I am using Insertion and Merge sort and trying to indicate the intersection point on the graph below. I found a possible solution on rosettacode, but a bit confusing for a newbie like me... Where should I look, do you guys know how to indicate it in Julia?
The output would is like
I am trying to show something like this
Your data has some noise, which means there might be more than one intersection point (all close). Assuming no noise, just two monotonically increasing functions, this should work:
using Plots
x = 1:5
y1 = x .+ 5
y2 = 3x
idx = argmin(abs.(y1-y2))
plot(x, [y1, y2], label=["y1", "y2"], markershape=:circle)
scatter!([x[idx]], [y1[idx]], label="intersection", markershape=:circle, markersize=10)
Note that this method gives you 1/4 possible points closest to the intersection. The intersection itself is not defined for functions only defined from discrete points. By changing the order of y1 and y2 in the code above you can get all 4 points. For fine enough discretization using one of those 4 points might be good enough for your needs.
A better approximation would be to use those 4 points to fit 2 lines, for y1 and y2, and find the intersection of those two lines analytically (e.g. with the equations y1=m1*x+b1, ...). But I leave this to you ;)
Related
I am back at college learning maths and I want to try and use some this knowledge to create some svg with d3.js.
If I have a function f(x) = x^3 - 3x^2 + 3x - 1
I would take the following steps:
Find the x intercepts for when y = 0
Find the y intercept when x = 0
Find the stationary points when dy\dx = 0
I would then have 2 x values from point 3 to plug into the original equation.
I would then draw a nature table do judge the flow of the graph or curve.
Plot the known points from the above and sketch the graph.
Translating what I would do on pen and paper into code instructions is what I really could do with any sort of advice on the following:
How can I programmatically factorise point 1 of the above to find the x-intercepts for when y = 0. I honestly do not know where to even start.
How would I programmatically find dy/dx and the values for the stationary points.
If I actually get this far then what should I use in d3 to join the points on the graph.
Your other "steps" have nothing to do with d3 or plotting.
Find the x intercepts for when y = 0
This is root finding. Look for algorithms to help with this.
Find the y intercept when x = 0
Easy: substitute to get y = 1.
Find the stationary points when dy\dx = 0
Take the first derivative to get 3x^2 - 12x + 9 and repeat the root finding step. Easy to get using quadratic equation.
I would then have 2 x values from point 3 to plug into the original
equation. I would then draw a nature table do judge the flow of the
graph or curve. Plot the known points from the above and sketch the
graph.
I would just draw the curve. Pick a range for x and go.
It's great to learn d3. You'll end up with something like this:
https://maurizzzio.github.io/function-plot/
For a cubic polynomial, there are closed formulas available to find all the particular points that you want (https://en.wikipedia.org/wiki/Cubic_function), and it is a sound approach to determine them.
Anyway, you will have to plot the smooth curve, which means that you will need to compute close enough points and draw a polyline that joins them.
Doing this, you are actually performing the first steps of numerical root isolation, with such an accuracy that the approximate and exact roots will be practically undistinguishable.
So an easy combined solution is to draw the curve as a polyline and find the intersections with the X axis as well as extrema using this polyline representation, rather than by means of more sophisticated methods.
This approach works for any continuous curve and is very easy to implement. So you actually draw the curve to find particular points rather than conversely as is done by analytical methods.
For best results on complicated curves, you can adapt the point density based on the local curvature, but this is another story.
I am working on a project of interpolating sample data {(x_i,y_i)} where the input domain for x_i locates in 4D space and output y_i locates in 3D space. I need generate two look up tables for both directions. I managed to generate the 4D -> 3D table. But the 3D -> 4D one is tricky. The sample data are not on regular grid points, and it is not one to one mapping. Is there any known method to treat this situation? I did some search online, but what I found is only for 3D -> 3D mapping, which are not suitable for this case. Thank you!
To answer the questions of Spektre:
X(3D) -> Y(4D) is the case 1X -> nY
I want to generate a table that for any given X, we can find the value for Y. The sample data is not occupy all the domain of X. But it's fine, we only need accuracy for point inside the domain of sample data. For example, we have sample data like {(x1,x2,x3) ->(y1,y2,y3,y4)}. It is possible we also have a sample data {(x1,x2,x3) -> (y1_1,y2_1,y3_1,y4_1)}. But it is OK. We need a table for any (a,b,c) in space X, it corresponds to ONE (e,f,g,h) in space Y. There might be more than one choice, but we only need one. (Sorry for the symbol confusing if any)
One possible way to deal with this: Since I have already established a smooth mapping from Y->X, I can use Newton's method or any other method to reverse search the point y for any given x. But it is not accurate enough, and time consuming. Because I need do search for each point in the table, and the error is the sum of the model error with the search error.
So I want to know it is possible to find a mapping directly to interpolate the sample data instead of doing such kind of search in 3.
You are looking for projections/mappings
as you mentioned you have projection X(3D) -> Y(4D) which is not one to one in your case so what case it is (1 X -> n Y) or (n X -> 1 Y) or (n X -> m Y) ?
you want to use look-up table
I assume you just want to generate all X for given Y the problem with non (1 to 1) mappings is that you can use lookup table only if it has
all valid points
or mapping has some geometric or mathematic symmetry (for example distance between points in X and Yspace is similar,and mapping is continuous)
You can not interpolate between generic mapped points so the question is what kind of mapping/projection you have in mind?
First the 1->1 projections/mappings interpolation
if your X->Y projection mapping is suitable for interpolation
then for 3D->4D use tri-linear interpolation. Find closest 8 points (each in its axis to form grid hypercube) and interpolate between them in all 4 dimensions
if your X<-Y projection mapping is suitable for interpolation
then for 4D->3D use quatro-linear interpolation. Find closest 16 points (each in its axis to form grid hypercube) and interpolate between them in all 3 dimensions.
Now what about 1->n or n->m projections/mappings
That solely depends on the projection/mapping properties which I know nothing of. Try to provide an example of your datasets and adding some image would be best.
[edit1] 1 X <- n Y
I still would use quatro-linear interpolation. You still will need to search your Y table but if you group it like 4D grid then it should be easy enough.
find 16 closest points in Y-table to your input Y point
These points should be the closest points to your Y in each +/- direction of all axises. In 3D it looks like this:
red point is your input Y point
blue points are the found closest points (grid) they do not need to be so symmetric as on image .
Please do not want me to draw 4D example that make sense :) (at least for sober mind)
interpolation
find corresponding X points. If there is more then one per point chose the closer one to the others ... Now you should have 16 X points and 16+1 Y points. Then from Y points you need just to calculate the distance along lines from your input Y point. These distances are used as parameter for linear interpolations. Normalize them to <0,1> where
0 means 'left' and 1 means 'right' point
0.5 means exact middle
You will need this scalar distance in each of Y-domain dimension. Now just compute all the X points along the linear interpolations until you get the corresponding red point in X-domain.
With tri-linear interpolation (3D) there are 4+2+1=7 linear interpolations (as on image). For quatro-linear interpolation (4D) there are 8+4+2+1=15 linear interpolations.
linear interpolation
X = X0 + (X1-X0)*t
X is interpolated point
X0,X1 are the 'left','right' points
t is the distance parameter <0,1>
I apologize in advance if my code looks very amateurish.
I'm trying to assign quadrants to 4 measurement stations approximately located on the edges of a town.
I have the coordinates of these 4 stations:
a <- c(13.2975,52.6556)
b <- c(14.0083,52.5583)
c <- c(13.3722,52.3997)
d <- c(12.7417,52.6917)
Now my idea was to create lines connecting the north-south and east-west stations:
line.1 <- matrix(c(d[1],b[1],d[2],b[2]),ncol=2)
line.2 <- matrix(c(a[1],c[1],a[2],c[2]),ncol=2)
Plotting all the stations the connecting lines looks allright, however not very helpful for analyzing it on a computer.
So I calculated the eucledian vectors for the two lines:
vec.1 <- as.vector(c((b[1]-d[1]),(b[2]-d[2])))
vec.2 <- as.vector(c((c[1]-a[1]),(c[2]-a[2])))
which allowed me to calculate the angle between the two lines in degrees:
alpha <- acos((vec.1%*%vec.2) / (sqrt(vec.1[1]^2+vec.1[2]^2)*
sqrt(vec.2[1]^2+vec.2[2]^2)))) * 180/pi
The angle I get for alpha is 67.7146°. This looks fairly good. From this angle I can easily calculate the other 3 angles of the intersection, however I need values relative to the grid so I can assign values from 0°-360° for the wind directions.
Now my next planned step was to find the point where the two lines intersect, add a horizontal and vertical abline through that point and then calculate the angle relative to the grid. However I can't find a proper example that does that and I don't think I have a nice linear equation system I could solve.
Is my code way off? Or maybe anyone knows of a package which could help me? It feels like my whole approach is a bit wrong.
Okay I managed to calculate the intersection point, using line equations. Here is how.
The basic equation for two points is like this:
y - y_1 = (y_2-y_1/x_2-x_1) * (x-x_1)
If you make one for each of the two lines, you can just substitute the fractions.
k.1 <- ((c[2]-a[2])/(c[1]-a[1]))
k.2 <- ((b[2]-d[2])/(b[1]-d[1]))
Reshaping the two functions you get a final form for y:
y <- (((-k.1/k.2)*d[2]+k.1*d[1]-k.1*c[1]+d[2])/(1-k.1/k.2))
This one you can now use to calculate the x-value:
x <- ((y-d[2])+d[1]*k.2)/k.2
In my case I get
y = 52.62319
x = 13.3922
I'm starting to really enjoy this program!
Wikipedia has a good article on finding the intersection between two line segments with an explicit formula. However, you don't need to know the point of intersection to calculate the angle to the grid (or axes of coordinate system.) Just compute the angles from your vec.1 and vec.2 to the basis vectors:
e1 <- c(1, 0)
e2 <- c(0, 1)
as you have done.
t = 0:%pi/50:10*%pi;
plot3d(sin(t),cos(t),t)
When I execute this code the plot is done but the line is not visible, only the box. Any ideas which property I have to change?
Thanks
The third argument should, in this case, be a matrix of the size (length arg1) x (length arg2).
You'd expect plot3d to behave like an extension of plot and plot2d but it isn't quite the case.
The 2d plot takes a vector of x and a vector of y and plots points at (x1,y1), (x2,y2) etc., joined with lines or not as per style settings. That fits the conceptual model we usually use for 2d plots - charting the relationship of one thing as a function of another, in most cases (y = f(x)). THere are other ways to use a 2d plot: scatter graphs are common but it's easy enough to produce one using the two-rows-of-data concept.
This doesn't extend smoothly to 3d though as there are many other ways you could use a 3d plot to represent data. If you gave it three vectors of coordinates and asked it to draw a line between them all what might we want to use that for? Is that the most useful way of using a 3d plot?
Most packages give you different visualisation types for the different kinds of data. Mathematica has a lot of 3d visualisation types and Python/Scipy/Mayavi2 has even more. Matlab has a number too but Scilab, while normally mirroring Matlab, in this case prefers to handle it all with the plot3d function.
I think of it like a contour plot: you give it a vector of x and a vector of y and it uses those to create a grid of (x,y) points. The third argument is then a matrix whose dimensions match those of the (x,y) grid holding the z-coordinates of each point. The first example in the docs does what I think you're after:
t=[0:0.3:2*%pi]';
z=sin(t)*cos(t');
plot3d(t,t,z);
The first line creates a column vector of length 21
-->size(t)
ans =
21. 1.
The second line computes a 21 x 21 matrix of products of the permutations of sin(t) with cos(t) - note the transpose in the cos(t') element.
-->size(z)
ans =
21. 21.
Then when it plots them it draws (x1,y1,z11), (x1,y2,x12), (x2,y2,z22) and so on. It draws lines between adjacent points in a mesh, or no lines, or just the surface.
I am capturing some points on an X, Y plane to represent some values. I ask the user to select a few points and I want the system to then generate a curve following the trend that the user creates. How do I calculate this? So say it is this:
Y = dollar amount
X = unit count
user input: (2500, 200), (4500, 500), (9500, 1000)
Is there a way I can calculate some sort of curve to follow those points so I would know based off that selection what Y = 100 would be on the same scale/trend?
EDIT: People keep asking for the nature of the curve, yes logarithmic. But I'd also like to check out some other options. It's for pricing the the restraint is that the as X increases Y should always be higher. However the rate of change of the curve should change related to the two adjacent points that the user selected, we could probably require a certain number of points. Does that help?
EDIT: Math is hard.
EDIT: Maybe a parabola then?
The problem is that there are multiple curves that you can fit to the same data. To borrow an example from my old stats book, here is the same data set (1, 1, 1, 10, 1, 1, 1) with four curves:
You need to specify the overall trend to get a meaningful result.
First, you are going to have to have an idea of what your line is or better said, what type of line fits your data the best. Is it linear (straight line) or does it curve (x-squared). Sounds like this is a curve.
If your curve is a parabola, then you will need to solve y = Ax(2) + Bx + c using your three points that the user has chosen. You will need at least 3 points to solve for 3 unknowns.
200 = A(2500)(2) + B(2500) + C
500 = A(4500)(2) + B(4500) + C
1000 = A(9500)(2) + B(9500) + C
Given these three equations, you should be able to solve for A, B and C, then use these to plot a new curve.
The Least Square Fit would give you a nice data matching curve.
This is a rather general extrapolation problem. In your case, fitting a quadric (parabola) is probably the most reasonable course of action. Depending on how well your data fits a quadric, you may want to fit it to more than 3 points (the noisier and weirder the data, the more points you should use).
Depending on the amount and type of data you have, you may want to try LOESS regression.
However, this may not be a good option if you only have 3 points as in your example (but keep in mind that you will not be able to have good extrapolation with 3 points no matter the algorithm you use)
Another option would be B-splines