Let A,B be matrixes of R^n space and b belong to R^n.Describe a fast algorithm to compute A^-2*B*A^-3*b.How many computations will the algorithm make?
This is an exam question I have for numerical analysis.I tried brute forcing an algorithm but i believe that the answer is more mathematical.
We haven't yet talked about Big O notations so the question asks for strictly the actions of the algorithm.How would you go about answering this question?
I would just work the problem from right to left, using a linear solver when dealing with the inverse of A, and matrix multiply when dealing with B:
x1 = linsolve(A, b)
x2 = linsolve(A, x1)
x3 = linsolve(A, x2)
y = B*x3
z1 = linsolve(A,y)
result = linsolve(A,z1)
You can reduce by a constant multiplier the number of operations by keeping the LU decomposition of A in memory, but unless you are given more structure on A and B, quadratic complexity seems to be the best you can aim for.
Related
This is part of a bigger question. Its actually a mathematical problem. So it would be really great if someone can direct me to any algorithm to obtain the solution of this problem or a pseudo code will be of help.
The question. Given an equation check if it has an integral solution.
For example:
(26a+5)/32=b
Here a is an integer. Is there an algorithm to predict or find if b can be an integer. I need a general solution not specific to this question. The equation can vary. Thanks
Your problem is an example of a linear Diophantine equation. About that, Wikipedia says:
This Diophantine equation [i.e., a x + b y = c] has a solution (where x and y are integers) if and only if c is a multiple of the greatest common divisor of a and b. Moreover, if (x, y) is a solution, then the other solutions have the form (x + k v, y - k u), where k is an arbitrary integer, and u and v are the quotients of a and b (respectively) by the greatest common divisor of a and b.
In this case, (26 a + 5)/32 = b is equivalent to 26 a - 32 b = -5. The gcd of the coefficients of the unknowns is gcd(26, -32) = 2. Since -5 is not a multiple of 2, there is no solution.
A general Diophantine equation is a polynomial in the unknowns, and can only be solved (if at all) by more complex methods. A web search might turn up specialized software for that problem.
Linear Diophantine equations take the form ax + by = c. If c is the greatest common divisor of a and b this means a=z'c and b=z''c then this is Bézout's identity of the form
with a=z' and b=z'' and the equation has an infinite number of solutions. So instead of trial searching method you can check if c is the greatest common divisor (GCD) of a and b
If indeed a and b are multiples of c then x and y can be computed using extended Euclidean algorithm which finds integers x and y (one of which is typically negative) that satisfy Bézout's identity
(as a side note: this holds also for any other Euclidean domain, i.e. polynomial ring & every Euclidean domain is unique factorization domain). You can use Iterative Method to find these solutions:
Integral solution to equation `a + bx = c + dy`
and thank you for the attention you're paying to my question :)
My question is about finding an (efficient enough) algorithm for finding orthogonal polynomials of a given weight function f.
I've tried to simply apply the Gram-Schmidt algorithm but this one is not efficient enough. Indeed, it requires O(n^2) integrals. But my goal is to use this algorithm in order to find Hankel determinants of a function f. So a "direct" computation wich consists in simply compute the matrix and take its determinants requires only 2*n - 1 integrals.
But I want to use the theorem stating that the Hankel determinant of order n of f is a product of the n first leading coefficients of the orthogonal polynomials of f. The reason is that when n gets larger (say about 20), Hankel determinant gets really big and my goal is to divided it by an other big constant (for n = 20, the constant is of order 10^103). My idea is then to "dilute" the computation of the constant in the product of the leading coefficients.
I hope there is a O(n) algorithm to compute the n first orthogonal polynomials :) I've done some digging and found nothing in that direction for general function f (f can be any smooth function, actually).
EDIT: I'll precise here what the objects I'm talking about are.
1) A Hankel determinant of order n is the determinant of a square matrix which is constant on the skew diagonals. Thus for example
a b c
b c d
c d e
is a Hankel matrix of size 3 by 3.
2) If you have a function f : R -> R, you can associate to f its "kth moment" which is defined as (I'll write it in tex) f_k := \int_{\mathbb{R}} f(x) x^k dx
With this, you can create a Hankel matrix A_n(f) whose entries are (A_n(f)){ij} = f{i+j-2}, that is something of the like
f_0 f_1 f_2
f_1 f_2 f_3
f_2 f_3 f_4
With this in mind, it is easy to define the Hankel determinant of f which is simply
H_n(f) := det(A_n(f)). (Of course, it is understood that f has sufficient decay at infinity, this means that all the moments are well defined. A typical choice for f could be the gaussian f(x) = exp(-x^2), or any continuous function on a compact set of R...)
3) What I call orthogonal polynomials of f is a set of polynomials (p_n) such that
\int_{\mathbb{R}} f(x) p_j(x) p_k(x) is 1 if j = k and 0 otherwize.
(They are called like that since they form an orthonormal basis of the vector space of polynomials with respect to the scalar product
(p|q) = \int_{\mathbb{R}} f(x) p(x) q(x) dx
4) Now, it is basic linear algebra that from any basis of a vector space equipped with a scalar product, you can built a orthonormal basis thanks to the Gram-Schmidt algorithm. This is where the n^2 integrations comes from. You start from the basis 1, x, x^2, ..., x^n. Then you need n(n-1) integrals for the family to be orthogonal, and you need n more in order to normalize them.
5) There is a theorem saying that if f : R -> R is a function having sufficient decay at infinity, then we have that its Hankel determinant H_n(f) is equal to
H_n(f) = \prod_{j = 0}^{n-1} \kappa_j^{-2}
where \kappa_j is the leading coefficient of the j+1th orthogonal polynomial of f.
Thank you for your answer!
(PS: I tagged octave because I work in octave so, with a bit of luck (but I doubt it), there is a built-in function or a package already done managing this kind of think)
Orthogonal polynomials obey a recurrence relation, which we can write as
P[n+1] = (X-a[n])*P[n] - b[n-1]*P[n-1]
P[0] = 1
P[1] = X-a[0]
and we can compute the a, b coefficients by
a[n] = <X*P[n]|P[n]> / c[n]
b[n-1] = c[n-1]/c[n]
where
c[n] = <P[n]|P[n]>
(Here < | > is your inner product).
However I cannot vouch for the stability of this process at large n.
I am working on algorithm to perform linear regression for one or more independent variables.
that is: (if I have m real world values and in the case of two independent variables a and b)
C + D*a1 + E* b1 = y1
C + D*a2 + E* b2 = y2
...
C + D*am + E* bm = ym
I would like to use the least squares solution to find best fitting straight line.
I will be using the matrix notation
so
where Beta is the vector [C, D, E] where these values will be the best fit line.
Question
What is the best way to solve this formula? Should I compute the inverse of
or should I use the LU factorization/decmposition of the matrix. What is the performance of each on large amount of data (i.e a big value of m , could be in order of 10^8 ...)
EDIT
If the answer was to use Cholesky decomposition or QR decomposition, are there any implementation hints/ simple libraries to use.
I am coding in C/ C++.
Two straightforward approaches spring to mind for solving a dense overdetermined system Ax=b:
Form A^T A x = A b, then Cholesky-factorise A^T A = L L^T, then do two back-solves. This usually gets you an answer precise to about sqrt(machine epsilon).
Compute the QR factorisation A = Q*R, where Q's columns are orthogonal and R is square and upper-triangular, using something like Householder elimination. Then solve Rx = Q^T b for x by back-substitution. This usually gets you an answer precise to about machine epsilon --- twice the precision as the Cholesky method, but it takes about twice as long.
For sparse systems, I'd usually prefer the Cholesky method because it takes better advantage of sparsity.
Your X^TX matrix should have a Cholesky decomposition. I'd look into this decomposition before LU. It is faster: http://en.wikipedia.org/wiki/Cholesky_decomposition
Imagine that I'm a bakery trying to maximize the number of pies I can produce with my limited quantities of ingredients.
Each of the following pie recipes A, B, C, and D produce exactly 1 pie:
A = i + j + k
B = t + z
C = 2z
D = 2j + 2k
*The recipes always have linear form, like above.
I have the following ingredients:
4 of i
5 of z
4 of j
2 of k
1 of t
I want an algorithm to maximize my pie production given my limited amount of ingredients.
The optimal solution of these example inputs would yield me the following quantities of pies:
2 x A
1 x B
2 x C
0 x D
= a total of 5 pies
I can solve this easily enough by taking the maximal producer of all combinations, but the number
of combos becomes prohibitive as the quantities of ingredients increases. I feel like there must
be generalizations of this type of optimization problem, I just don't know where to start.
While I can only bake whole pies, I would be still be interested in seeing a method which may produce non integer results.
You can define the linear programming problem. I'll show the usage on the example, but it can of course be generalized to any data.
Denote your pies as your variables (x1 = A, x2 = B, ...) and the LP problem will be as follows:
maximize x1 + x2 + x3 + x4
s.t. x1 <= 4 (needed i's)
x1 + 2x4 <= 4 (needed j's)
x1 + 2x4 <= 2 (needed k's)
x2 <= 1 (needed t's)
x2 + 2x3 <= 5 (needed z's)
and x1,x2,x3,x4 >= 0
The fractional solution to this problem is solveable polynomially, but the integer linear programming is NP-Complete.
The problem is indeed NP-Complete, because given an integer linear programming problem, you can reduce the problem to "maximize the number of pies" using the same approach, where each constraint is an ingredient in the pie and the variables are the number of pies.
For the integers problem - there are a lot of approximation techniques in the literature for the problem if you can do with "close up to a certain bound", (for example local ratio technique or primal-dual are often used) or if you need an exact solution - exponential solution is probably your best shot. (Unless of course, P=NP)
Since all your functions are linear, it sounds like you're looking for either linear programming (if continuous values are acceptable) or integer programming (if you require your variables to be integers).
Linear programming is a standard technique, and is efficiently solvable. A traditional algorithm for doing this is the simplex method.
Integer programming is intractable in general, because adding integral constraints allows it to describe intractable combinatorial problems. There seems to be a large number of approximation techniques (for example, you might try just using regular linear programming to see what that gets you), but of course they depend on the specific nature of your problem.
I've been trying to solve the codechef problem: http://www.codechef.com/MAY11/problems/TPRODUCT/
They have given the post-contest analysis here: http://www.codechef.com/wiki/may-2011-contest-problem-editorials
I need some help in understanding the logic discussed there:
They are talking about using logarithm in place of the function
Pi=max(Vi*PL, Vi*PR)
Math is not my strong area. [I've been trying to improve by participating in contests like this]. If someone can give a very dumbed down explanation for this problem, it would be helpful for mortals like me. Thanks.
One large problem with multiplication is that numbers get very large very fast, and there are issues with reaching the upper bounds of an int or long, and spilling over to the negatives. The logarithm allows us to keep the computations small, and then get the answer back modulo n.
In retracing the result found via dynamic programming, the naive solution is to multiply all the values together and then mod:
(x0 * x1 * x2 * ... * xk) (mod n)
this is replaced with a series of smaller computations, which avoid bound overflow:
z1 = e^(log(x0) + log(x1)) modulo n
z2 = e^(log(x2) + log(z1)) modulo n
...
zk = e^(log(xk) + log(z{k-1})) modulo n
and then zk contains the result.
Presumably, they are relying on the simple mathematical observation that if:
z = y * x
then:
log(z) = log(y) + log(x)
Thus turning multiplications into additions.