I have a recursive function f(n) with time complexicity
O(f(n)) = O(combination(n, n/2) * f(n/2)^2)
where combination(a, b) means combination nuber a above b.
I tried to simplify it, but don't have enough mathematical skills. The only thing that I foud out is that
combination(n,n/2) = 2^n * (gamma(n/2 + 1/2)/(sqrt(1/2) * gamma(n/2 + 1)))
I don't have experience with calculation of complexities but this seems to me to be of order n!. At least for calculation of special case with n=2^i. With integers combination(n, n/2) is equal to n!/((n/2)!)^2.
f(2^i) = (2^i)! / ((2^(i-1))!)^2 * f(2^(i-1))^2
= (2^i)! / ((2^(i-1))!)^2 * (2^(i-1))! / ((2^(i-2))!)^2)^2 * f(2^(i-2))^4
= (2^i)! / ((2^(i-2))!)^4 * f(2^(i-2))^4
...
= (2^i)! / (1!)^(2^i) = n!
I have allready solved it in this thread on satck exchange:
https://math.stackexchange.com/questions/2642848/time-complexicity-of-recursive-function?noredirect=1#comment5458208_2642848
Related
Refers to WolframAlpha and some other calculators:
1i / 1i = -1
i / i = 1
My questions are:
Why division between complex number to himself does'nt return the value 1?
The both expressions should be the same, so why the results are diffrent?
You need to be aware of what the first expression means. In particular, where are the hidden parentheses.
1i/1i is in fact 1 * i / 1 * i, which is (((1 * i) / 1) * i) when all the parentheses are added.
Then, step by step, you get ((i / 1) * i), then (i * i), which is -1.
This is very different from (1i)/(1i), which is effectively 1.
Assuming I have a spaceship (source); And an asteroid (target) is somewhere near it.
I know, in 3D space (XYZ vectors):
My ship's position (sourcePos) and velocity (sourceVel).
The asteroid's position (targetPos) and velocity (targetVel).
(eg. sourcePos = [30, 20, 10]; sourceVel = [30, 20, 10]; targetPos = [600, 400, 200]; targetVel = [300, 200, 100]`)
I also know that:
The ship's velocity is constant.
The asteroid's velocity is constant.
My ship's projectile speed (projSpd) is constant.
My ship's projectile trajectory, after being shot, is linear (/straight).
(eg. projSpd = 2000.00)
How can I calculate the interception coordinates I need to shoot at in order to hit the asteroid?
Notes:
This question is based on this Yahoo - Answers page.
I also searched for similar problems on Google and here on SO, but most of the answers are for 2D-space, and, of the few for 3D, neither the explanation nor the pseudo-codes explain what is doing what and/or why, so I couldn't really understand enough to apply them on my code successfully. Here are some of the pages I visited:
Danik Games Devlog, Blitz3D Forums thread, UnityAnswers, StackOverflow #1, StackOverflow #2
I really can't figure out the maths / execution-flow on the linked pages as they are, unless someone dissects it (further) into what is doing what, and why;
Provides a properly-commented pseudo-code for me to follow;
Or at least points me to links that actually explain how the equations work instead of just throwing even more random numbers and unfollowable equations in my already-confused psyche.
I find the easiest approach to these kind of problems to make sense of them first, and have a basic high school level of maths will help too.
Solving this problem is essentially solving 2 equations with 2 variables which are unknown to you:
The vector you want to find for your projectile (V)
The time of impact (t)
The variables you know are:
The target's position (P0)
The target's vector (V0)
The target's speed (s0)
The projectile's origin (P1)
The projectile's speed (s1)
Okay, so the 1st equation is basic. The impact point is the same for both the target and the projectile. It is equal to the starting point of both objects + a certain length along the line of both their vectors. This length is denoted by their respective speeds, and the time of impact. Here's the equation:
P0 + (t * s0 * V0) = P1 + (t * s0 * V)
Notice that there are two missing variables here - V & t, and so we won't be able to solve this equation right now. On to the 2nd equation.
The 2nd equation is also quite intuitive. The point of impact's distance from the origin of the projectile is equal to the speed of the projectile multiplied by the time passed:
We'll take a mathematical expression of the point of impact from the 1st equation:
P0 + (t * s0 * V0) <-- point of impact
The point of origin is P1
The distance between these two must be equal to the speed of the projectile multiplied by the time passed (distance = speed * time).
The formula for distance is: (x0 - x1)^2 + (y0 - y1)^2 = distance^2, and so the equation will look like this:
((P0.x + s0 * t * V0.x) - P1.x)^2 + ((P0.y + s0 * t * V0.y) - P1.y)^2 = (s1 * t)^2
(You can easily expand this for 3 dimensions)
Notice that here, you have an equation with only ONE unknown variable: t!. We can discover here what t is, then place it in the previous equation and find the vector V.
Let me solve you some pain by opening up this formula for you (if you really want to, you can do this yourself).
a = (V0.x * V0.x) + (V0.y * V0.y) - (s1 * s1)
b = 2 * ((P0.x * V0.x) + (P0.y * V0.y) - (P1.x * V0.x) - (P1.y * V0.y))
c = (P0.x * P0.x) + (P0.y * P0.y) + (P1.x * P1.x) + (P1.y * P1.y) - (2 * P1.x * P0.x) - (2 * P1.y * P0.y)
t1 = (-b + sqrt((b * b) - (4 * a * c))) / (2 * a)
t2 = (-b - sqrt((b * b) - (4 * a * c))) / (2 * a)
Now, notice - we will get 2 values for t here.
One or both may be negative or an invalid number. Obviously, since t denotes time, and time can't be invalid or negative, you'll need to discard these values of t.
It could very well be that both t's are bad (in which case, the projectile cannot hit the target since it's faster and out of range). It could also be that both t's are valid and positive, in which case you'll want to choose the smaller of the two (since it's preferable to hit the target sooner rather than later).
t = smallestWhichIsntNegativeOrNan(t1, t2)
Now that we've found the time of impact, let's find out what the direction the projectile should fly is. Back to our 1st equation:
P0 + (t * s0 * V0) = P1 + (t * s0 * V)
Now, t is no longer a missing variable, so we can solve this quite easily. Just tidy up the equation to isolate V:
V = (P0 - P1 + (t * s0 * V0)) / (t * s1)
V.x = (P0.x - P1.x + (t * s0 * V0.x)) / (t * s1)
V.y = (P0.y - P1.y + (t * s0 * V0.y)) / (t * s1)
And that's it, you're done!
Assign the vector V to the projectile and it will go to where the target will be rather than where it is now.
I really like this problem since it takes math equations we learnt in high school where everyone said "why are learning this?? we'll never use it in our lives!!", and gives them a pretty awesome and practical application.
I hope this helps you, or anyone else who's trying to solve this.
If you want a projectile to hit asteroid, it should be shoot at the point interceptionPos that satisfy the equation:
|interceptionPos - sourcePos| / |interceptionPos - targetPos| = projSpd / targetVel
where |x| is a length of vector x.
In other words, it would take equal amount of time for the target and the projectile to reach this point.
This problem would be solved by means of geometry and trigonometry, so let's draw it.
A will be asteroid position, S - ship, I - interception point.
Here we have:
AI = targetVel * t
SI = projSpd * t
AS = |targetPos - sourcePos|
vector AS and AI direction is defined, so you can easily calculate cosine of the SAI angle by means of simple vector math (take definitions from here and here). Then you should use the Law of cosines with the SAI angle. It will yield a quadratic equation with variable t that is easy to solve (no solutions = your projectile is slower than asteroid). Just pick the positive solution t, your point-to-shoot will be
targetPos + t * targetVel
I hope you can write a code to solve it by yourself. If you cannot get something please ask in comments.
I got a solution. Notice that the ship position, and the asteroid line (position and velocity) define a 3D plane where the intercept point lies. In my notation below | [x,y,z] | denotes the magnitude of the vector or Sqrt(x^2+y^2+z^2).
Notice that if the asteroid travels with targetSpd = |[300,200,100]| = 374.17 then to reach the intercept point (still unknown, called hitPos) will require time equal to t = |hitPos-targetPos|/targetSpd. This is the same time the projectile needs to reach the intercept point, or t = |hitPos - sourcePos|/projSpd. The two equations are used to solve for the time to intercept
t = |targetPos-sourcePos|/(projSpd - targetSpd)
= |[600,400,200]-[30,20,10]|/(2000 - |[300,200,100]|)
= 710.81 / ( 2000-374.17 ) = 0.4372
Now the location of the intetception point is found by
hitPos = targetPos + targetVel * t
= [600,400,200] + [300,200,100] * 0.4372
= [731.18, 487.45, 243.73 ]
Now that I know the hit position, I can calculate the direction of the projectile as
projDir = (hitPos-sourcePos)/|hitPos-sourcePos|
= [701.17, 467.45, 233.73]/874.52 = [0.8018, 0.5345, 0.2673]
Together the projDir and projSpd define the projectile velocity vector.
Credit to Gil Moshayof's answer, as it really was what I worked off of to build this. But they did two dimensions, and I did three, so I'll share my Unity code in case it helps anyone along. A little long winded and redundant. It helps me to read it and know what's going on.
Vector3 CalculateIntercept(Vector3 targetLocation, Vector3 targetVelocity, Vector3 interceptorLocation, float interceptorSpeed)
{
Vector3 A = targetLocation;
float Ax = targetLocation.x;
float Ay = targetLocation.y;
float Az = targetLocation.z;
float As = targetVelocity.magnitude;
Vector3 Av = Vector3.Normalize(targetVelocity);
float Avx = Av.x;
float Avy = Av.y;
float Avz = Av.z;
Vector3 B = interceptorLocation;
float Bx = interceptorLocation.x;
float By = interceptorLocation.y;
float Bz = interceptorLocation.z;
float Bs = interceptorSpeed;
float t = 0;
float a = (
Mathf.Pow(As, 2) * Mathf.Pow(Avx, 2) +
Mathf.Pow(As, 2) * Mathf.Pow(Avy, 2) +
Mathf.Pow(As, 2) * Mathf.Pow(Avz, 2) -
Mathf.Pow(Bs, 2)
);
if (a == 0)
{
Debug.Log("Quadratic formula not applicable");
return targetLocation;
}
float b = (
As * Avx * Ax +
As * Avy * Ay +
As * Avz * Az +
As * Avx * Bx +
As * Avy * By +
As * Avz * Bz
);
float c = (
Mathf.Pow(Ax, 2) +
Mathf.Pow(Ay, 2) +
Mathf.Pow(Az, 2) -
Ax * Bx -
Ay * By -
Az * Bz +
Mathf.Pow(Bx, 2) +
Mathf.Pow(By, 2) +
Mathf.Pow(Bz, 2)
);
float t1 = (-b + Mathf.Pow((Mathf.Pow(b, 2) - (4 * a * c)), (1 / 2))) / (2 * a);
float t2 = (-b - Mathf.Pow((Mathf.Pow(b, 2) - (4 * a * c)), (1 / 2))) / (2 * a);
Debug.Log("t1 = " + t1 + "; t2 = " + t2);
if (t1 <= 0 || t1 == Mathf.Infinity || float.IsNaN(t1))
if (t2 <= 0 || t2 == Mathf.Infinity || float.IsNaN(t2))
return targetLocation;
else
t = t2;
else if (t2 <= 0 || t2 == Mathf.Infinity || float.IsNaN(t2) || t2 > t1)
t = t1;
else
t = t2;
Debug.Log("t = " + t);
Debug.Log("Bs = " + Bs);
float Bvx = (Ax - Bx + (t * As + Avx)) / (t * Mathf.Pow(Bs, 2));
float Bvy = (Ay - By + (t * As + Avy)) / (t * Mathf.Pow(Bs, 2));
float Bvz = (Az - Bz + (t * As + Avz)) / (t * Mathf.Pow(Bs, 2));
Vector3 Bv = new Vector3(Bvx, Bvy, Bvz);
Debug.Log("||Bv|| = (Should be 1) " + Bv.magnitude);
return Bv * Bs;
}
I followed the problem formulation as described by Gil Moshayof's answer, but found that there was an error in the simplification of the quadratic formula. When I did the derivation by hand I got a different solution.
The following is what worked for me when finding the intersect in 2D:
std::pair<double, double> find_2D_intersect(Vector3 sourcePos, double projSpd, Vector3 targetPos, double targetSpd, double targetHeading)
{
double P0x = targetPos.x;
double P0y = targetPos.y;
double s0 = targetSpd;
double V0x = std::cos(targetHeading);
double V0y = std::sin(targetHeading);
double P1x = sourcePos.x;
double P1y = sourcePos.y;
double s1 = projSpd;
// quadratic formula
double a = (s0 * s0)*((V0x * V0x) + (V0y * V0y)) - (s1 * s1);
double b = 2 * s0 * ((P0x * V0x) + (P0y * V0y) - (P1x * V0x) - (P1y * V0y));
double c = (P0x * P0x) + (P0y * P0y) + (P1x * P1x) + (P1y * P1y) - (2 * P1x * P0x) - (2 * P1y * P0y);
double t1 = (-b + std::sqrt((b * b) - (4 * a * c))) / (2 * a);
double t2 = (-b - std::sqrt((b * b) - (4 * a * c))) / (2 * a);
double t = choose_best_time(t1, t2);
double intersect_x = P0x + t * s0 * V0x;
double intersect_y = P0y + t * s0 * V0y;
return std::make_pair(intersect_x, intersect_y);
}
How is the running time of a d-Ary heap simplified from O(logd n) to O( (log n) / (log d))?
A correct simplification would be:
logdn = log d * log n
How is the division simplification derived?
This uses the common identity to convert between logarithmic bases:
logx(z) = logm(z) / logm(x)
By multiplying both sides by logm(x), you get:
logm(z) = logx(z) * logm(x)
Which is equivalent to the answer in the question you site.
More information is available here.
Suppose x = logd(n) Equivalently we have, n = dx Then log2n = log2(dx) = x log2(d) Dividing through by log2(d) yields: log2(n) / log2(d) = x And so log2(n) / log2(d) = x = logd(n)
Of course, assuming d is fixed, then log2(d) is just a constant. And so O( logd(n) ) = O( 1 / log2(d) * log2(n) ) = O( log2(n) ) That is, so far as Big-Oh notation is concerned, you can change out any logarithm base (larger than 1) for any other (such) logarithm base. And so it's customary to just drop the base and write O( log(n) )
I need to round a number by 1/32 and THEN round it by 1/100th. I need to convert this into one single rounding rule though (using an archaic program...). I can multiply and divide the original number and all that, just can't round twice....
Is there a way to do this mathematically?
Thanks!
kcross
If whatever you're using lets you define functions, the most readable implementation would be this:
function round(x, interval){
//implementation left as an exercise to the reader
}
#rounds x by interval1, then by interval2
function doubleRound(x, interval1, interval2){
return round(round(x, interval1), interval2)
}
but if all you have is simple arithmetic, you can unroll everything into one statement.
To round a non-negative number x to the nearest interval of N, you can use this formula:
round(x,N) = floor((x + (N/2)) / N) * N
to round twice, you nest the function within itself:
round(round(x, N1), N2) = floor(((floor((x + (N1/2)) / N1) * N1) + (N2/2)) / N2) * N2
so to round by 1/32 and then 1/100, you use:
floor(((floor((x + ((1/32)/2)) / (1/32)) * (1/32)) + ((1/100)/2)) / (1/100)) * (1/100)
I came across a situation doing some advanced collision detection, where I needed to calculate the roots of a quartic function.
I wrote a function that seems to work fine using Ferrari's general solution as seen here: http://en.wikipedia.org/wiki/Quartic_function#Ferrari.27s_solution.
Here's my function:
private function solveQuartic(A:Number, B:Number, C:Number, D:Number, E:Number):Array{
// For paramters: Ax^4 + Bx^3 + Cx^2 + Dx + E
var solution:Array = new Array(4);
// Using Ferrari's formula: http://en.wikipedia.org/wiki/Quartic_function#Ferrari.27s_solution
var Alpha:Number = ((-3 * (B * B)) / (8 * (A * A))) + (C / A);
var Beta:Number = ((B * B * B) / (8 * A * A * A)) - ((B * C) / (2 * A * A)) + (D / A);
var Gamma:Number = ((-3 * B * B * B * B) / (256 * A * A * A * A)) + ((C * B * B) / (16 * A * A * A)) - ((B * D) / (4 * A * A)) + (E / A);
var P:Number = ((-1 * Alpha * Alpha) / 12) - Gamma;
var Q:Number = ((-1 * Alpha * Alpha * Alpha) / 108) + ((Alpha * Gamma) / 3) - ((Beta * Beta) / 8);
var PreRoot1:Number = ((Q * Q) / 4) + ((P * P * P) / 27);
var R:ComplexNumber = ComplexNumber.add(new ComplexNumber((-1 * Q) / 2), ComplexNumber.sqrt(new ComplexNumber(PreRoot1)));
var U:ComplexNumber = ComplexNumber.pow(R, 1/3);
var preY1:Number = (-5 / 6) * Alpha;
var RedundantY:ComplexNumber = ComplexNumber.add(new ComplexNumber(preY1), U);
var Y:ComplexNumber;
if(U.isZero()){
var preY2:ComplexNumber = ComplexNumber.pow(new ComplexNumber(Q), 1/3);
Y = ComplexNumber.subtract(RedundantY, preY2);
} else{
var preY3:ComplexNumber = ComplexNumber.multiply(new ComplexNumber(3), U);
var preY4:ComplexNumber = ComplexNumber.divide(new ComplexNumber(P), preY3);
Y = ComplexNumber.subtract(RedundantY, preY4);
}
var W:ComplexNumber = ComplexNumber.sqrt(ComplexNumber.add(new ComplexNumber(Alpha), ComplexNumber.multiply(new ComplexNumber(2), Y)));
var Two:ComplexNumber = new ComplexNumber(2);
var NegativeOne:ComplexNumber = new ComplexNumber(-1);
var NegativeBOverFourA:ComplexNumber = new ComplexNumber((-1 * B) / (4 * A));
var NegativeW:ComplexNumber = ComplexNumber.multiply(W, NegativeOne);
var ThreeAlphaPlusTwoY:ComplexNumber = ComplexNumber.add(new ComplexNumber(3 * Alpha), ComplexNumber.multiply(new ComplexNumber(2), Y));
var TwoBetaOverW:ComplexNumber = ComplexNumber.divide(new ComplexNumber(2 * Beta), W);
solution["root1"] = ComplexNumber.add(NegativeBOverFourA, ComplexNumber.divide(ComplexNumber.add(W, ComplexNumber.sqrt(ComplexNumber.multiply(NegativeOne, ComplexNumber.add(ThreeAlphaPlusTwoY, TwoBetaOverW)))), Two));
solution["root2"] = ComplexNumber.add(NegativeBOverFourA, ComplexNumber.divide(ComplexNumber.subtract(NegativeW, ComplexNumber.sqrt(ComplexNumber.multiply(NegativeOne, ComplexNumber.subtract(ThreeAlphaPlusTwoY, TwoBetaOverW)))), Two));
solution["root3"] = ComplexNumber.add(NegativeBOverFourA, ComplexNumber.divide(ComplexNumber.subtract(W, ComplexNumber.sqrt(ComplexNumber.multiply(NegativeOne, ComplexNumber.add(ThreeAlphaPlusTwoY, TwoBetaOverW)))), Two));
solution["root4"] = ComplexNumber.add(NegativeBOverFourA, ComplexNumber.divide(ComplexNumber.add(NegativeW, ComplexNumber.sqrt(ComplexNumber.multiply(NegativeOne, ComplexNumber.subtract(ThreeAlphaPlusTwoY, TwoBetaOverW)))), Two));
return solution;
}
The only issue is that I seem to get a few exceptions. Most notably when I have two real roots, and two imaginary roots.
For example, this equation:
y = 0.9604000000000001x^4 - 5.997600000000001x^3 + 13.951750054511718x^2 - 14.326264455924333x + 5.474214401412618
Returns the roots:
1.7820304835380467 + 0i
1.34041662585388 + 0i
1.3404185025061823 + 0i
1.7820323472855648 + 0i
If I graph that particular equation, I can see that the actual roots are closer to 1.2 and 2.9 (approximately). I can't dismiss the four incorrect roots as random, because they're actually two of the roots for the equation's first derivative:
y = 3.8416x^3 - 17.9928x^2 + 27.9035001x - 14.326264455924333
Keep in mind that I'm not actually looking for the specific roots to the equation I posted. My question is whether there's some sort of special case that I'm not taking into consideration.
Any ideas?
For finding roots of polynomials of degree >= 3, I've always had better results using Jenkins-Traub ( http://en.wikipedia.org/wiki/Jenkins-Traub_algorithm ) than explicit formulas.
I do not know why Ferrari's solution does not work, but I tried to use the standard numerical method (create a companion matrix and compute its eigenvalues), and I obtain the correct solution, i.e., two real roots at 1.2 and 1.9.
This method is not for the faint of heart. After constructing the companion matrix of the polynomial, you run the QR algorithm to find the eigenvalues of that matrix. Those are the zeroes of the polynomial.
I suggest you to use an existing implementation of the QR algorithm since a good deal of it is closer to kitchen recipe than algorithmics. But it is, I believe, the most widely used algorithm to compute eigenvalues, and thereby, roots of polynomials.
You can see my answer to a related question. I support the view of Olivier: the way to go may just be the companion matrix / eigenvalue approach (very stable, simple, reliable, and fast).
Edit
I guess it does not hurt if I reproduce the answer here, for convenience:
The numerical solution for doing this many times in a reliable, stable manner, involve: (1) Form the companion matrix, (2) find the eigenvalues of the companion matrix.
You may think this is a harder problem to solve than the original one, but this is how the solution is implemented in most production code (say, Matlab).
For the polynomial:
p(t) = c0 + c1 * t + c2 * t^2 + t^3
the companion matrix is:
[[0 0 -c0],[1 0 -c1],[0 1 -c2]]
Find the eigenvalues of such matrix; they correspond to the roots of the original polynomial.
For doing this very fast, download the singular value subroutines from LAPACK, compile them, and link them to your code. Do this in parallel if you have too many (say, about a million) sets of coefficients. You could use QR decomposition, or any other stable methodology for computing eigenvalues (see the Wikipedia entry on "matrix eigenvalues").
Notice that the coefficient of t^3 is one, if this is not the case in your polynomials, you will have to divide the whole thing by the coefficient and then proceed.
Good luck.
Edit: Numpy and octave also depend on this methodology for computing the roots of polynomials. See, for instance, this link.
The other answers are good and sound advice. However, recalling my experience with the implementation of Ferrari's method in Forth, I think your wrong results are probably caused by 1. wrong implementation of the necessary and rather tricky sign combinations, 2. not realizing yet that ".. == beta" in floating-point should become "abs(.. - beta) < eps, 3. not yet having found out that there are other square roots in the code that may return complex solutions.
For this particular problem my Forth code in diagnostic mode returns:
x1 = 1.5612244897959360787072371026316680470492e+0000 -1.6542769593216835969789894020584464029664e-0001 i
--> -4.2123274051525879873007970023884313331788e-0054 3.4544674220377778501545407451201598284464e-0077 i
x2 = 1.5612244897959360787072371026316680470492e+0000 1.6542769593216835969789894020584464029664e-0001 i
--> -4.2123274051525879873007970023884313331788e-0054 -3.4544674220377778501545407451201598284464e-0077 i
x3 = 1.2078440724224197532447709413299479764843e+0000 0.0000000000000000000000000000000000000000e-0001 i
--> -4.2123274051525879873010733597821943554068e-0054 0.0000000000000000000000000000000000000000e-0001 i
x4 = 1.9146049071693819497220585618954851525216e+0000 -0.0000000000000000000000000000000000000000e-0001 i
--> -4.2123274051525879873013497171759573776348e-0054 0.0000000000000000000000000000000000000000e-0001 i
The text after "-->" follows from backsubstituting the root into the original equation.
For reference, here are Mathematica/Alpha's results to the highest possible precision I managed to set it:
Mathematica:
x1 = 1.20784407242
x2 = 1.91460490717
x3 = 1.56122449 - 0.16542770 i
x4 = 1.56122449 + 0.16542770 i
A good alternative to the methods already mentioned is the TOMS Algorithm 326, which is based on the paper "Roots of Low Order Polynomials" by Terence R.F.Nonweiler CACM (Apr 1968).
This is an algebraic solution to 3rd and 4th order polynomials that is reasonably compact, fast, and quite accurate. It is much simpler than Jenkins Traub.
Be warned however that the TOMS code doesn't work all that well.
This Iowa Hills Root Solver page has code for a Quartic / Cubic root finder that is a bit more refined. It also has a Jenkins Traub type root finder.