Hello this semester we started a course for AI and we have a project, to calculate the best route to reach a target at a NxN board. This board can also randomnly contain obstacles that we cant pass through them and also we can move only vertically and horizontally. The cost of each vertical move is 1.0 and the cost of horizontal move is 0.5. We are asked to calculate it with the A* algorithm using the manhattan heuristic method.
My question is: When we are calculating the manhattan distance (i know that u have to add the horizontal and vertical "squares" until you reach the target) do we have to add the cost to each "square" ( 0.5 or 1.0) ?. Or Manhattan distance just counts the "squares" you need to reach the target?
You need to use the distance of your problem, i.e. 1 and 0.5 in your case.
When using A*, the distance from the start to the current node will be a sum of 1s and 0.5s along the real path to that point, and the estimate remaining distance to the end will be a sum of 1s and 0.5s along a 'perfect path', i.e. with no obstacles.
I hope that answers your question.
You should add the cost to each square as Manhattan does not simply counts the squares as it calculates the simplest cost which is same as counting squares in case of same weight vertical and horizontal moves.
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I have 5 {x,y} points randomly placed on a grid
Each of the points do not know the {x,y} coordinates of the other points
Each of the points do know the distance of each of the other points from their {x,y} position
Each of the points exchanges this distance information with every other point
So every point knows every distance of every other point
Using this distance information every point can calculate (by finding the angles) triangles for every other point using itself as a reference point
Example, point 1 can calculate the following triangles:
1-2-3,
1-2-4,
1-2-5,
1-3-4,
1-3-5,
1-4-5,
and using the distance data recieved from the other points it can also calculate
2-3-4,
2-3-5,
2-4-5,
3-4-5
I would like to build a map of the location of every other point relative to a single point
How should I go about doing this? I am asuming it would be some kind of triangulation algorithm but these mainly seem to compute the location of a point from three other points, not the other way around where the other points {x,y} coordinates are discovered based on only the distance information.
I have tried plotting the two possible triangles for every 3 triangle points and then rotating them on a fixed known point to try and align them, but I think this avenue will end up with too many possibilities and errors
Ultimately I would like every point to end up with {x,y} coordinates of every other point relative to itself
You know the distance from one point to every other, dij. Thus, point 2 lies in a circumference of center point 1 and radius = d12. Point 3 lies in a circumference of center point 1 and R=d13 and it also lies in another circumference of center point 2 and R=d23.
See this picture:
I've set point 2 in X-axis for simplicity.
As you see, point 3 is on the intersection of two cicrcumferences centered at P1 and P2. There is a second intersection, P3a. Let's choose the one that is upwards and continue.
For P4 we can use three circumferences, centered at P1, P2 and P3. Again we get two solutions.
The same process can be done with the rest of points. For Pn you have n-1 circumferences.
I'm sure you can find the maths for circle-circle intersection.
Some remarks must be observed:
1) The construction is simpler if you first sort the points by distance to P1.
2) Not all distances generate a solution. For example, increase d13 an there's no intersection between the two circumferences for P3. Or increase d14 and now the three circumferences don't intersect in just the two expected points 4 and 4a.
3) This fact can be overworked by considering the average of intersections and the distance from each solution to this average. You can set a tolerance in these distances and tell if the average is a solution or else some dij is wrong. Since two solutions are possible, you must consider two averages.
4) The two possible triangulations are symmetric, over X-axis in the case I've drawn.
The real solution is obtained by a rotation around P1. To calculate the angle of rotation you need the {x,y} coordinates of another point.
My requirement is to find the point closest to three circles. So lets say the three circles are C1, C2, C3. I want to find the point in the space such that the SUM of its distance from C1, C2 and C3 is MINIMUM.
The distance of a given point from a circle is the distance of the given point from the point that lies on the circle and is intersection of the circle with the line joining the given point with the center of the circle.
Is there a simple logic of find such a point?
Unless one of the distances will be zero, the circle radii are irrelevant: the sum of the distances to the circles will be the sum of the distances to the centers minus the sum of the radii. So in effect you are asking for the geometric median of the circle centers. You might want to iteratively compute an approximation. Or you make use of the fact that you have three circles, in which case the median is the Fermat point of the triangle formed by their centers.
If the point constructed as above lies within one of the circles, then you can move towards that circle while decreasing the sum. So you'd have to consider all the points on that given circle, which you can express as a one-parameter family. You could then compute the distance as a function of that parameter, and the derivative of the resulting formula, and setting that equal to zero will give the optimal solution for this case.
I have a set of 2D points, unorganized, and I want to find the "contour" of this set (not the convex hull). I can't use alpha shapes because I have a speed objective (less than 10ms on an average computer).
My first approach was to compute a grid and find the outline squares (squares which have an empty square as a neighbor). So I think I downsized efficiently my numbers of points (from 22000 to 3000 roughly). But I still need to refine this new set.
My question is : how do I find the real outlines points among my green points ?
After a weekend full of reflexions, I may have found a convenient solution.
So we need a grid, we need to fill it with our points, no difficulty here.
We have to decide which squares are considered as "Contour". Our criteria is : at least one empty neighbor and at least 3 non empty neighbors.
We lack connectivity information. So we choose a "Contour" square which as 2 "Contour" neighbors or less. We then pick one of the neighbor. From that, we can start the expansion. We just circle around the current square to find the next "Contour" square, knowing the previous "Contour" squares. Our contour criteria prevent us from a dead end.
We now have vectors of connected squares, and normally if our shape doesn't have a hole, only one vector of connected squares !
Now for each square, we need to find the best point for the contour. We select the one which is farther from the barycenter of our plane. It works for most of the shapes. Another technique is to compute the barycenter of the empty neighbors of the selected square and choose the nearest point.
The red points are the contour of the green one. The technique used is the plane barycenter one.
For a set of 28000 points, this techniques take 8 ms. CGAL's Alpha shapes would take an average 125 ms for 28000 points.
PS : I hope I made myself clear, English is not my mothertongue :s
You really should use the alpha shapes. Maybe use only green points as inputs of the alpha alpha algorithm.
Understanding that it would only be an estimate...
How what decimal constant would be able to be used to find a point X Miles away from a point in latitude and longitude to facilitate creating a lat long bounding box.
Unfortunately there is no such simple constant. As you go farther north, the "walking distance" between lines of longitude becomes smaller and smaller. If you were right next to the north pole, you could walk in a circle around it, covering almost no distance at all, and yet you'd still have touched every line of longtiude.
What you need is the great-circle distance between two points on a sphere.
Not sure what you are trying to do. You want a create a bounding box based on distance from a point? Are you looking for a way to calculate new lat/long from given lat/long using a distance in miles?
You can use the manhattan function to get an approximation of distance (realizing that lat/long are based on a spheroid approximating the earth, and truly calculating this requires more math), calculating x and y values with a forumla to follow
Manhattan function:
sqrt(x*x + y*y)
X and Y from Lat/Long:
x = 69.1 * (lat2 - lat1)
y = 53 * (lon2 - lon1)
Still, that method is pretty error-prone.
There's the great circle distance formula too, gotta use some trig for it, but it's probably worth it since you can get pretty good error in approximation depending upon what part of the spheroid (e.g. the lat/long) you start at.
Check out this page:
http://www.meridianworlddata.com/Distance-Calculation.asp
If i have a position specified in latitude and longitude, then it can cover a box, depending on how many digits of accuracy are in the position.
For example, given a position of 55.0° N, 3.0° W (i.e. to 1 decimal place), and assuming a truncation (as opposed to rounding), this could cover anything that's 55.01° to 55.09°. This would cover the area in this box: http://www.openstreetmap.org/?minlat=55.0&maxlat=55.1&maxlon=-3.0&minlon=-3.1&box=yes
Is there anyway to calculate the area of that box? I only want to do this once, so a simple website that provides this calculation would suffice.
The main reason I want to do this is because I have a position to a very high number of decimal places, and I want to see how precise it is.
While the Earth isn't exactly spherical you can treat it as such for these calculations.
The North/South calculation is relatively simple as there are 180° from pole to pole and the distance is 20,014 km (Source) so one degree == 20014/180 = 111.19 km.
The East/West calculation is more difficult as it depends on the latitude. The Equatorial distance is 40,076 km (Source) so one degree = 40076/360 = 111.32 km. The circumference at the poles is (by definition) 0 km. So you can calculate the circumference at any latitude by trigonometry (circumference * sin(latitude)).
I have written an online geodesic area calculator which is available at
http://geographiclib.sf.net/cgi-bin/Planimeter. Enter in the four
corners of your box (counter clockwise) and it will give its area.