I am integrating a differential equation with Runge-Kutta 2 method in order to obtain an approximate solution y_n(t), where n is a varying initial parameter. Then, for a sample of n's in a chosen range I want to find the first root of y_n(t) by recording the first change of sign: the result is a function R(n) that has a vertical asymptote at some value of n, where R(n) is the first root of y_n(t).
My question is then: how can I estimate the value of the vertical asymptote?
I do not see any way of doing that for the moment. I have thought about checking that for small changes of n the value of R(n) should not change too much at each step, but that is not true especially if there is an exponential or worse growth. I have also thought about just choosing an "upper bound", meaning that if R(n) is greater than some limit then I could approximate the vertical asymptote by that value of n, but I do not like this solution because I have no control over the error I am making.
Are there any more clever ideas?
Any comment or answer is much appreciated and let me know if I can explain myself clearer!
Related
The Matlab function bvp4c solves boundary value problems. It takes a differential equation, boundary conditions and an initial guess as input, and returns a structure array containing arrays of x, y and yp (which stands for "y prime", or y').
The length of the output arrays should be the same as that of the initial guess, but I found that it isn't always. I have checked the dimensions of the input (the initial guess, always 1x101 double for x and 16x101 double for y) and the output (sometimes 1x101 double for x and 16x101 double for y and yp as it should be, but often different values, such as 1x91 double and 16x91 double or 1x175 double and 16x175 double).
Looking at the output array x when its length is off, some extra values are squeezed in, or some are taken out. For example, the initial guess has 100 positions between x=0 and x=1, and the x array should be [0 0.01 0.02 ... 1], but sometimes a new position like 0.015 shows up.
Question: Why does this happen, and how can this be solved?
"The length of the output arrays should be the same as that of the initial guess ...." This is incorrect.
As described in the bvp4c documentation, sol.x contains a "[mesh] selected by bvp4c" with an "[approximation] to y(x) at the mesh points of sol.x". In order to evaluate bvp4c's solution on your mesh, use deval.
Why does bvp4c choose a mesh? Quoting from the cited paper1, which you can get in full here if you have a MathWorks account:
Because BVPs can have more than one solution, BVP codes require users to supply a guess for the solution desired. The guess includes a guess for an initial mesh that reveals the behavior of the desired solution. The codes then adapt the mesh so as to obtain an accurate numerical solution with a modest number of mesh points.
Because a steady BVP generally has a global behavior strongly dependent on its boundary values, the spatial mesh between the two boundaries may need to be refined in order to properly approximate the desired solution with the locally chosen basis functions for the method. However, there may also be portions of the mesh that do not need to be refined and can even be coarsened in some cases to maintain a reasonably small residual and accurate approximation. Therefore, for general efficiency, the guess mesh is adaptively refined or coarsened depending on some locally chosen metric (since bvp4c is collocation based, the metric is probably point-based or division-integrated based) such that the mesh returned by bvp4c is, in some sense, adequate enough for generic interpolation within the boundaries.
I'll also note that this is different from numerically solving IVPs since their state is not global across the entire time integration locus and only depends on the current state to the next time-step, and possibly previous time steps if using a multi-step method or solving a delay differential equation, which makes the refinement inherently local. This local behavior of IVPs is what allows functions like ode45 to return a solution at pre-selected time values because it can locally refine the solution at the selected point while performing the time march (this is known as dense output).
1 Shampine, L.F., M.W. Reichelt, and J. Kierzenka, "Solving Boundary Value Problems for Ordinary Differential Equations in MATLAB with bvp4c".
I am trying to find the optimal solution to a Sliding Block Puzzle of any length using the A* algorithm.
The Sliding Block Puzzle is a game with white (W) and black tiles (B) arranged on a linear game board with a single empty space(-). Given the initial state of the board, the aim of the game is to arrange the tiles into a target pattern.
For example my current state on the board is BBW-WWB and I have to achieve BBB-WWW state.
Tiles can move in these ways :
1. slide into an adjacent empty space with a cost of 1.
2. hop over another tile into the empty space with a cost of 1.
3. hop over 2 tiles into the empty space with a cost of 2.
I have everything implemented, but I am not sure about the heuristic function. It computes the shortest distance (minimal cost) possible for a misplaced tile in current state to a closest placed same color tile in goal state.
Considering the given problem for the current state BWB-W and goal state BB-WW the heuristic function gives me a result of 3. (according to minimal distance: B=0 + W=2 + B=1 + W=0). But the actual cost of reaching the goal is not 3 (moving the misplaced W => cost 1 then the misplaced B => cost 1) but 2.
My question is: should I compute the minimal distance this way and don't care about the overestimation, or should I divide it by 2? According to the ways tiles can move, one tile can for the same cost overcome twice as much(see moves 1 and 2).
I tried both versions. While the divided distance gives better final path cost to the achieved goal, it visits more nodes => takes more time than the not divided one. What is the proper way to compute it? Which one should I use?
It is not obvious to me what an admissible heuristic function for this problem looks like, so I won't commit to saying, "Use the divided by two function." But I will tell you that the naive function you came up with is not admissible, and therefore will not give you good performance. In order for A* to work properly, the heuristic used must be admissible; in order to be admissible, the heuristic must absolutely always give an optimistic estimate. This one doesn't, for exactly the reason you highlight in your example.
(Although now that I think about it, dividing by two does seem like a reasonable way to force admissibility. I'm just not going to commit to it.)
Your heuristic is not admissible, so your A* is not guaranteed to find the optimal answer every time. An admissible heuristic must never overestimate the cost.
A better heuristic than dividing your heuristic cost by 3, would be: instead of adding the distance D of each letter to its final position, add ceil(D/2). This way, a letter 1 or 2 away, gets a 1 value, 3 or 4 away, gets a 2 value, an so on.
I'm looking for a fast solution for the following problem:
I have a fixed point (let's say the upper right on the white measurement line) and need to find the closest point on a curve made of equally spaced points (the lower curve). Additionally, I do this for every point on the upper curve to draw the distances between the curves with different colours (three levels: below minimum [red], between minimum and maximum [orange] and above maximum [green]).
My current solution is a tradeoff: I take the fixed point, iterate through an arbitrary interval (e. g. 50 units to the left and right of the fixed point) and calculate the distance of each pair. This saves some CPU power, but it is neither elegant nor accurate, since I could miss a minimum distance outside my chosen interval.
Any proposals for a faster algorithm?
Edit: Equally spaced means all points have the same distance on the x-axis, this is true for both curves. Also I do not need to interpolate between the points, this would be too time consuming.
Rather than an arbitrary distance, you could perhaps iterate until "out of range".
In your example, suppose you start with the point on the upper curve at the top-right of your line. Then drop vertically downwards, you get a distance of (by my eye) about 200um.
Now you can move right from here testing points until the horizontal distance is 200um. Beyond that, it's impossible to get a distance less than 200um.
Moving left, the distance goes down until you find the 150um minimum, then starts rising again. Once you're 150um to the left of your upper point, again, it's impossible to beat the minimum you've found.
If you'd gone left first, you wouldn't have had to go so far right, so as an optimization either follow the direction in which the distance falls, or else work out from the middle in both directions at once.
I don't know how many um 50 units is, so this might be slower or faster than what you have. It does avoid the risk of missing a lower value, though.
Since you're doing lots of tests against the same set of points on the lower curve, you can proably improve on this by ignoring the fact that the points form a curve at all. Stick them all in a k-d tree or similar, and search that repeatedly. It's called a Nearest neighbor search.
It may help to identify this problem as a nearest neighbour search problem. That link includes a good discussion about the various algorithms that are used for this. If you are OK with using C++ rather than straight C, ANN looks like a good library for this.
It also looks as though this question has been asked before.
We can label the top curve y=t(x) and the bottom curve y=b(x). Label the closest-function x_b=c(x_t). We know that the closest-function is weakly monotone non-decreasing as two shortest paths never cross each other.
If you know that the distance function d(x_t,x_b) has only one local minimum for every fixed x_t (this happens if the curve is "smooth enough"), then you can save time by "walking" the curve:
- start with x_t=0, x_b=0
- while x_t <= x_max
-- find the closest x_b by local search
(increment x_b while the distance is decreasing)
-- add {x_t, x_b} to the result set
-- increment x_t
If you expect x_b to be smooth enough, but you cannot assume that and you want an exact result,
Walk the curve in both directions. Where the results agree, they are correct. Where they disagree, run a complete search betwen the two results (the leftmost and the rightmost local maxima). Sample the "ambiguous block" in such an order (binary division) to allow the most pruning due to the monotonicity.
As a middle ground:
Walk the curve in both directions. If the results disagree, choose among the two. If you can guarantee at most two local maxima for each fixed x_t, this produces the optimal solution. There are still some pathological cases where the optimal solution is not found, and contain a local minimum that is flanked by two other local minima that are both worse than this one. I dare say it is uncommon to find a case where the solution is far from optimal (assuming smooth y=b(x)).
I have a simple machine learning question:
I have n (~110) elements, and a matrix of all the pairwise distances. I would like to choose the 10 elements that are most far apart. That is, I want to
Maximize:
Choose 10 different elements.
Return min distance over (all pairings within the 10).
My distance metric is symmetric and respects the triangle inequality.
What kind of algorithm can I use? My first instinct is to do the following:
Cluster the n elements into 20
clusters.
Replace each cluster with just the
element of that cluster that is
furthest from the mean element of
the original n.
Use brute force to solve the
problem on the remaining 20
candidates. Luckily, 20 choose 10 is
only 184,756.
Edit: thanks to etarion's insightful comment, changed "Return sum of (distances)" to "Return min distance" in the optimization problem statement.
Here's how you might approach this combinatorial optimization problem by taking the convex relaxation.
Let D be an upper triangular matrix with your distances on the upper triangle. I.e. where i < j, D_i,j is the distance between elements i and j. (Presumably, you'll have zeros on the diagonal, as well.)
Then your objective is to maximize x'*D*x, where x is binary valued with 10 elements set to 1 and the rest to 0. (Setting the ith entry in x to 1 is analogous to selecting the ith element as one of your 10 elements.)
The "standard" convex optimization thing to do with a combinatorial problem like this is to relax the constraints such that x need not be discrete valued. Doing so gives us the following problem:
maximize y'*D*y
subject to: 0 <= y_i <= 1 for all i, 1'*y = 10
This is (morally) a quadratic program. (If we replace D with D + D', it'll become a bona fide quadratic program and the y you get out should be no different.) You can use an off-the-shelf QP solver, or just plug it in to the convex optimization solver of your choice (e.g. cvx).
The y you get out need not be (and probably won't be) a binary vector, but you can convert the scalar values to discrete ones in a bunch of ways. (The simplest is probably to let x be 1 in the 10 entries where y_i is highest, but you might need to do something a little more complicated.) In any case, y'*D*y with the y you get out does give you an upper bound for the optimal value of x'*D*x, so if the x you construct from y has x'*D*x very close to y'*D*y, you can be pretty happy with your approximation.
Let me know if any of this is unclear, notation or otherwise.
Nice question.
I'm not sure if it can be solved exactly in an efficient manner, and your clustering based solution seems reasonable. Another direction to look at would be local search method such as simulated annealing and hill climbing.
Here's an obvious baseline I would compare any other solution against:
Repeat 100 times:
Greedily select the datapoint that whose removal decreases the objective function the least and remove it.
Given is an array of 320 elements (int16), which represent an audio signal (16-bit LPCM) of 20 ms duration. I am looking for a most simple and very fast method which should decide whether this array contains active audio (like speech or music), but not noise or silence. I don't need a very high quality of the decision, but it must be very fast.
It occurred to me first to add all squares or absolute values of the elements and compare their sum with a threshold, but such a method is very slow on my system, even if it is O(n).
You're not going to get much faster than a sum-of-squares approach.
One optimization that you may not be doing so far is to use a running total. That is, in each time step, instead of summing the squares of the last n samples, keep a running total and update that with the square of the most recent sample. To avoid your running total from growing and growing over time, add an exponential decay. In pseudocode:
decay_constant=0.999; // Some suitable value smaller than 1
total=0;
for t=1,...
// Exponential decay
total=total*decay_constant;
// Add in latest sample
total+=current_sample;
if total>threshold
// do something
end
end
Of course, you'll have to tune the decay constant and threshold to suit your application. If this isn't fast enough to run in real time, you have a seriously underpowered DSP...
You might try calculating two simple "statistics" - first would be spread (max-min). Silence will have very low spread. Second would be variety - divide the range of possible values into say 16 brackets (= value range) and as you go through the elements, determine in which bracket that element goes. Noise will have similar numbers for all brackets, whereas music or speech should prefer some of them while neglecting others.
This should be possible to do in just one pass through the array and you do not need complicated arithmetics, just some addition and comparison of values.
Also consider some approximation, for example take only each fourth value, thus reducing the number of checked elements to 80. For audio signal, this should be okay.
I did something like this a while back. After some experimentation I arrived at a solution that worked sufficiently well in my case.
I used the rate of change in the cube of the running average over about 120ms. When there is silence (only noise that is) the expression should be hovering around zero. As soon as the rate starts increasing over a couple of runs, you probably have some action going on.
rate = cur_avg^3 - prev_avg^3
I used a cube because the square just wasn't agressive enough. If the cube is to slow for you, try using the square and a bitshift instead. Hope this helps.
It is clearly that the complexity should be at least O(n). Probably some simple algorithms that calculate some value range are good for the moment but I would look for Voice Activity Detection on web and for related code samples.