.
Problem 2-2. Assume that the concept of "young" is represented by a fuzzy set Young, whose membership function is given by the following fuzzy set.
Young= FuzzyTrapeZoid [0 ,0 ,25 ,40]
All i want to understand is how i can get the Complement[Young]
The quick answer is that Complement[Young] = FuzzyTrapeZoid[25,40,100,100]. Here is an image to show (in red) Young, and the complement in green.
Were you looking for an algorithm to solve this?
edit: adding more:
A generic fuzzy trapezoid is: FuzzyTrapeZoid[A,B,C,D]
The membership value is 0 up to A, then ramps from 0 to 1 between A and B, stays at 1 from B to C, then ramps from 1 to 0 between C and D. see page 3 of this intro (warning! pdf)
Since the complement of a fuzzy set = 1 - the membership function, then you can pretty much see the values by inspection. For the original problem (which comes from Mathematica), the complement is a single function. For the generic one FuzzyTrapeZoid[A,B,C,D] you will need 2 trapeziods to make the complement: FuzzyTrapeZoid[0,0,A,B] + FuzzyTrapeZoid[C,D,100,100]
For the Young membership function, it is 1 up to 25, so the complement will be 0 up to 25 (this yields [25,x,x,x] where x is yet to be determined). Since the Young membership function ramps to 0 between 25 and 40, it is clear that the complement will ramp from 0 to 1 in the same range (this yields the [25,40,x,x] where x is yet to be determined). Finally, since the Young membership function is 0 from 40 to 100, the complement will be 1 in the same range, this gives [x,40,100,100] (we knew from before that x = 25).
If you were looking for some more formal proof, I'm sorry, I do proofs poorly as I come from the Captain Kirk school of math: I can see it, and I can jump to the right answer, but I can't tell you exactly how I did it.
I have come across this problem in a calculation I do in my code, where the divisor is 0 if the divident is 0 too. In my code I return 0 for that case. I am wondering, while division by zero is generally undefined, why not make an exception for this case? My understanding why division by zero is undefined is basically that it cannot be reversed. However, I do not see this problem in the case 0/0.
EDIT OK, so this question spawned a lot of discussion. I made the mistake of over-eagerly accepting an answer based on the fact that it received a lot of votes. I now accepted AakashM's answer, because it provides an idea on how to analyze the problem.
Let's say:
0/0 = x
Now, rearranging the equation (multiplying both sides by 0) gives:
x * 0 = 0
Now do you see the problem? There are an infinite number of values for x as anything multiplied by 0 is 0.
This is maths rather than programming, but briefly:
It's in some sense justifiable to assign a 'value' of positive-infinity to some-strictly-positive-quantity / 0, because the limit is well-defined
However, the limit of x / y as x and y both tend to zero depends on the path they take. For example, lim (x -> 0) 2x / x is clearly 2, whereas lim (x -> 0) x / 5x is clearly 1/5. The mathematical definition of a limit requires that it is the same whatever path is followed to the limit.
(Was inspired by Tony Breyal's rather good answer to post one of my own)
Zero is a tricky and subtle beast - it does not conform to the usual laws of algebra as we know them.
Zero divided by any number (except zero itself) is zero. Put more mathematically:
0/n = 0 for all non-zero numbers n.
You get into the tricky realms when you try to divide by zero itself. It's not true that a number divided by 0 is always undefined. It depends on the problem. I'm going to give you an example from calculus where the number 0/0 is defined.
Say we have two functions, f(x) and g(x). If you take their quotient, f(x)/g(x), you get another function. Let's call this h(x).
You can also take limits of functions. For example, the limit of a function f(x) as x goes to 2 is the value that the function gets closest to as it takes on x's that approach 2. We would write this limit as:
lim{x->2} f(x)
This is a pretty intuitive notion. Just draw a graph of your function, and move your pencil along it. As the x values approach 2, see where the function goes.
Now for our example. Let:
f(x) = 2x - 2
g(x) = x - 1
and consider their quotient:
h(x) = f(x)/g(x)
What if we want the lim{x->1} h(x)? There are theorems that say that
lim{x->1} h(x) = lim{x->1} f(x) / g(x)
= (lim{x->1} f(x)) / (lim{x->1} g(x))
= (lim{x->1} 2x-2) / (lim{x->1} x-1)
=~ [2*(1) - 2] / [(1) - 1] # informally speaking...
= 0 / 0
(!!!)
So we now have:
lim{x->1} h(x) = 0/0
But I can employ another theorem, called l'Hopital's rule, that tells me that this limit is also equal to 2. So in this case, 0/0 = 2 (didn't I tell you it was a strange beast?)
Here's another bit of weirdness with 0. Let's say that 0/0 followed that old algebraic rule that anything divided by itself is 1. Then you can do the following proof:
We're given that:
0/0 = 1
Now multiply both sides by any number n.
n * (0/0) = n * 1
Simplify both sides:
(n*0)/0 = n
(0/0) = n
Again, use the assumption that 0/0 = 1:
1 = n
So we just proved that all other numbers n are equal to 1! So 0/0 can't be equal to 1.
walks on back to her home over at mathoverflow.com
Here's a full explanation:
http://en.wikipedia.org/wiki/Division_by_zero
( Including the proof that 1 = 2 :-) )
You normally deal with this in programming by using an if statement to get the desired behaviour for your application.
The problem is with the denominator. The numerator is effectively irrelevant.
10 / n
10 / 1 = 10
10 / 0.1 = 100
10 / 0.001 = 1,000
10 / 0.0001 = 10,000
Therefore: 10 / 0 = infinity (in the limit as n reaches 0)
The Pattern is that as n gets smaller, the results gets bigger. At n = 0, the result is infinity, which is a unstable or non-fixed point. You can't write infinity down as a number, because it isn't, it's a concept of an ever increasing number.
Otherwise, you could think of it mathematically using the laws on logarithms and thus take division out of the equation altogther:
log(0/0) = log(0) - log(0)
BUT
log(0) = -infinity
Again, the problem is the the result is undefined because it's a concept and not a numerical number you can input.
Having said all this, if you're interested in how to turn an indeterminate form into a determinate form, look up l'Hopital's rule, which effectively says:
f(x) / g(x) = f'(x) / g'(x)
assuming the limit exists, and therefore you can get a result which is a fixed point instead of a unstable point.
Hope that helps a little,
Tony Breyal
P.S. using the rules of logs is often a good computational way to get around the problems of performing operations which result in numbers which are so infinitesimal small that given the precision of a machine’s floating point values, is indistinguishable from zero. Practical programming example is 'maximum likelihood' which generally has to make use of logs in order to keep solutions stable
Look at division in reverse: if a/b = c then c*b = a. Now, if you substitute a=b=0, you end up with c*0 = 0. But ANYTHING multiplied by zero equals zero, so the result can be anything at all. You would like 0/0 to be 0, someone else might like it to be 1 (for example, the limiting value of sin(x)/x is 1 when x approaches 0). So the best solution is to leave it undefined and report an error.
You may want to look at Dr. James Anderson's work on Transarithmetic. It isn't widely accepted.
Transarithmetic introduces the term/number 'Nullity' to take the value of 0/0, which James likens to the introduction 'i' and 'j'.
The structure of modern math is set by mathematicians who think in terms of axioms.
Having additional axioms that aren't productive and don't really allow one to do more stuff is against the ideal of having clear math.
How many times does 0 go into 0? 5. Yes - 5 * 0 = 0, 11. Yes - 11 * 0 = 0, 43. Yes - 43 * 0 = 0. Perhaps you can see why it's undefined now? :)
Since x/y=z should be equivalent to x=yz, and any z would satisfy 0=0z, how useful would such an 'exception' be?
Another explanation of why 0/0 is undefined is that you could write:
0/0 = (4 - 4)/0 = 4/0 - 4/0
And 4/0 is undefined.
If a/b = c, then a = b * c.
In the case of a=0 and b=0, c can be anything because 0 * c = 0 will be true for all possible values of c. Therefore, 0/0 is undefined.
This is only a Logical answer not a mathamatical one,
imagine Zero as empty how can you Divide an empty by an empty this is the case in division by zero also how can you divide by something which is empty.
0 means nothing, so if you have nothing, it does not imply towards anything to distribute to anything. Some Transit Facilities when they list out the number of trips of a particular line, trip number 0 is usually the special route that is routed in a different way. Typically, a good example would be in the Torrance Transit Systems where Line 2 has a trip before the first trip known as trip number 0 that operates on weekdays only, that trip in particular is trip number 0 because it is a specialized route that is routed differently from all the other routes.
See the following web pages for details:
http://transit.torrnet.com/PDF/Line-2_MAP.pdf
http://transit.torrnet.com/PDF/Line-2_Time_PDF.pdf
On the map, trip number 0 is the trip that is mapped in dotted line, the solid line maps the regular routing.
Sometimes 0 can be used on numbering the trips a route takes where it is considered the "Express Service" route.
why not make an exception for this
case?
Because:
as others said, it's not that easy;)
there's no application for defining 0/0 - adding exception would complicate mathematics for no gains.
This is what I'd do:
function div(a, b) {
if(b === 0 && a !== 0) {
return undefined;
}
if(b === 0 && a === 0) {
return Math.random;
}
return a/b;
}
When you type in zero divided by zero, there's an error because whatever you multiply zero from will be zero so it could be any number.
As Andrzej Doyle said:
Anything dived by zero is infinity. 0/0 is also infinity. You can't get 0/0 = 1. That's the basic principle of maths. That's how the whole world goes round. But you can sure edit a program to say "0/0 is not possible" or "Cannot divide by zero" as they say in cell phones.
I've never been much for math and I'm hoping that someone can help me out with the following.
I have 5 boxes:
1 2 3 4 5
[ ] [ ] [ ] [ ] [ ]
The boxes can either be white, gray, or black (or think of it as 0, 1, 2)
How many possible states can the box set be in?
What is the pseudocode (or in any language) to generate all the possible outcomes??
ie...
00000
00001
00011
00111
etc, etc...
I really appreciate any help anyone can give me with this.
the answer for the number of combinations is: 3x3x3x3x3 (3^5) since each box can have 3 possible colors.
As for generating the outcomes, see if you can figure it out using this matrix with 0, 1, or 2 to represent the color of the box. On a smaller scale (lets assume 3 boxes) it would look like this:
0 0 0
0 0 1
0 0 2
0 1 0
0 1 1
0 1 2
0 2 0
0 2 1
0 2 2
1 0 0
1 0 1
1 0 2
1 1 0
1 1 1
1 1 2
1 2 0
1 2 1
1 2 2
2 0 0
2 0 1
2 0 2
2 1 0
2 1 1
2 1 2
2 2 0
2 2 1
2 2 2
This is a classic permutation generation problem. You have 3 possibilities for each position, and 5 positions. The total number of generated string is 3^5 = 243.
You need recursion if you want a general solution (a simple iterative loop only works for a single instance of the problem).
Here's a quick example:
public static void Main(string[] args){
Generate("", 5);
}
private void Generate(string s, int limit)
{
if (s.Length == limit)
Console.WriteLine(s);
else
{
Generate(s+"0", limit);
Generate(s+"1", limit);
Generate(s+"2", limit);
}
}
To answer your first question, what would the answer be if the boxes could contain only one of two values? So, what's the answer if the boxes contain one of three values?
To answer your second question, what pseudocode generates all possible outcomes of one box? Now, pseudocode generates all possible outcomes of two boxes?
I'd recommend solving the problem on paper first. Try to solve it with a smaller number of boxes (maybe three), and list all possibilities. Then, think of how your reasoning went, or how you'd explain what you did to a small child.
This seems like a homework problem. I'll just give you some help as to the solution then.
What you are saying is that each box has three states, which are all independent. One box would have 3 solutions, and two boxes would have 3 * 3 solutions - for each state of the first box the second box would have three states as well. Extend that to 5 boxes.
To generate each solution, you can just cycle through it. It is easy to make nested for loops for each box, and multiplying by powers of 10 can let you show the number at once.
You can generalize the code for multiple boxes in a similar way.
Thank you all for your answers, at least those of you who actually gave me one.
While I can appreciate that the question sounded like it was pulled straight out of Computer Science 101, it wasn't. The irony of the matter is that it was for real life on a real deadline and I didn't have time to hearken back to when I was being taught this stuff and said to myself, "when am I ever going to need this crap"
If I wanted to be patronized and treated like a school boy I would go back to my elementary school and ask my 5th grade teacher if I can go to the bathroom
Thanks again
the number of states is 3^5.
pseudocode is
for value from 0 to 3^5-1
print base3(value)
where base3 is a function that repeatedly takes modulo 3 to get a digit, then removes that digit (by dividing by 3)
Hint: imagine that each box is a position in a number and each colour is a different digit. In the real world, how many combinations (including zero) do you get with 2 positions and 10 possible digits? What about 3 positions? What's the relationship between adding an extra position and the number of combinations, given the number of digits you have available?
Unique number of combinations: 3^5=243
Code:
n = 0
for i = 0 to 3^5-1
{
s = ""
for j = 1 to 5
{
d = n mod 3
s = toascii(d) . s
n = n / 3
}
println s
i = i + 1
}
Here's how I first learned to do this: first think about how many choices you are making. You are making five choices, one for each box. So write down five blank lines with multiplication signs:
__ x __ x __ x __ x __ = ?
In each blank, write the number of objects you have to choose from for that box. Since you have 3 numbers to choose from for each box, you write a 3 in every blank:
3 x 3 x 3 x 3 x 3 = 243
This gives you the total number of permutations for those choices.
The number of possibilities is 3 to the power 5
If you loop from 0 to that number minus 1 and express it in base 3 you will have all the possibilities (remember to prepend 0s where necessary)
In Ruby:
number_of_possibilities = 3**5-1
for i in (0..number_of_possibilities)
base_3_number = i.to_s(3)
puts "%05d" % base_3_number # number formatting used to prepend 0s where necessary
end
Can I ask what about this you don't understand or whats tripping you up? I see that everyone here has simply answered the question, but if you've copied their answers, you've learned nothing, and thus completely missed the point of the homework. Assuming your next lesson builds upon this one, you're just going to fall further behind.
If you either worked for me or were in my class I'd simply ask the following...
"How do you think the problem should be solved?" The answer to which might reveal where you're getting hung up. A wise professor of mine at CMU once said "I can't help you understand this until you know what you don't understand" I never did figure out what I didn't understand and I dropped his class, but the lesson stuck with me.
I know its probably too late, but for these homework questions I really think we should be helping the person learn as opposed to simply providing an answer and doing their homework for them.
Your problem needs nothing more than the rule of product in combinatorics.
You can choose the state of the first box in 3 ways, and the state of the second box in 3 ways, and ... and the state of the 5th box in 3 ways. The number of ways in which you can set the state of all the boxes is the product of all the five (equal) numbers of ways, i.e. 3x3x3x3x3 = 35.
Similar question: how many numbers can you form with 5 digits in the decimal system, counting the leading zeros? That is, how many numbers are there from 00000 to 99999? You can choose the first digit in 10 ways (0...9), and so on and so on, and the answer is 10x10x10x10x10 = 100000, as you already know.
Don't even try to write code to answer this! The reason is that you need some very large numbers (factorials) to calculate it. These create numbers much larger than any base type in the CLR. You can use this opensource library to do the calculation.
void solve(int p=0,int n=5,int d=0)
{
if (n==p)
{
int rev=d;
int i=0;
while (i<5) {
cout << rev%10;
rev /= 10;
i++;## Heading ##
}
cout << endl;
return;
}
for(int i=0; i<3 ; i++)
{
solve(p+1,n, d*10 + i);
}
}