How can I get the not-normalized output refracted vector, with an also not-normalized incident vector?
I'm following that formulas, work with normalized input, but if I pass not-normalized doesn't. Tried to divide the dot product by input vector length but also nothing.
Wikipedia Snell's Law Vector form
If you divide the dot product by the incident vector length, then your thetas will be correct.
After that, if you multiply n by the incident vector length, then your vreflected and vrefracted vectors will be correct.
Related
Suppose the problem posed is as follows:
On Mars there lives a colony of worms. Each worm is represented as elements in an 1D array. Worms decide to eat each other but any worm can eat only its nearest neighbour. Each worm has a preset amount of energy(i.e the value of the element). On Mars, the laws dictate that when a worm i with energy x eats another worm with energy y, the i-th worm’s final energy becomes x-y. A worm is allowed to have negative energy levels.
Find the maximum value of energy of the last standing worm.
Sample data:
0,-1,-1,-1,-1 has answer 4.
2,1,2,1 has answer 4.
What will be the suitable logic to address this problem?
This problem has a surprisingly simple O(N) solution.
If any two members in the array have different signs, the answer is then sum of absolute values of all elements.
To see why, imagine a single positive value in the array, all other elements are negative (Example 1). Now the best strategy would be keeping this value positive and gradually eating all neighbors away to increase this positive value. The position of the positive value doesn't matter. The strategy is same in case of a single negative element.
In more general case, if an array of size N have values of different signs, we can always find an array of size N-1 with different signs, because there must be a pair of neighbors with different sign, which we can combine to form a number of any sign we prefer.
For example with this array : [1,2,-5,4,-10]
we can combine either (2,-5) or (4,-10). Lets combine (4,-10) to get [1,2,-5,-14]
We can only take (2,-5) now. So our array now is : [1,-7,-14]
Again only (1,-7) possible. But this time we have to keep combined value positive. So we are left with: [8,-14]
Final combining gives us 22, sum of all absolute values.
In case of all values with same sign, our first move would be to produce an opposite sign combining a neighbor pair with as little "cost" as possible. Intuitively, we don't want to waste two big numbers on this conversion. If we take x,y neighbor pair, when combined the new value (of opposite sign) will be abs(x-y). Since result is simply sum of absolute values, we can interpret it as - "loosing" abs(x) and abs(y) from maximum possible output and "gaining" abs(x-y) instead. So the "cost" for using this pair for sign conversion is abs(x)+abs(y)-abs(x-y). Since we need to minimise this cost, we choose from initial array neighbor pair that have lowest such value.
So if we take the above array but now all values are positive [1,2,5,4,10]:
"cost" of converting (1,2) to -1 is 1+2-abs(-1)=2.
"cost" of converting (2,5) to -3 is 2+5-abs(-3)=4.
"cost" of converting (5,4) to -1 is 5+4-abs(-1)=8.
"cost" of converting (4,10) to -6 is 4+10-abs(-6)=8.
So, we take and convert pair (1,2) to -1. Then just sum absolute values of resultant array to get 20. Notice that this value is exactly 2 less than our previous example.
I have two tensors of rank 3 each, in other words two 3D matrix. I want to take dot product of these two matrix. I am confused to continue with this problem. Help me out with formula to do so.
A 3-way tensor (or equivalently 3D array or 3-order array) need not necessarily be of rank-3; Here, "rank of a tensor" means the minimum number of rank-1 tensors (i.e. outer product of vectors; For N-way tensor, it's the outer product of N vectors) needed to get your original tensor. This is explained in the below figure of so-called CP decomposition.
In the above figure, the original tensor(x) can be written as a sum of R rank-1 tensors, where R is a positive integer. In CP decomposition, we aim to find a minimum R that yields our original tensor X. And this minimum R is called the rank of our original tensor.
For a 3-way tensor, it is the minimum number of (a1,a2,a3...aR; b1,b2,b3...bR; c1,c2,c3...cR) vectors (where each of the vectors is n dimensional) required to obtain the original tensor. The tensor can be written as the outer product of these vectors as:
In terms of element-wise, we can write the 3-way tensor as:
Now, with that background, to answer your specific question, to take the dot product (also called tensor inner product), both tensors must be of same shape (for e.g. 3x2x5 and 3x2x5), then the inner product is defined as the sum of the element-wise product of their values.
where the script X and Y are the same-shape tensors.
P.S.: The tilde in the above formulae should not be interpreted as an approximation.
The vector inner product sum the elementwise products. The tensor inner product follows the same idea. Match the elements, multiply them, and add them all .
I have a double vector:
r = -50 + (50+50)*rand(10,1)
Now i want to ideally have all the numbers in the vector equal upto a tolerance of say 1e-4. I want to represent each r with a scalar say s(r) such that its value gives an idea of the quality of the vector. The vector is high quality if all elements in the vector are equal-like. I can easily run a for loop like
for i=1:10
for j=i+1:10
check equality upto the tolerance
end
end
But even then i cannot figure what computation to do inside the nested for loops to assign a scalar representing the quality . Is there a better way such that given any vector r length n, i can quickly calculate a scalar representing the quality of the vector.
Your double-loop algorithm is somewhat slow, of order O(n**2) where n is the number of dimensions of the vector. Here is a quick way to find the closeness of the vector elements, which can be done in order O(n), just one pass through the elements.
Find the maximum and the minimum of the vector elements. Just use two variables to store the maximum and minimum so far and run once through all the elements. The difference between the maximum and the minimum is called the range of the values, a commonly accepted measure of dispersion of the values. If the values are exactly equal, the range is zero which shows perfect quality. If the range is below 1e-4 then the vector is of acceptable quality. The bigger the range, the worse the equality.
The code is obvious for just about any given language, so I'll leave that to you. If the fact that the range only really considers the two extreme values of the vector bothers you, you could use other measures of variation such as the interquartile range, variance, or standard deviation. But the range seems to best fit what you request.
I have records (rows) in a database and I want to identify similar records. I have a constraint to use cosine similarity. If the variables (attributes, columns) vary in type and come in this form:
[number] [number] [boolean] [20 words string]
how can I proceed to the vectorization to apply the cosine similarity? For the string I can take the simple tf-idf. But for numbers and boolean values?. And how can this be combined? My thought is that the vector would be of 1+1+1+20 length. But is it semantically "efficient" to just transform the numbers of the record to coefficients in my vector and to concatenate them with the tf-idf of the string to compute the cosine similarity? Or i can treat numbers as words and apply tf-idf to numbers as well. Is there another technique?
Each positional element of the vectors must measure a particular attribute/feature of the entities of interest. Frequently, when words are involved, there is a vector element for the count of each word that may appear. Thus, your vector might have the size of 1 + 1 + 1 + (vocabulary size).
Because cosine similarity calculates based on numbers, you might have to convert non-numbers to numbers. For example, you might use 0, 1 for booleans.
You don't mention whether your numeric fields represent measurements or discrete values (e.g., keys). If the numeric values are measurements, then cosine similarity is well-suited (although if there are different scales of the numbers of the different attributes, it can bias your results). However, if the numbers represent keys, then using a single attribute for each field will give poor results, because a key of 5 is no closer to 6 than it is to 200. But cosine similarity doesn't know that. In the case where a database field contains keys, you might want to have a boolean (0, 1) vector element for each possible value.
I have 2 questions,
I've made a vector from a document by finding out how many times each word appeared in a document. Is this the right way of making the vector? Or do I have to do something else also?
Using the above method I've created vectors of 16 documents, which are of different sizes. Now i want to apply cosine similarity to find out how similar each document is. The problem I'm having is getting the dot product of two vectors because they are of different sizes. How would i do this?
Sounds reasonable, as long as it means you have a list/map/dict/hash of (word, count) pairs as your vector representation.
You should pretend that you have zero values for the words that do not occur in some vector, without storing these zeros anywhere. Then, you can use the following algorithm to compute the dot product of these vectors (pseudocode):
algorithm dot_product(a : WordVector, b : WordVector):
dot = 0
for word, x in a do
y = lookup(word, b)
dot += x * y
return dot
The lookup part can be anything, but for speed, I'd use hashtables as the vector representation (e.g. Python's dict).