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  • Finding cities close to one another using longitude and latitude

    - by Jamie
    Each user in my db is associated to a city (with it's longitude and latitude) How would I go about finding out which cities are close to one another? i.e. in England, Cambridge is fairly close to London. So If I have a user who lives in Cambridge. Users close to them would be users living in close surrounding cities, such as London, Hertford etc. Any ideas how I could go about this? And also, how would I define what is close? i.e. in the UK close would be much closer than if it were in the US as the US is far more spread out. Ideas and suggestions. Also, do you know any services that provide this sort of functionality? Thanks

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  • Vector deltas and moving in unknown areas

    - by dekz
    Hi All, I was in need of a little math help that I can't seem to find the answer to, any links to documentation would be greatly appreciated. Heres my situation, I have no idea where I am in this maze, but I need to move around and find my way back to the start. I was thinking of implementing a waypoint list of places i've been offset from my start at 0,0. This is a 2D cartesian plane. I've been given 2 properties, my translation speed from 0-1 and my rotation speed from -1 to 1. -1 is very left and +1 is very right. These are speed and not angles so thats where my problem lies. If I'm given 0 as a translation speed and 0.2 I will continually turn to my right at a slow speed. How do I figure out the offsets given these 2 variables? I can store it every time I take a 'step'. I just need to figure out the offsets in x and y terms given the translations and rotation speeds. And the rotation to get to those points. Any help is appreciated.

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  • Find a node in a Graph that minimizes the distance between two other nodes

    - by Andrés
    Here is the thing. I have a directed weighted graph G, with V vertices and E edges. Given two nodes in the graph, let's say A, and B, and given the weight of an edge A-B denoted as w(A, B), I need to find a node C so that max(w(A, C), w(B, C)) is minimal among all possibilities. By possibilities I mean all the values C can take. I don't know if it is completely clear, if it's not, I'll try to be more precise. Thanks in advance.

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  • Connecting grouped dots/points on a scatter plot based on distance

    - by ToNoY
    I have 2 sets of depth point measurements, for example: > a depth value 1 2 2 2 4 3 3 6 4 4 8 5 5 16 40 6 18 45 7 20 58 > b depth value 1 10 10 2 12 20 3 14 35 I want to show both groups in one figure plotted with depth and with different symbols as you can see here plot(a$value, a$depth, type='b', col='green', pch=15) points(b$value, b$depth, type='b', col='red', pch=14) The plot seems okay, but the annoying part is that the green symbols are all connected (though I want connected lines also). I want connection only when one group has a continued data points at 2 m interval i.e. the symbols should be connected with a line from 2 to 8 m (green) and then group B symbols should be connected from 10-14 m (red) and again group A symbols should be connected (green), which means I do NOT want to see the connection between 8 m sample with the 16 m for group A. An easy solution may be dividing the group A into two parts (say, A-shallow and A-deep) and then plotting A-shallow, B, and A-deep separately. But this is completely impractical because I have thousands of data points with hundreds of groups i.e. I have to produce many depth profiles. Therefore, there has to be a way to program so that dots are NOT connected beyond a prescribed frequency/depth interval (e.g. 2 m in this case) for a particular group of samples. Any idea?

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  • Am I right about the differences between Floyd-Warshall, Dijkstra's and Bellman-Ford algorithms?

    - by Programming Noob
    I've been studying the three and I'm stating my inferences from them below. Could someone tell me if I have understood them accurately enough or not? Thank you. Dijkstra's algorithm is used only when you have a single source and you want to know the smallest path from one node to another, but fails in cases like this Floyd-Warshall's algorithm is used when any of all the nodes can be a source, so you want the shortest distance to reach any destination node from any source node. This only fails when there are negative cycles (this is the most important one. I mean, this is the one I'm least sure about:) 3.Bellman-Ford is used like Dijkstra's, when there is only one source. This can handle negative weights and its working is the same as Floyd-Warshall's except for one source, right? If you need to have a look, the corresponding algorithms are (courtesy Wikipedia): Bellman-Ford: procedure BellmanFord(list vertices, list edges, vertex source) // This implementation takes in a graph, represented as lists of vertices // and edges, and modifies the vertices so that their distance and // predecessor attributes store the shortest paths. // Step 1: initialize graph for each vertex v in vertices: if v is source then v.distance := 0 else v.distance := infinity v.predecessor := null // Step 2: relax edges repeatedly for i from 1 to size(vertices)-1: for each edge uv in edges: // uv is the edge from u to v u := uv.source v := uv.destination if u.distance + uv.weight < v.distance: v.distance := u.distance + uv.weight v.predecessor := u // Step 3: check for negative-weight cycles for each edge uv in edges: u := uv.source v := uv.destination if u.distance + uv.weight < v.distance: error "Graph contains a negative-weight cycle" Dijkstra: 1 function Dijkstra(Graph, source): 2 for each vertex v in Graph: // Initializations 3 dist[v] := infinity ; // Unknown distance function from 4 // source to v 5 previous[v] := undefined ; // Previous node in optimal path 6 // from source 7 8 dist[source] := 0 ; // Distance from source to source 9 Q := the set of all nodes in Graph ; // All nodes in the graph are 10 // unoptimized - thus are in Q 11 while Q is not empty: // The main loop 12 u := vertex in Q with smallest distance in dist[] ; // Start node in first case 13 if dist[u] = infinity: 14 break ; // all remaining vertices are 15 // inaccessible from source 16 17 remove u from Q ; 18 for each neighbor v of u: // where v has not yet been 19 removed from Q. 20 alt := dist[u] + dist_between(u, v) ; 21 if alt < dist[v]: // Relax (u,v,a) 22 dist[v] := alt ; 23 previous[v] := u ; 24 decrease-key v in Q; // Reorder v in the Queue 25 return dist; Floyd-Warshall: 1 /* Assume a function edgeCost(i,j) which returns the cost of the edge from i to j 2 (infinity if there is none). 3 Also assume that n is the number of vertices and edgeCost(i,i) = 0 4 */ 5 6 int path[][]; 7 /* A 2-dimensional matrix. At each step in the algorithm, path[i][j] is the shortest path 8 from i to j using intermediate vertices (1..k-1). Each path[i][j] is initialized to 9 edgeCost(i,j). 10 */ 11 12 procedure FloydWarshall () 13 for k := 1 to n 14 for i := 1 to n 15 for j := 1 to n 16 path[i][j] = min ( path[i][j], path[i][k]+path[k][j] );

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  • Intune : administrez vos PC à distance via le Cloud et sécurisez-les contre les menaces, Microsoft offre un test gratuit de 30 jours

    Intune : administrez vos PC à distance via le Cloud et sécurisez-les contre les logiciels et les menaces, Microsoft offre un test gratuit de 30 jours [IMG]http://rdonfack.developpez.com/images/intune.PNG[/IMG] Windows Intune est une solution d'administration de postes de travail dans le Cloud qui permet également de les faire évoluer vers Windows 7 Entreprise. Le service permet aux responsables IT de gérer à distance l'ensemble de leur parc informatique sous Windows. Il intègre des services de gestion pour les parcs de PCs, le suivi des protections installées contre les logiciels malveillants, et le suivi des mises à jour Windows. Windows InTune perme...

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  • Faille zero-day découverte dans Internet Explorer, permettant de prendre à distance le contrôle d'un PC, Microsoft recommande EMET

    Faille zero-day découverte dans Internet Explorer permettant de prendre à distance le contrôle d'un ordinateur, Microsoft recommande EMET Une faille de sécurité critique dans Internet Explorer 6, 7, 8 et 9 vient d'être découverte par des experts en sécurité. Dans un billet de blog sur ZATAZ.com, Eric Romang, un conseiller en sécurité luxembourgeois, déclare avoir découvert la vulnérabilité lorsque son PC a été infecté par le cheval de troie Poison Ivy, qui est utilisé pour voler des données ou prendre le contrôle à distance d'un ordinateur. La faille pourrait être exploitée par des pirates distants qui peuvent obtenir les mêmes privilèges que l'utilisateur local, pour installer de...

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  • search all paths and the shortest path for a graph - Prolog

    - by prologian
    Hi , I have a problem in my code with turbo prolog wich search all paths and the shortest path for a graph between 2 nodes the problem that i have is to test if the node is on the list or not exactly in the clause of member and this is my code : /* 1 ---- b ---- 3 --- | --- --- | ----- a |5 d --- | ----- --- | --- 2 --- | --- 4 -- c -- for example we have for b--->c ([b,c],5) , ([b,a,c],3) and ([b,d,c],7) : possible paths. ([b,a,c],3) : the shortest path. */ DOMAINS list=Symbol * PREDICATES distance(Symbol,Symbol) path1(Symbol,Symbol,list,integer) path(Symbol,Symbol,list,list,integer) distance(Symbol,list,integer) member(Symbol,list) shortest(Symbol,Symbol,list,integer) CLAUSES distance(a,b,1). distance(a,c,2). distance(b,d,3). distance(c,d,4). distance(b,c,5). distance(b,a,1). distance(c,a,2). distance(d,b,3). distance(d,c,4). distance(c,b,5). member(X, [Y|T]) :- X = Y; member(X, T). absent(X,L) :-member(X, L),!,fail. absent(_,_). /*find all paths*/ path1(X, Y, L, C):- path(X, Y, L, I, C). path(X, X, [X], I, C) :- absent(X, I). path(X, Y, [X|R], I, C) :- distance(X, Z, A) , absent(Z, I), path(Z, Y, R, [X|I] ,C1) , C=C1+A . /*to find the shortest path*/ shortest(X, Y, L, C):-path(X, Y, L, C),path(X, Y, L1, C1),C<C1.

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  • Zoom Layer centered on a Sprite

    - by clops
    I am in process of developing a small game where a space-ship travels through a layer (doh!), in some situations the spaceship comes close to an enemy space ship, and the whole layer is zoomed in on the two with the zoom level being dependent on the distance between the ship and the enemy. All of this works fine. The main question, however, is how do I keep the zoom being centered on the center point between the two space-ships and make sure that the two are not off-screen? Currently I control the zooming in the GameLayer object through the update method, here is the code (there is no layer repositioning here yet): -(void) prepareLayerZoomBetweenSpaceship{ CGPoint mainSpaceShipPosition = [mainSpaceShip position]; CGPoint enemySpaceShipPosition = [enemySpaceShip position]; float distance = powf(mainSpaceShipPosition.x - enemySpaceShipPosition.x, 2) + powf(mainSpaceShipPosition.y - enemySpaceShipPosition.y,2); distance = sqrtf(distance); /* Distance > 250 --> no zoom Distance < 100 --> maximum zoom */ float myZoomLevel = 0.5f; if(distance < 100){ //maximum zoom in myZoomLevel = 1.0f; }else if(distance > 250){ myZoomLevel = 0.5f; }else{ myZoomLevel = 1.0f - (distance-100)*0.0033f; } [self zoomTo:myZoomLevel]; } -(void) zoomTo:(float)zoom { if(zoom > 1){ zoom = 1; } // Set the scale. if(self.scale != zoom){ self.scale = zoom; } } Basically my question is: How do I zoom the layer and center it exactly between the two ships? I guess this is like a pinch zoom with two fingers!

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  • compareTo() method java is acting weird

    - by Ron Paul
    hi im having trouble getting this to work im getting an error here with my object comparison...how could I cast the inches to a string ( i never used compare to with anything other than strings) , or use comparison operators to compare the intigers, Object comparison = this.inches.compareTo(obj.inches); here is my code so far import java.io.*; import java.util.*; import java.lang.Integer; import java.lang.reflect.Array; public class Distance implements Comparable<Distance> { private static final String HashCodeUtil = null; private int feet; private int inches; private final int DEFAULT_FT = 1; private final int DEFAULT_IN = 1; public Distance(){ feet = DEFAULT_FT; inches = DEFAULT_IN; } public Distance(int ft, int in){ feet = ft; inches = in; } public void setFeet(int ft){ try { if(ft<0){ throw new CustomException("Distance is not negative"); } } catch(CustomException c){ System.err.println(c); feet =ft; } } public int getFeet(){ return feet; } public void setInches(int in){ try { if (in<0) throw new CustomException("Distance is not negative"); //inches = in; } catch(CustomException c) { System.err.println(c); inches = in; } } public int getInches(){ return inches; } public String toString (){ return "<" + feet + ":" + inches + ">"; } public Distance add(Distance m){ Distance n = new Distance(); n.inches = this.inches + m.inches; n.feet = this.feet + m.feet; while(n.inches>12){ n.inches = n.inches - 12; n.feet++; } return n; } public Distance subtract(Distance f){ Distance m = new Distance(); m.inches = this.inches - f.inches; m.feet = this.feet - f.feet; while(m.inches<0){ m.inches = m.inches - 12; feet--; } return m; } @Override public int compareTo(Distance obj) { // TODO Auto-generated method stub final int BEFORE = -1; final int EQUAL = 0; final int AFTER = 1; if (this == obj) return EQUAL; if(this.DEFAULT_IN < obj.DEFAULT_FT) return BEFORE; if(this.DEFAULT_IN > obj.DEFAULT_FT) return AFTER; Object comparison = this.inches.compareTo(obj.inches); if (this.inches == obj.inches) return compareTo(null); assert this.equals(obj) : "compareTo inconsistent with equals"; return EQUAL; } @Override public boolean equals( Object obj){ if (obj != null) return false; if (!(obj intanceof Distance)) return false; Distance that = (Distance)obj; ( this.feet == that.feet && this.inches == that.inches); return true; else return false; } @Override public int hashCode(int, int) { int result = HashCodeUtil.inches; result = HashCodeUtil.hash(result, inches ); result = HashCodeUtil.hash(result, feet); ruturn result; }

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  • Find cosine similarity in R

    - by Derek
    I'm wondering if there is a built in function in R that can find the cosine similarity (or cosine distance) between two arrays? Currently, I implemented my own function, but I can't help but think that R should already come with one :) Thanks, Derek

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  • Using base64 encoding as a mechanism to detect changes

    - by Mikos
    Is it possible to detect changes in the base64 encoding of an object to detect the degree of changes in the object. Suppose I send a document attachment to several users and each makes changes to it and emails back to me, can I use the string distance between original base64 and the received base64s to detect which version has the most changes. Would that be a valid metric? If not, would there be any other metrics to quantify the deltas?

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  • I’m new to c++. can anyone help me in the following function [migrated]

    - by Laian Alsabbagh
    In this code I’m using a function Distance{to calculate the distance between two nodes } I declare the function like this : int Distance( int x1 , int y1 ,int x2 , y2){ int distance_x = x1-x2; int distance_y = y1- y2; int distance =sqrt( (distance_x * distance_x) + (distance_y * distance_y)); return distance ; } and in the main memory I have 2 for loops what iam asking for ,can I pass the values like this?? Distance (i, j , i+1 ,j+1) for(int i=0;i< No_Max;i++) { for(int j=0;j<No_Max;j++) { if( Distance (i, j , i+1 ,j+1)<=Radio_Range)//the function node_degree[i]=node_degree[i]+1; cout<<node_degree[i]<<endl; } }

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  • Restart program from a certain line with an if statement?

    - by user1744093
    could anyone help me restart my program from line 46 if the user enters 1 (just after the comment where it states that the next code is going to ask the user for 2 inputs) and if the user enters -1 end it. I cannot think how to do it. I'm new to C# any help you could give would be great! class Program { static void Main(string[] args) { //Displays data in correct Format List<float> inputList = new List<float>(); TextReader tr = new StreamReader("c:/users/tom/documents/visual studio 2010/Projects/DistanceCalculator3/DistanceCalculator3/TextFile1.txt"); String input = Convert.ToString(tr.ReadToEnd()); String[] items = input.Split(','); Console.WriteLine("Point Latitude Longtitude Elevation"); for (int i = 0; i < items.Length; i++) { if (i % 3 == 0) { Console.Write((i / 3) + "\t\t"); } Console.Write(items[i]); Console.Write("\t\t"); if (((i - 2) % 3) == 0) { Console.WriteLine(); } } Console.WriteLine(); Console.WriteLine(); // Ask for two inputs from the user which is then converted into 6 floats and transfered in class Coordinates Console.WriteLine("Please enter the two points that you wish to know the distance between:"); string point = Console.ReadLine(); string[] pointInput = point.Split(' '); int pointNumber = Convert.ToInt16(pointInput[0]); int pointNumber2 = Convert.ToInt16(pointInput[1]); Coordinates distance = new Coordinates(); distance.latitude = (Convert.ToDouble(items[pointNumber * 3])); distance.longtitude = (Convert.ToDouble(items[(pointNumber * 3) + 1])); distance.elevation = (Convert.ToDouble(items[(pointNumber * 3) + 2])); distance.latitude2 = (Convert.ToDouble(items[pointNumber2 * 3])); distance.longtitude2 = (Convert.ToDouble(items[(pointNumber2 * 3) + 1])); distance.elevation2 = (Convert.ToDouble(items[(pointNumber2 * 3) + 2])); //Calculate the distance between two points const double PIx = 3.141592653589793; const double RADIO = 6371; double dlat = ((distance.latitude2) * (PIx / 180)) - ((distance.latitude) * (PIx / 180)); double dlon = ((distance.longtitude2) * (PIx / 180)) - ((distance.longtitude) * (PIx / 180)); double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos((distance.latitude) * (PIx / 180)) * Math.Cos((distance.latitude2) * (PIx / 180)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2)); double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a)); double ultimateDistance = (angle * RADIO); Console.WriteLine("The distance between your two points is..."); Console.WriteLine(ultimateDistance); //Repeat the program if the user enters 1, end the program if the user enters -1 Console.WriteLine("If you wish to calculate another distance type 1 and return, if you wish to end the program, type -1."); Console.ReadLine(); if (Convert.ToInt16(Console.ReadLine()) == 1); { //here is where I need it to repeat }

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  • Objective c string formatter for distances

    - by nevan
    I have a distance as a float and I'm looking for a way to format it nicely for human readers. Ideally, I'd like it to change from m to km as it gets bigger, and to round the number nicely. Converting to miles would be a bonus. I'm sure many people have had a need for one of these and I'm hoping that there's some code floating around somewhere. Here's how I'd like the formats: 0-100m: 47m (as a whole number) 100-1000m: 325m or 320m (round to the nearest 5 or 10 meters) 1000-10000m: 1.2km (round to nearest with one decimal place) 10000m +: 21km If there's no code available, how can I write my own formatter? Thanks

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  • How to get nearby POIs

    - by balexandre
    I have a database with Points of Interest that all have an address. I want to know what is the method/name/call to get all nearby POIs from a given position. I understand that I need to convert all my addresses to LAT / LON coordinates at least, but my question is: for a given LAT / LONG how do I get from the database/array what POIs are nearby by distance, for example: You are here 0,0 nearest POIs in a 2km radius are: POI A (at 1.1 Km) POI C (at 1.3 Km) POI F (at 1.9 Km) I have no idea what should I look into to get what I want :-( Any help is greatly appreciated. Thank you

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  • Writing a post search algorithm.

    - by MdaG
    I'm trying to write a free text search algorithm for finding specific posts on a wall (similar kind of wall as Facebook uses). A user is suppose to be able to write some words in a search field and get hits on posts that contain the words; with the best match on top and then other posts in decreasing order according to match score. I'm using the edit distance (Levenshtein) "e(x, y) = e" to calculate the score for each post when compared to the query word "x" and post word "y" according to: score(x, y) = 2^(2 - e)(1 - min(e, |x|) / |x|) Each word in a post contributes to the total score for that specific post. This approach seems to work well when the posts are of roughly the same size, but sometime certain large posts manages to rack up score solely on having a lot of words in them while in practice not being relevant to the query. Am I approaching this problem in the wrong way or is there some way to normalize the score that I haven't thought of?

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  • Fastest way to find the closest point to a given point in 3D, in Python.

    - by Saebin
    So lets say I have 10,000 points in A and 10,000 points in B and want to find out the closest point in A for every B point. Currently, I simply loop through every point in B and A to find which one is closest in distance. ie. B = [(.5, 1, 1), (1, .1, 1), (1, 1, .2)] A = [(1, 1, .3), (1, 0, 1), (.4, 1, 1)] C = {} for bp in B: closestDist = -1 for ap in A: dist = sum(((bp[0]-ap[0])**2, (bp[1]-ap[1])**2, (bp[2]-ap[2])**2)) if(closestDist > dist or closestDist == -1): C[bp] = ap closestDist = dist print C However, I am sure there is a faster way to do this... any ideas?

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  • Calculating similarites between sentences

    - by codecreator
    I have datbase with thousands of rows of error logs and their description.This error log is for an application that running 24/7. I want to create a dashboard/UI to view the current common errors happening for prodcution support. The problem I am having is that even though there are lot of common errors, the error description differs by the transcation ID or user ID or things that are unique for that sigle prcoess. e.g Error trasaction XYz failed for user 233 e.g 2. Error trasaction XYz failed for user 567 I consider these two erros to be same. So I want to a program that will go through the new error logs and classify them into groups. I am trying to use "edit distance" but its very slow.Since I alraedy have old error logs, i am trying to think of solutions using that information too. Any thoughts?

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  • What Java data structure/design pattern best models this object, considering it would perform these methods?

    - by zundarz
    Methods: 1. getDistance(CityA,CityB) // Returns distance between two cities 2. getCitiesInRadius(CityA,integer) // Returns cities within a given distance of another city 3. getCitiesBeyondRadius(CityA,integer) //Returns cities beyond a given distance of another city 4. getRemoteDestinations(integer) // Returns all city pairs greater than x distance of each other 5. getLocalDestinations(integer) //Returns all city pairs within x distance of each other

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  • What OO Design to use ( is there a Design Pattern )?

    - by Blundell
    I have two objects that represent a 'Bar/Club' ( a place where you drink/socialise). In one scenario I need the bar name, address, distance, slogon In another scenario I need the bar name, address, website url, logo So I've got two objects representing the same thing but with different fields. I like to use immutable objects, so all the fields are set from the constructor. One option is to have two constructors and null the other fields i.e: class Bar { private final String name; private final Distance distance; private final Url url; public Bar(String name, Distance distance){ this.name = name; this.distance = distance; this.url = null; } public Bar(String name, Url url){ this.name = name; this.distance = null; this.url = url; } // getters } I don't like this as you would have to null check when you use the getters In my real example the first scenario has 3 fields and the second scenario has about 10, so it would be a real pain having two constructors, the amount of fields I would have to declare null and then when the object are in use you wouldn't know which Bar you where using and so what fields would be null and what wouldn't. What other options do I have? Two classes called BarPreview and Bar? Some type of inheritance / interface? Something else that is awesome?

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  • How to get distance from point to line with distinction between side of line?

    - by tesselode
    I'm making a 2d racing game. I'm taking the nice standard approach of having a set of points defining the center of the track and detecting whether the car is off the track by detecting its distance from the nearest point. The nicest way I've found of doing this is using the formula: d = |Am + Bn + C| / sqrt(A^2 + B^2) Unfortunately, to have proper collision resolution, I need to know which side of the line the car is hitting, but I can't do that with this formula because it only returns positive numbers. So my question is: is there a formula that will give me positive or negative numbers based on which side of the line the point is on? Can I just get rid of the absolute value in the formula or do I need to do something else?

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