I want to get the point "Search".
All other points are known.
A, B, C, D are always Z=0
F, E, G, H 's Z could be anything above 0 (Z-Axis is representing a surface)
"I" is also known (X, Y, Z=0)
"Search" = (I.X, I.Y and Z =?)
So I need something to calculate "Search.Z"
I guess there is already something like this in 3D libraries.
Roughly speaking, the choice of interpolation method depends on whether your data fall on a regular grid or not.
If you have a regular grid then you should use recursive interpolation. From your 8 points you would do 4 interpolations, say along the y axis. From the resulting rectangle you would do 2 interpolations, say along the x axis. Then you would do an interpolation along the z axis. If you just have the 8 points you describe then you can only realistically perform linear interpolation (unless you have some other constraints like known values of the slope at the edges). If you have more points [at least 4 for each desired dimension] then you can use cubic interpolation.
If you have an irregular grid then there are several techniques.
Related
I get the positions of 5 enemies in the game in vectors. Depending on the distance I choose, the number of enemies can vary from 0 to 5. I need to know their vectors each time to check whether it is possible to draw a straight line through a certain number of heroes (vectors).
After that, my hero will have to use his ability called wall. It consists of 2 start and end vectors. Thus, check whether my hero can put a wall on the enemies in the line to catch them
Let's say there are 3 enemy heroes whose positions I can get. I need to find out if I can pass through them directly, in order to use the ability on them.
Here's what using the ability looks like in the game
Here is getting the vector of one of the heroes
The ability itself can be twisted at a certain point. But anyway, it is necessary that the wall would touch several heroes
Wherever I move the mouse, I can put it in the desired position. But unfortunately it takes a lot of time, so I would like to automate
The coordinates of the wall itself, or rather its two edges, I can also get, but only after the ability has been used
If one prefers geometry to linear algebra...
Then One can compute the dot product of (unit-vector1. Unit-Vector2). That is equal to the SIN of the angle between them.
So if unit vector is the shooter position to target1, unit vector2 is the shooter to target2, etc... then when DOTPRODUCT(Vector1,vector2) = 1 and DOTPRODUCT(Vector1,vector3) = 1, then the three points are in syzygy.
And repeat from shooter to as many targets as you have to determine whether some or all of the points are in syzygy.
From your statement that there is a start and an endpoint I take that you select two enemys and want to trap anything in between.
So you're actually not looking for a straight line that can be drawn through your enemy positions but if they are withn a rectangle. It would be very unlikely and for more points nearly impossible that they are all collinear anyway.
So it becomes quite trivial. You draw a line through start and end enemy. Then you check the remaining enemies distance to that line vs the width of your AoE. Maye you want to also handle some body width in that calculation.
https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
You can describe all points belonging to line (x0, y0) + (dx, dy)t = (x1, y1). Chose any two points and t as 1 and you will get (dx, dy) for line connecting two dots. Now you will need to find distance between this line and (x2, y2). it is distance between (x2, y2) and (xd, yd), where on one hand (xd, yd) = (x0, y0) + t1(dx, dy) and on other hand (xd, yd) = (x2, y2) + t2*(-dy, dx). Solving this two equations you will find t1, t2, (xd, yd) and distance between (x2, y2) and (xd, yd), which is distance between (x2, y2) and line, connecting (x0, y0) and (x1, y1).
Knowing this, you select dots with min_x and max_x and calculate ditance between line, connecting said dots and rest of the dots. If distance is lesser then some threshold of your choice, then you can assume that you can have line passing through all dots.
Any line in the plane can be described by an equation a*x + b*y + c = 0 with (a, b) ≠ (0, 0). Note that if you have an equation of this form, then multiplying each coefficient a, b, c with the same number yields an equation describing the same line. That's the reason (a, b, c) is called a homogeneous coordinate vector for that line.
How do you find a, b, c? One simple approach would be treating this as three linear equations in three unknowns. You plug in the x and y coordinates for all your three points, and get tree equations for a thorough c. However, there is a catch. Since the right hand side of each equation is zero, a = b = c = 0 is always a solution. In those cases where there is only one solution, that will be it. So in order for there to be a line, you need more than one solution. The mathematical tool to determine whether a set of equations had more than one solution is the determinant. It is zero if the system has no single unique solution.
Long story short: three points are collinear (on a line) if
⎛x1 y1 1⎞
det ⎜x2 y2 1⎟ = 0
⎝x3 y3 1⎠
The homogeneous coordinate vector describing the line world correspond to the kernel of that matrix.
Of course, if your input coordinates are floating point numbers, exact zero is unlikely. Presumably that wall does allow for some error in some way, and you'd need to tell us about that in order to get an answer that models this aspect correctly. In the mean time, know that the absolute value of the determinant above is proportional to the area of the triangle created by these three points. So if your were to pick a constant threshold value, the farther your enemies are apart along the direction of the wall, the less they could deviate from the straight line without violating that threshold.
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>
Given a point p exterior to an axially aligned, origin centered ellipse E, find the (upto) four unique normals to E passing through p.
This is not a Mathematica question. Direct computation is too slow; I am willing to sacrifice precision and accuracy for speed.
I have searched the web, but all I found involved overly complex calculations which if implemented directly appear to lack the performance I need. Is there a more "programmatical" way to do this, like using matrices or scaling the ellipse into a circle?
Let's assume the ellipse E is in "standard position", center at the origin and axes parallel to the coordinate axes:
(x/a)^2 + (y/b)^2 = 1 where a > b > 0
The boundary cases a=b are circles, where the normal lines are simply ones that pass through the center (origin) and are thus easy to find. So we omit discussion of these cases.
The slope of the tangent to the ellipse at any point (x,y) may be found by implicit differentiation:
dy/dx = -(b^2 x)/(a^2 y)
For the line passing through (x,y) and a specified point p = (u,v) not on the ellipse, that is normal to ellipse E when its slope is the negative reciprocal of dy/dx:
(y-v)/(x-u) * (-b^2 x)/(a^2 y) = -1 (N)
which simplifies to:
(x - (1+g)u) * (y + gv) = -g(1+g)uv where g = b^2/(a^2 - b^2)
In this form we recognize it is the equation for a right rectangular hyperbola. Depending on how many points of intersection there are between the ellipse and the hyperbola (2,3,4), we have that many normals to E passing through p.
By reflected symmetry, if p is assumed exterior to E, we may take p to be in the first quadrant:
(u/a)^2 + (v/b)^2 > 1 (exterior to E)
u,v > 0 (1'st quadrant)
We could have boundary cases where u=0 or v=0, i.e. point p lies on an axis of E, but these cases may be reduced to solving a quadratic, because two normals are the (coinciding) lines through the endpoints of that axis. We defer further discussion of these special cases for the moment.
Here's an illustration with a=u=5,b=v=3 in which only one branch of the hyperbola intersects E, and there will be only two normals:
If the system of two equations in two unknowns (x,y) is reduced to one equation in one unknown, the simplest root-finding method to code is a bisection method, but knowing something about the possible locations of roots/intersections will expedite our search. The intersection in the first quadrant is the nearest point of E to p, and likewise the intersection in the third quadrant is the farthest point of E from p. If the point p were a good bit closer to the upper endpoint of the minor axis, the branches of the hyperbola would shift together enough to create up to two more points of intersection in the fourth quadrant.
One approach would be to parameterize E by points of intersection with the x-axis. The lines from p normal to the ellipse must intersect the major axis which is a finite interval [-a,+a]. We can test both the upper and lower points of intersection q=(x,y) of a line passing through p=(u,v) and (z,0) as z sweeps from -a to +a, looking for places where the ellipse and hyperbola intersect.
In more detail:
1. Find the upper and lower points `q` of intersection of E with the
line through `p` and `(z,0)` (amounts to solving a quadratic)
3. Check the sign of a^2 y(x-u) - b^2 x(y-v) at `q=(x,y)`, because it
is zero if and only `q` is a point of normal intersection
Once a subinterval is detected (either for upper or lower portion) where the sign changes, it can be refined to get the desired accuracy. If only modest accuracy is needed, there may be no need to use faster root finding methods, but even if they are needed, having a short subinterval that isolates a root (or root pair in the fourth quadrant) will be useful.
** more to come comparing convergence of various methods **
I had to solve a problem similar to this, for GPS initialization. The question is: what is the latitude of a point interior to the Earth, especially near the center, and is it single-valued? There are lots of methods for converting ECEF cartesian coordinates to geodetic latitude, longitude and altitude (look up "ECEF to Geodetic"). We use a fast one with only one divide and sqrt per iteration, instead of several trig evaluations like most methods, but since I can't find it in the wild, I can't give it to you here. I would start with Lin and Wang's method, since it only uses divisions in its iterations. Here is a plot of the ellipsoid surface normals to points within 100 km of Earth's center (North is up in the diagram, which is really ECEF Z, not Y):
The star-shaped "caustic" in the figure center traces the center of curvature of the WGS-84 ellipsoid as latitude is varied from pole to equator. Note that the center of curvature at the poles is on the opposite side of the equator, due to polar flattening, and that the center of curvature at the equator is nearer to the surface than the axis of rotation.
Wherever lines cross, there is more than one latitude for that cartesian position. The green circle shows where our algorithm was struggling. If you consider that I cut off these normal vectors where they reach the axis, you would have even more normals for a given position for the problem considered in this SO thread. You would have 4 latitudes / normals inside the caustic, and 2 outside.
The problem can be expressed as the solution of a cubic equation which
gives 1, 2, or 3 real roots. For the derivation and closed form
solution see Appendix B of Geodesics on an ellipsoid of revolution. The boundary between 1 and 3 solutions is an astroid.
Imagine you have a grid of sample points of a function z = f(x, y) where 1 < x < N and 1 < y < N. The formula is not given, but just the raw data, that could be for example the grey level of an image.
I would like to find, given a point A, whose x and y coordinates are given (and z is known from the data, so A is a vertex of the surface) a number M of points that lie on the circumference of the circle with center in A and radius R that are a good approximation of a circular "cloth" draped on the imaginary surface described by the data points. Imagine also that the edges of the surface are a triangle mesh.
The biggest constraint in the approximation is that the sum of the length of the edges of the resulting polygon is constantly R * 2 * PI, so that moving the A point across the surface would just change the M points but never the sum of their reciprocal distances. The draping doesn't need to be perfect, it would be nice though to be as close as possible to the surface., or always on one side of the surface, above or below.
Could anybody give me a pointer to something to read about this? Is this a known problem?
I feel that the problem is not completely formulated, I'd already like some help to give a complete description of it.