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  • How can I draw an arrow at the edge of the screen pointing to an object that is off screen?

    - by Adam Henderson
    I am wishing to do what is described in this topic: http://www.allegro.cc/forums/print-thread/283220 I have attempted a variety of the methods mentioned here. First I tried to use the method described by Carrus85: Just take the ratio of the two triangle hypontenuses (doesn't matter which triagle you use for the other, I suggest point 1 and point 2 as the distance you calculate). This will give you the aspect ratio percentage of the triangle in the corner from the larger triangle. Then you simply multiply deltax by that value to get the x-coordinate offset, and deltay by that value to get the y-coordinate offset. But I could not find a way to calculate how far the object is away from the edge of the screen. I then tried using ray casting (which I have never done before) suggested by 23yrold3yrold: Fire a ray from the center of the screen to the offscreen object. Calculate where on the rectangle the ray intersects. There's your coordinates. I first calculated the hypotenuse of the triangle formed by the difference in x and y positions of the two points. I used this to create a unit vector along that line. I looped through that vector until either the x coordinate or the y coordinate was off the screen. The two current x and y values then form the x and y of the arrow. Here is the code for my ray casting method (written in C++ and Allegro 5) void renderArrows(Object* i) { float x1 = i->getX() + (i->getWidth() / 2); float y1 = i->getY() + (i->getHeight() / 2); float x2 = screenCentreX; float y2 = ScreenCentreY; float dx = x2 - x1; float dy = y2 - y1; float hypotSquared = (dx * dx) + (dy * dy); float hypot = sqrt(hypotSquared); float unitX = dx / hypot; float unitY = dy / hypot; float rayX = x2 - view->getViewportX(); float rayY = y2 - view->getViewportY(); float arrowX = 0; float arrowY = 0; bool posFound = false; while(posFound == false) { rayX += unitX; rayY += unitY; if(rayX <= 0 || rayX >= screenWidth || rayY <= 0 || rayY >= screenHeight) { arrowX = rayX; arrowY = rayY; posFound = true; } } al_draw_bitmap(sprite, arrowX - spriteWidth, arrowY - spriteHeight, 0); } This was relatively successful. Arrows are displayed in the bottom right section of the screen when objects are located above and left of the screen as if the locations of the where the arrows are drawn have been rotated 180 degrees around the center of the screen. I assumed this was due to the fact that when I was calculating the hypotenuse of the triangle, it would always be positive regardless of whether or not the difference in x or difference in y is negative. Thinking about it, ray casting does not seem like a good way of solving the problem (due to the fact that it involves using sqrt() and a large for loop). Any help finding a suitable solution would be greatly appreciated, Thanks Adam

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  • Cutting edge technology, a lone Movember ranger and a 5-a-side football club ...meet the team at Oracle’s Belfast Offices.

    - by user10729410
    Normal 0 false false false EN-IE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Normal 0 false false false EN-IE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} By Olivia O’Connell To see what’s in store at Oracle’s next Open Day which comes to Belfast this week, I visited the offices with some colleagues to meet the team and get a feel for what‘s in store on November 29th. After being warmly greeted by Frances and Francesca, who make sure Front of House and Facilities run smoothly, we embarked on a quick tour of the 2 floors Oracle occupies, led by VP Bo, it was time to seek out some willing volunteers to be interviewed/photographed - what a shy bunch! A bit of coaxing from the social media team was needed here! In a male-dominated environment, the few women on the team caught my eye immediately. I got chatting to Susan, a business analyst and Bronagh, a tech writer. It becomes clear during our chat that the male/female divide is not an issue – “everyone here just gets on with the job,” says Suzanne, “We’re all around the same age and have similar priorities and luckily everyone is really friendly so there are no problems. ” A graduate of Queen’s University in Belfast majoring in maths & computer science, Susan works closely with product management and the development teams to ensure that the final project delivered to clients meets and exceeds their expectations. Bronagh, who joined us following working for a tech company in Montreal and gaining her post-grad degree at University of Ulster agrees that the work is challenging but “the environment is so relaxed and friendly”. Normal 0 false false false EN-IE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Software developer David is taking the Movember challenge for the first time to raise vital funds and awareness for men’s health. Like other colleagues in the office, he is a University of Ulster graduate and works on Reference applications and Merchandising Tools which enable customers to establish e-shops using Oracle technologies. The social activities are headed up by Gordon, a software engineer on the commerce team who joined the team 4 years ago after graduating from the University of Strathclyde at Glasgow with a degree in Computer Science. Everyone is unanimous that the best things about working at Oracle’s Belfast offices are the casual friendly environment and the opportunity to be at the cutting edge of technology. We’re looking forward to our next trip to Belfast for some cool demos and meet candidates. And as for the camera-shyness? Look who came out to have their picture taken at the end of the day! Normal 0 false false false EN-IE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} The Oracle offices in Belfast are located on the 6th floor, Victoria House, Gloucester Street, Belfast BT1 4LS, UK View Larger Map Normal 0 false false false EN-IE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Open day takes place on Thursday, 29th November 4pm – 8pm. Visit the 5 Demo Stations to find out more about each teams' activities and projects to date. See live demos including "Engaging the Customer", "Managing Your Store", "Helping the Customer", "Shopping on-line" and "The Commerce Experience" processes. The "Working @Oracle" stand will give you the chance to connect with our recruitment team and get information about the Recruitment process and making your career path in Oracle. Register here.

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  • AABB vs OBB Collision Resolution jitter on corners

    - by patt4179
    I've implemented a collision library for a character who is an AABB and am resolving collisions between AABB vs AABB and AABB vs OBB. I wanted slopes for certain sections, so I've toyed around with using several OBBs to make one, and it's working great except for one glaring issue; The collision resolution on the corner of an OBB makes the player's AABB jitter up and down constantly. I've tried a few things I've thought of, but I just can't wrap my head around what's going on exactly. Here's a video of what's happening as well as my code: Here's the function to get the collision resolution (I'm likely not doing this the right way, so this may be where the issue lies): public Vector2 GetCollisionResolveAmount(RectangleCollisionObject resolvedObject, OrientedRectangleCollisionObject b) { Vector2 overlap = Vector2.Zero; LineSegment edge = GetOrientedRectangleEdge(b, 0); if (!SeparatingAxisForRectangle(edge, resolvedObject)) { LineSegment rEdgeA = new LineSegment(), rEdgeB = new LineSegment(); Range axisRange = new Range(), rEdgeARange = new Range(), rEdgeBRange = new Range(), rProjection = new Range(); Vector2 n = edge.PointA - edge.PointB; rEdgeA.PointA = RectangleCorner(resolvedObject, 0); rEdgeA.PointB = RectangleCorner(resolvedObject, 1); rEdgeB.PointA = RectangleCorner(resolvedObject, 2); rEdgeB.PointB = RectangleCorner(resolvedObject, 3); rEdgeARange = ProjectLineSegment(rEdgeA, n); rEdgeBRange = ProjectLineSegment(rEdgeB, n); rProjection = GetRangeHull(rEdgeARange, rEdgeBRange); axisRange = ProjectLineSegment(edge, n); float axisMid = (axisRange.Maximum + axisRange.Minimum) / 2; float projectionMid = (rProjection.Maximum + rProjection.Minimum) / 2; if (projectionMid > axisMid) { overlap.X = axisRange.Maximum - rProjection.Minimum; } else { overlap.X = rProjection.Maximum - axisRange.Minimum; overlap.X = -overlap.X; } } edge = GetOrientedRectangleEdge(b, 1); if (!SeparatingAxisForRectangle(edge, resolvedObject)) { LineSegment rEdgeA = new LineSegment(), rEdgeB = new LineSegment(); Range axisRange = new Range(), rEdgeARange = new Range(), rEdgeBRange = new Range(), rProjection = new Range(); Vector2 n = edge.PointA - edge.PointB; rEdgeA.PointA = RectangleCorner(resolvedObject, 0); rEdgeA.PointB = RectangleCorner(resolvedObject, 1); rEdgeB.PointA = RectangleCorner(resolvedObject, 2); rEdgeB.PointB = RectangleCorner(resolvedObject, 3); rEdgeARange = ProjectLineSegment(rEdgeA, n); rEdgeBRange = ProjectLineSegment(rEdgeB, n); rProjection = GetRangeHull(rEdgeARange, rEdgeBRange); axisRange = ProjectLineSegment(edge, n); float axisMid = (axisRange.Maximum + axisRange.Minimum) / 2; float projectionMid = (rProjection.Maximum + rProjection.Minimum) / 2; if (projectionMid > axisMid) { overlap.Y = axisRange.Maximum - rProjection.Minimum; overlap.Y = -overlap.Y; } else { overlap.Y = rProjection.Maximum - axisRange.Minimum; } } return overlap; } And here is what I'm doing to resolve it right now: if (collisionDetection.OrientedRectangleAndRectangleCollide(obb, player.PlayerCollision)) { var resolveAmount = collisionDetection.GetCollisionResolveAmount(player.PlayerCollision, obb); if (Math.Abs(resolveAmount.Y) < Math.Abs(resolveAmount.X)) { var roundedAmount = (float)Math.Floor(resolveAmount.Y); player.PlayerCollision._position.Y -= roundedAmount; } else if (Math.Abs(resolveAmount.Y) <= 30.0f) //Catch cases where the player should be able to step over the top of something { var roundedAmount = (float)Math.Floor(resolveAmount.Y); player.PlayerCollision._position.Y -= roundedAmount; } else { var roundedAmount = (float)Math.Floor(resolveAmount.X); player.PlayerCollision._position.X -= roundedAmount; } } Can anyone see what might be the issue here, or has anyone experienced this before that knows a possible solution? I've tried for a few days to figure this out on my own, but I'm just stumped.

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  • Top/left edges of screen, in Virtualbox, act like bottom/right edges

    - by Ken
    I have Virtualbox running on Windows Vista, and Debian running inside Virtualbox. Everything's running great, for the most part. Everything looks correct. But when I'm in full-screen mode, the top edge seems to act (to the mouse) like it's the bottom edge, and the left edge seems to act like the right edge. For example, if I click in the middle of the desktop and drag left, as if to select some icons, when I hit the very leftmost pixel of the screen, the selection (but not the mouse pointer) jumps to the far right edge of the screen). For the left edge, it's not such a big deal, but not having the top edge is kind of annoying: it means I can't select things from the menu in my top panel by slamming the mouse against the top of the screen. Anyone seen this before? Is there some way to make this work? Thanks!

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  • How to scan convert right edges and slopes less than one?

    - by Zachary
    I'm writing a program which will use scan conversion on triangles to fill in the pixels contained within the triangle. One thing that has me confused is how to determine the x increment for the right edge of the triangle, or for slopes less than or equal to one. Here is the code I have to handle left edges with a slope greater than one (obtained from Computer Graphics: Principles and Practice second edition): for(y=ymin;y<=ymax;y++) { edge.increment+=edge.numerator; if(edge.increment>edge.denominator) { edge.x++; edge.increment -= edge.denominator; } } The numerator is set from (xMax-xMin), and the denominator is set from (yMax-yMin)...which makes sense as it represents the slope of the line. As you move up the scan lines (represented by the y values). X is incremented by 1/(denomniator/numerator) ...which results in x having a whole part and a fractional part. If the fractional part is greater than one, then the x value has to be incremented by 1 (as shown in edge.incrementedge.denominator). This works fine for any left handed lines with a slope greater than one, but I'm having trouble generalizing it for any edge, and google-ing has proved fruitless. Does anyone know the algorithm for that?

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  • Concurrency problem with arrays (Java)

    - by Johannes
    For an algorithm I'm working on I tried to develop a blacklisting mechanism that can blacklist arrays in a specific way: If "1, 2, 3" is blacklisted "1, 2, 3, 4, 5" is also considered blacklisted. I'm quite happy with the solution I've come up with so far. But there seem to be some serious problems when I access a blacklist from multiple threads. The method "contains" (see code below) sometimes returns true, even if an array is not blacklisted. This problem does not occur if I only use one thread, so it most likely is a concurrency problem. I've tried adding some synchronization, but it didn't change anything. I also tried some slightly different implementations using java.util.concurrent classes. Any ideas on how to fix this? public class Blacklist { private static final int ARRAY_GROWTH = 10; private final Node root = new Node(); private static class Node{ private volatile Node[] childNodes = new Node[ARRAY_GROWTH]; private volatile boolean blacklisted = false; public void blacklist(){ this.blacklisted = true; this.childNodes = null; } } public void add(final int[] array){ synchronized (root) { Node currentNode = this.root; for(final int edge : array){ if(currentNode.blacklisted) return; else if(currentNode.childNodes.length <= edge) { currentNode.childNodes = Arrays.copyOf(currentNode.childNodes, edge + ARRAY_GROWTH); } if(currentNode.childNodes[edge] == null) { currentNode.childNodes[edge] = new Node(); } currentNode = currentNode.childNodes[edge]; } currentNode.blacklist(); } } public boolean contains(final int[] array){ synchronized (root) { Node currentNode = this.root; for(final int edge : array){ if(currentNode.blacklisted) return true; else if(currentNode.childNodes.length <= edge || currentNode.childNodes[edge] == null) return false; currentNode = currentNode.childNodes[edge]; } return currentNode.blacklisted; } } }

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  • Drawing Directed Acyclic Graphs: Using DAG property to improve layout/edge routing?

    - by Robert Fraser
    Hi, Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of:

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  • LINQ, creating unique collection of a collection

    - by Wish
    I have class Vertex and a class Edge (Edge holds 2 properties - Vertex Source and Vertex Target); Edges and Vertexes are collected into lists Some example: A-->B // edge from vertex A to B B-->C // edge from vertex B to C C-->A // edge from vertex C to A A-->C // edge from vertex A to C -- this is two way edge So I would like to make IDictionary<Edge, bool> which would hold edges (A--B and B--A would be like 1), and bool - if it is two way or no. I need it because when I draw them now, it draws 2 arrows under one another. I would better make 1 arrow. So I'm pretty stuck right here... May anybody help me a bit ?

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  • How do I stop that launcher from popping up when mouse goes to the left edge?

    - by Michael
    This is what it says on the help.ubuntu.com website, but when i go to system settings, it does not have a Launcher & Menus option. I wish I could revert back to the old Ubuntu, I hate this launcher. Does any know a fix plz? "Stop the launcher from showing when I point to the left side of the screen The launcher unhides when you move your mouse or touchpad pointer to the left side of the screen. If you prefer, you can tell Unity to only show the launcher when you click the top left Ubuntu button instead. Click the icon at the very right of the top bar and select System Settings. In the Personal section, click Launcher & Menus. Choose Touches the top left corner of the screen"

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  • How can I solve this SAT direct corner intersection edge case?

    - by ssb
    I have a working SAT implementation, but I am running into a problem where direct collisions at a corner do not work for tiled surfaces. That is, it clips on the surface when going in a certain direction because it gets hung up on one of the tiles, and so, for example, if I walk across a floor while holding both down and left, the player will stop when meeting the next shape because the player will be colliding with the right side rather than with the top of the floor tile. This illustration shows what I mean: The top block will translate right first and then up. I have checked here and here which are helpful, but this does not address what I should do in a situation where I don't have a tile-based world. My usage of the term "tile" before isn't really accurate since what I'm doing here is manually placing square obstacles next to each other, not assigning them spots on a grid. What can I do to fix this?

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  • What would be a good topic for research on "edge of multiple processors / computers programming" topic?

    - by Kabumbus
    This is a subjective discussion so we can express our dreams and hopes here. A "topic" must be like a task with point to have as end result a software poduct. A "topic" must be mainly about "Software engineering", "Algorithm and data structure concepts" and perhaps "Design patterns". I mean let us try to look what is not already there? What can be developed in fiew month and give a breakthrue / start a new leap / show somethig not realized before in science of f multiple computers programming? What i see is already there: LAN / wire and other infrastractural programms for connecting on device level MPI/ Bit torrent/Jabber protocols / APIs / servers for messaging on top Boost and analogs on evry OS in most languages for multithreading there are lots of CUDA like on computer frameworks for fast calculating on computers GPUs What I personally do not see out there is a crossplatform framework for multiple processes interaction. Meaning one that would allow easy creation of multyple processes running in paralell inside one hoster app on one machine. In level not harder than needed for threads creation (so no seprate server apps - just one lib doing it all) Is there ny such lib and what can you propose for research topic?

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  • How can I get textures on edge of walls like in Super Metroid and Aquaria?

    - by meds
    Games like Super Metroid and Aquaria present the terrain with the other facing parts having rocks and stuff while deeper behind them (i.e. underground) there's different detail or just black. I would like to do something similar using polygons. Terrain is created in my current level as a set of overlapping square boxes. I'm not sure if this rendering method will work such a system for creating terrain but if anyone has ideas I'd love to hear them. Otherwise I'd like to know how I should re-write the terrain rendering system so it actually works to draw terrain in this manner...

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  • Windows 7: How to disable auto-maximize/resize window (aero-snap) when near screen edge?

    - by glenneroo
    Whenever I drag or resize a window near any edge or corner of either monitor, Windows wants to maximize or resize the window for me in several different ways: dragging a window near a corner offers resize to full-screen dragging directly to the corner offers half-screen maximize resizing to top or bottom edges offers a vertical maximize (one long strip from top to bottom) Actually, now that I think about it the 3rd one isn't so bad, its just the full and half-screen maximize features that drives me mad. Is there a registry hack to disable these settings, preferably independently?

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  • How do I balance program CPU reverse compatibility whist still being able to use cutting edge features?

    - by TheLQ
    As I learn more about C and C++ I'm starting to wonder: How can a compiler use newer features of processors without limiting it just to people with, for example, Intel Core i7's? Think about it: new processors come out every year with lots of new technologies. However you can't just only target them since a significant portion of the market will not upgrade to the latest and greatest processors for a long time. I'm more or less wondering how this is handled in general by C and C++ devs and compilers. Do compilers make code similar to if SSE is supported, do this using it, else do that using the slower way or do developers have to implement their algorithm twice, or what? More or less how do you release software that takes advantage of newer processor technologies while still keeping a low common denominator?

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  • New Features, Fresh Competitive Edge: Help Your Customers Get the Most from Oracle RightNow Cloud Service by Sandi Main

    - by Tuula Fai
    Are your customers taking full advantage of their customer service solution? If they’re not up-to-speed with the new features of Oracle RightNow Cloud Service, this is their chance to catch up—and gain competitive advantage. Invite customers to join our live Webcast on Thursday, July 16, for a rapid overview of Oracle RightNow Cloud Service's most powerful new capabilities and what they mean for their organization. They’ll learn how they can: Take full advantage of RightNow's evolution under Oracle Stay ahead of the competition with key innovations and enhancements Upgrade with confidence based on other user experiences Don't miss the chance to help your customers deliver an even smarter customer experience.  Click here for the Evite to send to your customers.

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  • solve TOR edge node problem by using .onion proxy?

    - by rd.
    I would like to improve the TOR network, where the exit nodes are a vulnerability to concealing traffic. From my understanding, traffic to .onion sites are not decrypted by exit nodes, so therefore - in theory - a .onion site web proxy could be used to further anonymize traffic. Yes/no? perhaps you have insight into the coding and routing behind these concepts to elaborate on why this is a good/not good idea.

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  • Is it possible to prevent windows in Mac OS from opening below the bottom edge of the screen?

    - by user31262
    When opening a new window in finder, the new window will go below the screen. This is not a big deal, but when using macvim or terminal it can be annoying since your command prompt is now hidden. Firefox seems to be smarter, it never overlaps below the screen. Same deal for the right hand side of the screen, windows often open with their right edge cut off. Is there any way to prevent this?

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  • What would be topic for research in on edge of multiple processors / computers programming?

    - by Kabumbus
    I mean what is not already there? What can be developed in fiew month and give a breakthrue/ start a new leap in science of f multiple computers programming? What i see is already there MPI/ Bit torrent/Jabber protocols / APIs / servers for messaging LAN / wire and other infrastractural cabels for connecting Boost and analogs on evry OS in most languages for multithreading there are lots of CUDA like on computer frameworks for fast calculating on computers GPUs What I personally do not see out there is a crossplatform framework for multiple processes interaction. Meaning one that would allow easy creation of multyple processes running in paralell inside one hoster app on one machine. In level not harder than needed for threads creation (so no seprate server apps - just one lib doing it all) Is there ny such lib and what can you propose for research topic?

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  • Prim's MST algorithm implementation with Java

    - by user1290164
    I'm trying to write a program that'll find the MST of a given undirected weighted graph with Kruskal's and Prim's algorithms. I've successfully implemented Kruskal's algorithm in the program, but I'm having trouble with Prim's. To be more precise, I can't figure out how to actually build the Prim function so that it'll iterate through all the vertices in the graph. I'm getting some IndexOutOfBoundsException errors during program execution. I'm not sure how much information is needed for others to get the idea of what I have done so far, but hopefully there won't be too much useless information. This is what I have so far: I have a Graph, Edge and a Vertex class. Vertex class mostly just an information storage that contains the name (number) of the vertex. Edge class can create a new Edge that has gets parameters (Vertex start, Vertex end, int edgeWeight). The class has methods to return the usual info like start vertex, end vertex and the weight. Graph class reads data from a text file and adds new Edges to an ArrayList. The text file also tells us how many vertecis the graph has, and that gets stored too. In the Graph class, I have a Prim() -method that's supposed to calculate the MST: public ArrayList<Edge> Prim(Graph G) { ArrayList<Edge> edges = G.graph; // Copies the ArrayList with all edges in it. ArrayList<Edge> MST = new ArrayList<Edge>(); Random rnd = new Random(); Vertex startingVertex = edges.get(rnd.nextInt(G.returnVertexCount())).returnStartingVertex(); // This is just to randomize the starting vertex. // This is supposed to be the main loop to find the MST, but this is probably horribly wrong.. while (MST.size() < returnVertexCount()) { Edge e = findClosestNeighbour(startingVertex); MST.add(e); visited.add(e.returnStartingVertex()); visited.add(e.returnEndingVertex()); edges.remove(e); } return MST; } The method findClosesNeighbour() looks like this: public Edge findClosestNeighbour(Vertex v) { ArrayList<Edge> neighbours = new ArrayList<Edge>(); ArrayList<Edge> edges = graph; for (int i = 0; i < edges.size() -1; ++i) { if (edges.get(i).endPoint() == s.returnVertexID() && !visited(edges.get(i).returnEndingVertex())) { neighbours.add(edges.get(i)); } } return neighbours.get(0); // This is the minimum weight edge in the list. } ArrayList<Vertex> visited and ArrayList<Edges> graph get constructed when creating a new graph. Visited() -method is simply a boolean check to see if ArrayList visited contains the Vertex we're thinking about moving to. I tested the findClosestNeighbour() independantly and it seemed to be working but if someone finds something wrong with it then that feedback is welcome also. Mainly though as I mentioned my problem is with actually building the main loop in the Prim() -method, and if there's any additional info needed I'm happy to provide it. Thank you. Edit: To clarify what my train of thought with the Prim() method is. What I want to do is first randomize the starting point in the graph. After that, I will find the closest neighbor to that starting point. Then we'll add the edge connecting those two points to the MST, and also add the vertices to the visited list for checking later, so that we won't form any loops in the graph. Here's the error that gets thrown: Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 0, Size: 0 at java.util.ArrayList.rangeCheck(Unknown Source) at java.util.ArrayList.get(Unknown Source) at Graph.findClosestNeighbour(graph.java:203) at Graph.Prim(graph.java:179) at MST.main(MST.java:49) Line 203: return neighbour.get(0); in findClosestNeighbour() Line 179: Edge e = findClosestNeighbour(startingVertex); in Prim()

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  • Scala: Recursively building all pathes in a graph?

    - by DarqMoth
    Trying to build all existing paths for an udirected graph defined as a map of edges using the following algorithm: Start: with a given vertice A Find an edge (X.A, X.B) or (X.B, X.A), add this edge to path Find all edges Ys fpr which either (Y.C, Y.B) or (Y.B, Y.C) is true For each Ys: A=B, goto Start Providing edges are defined as the following map, where keys are tuples consisting of two vertices: val edges = Map( ("n1", "n2") -> "n1n2", ("n1", "n3") -> "n1n3", ("n3", "n4") -> "n3n4", ("n5", "n1") -> "n5n1", ("n5", "n4") -> "n5n4") As an output I need to get a list of ALL pathes where each path is a list of adjecent edges like this: val allPaths = List( List(("n1", "n2") -> "n1n2"), List(("n1", "n3") -> "n1n3", ("n3", "n4") -> "n3n4"), List(("n5", "n1") -> "n5n1"), List(("n5", "n4") -> "n5n4"), List(("n2", "n1") -> "n1n2", ("n1", "n3") -> "n1n3", ("n3", "n4") -> "n3n4", ("n5", "n4") -> "n5n4")) //... //... more pathes to go } Note: Edge XY = (x,y) - "xy" and YX = (y,x) - "yx" exist as one instance only, either as XY or YX So far I have managed to implement code that duplicates edges in the path, which is wrong and I can not find the error: object Graph2 { type Vertice = String type Edge = ((String, String), String) type Path = List[((String, String), String)] val edges = Map( //(("v1", "v2") , "v1v2"), (("v1", "v3") , "v1v3"), (("v3", "v4") , "v3v4") //(("v5", "v1") , "v5v1"), //(("v5", "v4") , "v5v4") ) def main(args: Array[String]): Unit = { val processedVerticies: Map[Vertice, Vertice] = Map() val processedEdges: Map[(Vertice, Vertice), (Vertice, Vertice)] = Map() val path: Path = List() println(buildPath(path, "v1", processedVerticies, processedEdges)) } /** * Builds path from connected by edges vertices starting from given vertice * Input: map of edges * Output: list of connected edges like: List(("n1", "n2") -> "n1n2"), List(("n1", "n3") -> "n1n3", ("n3", "n4") -> "n3n4"), List(("n5", "n1") -> "n5n1"), List(("n5", "n4") -> "n5n4"), List(("n2", "n1") -> "n1n2", ("n1", "n3") -> "n1n3", ("n3", "n4") -> "n3n4", ("n5", "n4") -> "n5n4")) */ def buildPath(path: Path, vertice: Vertice, processedVerticies: Map[Vertice, Vertice], processedEdges: Map[(Vertice, Vertice), (Vertice, Vertice)]): List[Path] = { println("V: " + vertice + " VM: " + processedVerticies + " EM: " + processedEdges) if (!processedVerticies.contains(vertice)) { val edges = children(vertice) println("Edges: " + edges) val x = edges.map(edge => { if (!processedEdges.contains(edge._1)) { addToPath(vertice, processedVerticies.++(Map(vertice -> vertice)), processedEdges, path, edge) } else { println("ALready have edge: "+edge+" Return path:"+path) path } }) val y = x.toList y } else { List(path) } } def addToPath( vertice: Vertice, processedVerticies: Map[Vertice, Vertice], processedEdges: Map[(Vertice, Vertice), (Vertice, Vertice)], path: Path, edge: Edge): Path = { val newPath: Path = path ::: List(edge) val key = edge._1 val nextVertice = neighbor(vertice, key) val x = buildPath (newPath, nextVertice, processedVerticies, processedEdges ++ (Map((vertice, nextVertice) -> (vertice, nextVertice))) ).flatten // need define buidPath type x } def children(vertice: Vertice) = { edges.filter(p => (p._1)._1 == vertice || (p._1)._2 == vertice) } def containsPair(x: (Vertice, Vertice), m: Map[(Vertice, Vertice), (Vertice, Vertice)]): Boolean = { m.contains((x._1, x._2)) || m.contains((x._2, x._1)) } def neighbor(vertice: String, key: (String, String)): String = key match { case (`vertice`, x) => x case (x, `vertice`) => x } } Running this results in: List(List(((v1,v3),v1v3), ((v1,v3),v1v3), ((v3,v4),v3v4))) Why is that?

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  • how do I make my submenu position dynamic based on the distance to the edge of the window?

    - by Mario Antoci
    I'm trying to write a jQuery script that will find the distance to the right edge of the browser window from my css class element and then position the child submenu dropdowns to the right or left depending on the available space to the right. Also it needs to revert to the default settings on hoverout. Here is what I have so far but it's not calculating properly. $(document).ready(function(){ $('#dnnMenu .subLevel').hover(function(){ if ($(window).width() - $('#dnnMenu .subLevel').offset().left - '540' >= '270') { $('#dnnMenu .subLevelRight').css('left', '270px');} else {$('#dnnMenu .subLevelRight').css('left', '-270px');} }); $(document).ready(function () { function HoverOver() { $(this).addClass('hover'); } function HoverOut() { $(this).removeClass('hover'); } var config = { sensitivity: 2, interval: 100, over: HoverOver, timeout: 100, out: HoverOut }; $("#dnnMenu .topLevel > li.haschild").hoverIntent(config); $(".subLevel li.haschild").hover(HoverOver, HoverOut); }); Basically I tried to take the width of the current window, minus the distance to the left edge of the browser of the first level submenu, minus the width of both elements together which would equal 540px, to calculate the distance to the right edge of the window when the first level submenu is hovered over. if the distance to the right of my first level submenu element is less than 540px then the second level sub menu css property is changed to position to the left instead of right. I also know that it needs to revert back to default after hover out so it can recalculate the distance from other positions within the menu structure and still have those second level submenus with enough room to still display on the right of the first level. here is css for the elements in question. #dnnMenu .subLevel{ display: none; position: absolute; margin: 0; z-index: 1210; background: #639ec8; text-transform: none;} #dnnMenu .subLevelRight{ position: absolute; display: none; left: 270px; top: 0px;} The site's not live yet and I tried to create a jsfiddle but it doesn't look right. Any help would be greatly appreciated! Best Regards, Mario

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  • How to switch off the monitor when mouse reaches the edge of the screen?

    - by evgeny9
    I have 2 computers at home (Windows XP and Windows 7), but one monitor for both of them. They are connected to this monitor using different interfaces: DVI and VGA. I'm also using one keyboard and one mouse to control both PCs with the help of Synergy or Input Director. But I still need to manually switch between monitor interfaces. I wonder, if there's some way (software) that will switch this interfaces (turn off the monitor), when reach the edge of the screen with the mouse. Until now I found several answers, which help to avoid pressing hardware buttons, but still can not do the job automatically based on mouse pointer coordinates. Thank you.

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