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So I'm attempting to make a wall system like the Sims. While it is 3D, it can be simplified to 2D (think a bird's eye view). A wall is defined by two points: a starting point an an ending point and the walls themselves have directions and can be simplified to unit vectors. Walls are also made with rectangular prisms, each of these prisms start at one point and end at another. So, bearing this in mind, sometimes walls have to be extended, otherwise there will be a small gap in the convex corner of the wall
if one of the points is shared between the walls. Any number of walls can be placed in a single position, so all of this in mind, a grid point's data could contain something like this (using Lua), each table represents a vector2, using a table for ease of visualization, this code won't actually work:
local gridPoint = {
{x = 1, y = 0},
{x = 0.7071, y = 0.7071},
{x = 0, y = 1},
{x = 0.7071, y = -0.7071},
}
Here's a visual of what the points might look like, where the intersection is the grid point and the walls are the vectors extending from the intersection, assuming each of the vectors have a length of 1 (excuse my terrible drawing skills). Since dot products flip when greater than 180 degrees, the largest angle would always be <=180.
So now, to get the angle between any two of points, we can just do math.deg(math.acos(v1:dot(v2))) to get the angle between these two points in degrees, where v1 is vector 1, v2 is vector 2, and dot is a function that returns the dot product of v1 and v2.
So up to this point, everything is fine. The issue is that I have to create two loops which goes through every single combination of possible dot products which I'm not sure is the best method of finding the largest angle, this is #gridPoint^2 possible combinations which is fine at lower numbers, but this number of possible combination increases exponentially with each new wall.
Is there an easier way of doing this?
This problem is equivalent to finding of the farthest pair of points, where all the points lie at unit circle.
Sort points - it is simpler to separate points with positive y (angles 0..Pi) into the first list, and points with negative y into the second list,sort them by X-coordinate, then join the first list and reversed second one. This stage takes O(nlogn) time
Then perform searching of the farthest points - fix the first point index (A[i]), walk through the list until squared distance A[i]-A[j+1] becomes less than squared distance A[i]-A[j]. So we find the farthest point A[j] for A[i]. Now increment i and search the farthest point for it - starting from j.
Repeat until j exceeds n.
This stage is linear because we move i and j only in forward direction. Essentially this is a method of rotating calipers, so you can get some implementation elsewhere.
Can anyone figure out a function that can perform a mapping from a finite set of N numbers X = {x0, x1, x2, ..., xN} where each x can be valued 0 to 999999999 and N < 999999999, to a set Y = {0, 1, 2, 3, ..., N}.
In my case, i have about 24000000 element in the first set whose values can range as X. This elements have continuous block (for example 53000 to 1234500, then 8000000 to 9000000 and so on) and i have to remap this elements from 0 to 2400000. I don't require to maintain order.
I need a (possibly simple and rapid) math function, or a bitwise transformation, not something like put it ordered into an array and then binary search for their position.
Really thank to whom that can figure out a way to solve this!
Luca
If you don't want to keep some gigabytes of straight map, then augmented segment tree is reasonable approach. Tree should contain intervals and shift of every interval (sum of left intervals). Of course, finding appropriate interval (and shift) in this method is close to the binary search.
For example, you get X=80000015. Find interval for this value - it is 8000000 to 9000000. Rank of this interval is 175501 (1234500-53000 + 1). So X maps to
X => 175501 + 80000015 - 80000000 = 175516
For sparse elements make counting stage - find what is rank R for every number M and put (key=M, value=R) pair in hash table.
X = (3, 19, 20, 101)
table: [(3:0), (19:1), (20:2), (101:3)]
Note that one should keep balance between speed and space - for long filled intervals it is better to store only interval ends.
I have started to develop unit trajectories for a game server and for now I'm trying to retrieve the position of a unit at a given time. It is easy when the trajectory is just a straight line, but it is far more complicated when unit chases another unit.
I've done flash app to illustrate the problem. Black trajectory is for unit which travels in a single direction. Blue chases black and red chases blue. What I want is to precalculate whole trajectory for blue and red to be able to retrieve their position in a constant time.
Is it possible? Thanks for any help!!
Here's a paper A classic chase problem solved from a
physics perspective by Carl E. Mungan that solves a particular version in which the chaser is initially perpendicular to the chased object's trajectory. I believe this is an inessential element of the solution since that perpendicularity disappears along the rest of the trajectory.
It is an autonomous system of differential equations in the sense that time does not appear explicitly in the coefficients or terms of the equations. This supports the idea that the family of solutions given in the paper is general enough to provide for non-perpendicular initial conditions.
The paper provides further links and references, as well as a useful search term, "curves of pursuit".
Let's state a slight different, slightly more general initial condition than Mungan's. Suppose the chased object ("ship") is initially located at the origin and travels up the positive y-axis (x=0) with constant speed V. The chasing object ("torpedo") is initially located at (x0,y0), and although instantaneous reorienting directly at the "ship", also travels at some constant speed v.
The special case where x0 is zero results in a linear pursuit curve, i.e. a head-on collision or a trailing chase accordingly as y0 is positive or negative. Otherwise by reflection in the y-axis one may assume without loss of generality that x0 > 0. Thus rational powers of x-coordinates will be well-defined.
Assume for our immediate purpose that speeds V,v are unequal, so that ratio r = V/v is not 1. The following is a closed-form solution (1) for the "torpedo" curve similar to Mungan's equation (10):
(1+r) (1-r)
[ (x/H) (x/H) ]
(y/H) = (1/2) [ ----- - ----- ] + C (1)
[ (1+r) (1-r) ]
in which the constants H,C can be determined by the initial conditions.
Applying the condition that initially the torpedo moves toward the ship's location at the origin, we take the derivative with respect to x in (1) and cancel a factor 1/H from both sides:
r -r
dy/dx = (1/2) [ (x/H) - (x/H) ] (2)
Now equate the curve's slope dy/dx at initial point (x0,y0) with that of its line passing through the origin:
r -r
(x0/H) - (x0/H) = 2y0/x0 = K (3)
This amounts to a quadratic equation for positive B = (x0/H)^r:
B^2 - K*B - 1 = 0 (4)
namely B = [K + sqrt(K^2 + 4)]/2 (but use the alternative form if K < 0 to avoid cancellation error), which allows H to be determined from our knowledge of x0 and r:
H = x0/(B^(1/r)) (5)
Knowing H makes it a simple matter to determine the additive constant C in (1) by substituting the initial point (x0,y0) there.
The tricky part will be to determine which point on the "torpedo" trajectory corresponds to a given time t > 0. The inverse of that problem is solved fairly simply. Given a point on the trajectory, find the tangent line at that point using derivative formula (2) and deduce time t from the y-intercept b of that line (i.e. from the current "ship" position):
t = b/V (6)
Therefore determining (x(t),y(t)) where the "torpedo" is located at a given time t > 0 is essentially a root-finding exercise. One readily brackets the desired x(t) between two x-coordinates x1 and x2 that correspond to times t1 and t2 such that t1 < t < t2. A root-finding method can be used to refine this interval until the desired accuracy is achieved. Once a fairly small interval has been refined, Newton's method will provide rapid convergence. We can look at the details of such a procedure in a next installment!
I can set up the problem for you but not solve it.
The black curve is moving at a constant velocity v0, and in a straight line.
The blue curve moves at a constant velocity v1 in the direction of black.
For simplicity, choose coordinates so that at time t=0 the black curve starts at (x=0, y=0) and is moving in the direction x.
Thus, at time t >= 0, the position of the black curve is (v0 t, 0).
Problem statement
The goal is to find x, y of the blue curve for times t >= 0 given the initial position (x(t=0), y(t=0)). The differential equations of motion are
dx / dt = v1 (v0 t - x) / a(t)
dy / dt = v1 (- y) / a(t)
where a(t) = sqrt((v0 t - x)^2 + (y^2)) is the distance between blue and black at time t.
This is a system of two nonlinear coupled differential equations. It seems likely that there is no complete anaytical solution. Wolfram Alpha gives up without trying for the input
D[y[t],t] = -y[t] / sqrt[(t-x[t])^2 + y[t]^2], D[x[t],t] = (t-x[t]) / sqrt[(t-x[t])^2 + y[t]^2]
You could try asking on math.stackexchange. Good luck!
I am new to this forum and not a native english speaker, so please be nice! :)
Here is the challenge I face at the moment:
I want to calculate the (approximate) relative coordinates of yet unknown points in a 3D euclidean space based on a set of given distances between 2 points.
In my first approach I want to ignore possible multiple solutions, just taking the first one by random.
e.g.:
given set of distances: (I think its creating a pyramid with a right-angled triangle as a base)
P1-P2-Distance
1-2-30
2-3-40
1-3-50
1-4-60
2-4-60
3-4-60
Step1:
Now, how do I calculate the relative coordinates for those points?
I figured that the first point goes to 0,0,0 so the second one is 30,0,0.
After that the third points can be calculated by finding the crossing of the 2 circles from points 1 and 2 with their distances to point 3 (50 and 40 respectively). How do I do that mathematically? (though I took these simple numbers for an easy representation of the situation in my mind). Besides I do not know how to get to the answer in a correct mathematical way the third point is at 30,40,0 (or 30,0,40 but i will ignore that).
But getting the fourth point is not as easy as that. I thought I have to use 3 spheres in calculate the crossing to get the point, but how do I do that?
Step2:
After I figured out how to calculate this "simple" example I want to use more unknown points... For each point there is minimum 1 given distance to another point to "link" it to the others. If the coords can not be calculated because of its degrees of freedom I want to ignore all possibilities except one I choose randomly, but with respect to the known distances.
Step3:
Now the final stage should be this: Each measured distance is a bit incorrect due to real life situation. So if there are more then 1 distances for a given pair of points the distances are averaged. But due to the imprecise distances there can be a difficulty when determining the exact (relative) location of a point. So I want to average the different possible locations to the "optimal" one.
Can you help me going through my challenge step by step?
You need to use trigonometry - specifically, the 'cosine rule'. This will give you the angles of the triangle, which lets you solve the 3rd and 4th points.
The rules states that
c^2 = a^2 + b^2 - 2abCosC
where a, b and c are the lengths of the sides, and C is the angle opposite side c.
In your case, we want the angle between 1-2 and 1-3 - the angle between the two lines crossing at (0,0,0). It's going to be 90 degrees because you have the 3-4-5 triangle, but let's prove:
50^2 = 30^2 + 40^2 - 2*30*40*CosC
CosC = 0
C = 90 degrees
This is the angle between the lines (0,0,0)-(30,0,0) and (0,0,0)- point 3; extend along that line the length of side 1-3 (which is 50) and you'll get your second point (0,50,0).
Finding your 4th point is slightly trickier. The most straightforward algorithm that I can think of is to firstly find the (x,y) component of the point, and from there the z component is straightforward using Pythagoras'.
Consider that there is a point on the (x,y,0) plane which sits directly 'below' your point 4 - call this point 5. You can now create 3 right-angled triangles 1-5-4, 2-5-4, and 3-5-4.
You know the lengths of 1-4, 2-4 and 3-4. Because these are right triangles, the ratio 1-4 : 2-4 : 3-4 is equal to 1-5 : 2-5 : 3-5. Find the point 5 using trigonometric methods - the 'sine rule' will give you the angles between 1-2 & 1-4, 2-1 and 2-4 etc.
The 'sine rule' states that (in a right triangle)
a / SinA = b / SinB = c / SinC
So for triangle 1-2-4, although you don't know lengths 1-4 and 2-4, you do know the ratio 1-4 : 2-4. Similarly you know the ratios 2-4 : 3-4 and 1-4 : 3-4 in the other triangles.
I'll leave you to solve point 4. Once you have this point, you can easily solve the z component of 4 using pythagoras' - you'll have the sides 1-4, 1-5 and the length 4-5 will be the z component.
I'll initially assume you know the distances between all pairs of points.
As you say, you can choose one point (A) as the origin, orient a second point (B) along the x-axis, and place a third point (C) along the xy-plane. You can solve for the coordinates of C as follows:
given: distances ab, ac, bc
assume
A = (0,0)
B = (ab,0)
C = (x,y) <- solve for x and y, where:
ac^2 = (A-C)^2 = (0-x)^2 + (0-y)^2 = x^2 + y^2
bc^2 = (B-C)^2 = (ab-x)^2 + (0-y)^2 = ab^2 - 2*ab*x + x^2 + y^2
-> bc^2 - ac^2 = ab^2 - 2*ab*x
-> x = (ab^2 + ac^2 - bc^2)/2*ab
-> y = +/- sqrt(ac^2 - x^2)
For this to work accurately, you will want to avoid cases where the points {A,B,C} are in a straight line, or close to it.
Solving for additional points in 3-space is similar -- you can expand the Pythagorean formula for the distance, cancel the quadratic elements, and solve the resulting linear system. However, this does not directly help you with your steps 2 and 3...
Unfortunately, I don't know a well-behaved exact solution for steps 2 and 3, either. Your overall problem will generally be both over-constrained (due to conflicting noisy distances) and under-constrained (due to missing distances).
You could try an iterative solver: start with a random placement of all your points, compare the current distances with the given ones, and use that to adjust your points in such a way as to improve the match. This is an optimization technique, so I would look up books on numerical optimization.
If you know the distance between the nodes (fixed part of system) and the distance to the tag (mobile) you can use trilateration to find the x,y postion.
I have done this using the Nanotron radio modules which have a ranging capability.
I am writing a physics engine/simulator which incorporates 3D space flight, planetary/stellar gravitation, ship thrust and relativistic effects. So far, it is going very well, however, one thing that I need help with is the math of the collision detection algorithm.
The iterative simulation of movement that I am using is basically as follows:
(Note: 3D Vectors are ALL CAPS.)
For each obj
obj.ACC = Sum(all acceleration influences)
obj.POS = obj.POS + (obj.VEL * dT) + (obj.ACC * dT^2)/2 (*EQ.2*)
obj.VEL = obj.VEL + (obj.ACC * dT)
Next
Where:
obj.ACC is the acceleration vector of the object
obj.POS is the position or location vector of the object
obj.VEL is the velocity vector of the object
obj.Radius is the radius (scalar) of the object
dT is the time delta or increment
What I basically need to do is to find some efficient formula that derives from (EQ.2) above for two objects (obj1, obj2) and tell if they ever collide, and if so, at what time. I need the exact time both so that I can determine if it is in this particular time increment (because acceleration will be different at different time increments) and also so that I can locate the exact position (which I know how to do, given the time)
For this engine, I am modelling all objects as spheres, all this formula/algorithm needs to do is to figure out at what points:
(obj1.POS - obj2.POS).Distance = (obj1.Radius + obj2.Radius)
where .Distance is a positive scalar value. (You can also square both sides if this is easier, to avoid the square root function implicit in the .Distance calculation).
(yes, I am aware of many, many other collision detection questions, however, their solutions all seem to be very particular to their engine and assumptions, and none appear to match my conditions: 3D, spheres, and acceleration applied within the simulation increments. Let me know if I am wrong.)
Some Clarifications:
1) It is not sufficient for me to check for Intersection of the two spheres before and after the time increment. In many cases their velocities and position changes will far exceed their radii.
2) RE: efficiency, I do not need help (at this point anyway) with respect to determine likely candidates for collisions, I think that I have that covered.
Another clarification, which seems to be coming up a lot:
3) My equation (EQ.2) of incremental movement is a quadratic equation that applies both Velocity and Acceleration:
obj.POS = obj.POS + (obj.VEL * dT) + (obj.ACC * dT^2)/2
In the physics engines that I have seen, (and certainly every game engine that I ever heard of) only linear equations of incremental movement that apply only Velocity:
obj.POS = obj.POS + (obj.VEL * dT)
This is why I cannot use the commonly published solutions for collision detection found on StackOverflow, on Wikipedia and all over the Web, such as finding the intersection/closest approach of two line segments. My simulation deals with variable accelerations that are fundamental to the results, so what I need is the intersection/closest approach of two parabolic segments.
On the webpage AShelley referred to, the Closest Point of Approach method is developed for the case of two objects moving at constant velocity. However, I believe the same vector-calculus method can be used to derive a result in the case of two objects both moving with constant non-zero acceleration (quadratic time dependence).
In this case, the time derivative of the distance-squared function is 3rd order (cubic) instead of 1st order. Therefore there will be 3 solutions to the Time of Closest Approach, which is not surprising since the path of both objects is curved so multiple intersections are possible. For this application, you would probably want to use the earliest value of t which is within the interval defined by the current simulation step (if such a time exists).
I worked out the derivative equation which should give the times of closest approach:
0 = |D_ACC|^2 * t^3 + 3 * dot(D_ACC, D_VEL) * t^2 + 2 * [ |D_VEL|^2 + dot(D_POS, D_ACC) ] * t + 2 * dot(D_POS, D_VEL)
where:
D_ACC = ob1.ACC-obj2.ACC
D_VEL = ob1.VEL-obj2.VEL (before update)
D_POS = ob1.POS-obj2.POS (also before update)
and dot(A, B) = A.x*B.x + A.y*B.y + A.z*B.z
(Note that the square of the magnitude |A|^2 can be computed using dot(A, A))
To solve this for t, you'll probably need to use formulas like the ones found on Wikipedia.
Of course, this will only give you the moment of closest approach. You will need to test the distance at this moment (using something like Eq. 2). If it is greater than (obj1.Radius + obj2.Radius), it can be disregarded (i.e. no collision). However, if the distance is less, that means the spheres collide before this moment. You could then use an iterative search to test the distance at earlier times. It might also be possible to come up with another (even more complicated) derivation which takes the size into account, or possible to find some other analytic solution, without resorting to iterative solving.
Edit: because of the higher order, some of the solutions to the equation are actually moments of farthest separation. I believe in all cases either 1 of the 3 solutions or 2 of the 3 solutions will be a time of farthest separation. You can test analytically whether you're at a min or a max by evaluating the second derivative with respect to time (at the values of t which you found by setting the first derivative to zero):
D''(t) = 3 * |D_ACC|^2 * t^2 + 6 * dot(D_ACC, D_VEL) * t + 2 * [ |D_VEL|^2 + dot(D_POS, D_ACC) ]
If the second derivative evaluates to a positive number, then you know the distance is at a minimum, not a maximum, for the given time t.
Draw a line between the start location and end location of each sphere. If the resulting line segments intersect the spheres definitely collided at some point and some clever math can find at what time the collision occurred. Also make sure to check if the minimum distance between the segments (if they don't intersect) is ever less than 2*radius. This will also indicate a collision.
From there you can backstep your delta time to happen exactly at collision so you can correctly calculate the forces.
Have you considered using a physics library which already does this work? Many libraries use far more advanced and more stable (better integrators) systems for solving the systems of equations you're working with. Bullet Physics comes to mind.
op asked for time of collision. A slightly different approach will compute it exactly...
Remember that the position projection equation is:
NEW_POS=POS+VEL*t+(ACC*t^2)/2
If we replace POS with D_POS=POS_A-POS_B, VEL with D_VEL=VEL_A-VEL_B, and ACC=ACC_A-ACC_B for objects A and B we get:
$D_NEW_POS=D_POS+D_VEL*t+(D_ACC*t^2)/2
This is the formula for vectored distance between the objects. In order to get the squared scalar distance between them, we can take the square of this equation, which after expansion looks like:
distsq(t) = D_POS^2+2*dot(D_POS,D_VEL)*t + (dot(D_POS, D_ACC)+D_VEL^2)*t^2 + dot(D_VEL,D_ACC)*t^3 + D_ACC^2*t^4/4
In order to find the time where collision occurs, we can set the equation equal to the square of the sum of radii and solve for t:
0 = D_POS^2-(r_A+r_B)^2 + 2*dot(D_POS,D_VEL)*t + (dot(D_POS, D_ACC)+D_VEL^2)*t^2 + dot(D_VEL,D_ACC)*t^3 + D_ACC^2*t^4/4
Now, we can solve for the equation using the quartic formula.
The quartic formula will yield 4 roots, but we are only interested in real roots. If there is a double real root, then the two objects touch edges at exactly one point in time. If there are two real roots, then the objects continuously overlap between root 1 and root 2 (i.e. root 1 is the time when collision starts and root 2 is the time when collision stops). Four real roots means that the objects collide twice, continuously between root pairs 1,2 and 3,4.
In R, I used polyroot() to solve as follows:
# initial positions
POS_A=matrix(c(0,0),2,1)
POS_B=matrix(c(2,0),2,1)
# initial velocities
VEL_A=matrix(c(sqrt(2)/2,sqrt(2)/2),2,1)
VEL_B=matrix(c(-sqrt(2)/2,sqrt(2)/2),2,1)
# acceleration
ACC_A=matrix(c(sqrt(2)/2,sqrt(2)/2),2,1)
ACC_B=matrix(c(0,0),2,1)
# radii
r_A=.25
r_B=.25
# deltas
D_POS=POS_B-POS_A
D_VEL=VEL_B-VEL_A
D_ACC=ACC_B-ACC_A
# quartic coefficients
z=c(t(D_POS)%*%D_POS-r*r, 2*t(D_POS)%*%D_VEL, t(D_VEL)%*%D_VEL+t(D_POS)%*%D_ACC, t(D_ACC)%*%D_VEL, .25*(t(D_ACC)%*%D_ACC))
# get roots
roots=polyroot(z)
# In this case there are only two real roots...
root1=as.numeric(roots[1])
root2=as.numeric(roots[2])
# trajectory over time
pos=function(p,v,a,t){
T=t(matrix(t,length(t),2))
return(t(matrix(p,2,length(t))+matrix(v,2,length(t))*T+.5*matrix(a,2,length(t))*T*T))
}
# plot A in red and B in blue
t=seq(0,2,by=.1) # from 0 to 2 seconds.
a1=pos(POS_A,VEL_A,ACC_A,t)
a2=pos(POS_B,VEL_B,ACC_B,t)
plot(a1,type='o',col='red')
lines(a2,type='o',col='blue')
# points of a circle with center 'p' and radius 'r'
circle=function(p,r,s=36){
e=matrix(0,s+1,2)
for(i in 1:s){
e[i,1]=cos(2*pi*(1/s)*i)*r+p[1]
e[i,2]=sin(2*pi*(1/s)*i)*r+p[2]
}
e[s+1,]=e[1,]
return(e)
}
# plot circles with radius r_A and r_B at time of collision start in black
lines(circle(pos(POS_A,VEL_A,ACC_A,root1),r_A))
lines(circle(pos(POS_B,VEL_B,ACC_B,root1),r_B))
# plot circles with radius r_A and r_B at time of collision stop in gray
lines(circle(pos(POS_A,VEL_A,ACC_A,root2),r_A),col='gray')
lines(circle(pos(POS_B,VEL_B,ACC_B,root2),r_B),col='gray')
Object A follows the red trajectory from the lower left to the upper right. Object B follows the blue trajectory from the lower right to the upper left. The two objects collide continuously between time 0.9194381 and time 1.167549. The two black circles just touch, showing the beginning of overlap - and overlap continues in time until the objects reach the location of the gray circles.
Seems like you want the Closest Point of Approach (CPA). If it is less than the sum of the radiuses, you have a collision. There is example code in the link. You can calculate each frame with the current velocity, and check if the CPA time is less than your tick size. You could even cache the cpa time, and only update when acceleration was applied to either item.