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I have for example 4 points: A (latitute1, longitude1), B (latitute2, longitude2), C (latitute3, longitude3), D (latitute4, longitude4).
If I am a driver and I go from point A, I need an algorithm that calculates the most efficient way for me to visit all the points B, C, D starting from A. So that the distance is the smallest possible.
The algorithm should tell me the most effective order: A --> C --> B --> D (for example).
What matters is the total distance traveled is the lowest possible.
Thanks so much!!! :)
Maybe look into Dijkstras algorithm?
https://en.m.wikipedia.org/wiki/Dijkstra%27s_algorithm
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I'm developing a little 2D game and i have to predict when and where things will collide.
So, i've got four Vector2 :
A position
B position
A linear velocity
B linear velocity
I have to find if they intersect, where they intersect and at what time from now.
I've found many math solutions but i could't translate them into code.
The visualization of the problem, numbers are velocities
You want to compute the minimum of
norm((A+t*vA)-(B+t*vB))=norm((A-B)+t*(vA-vB))
Taking the square of these Euclidean norms
norm((A-B)+t*(vA-vB))^2 = norm(A-B)^2 + 2*t*dot(A-B,vA-vB) + t^2*norm(vA-vB)^2
gives you a simple quadratic function in t where the minimum has the value
min_dist =norm(A-B)^2 - dot(A-B,vA-vB)^2/norm(vA-vB)^2
at time
t = -dot(A-B,vA-vB)/norm(vA-vB)^2
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An input sequence is given. Each stage of the iteration finds another sequence by calculating difference between n-i and n-i-1 number. We continue the process and at the end of the last iteration (iteration: n-1) we find only 1 number. What is the mathematical formulation for finding the last number as shown in the image?
Basically, the mathematical formulation is finding the n-1'th derivative of the degree-n-1 polynomial passing through all points (i,arr[i]). That derivative is guaranteed to be a constant. This is equivalent to the coefficient of the term with exponent n-1, divided by (n-1)!.
This method is a special case of what is known as Neville's Algorithm.
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Number of digits that are present in the maximum number that is formed using three digits?
Maximum factorial that is formed by three digits?
This was a question asked on a site.
I am not able to understand is there any thing tricky i am not getting?
i have tried 3 and 720 but it is incorrect
The maximum factorial which can be formed using 3 digits is 999!.
The answer can be easily obtained from wolfram alpha.
Number of digits in 999!.
999!=Answer
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I would like to implement a revolve tool for 2d splines in 3d. The geometry computation already works but the normals are a bit tricky.
The problem is the angle between 2 points like the following image:
Here P1 is the previous point P2 the current point and P3 the next point.
How would I calculate the vector N.
Are you looking for the inverse angle bisector?
dir1 = normalize(p1 - p2)
dir2 = normalize(p3 - p2)
n = normalize(-dir1 - dir2)
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For a connected undirected graph, G = (V, E)
And a desired vertex set D, D is a subset of V (i.e. D \in V)
Is it NP-hard to find a minimum dominating set, containing the desired vertex set D?
Yes, it is a NP-hard problem. Please refer to the following document to read the reduction. Feel free to ask if you have problems in understanding the proof.
http://www.cs.iastate.edu/~chaudhur/cs611/Sp07/notes/lec22.pdf
To explain a bit more on your problem, i.e. adding the restriction D is a subset of V.....think like this -- when you are trying to prove your problem is NP, you reduce a known NP problem to a specific instance of your problem. Your specific instance of the problem can be a case when D=V...and you can prove your problem is also NP. Hope this helps.