I need to calculate (a/b) mod m where a and b are very large numbers.
What I am trying to do is to calculate (a mod m) * (x mod m), where x is the modular inverse of b.
I tried using Extended Euclidean algorithm, but what to do when b and m are not co-prime?
It is specifically mentioned that b and m need to be co-prime.
I tried using the code here, and realized that for example:
3 * x mod 12 is not at all possible for any value of x, it does not exist!
What should I do? Can the algorithm be modified somehow?
Yep, you are in trouble. x has no solution in b*x = 1 mod m if b and m have a common divisor. Similarly, in your original problem a/b = y mod m, you are looking for y such that a=by mod m. If a is divisible by gcd(b,m), then you can divide out by that factor and solve for y. If not, then there is no y that can solve the equation (i.e. a/b mod m is not defined).
The reason that b and m have to be coprime is because of the Chinese Remainder Theorem. Basically the problem:
3 * x mod 12
Can be thought of as a compound problem involving
3*x mod 3 and 3*x mod 4 = 2^2
Now if b is not coprime to 12, this is like trying to divide by zero. Thus the answer doesn't exist!
This is due to field theory in abstract algebra. A field is basically a set which has addition, subtraction, multiplication, and division well-defined. A finite field is always of the form GF(p^n), where p is prime and n is a positive integer, and the operations are addition and multiplication modulo p^n. Now, 12 is not a prime power, so your ring is not a field. Thus this problem can't be solved for any b which is not coprime to m.
Check this: http://www.math.harvard.edu/~sarah/magic/topics/division
It might help.
It explains methods of modular division.
Related
Let assume an equation A x = b, where A is a real n x n matrix, b is a real valued vector of length n and x is the solution vector of this linear system.
We can find the solution of x by finding the inverse of A.
B= inv(A)
And therefore
x =A^{-1}b.
x= B*b
Can I apply the same solver if A and b are complex?
EDIT: I'm looking for explanation why it should work. Thanks :)
You can do that. Better in Matlab would be x = A\b. That will usually get the answer faster and more accurately.
In short yes, it works for any matrices irrespective of the field. (Well, at least real or complex field works.)
I think your question is whether B exists or not. It does as long as the determinant of B is non-zero. Vice versa.
The title of this post may be a bit confusing. Please allow me to provide a bit of context and then elaborate on what I'm asking. For your reference, the question I'm asking is toward the end and is denoted by bold letters. I provide some code, outlining where I'm currently at in solving the problem, immediately beforehand.
Essentially what I'm trying to do is Kernel Regression, which is usually done using a single test point x and a set of training instances . A reference to this can be found on wikipedia here. The kernel I'm using is the RBF kernel, a Wikipedia reference for which can be found here.
Anyway, I have some code written in Matlab so that this can be done quickly for a single instance of x, which is 1 x p in size. What I'd like to do is make it so I can estimate for numerous points very quickly, say m x p.
For the sake of avoiding notational mixups, I'll let the training instances be denoted Train and the instances I want estimates for as Test: and . It also needs to be mentioned that I want to estimate a vector of numbers for each of the m points. For a single point this vector would be 1 x v in size. Now I need it to be m x v. Therefore, Train will also have a vector of these know values associated with it called TS: . Lastly, we need a vector of sigmas that is 1 x v in size. This is denoted as Sig.
Here's the code I have so far:
%First, we have to get the matrices to equivalent size so we can subtract Train from Test
tm0 = kron(ones(size(Train,1),1),Test) - kron(ones(size(Test,1),1),Train);
%Secondly, we apply the Euclidean norm sq by row and then multiply each of these results by each element (j) in Sig times 1/2j^2
tm3 = exp(-kron(sum((tm0).^2,2),1/2./(Sig.^2)));
Now, at this point tm3 is an (m*n) x v matrix. This is where my question is: I now need to multiply TS' (TS transpose) times each of the n x v-sized segments in tm3 (there are m of these segments), get the diagonal elements of each of these resulting segments (after multiplication one of the m segments will be v x v, so each chunk of diagonal elements will be 1 x v meaning the resulting matrix is m x v) and sum these diagonal elements together to produce an m x 1 sized matrix. Lastly, I will need to divide each entry i in this m x 1 matrix by each of the v elements in the ith row of the diagonal-holding m x v-sized matrix, producing an m x v-sized result matrix.
I hope all of that makes sense. I'm sure there's some kind of trick that can be employed, but I'm just not coming up with it. Any help is greatly appreciated.
Edit 1: I was asked to provide more of an example to help demonstrate what it is that I would like done. The following represent that two matrices I'm talking about, TS and tm3:
As you can see, TS'(TS transpose) is v x n and tm3 is mn x v. In tm3 there are blocks that are of size n x v -- there are m blocks of this size. Notice that the size of TS' is of size v x n. This means that I can multiply TS' by a single block of tm3, which again is of size n x v. This would result in a matrix that is v x v in size. I would like to do this operation -- individually multiplying TS' by each of the n x v-sized blocks of tm3, which would produce m v x v matrices.
From here, though, I would like to obtain the diagonal elements from each of these v x v matrices. So, for a single v x v matrix, denoted using a:
Ultimately, I would to do this for each of the m v x v matrices giving me something that looks like the following, where s is the mth v x v matrix:
If I denote this last matrix as Q, which is m x v in size, it is trivial to sum the elements across the rows to produce the m x 1 vector I was looking for. I will refer to this vector as C. However, I would then like to divide each of these m scalar values by the corresponding row of matrix Q, to produce another m x v matrix:
This is the final matrix I'm looking for. Hopefully this helps make it clear what I'm looking for. Thanks for taking the time to read this!
Thought: I'm pretty sure I could accomplish this by converting tm3 to a cell matrix by doing tc1 = mat2cell(tm3,repmat(length(Train),1,m),length(Sig)), and then put replicate TS m times in another cell matrix tc2 = mat2cell(TS',length(indirectSigma),repmat(length(Train),1,m))'. Finally, I could do operations like tc3 = cellfun(#(a,b) a*b, tc2,tc1,'UniformOutput',false), which would give me m cells filled with the v x v matrices I was looking for. I could proceed from there. However, I'm not sure how fast these cell operations are. Can anybody comment? I'm afraid they might be slow, so I would prefer operations be performed on normal matrices, which I know to be fast. Thanks!
I have 37 linear equations and 36 variables in the form of a matrix equation; A*X=B . The equations don't have an exact answer. I want to use Matlab least square method to find the answers with the least error. I am new to Matlab so any comments will help. Thank you
If A is of full rank, i.e. the columns of A are linearly independent, the least-squares solution of an overdetermined system of linear equations
A * x = b
can be found by inverting the normal equations (see Linear Least Squares):
x = inv(A' * A) * A' * b
If A is not of full rank, A' * A is not invertible. Instead, one can use the pseudoinverse of A
x = pinv(A) * b
or Matlab's left-division operator
x = A \ b
Both give the same solution, but the left division is more computationally efficient.
The two latter computation methods can also deal with underdetermined systems of linear equations, but they give different solutions in that case: The pseudoinverse gives the solution where x has the smallest sum of squares, while the left-division operator gives a solution with as many 0 coefficients as possible.
The most general way to solve this is to use the pseudoinverse:
X = pinv(A) * B;
You can calculate x by:
x = (A'*A)\A'*B
I wanna to try calculate multiply of three matrix in matlab.
The formation of matrices described below:
L = D^(-1/2) * A * D^(-1/2);
D, A and L are a n*n matrices. A and L are not diagonal or sparse but D is diagonal. In this case n = 16900. When I calculate L in matlab, it takes a long time, about 4 hours!
My question is: Is there a more efficient way to calculate L?
You can use bsxfun twice. I'm not sure it will be faster or not:
v = diag(D).^(-1/2); %// this is the same as diag(D.^(-1/2))
L = bsxfun(#times, v.', bsxfun(#times, A, v));
Instead of using naive matrix multiplication, you can specialised asymptotically faster ones. Strassen's algorithm comes to mind but if I recall correctly it typically has a high constant, despite it's better asymptotic complexity. If you have a very limited set of possible values in your matrices, you can use a variation of the "four Russians" method.
This is the equation I am trying to solve:
h = (X'*X)^-1*X'*y
where X is a matrix and y is a vector ((X'X)^-1 is the inverse of X-transpose times X). I have coded this in Matlab as:
h = (X'*X)\X'*y
which I believe is correct. The problem is that X is around 10000x10000, and trying to calculate that inverse is crashing Matlab on even the most powerful computer I can find (16 cores, 24GB RAM). Is there any way to split this up, or a library designed for doing such large inversions?
Thank you.
That looks like a pseudo inverse. Are you perhaps looking for just
h = X \ y;
I generated a random 10,000 by 10,000 matrix X and a random 10,000 by 1 vector y.
I just broke up my computation step by step. (Code shown below)
Computed the transpose and held it in matrix K
Then I computed Matrix A by multiplying K by X
Computed vector b by multiplying K by vector y
Lastly, I used the backslash operator on A and b to solve
I didn't have a problem with the computation. It took a while, but breaking up the operations into the smallest groups possible helped to prevent the computer from being overwhelmed. However, it could be the composition of the matrix that you are using (ie. Sparse, decimals, etc.).
X = randi(2000, [10000, 10000]);
y = randi(2000, 10000, 1);
K = X';
A = K*X;
b = K*y;
S = A\b;
If you have multiple machines at your disposal, and you can recast your problem into the form h = X\y as proposed by #Ben, then you could use distributed arrays. This demo shows how you can do that.
Jordan,
Your equation is exactly the definition for "Moore-Penrose Matrix Inverse".
Check: http://mathworld.wolfram.com/Moore-PenroseMatrixInverse.html
Directly using h = X \ y; should help.
Or check Matlab pinv(X)*y