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  • XNA - Use Mouse To Rotate & Arrow Keys To Scroll A Linearly Wrapped Texture:

    - by The Thing
    Using XNA I'm working on my first, relatively simple, videogame for the PC. At the moment my game window is 1024 X 768 and I have a 'Starfield' linearly wrapped background texture 1280 X 1280 in size whose origin has been set to its center point (width / 2, height / 2). This texture is drawn onscreen using (graphics.PreferredBackBufferWidth / 2, graphics.PreferredBackBufferHeight / 2) to place the origin in the center of the window. I want to be able to use the horizontal movement of the mouse to rotate my texture left or right and use the arrow keys to scroll the texture in four directions. From my own related coding experiments I have found that once I rotate the texture it no longer scrolls in the direction I want, it's as if somehow the XNA framework's 'sense of direction' has been 'rotated' along with the texture. As an example of what I've described above lets say I rotate the texture 45 degrees to the right, then pressing the up arrow key results in the texture scrolling diagonally from top-right to bottom-left. This is not what I want, regardless of the degree or direction of rotation I want my texture to scroll straight up, straight down, or to the left or right depending on which arrow key was pressed. How do I go about accomplishing this? Any help or guidance is appreciated. To finish up there are two points I'd like to clarify: [1] The reason I'm using linear wrapping on my starfield texture is that it gives a nice impression of an endless starfield. [2] Using a texture at least 1280 X 1280 in conjunction with a game window of 1024 X 768 means that at no point in it's rotation will the edges of the texture become visible. Thanks for reading..... Update # 1 - as requested by RCIX: The code below is what I was referring to earlier when I mentioned 'related coding experiments'. As you can see I am scrolling a linearly wrapped texture in the direction I've moved the mouse relative to the center of the screen. This works perfectly if I don't rotate the texture, but once I do rotate it the direction of the scrolling gets messed up for some reason. public class Game1 : Microsoft.Xna.Framework.Game { GraphicsDeviceManager graphics; SpriteBatch spriteBatch; int x; int y; float z = 250f; Texture2D Overlay; Texture2D RotatingBackground; Rectangle? sourceRectangle; Color color; float rotation; Vector2 ScreenCenter; Vector2 Origin; Vector2 scale; Vector2 Direction; SpriteEffects effects; float layerDepth; public Game1() { graphics = new GraphicsDeviceManager(this); Content.RootDirectory = "Content"; } protected override void Initialize() { graphics.PreferredBackBufferWidth = 1024; graphics.PreferredBackBufferHeight = 768; graphics.ApplyChanges(); Direction = Vector2.Zero; IsMouseVisible = true; ScreenCenter = new Vector2(graphics.PreferredBackBufferWidth / 2, graphics.PreferredBackBufferHeight / 2); Mouse.SetPosition((int)graphics.PreferredBackBufferWidth / 2, (int)graphics.PreferredBackBufferHeight / 2); sourceRectangle = null; color = Color.White; rotation = 0.0f; scale = new Vector2(1.0f, 1.0f); effects = SpriteEffects.None; layerDepth = 1.0f; base.Initialize(); } protected override void LoadContent() { spriteBatch = new SpriteBatch(GraphicsDevice); Overlay = Content.Load<Texture2D>("Overlay"); RotatingBackground = Content.Load<Texture2D>("Background"); Origin = new Vector2((int)RotatingBackground.Width / 2, (int)RotatingBackground.Height / 2); } protected override void UnloadContent() { } protected override void Update(GameTime gameTime) { float timePassed = (float)gameTime.ElapsedGameTime.TotalSeconds; MouseState ms = Mouse.GetState(); Vector2 MousePosition = new Vector2(ms.X, ms.Y); Direction = ScreenCenter - MousePosition; if (Direction != Vector2.Zero) { Direction.Normalize(); } x += (int)(Direction.X * z * timePassed); y += (int)(Direction.Y * z * timePassed); //No rotation = texture scrolls as intended, With rotation = texture no longer scrolls in the direction of the mouse. My update method needs to somehow compensate for this. //rotation += 0.01f; base.Update(gameTime); } protected override void Draw(GameTime gameTime) { spriteBatch.Begin(SpriteSortMode.Deferred, null, SamplerState.LinearWrap, null, null); spriteBatch.Draw(RotatingBackground, ScreenCenter, new Rectangle(x, y, RotatingBackground.Width, RotatingBackground.Height), color, rotation, Origin, scale, effects, layerDepth); spriteBatch.Draw(Overlay, Vector2.Zero, Color.White); spriteBatch.End(); base.Draw(gameTime); } }

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  • UIScrollView image/photo viewer with paging enabled and zooming

    - by Mike Weller
    OK, I think it's time to make an official place on the internet for this problem: How to make a UIScrollView photoviewer with paging and zooming. Welcome my fellow UIScrollView hackers. I have a UIScrollView with paging enabled, and I'm displaying UIImageViews like the built-in photos app. (Does this sound familiar yet?) I found the following project on github: http://wiki.github.com/andreyvit/ScrollingMadness Which shows how to implement zooming in a scroll view while paging is enabled. If anyone else tries this out, I actually had to remove the UIScrollView subclass and use the native class otherwise it doesn't work. I think it's because of changes in the 3.0 SDK relating to how the scroll view intercepts touch events. So the the idea is to remove all the other views when you start zooming, and move the current view to (0, 0) in the scrollview, updating the contentsize etc. Then when you zoom back to 1.0f it adds the other views back and puts things all back in order. Anyway, that project works perfectly in the simulator, but on the device there is some nasty movement of the view you are resizing, which looks like it's caused by the fact we are changing the contentsize/offset etc. for the view being resized. You have to do this view moving otherwise you can pan left through the whitespace left by the other views. I found one interesting note in the "Known Issues" of the 3.0 SDK release notes: UIScrollView: After zooming, content inset is ignored and content is left in the wrong position. This kind of sounds like what is happening here. After zooming in, the view will shift offscreen because you have changed the offset etc. I've spent hours on this already and I'm slowing coming to the sad realization that this just isn't going to work. Three20's photo viewer is out of the question: it's too heavy weight and there is too much unnecessary UI and other behaviour. The built in Photo app seems to do some magic. If you zoom in on an image and pan to the far edges, the current photo moves independently of the photo next to it which isn't what you get when trying this with a standard UIScrollView. I've seen discussion about nesting the UIScrollView's but I really don't want to go there. Has anybody managed this with the standard UIScrollView (and works in the 2.2 and 3.0 SDK)? I don't fancy rolling my own zoom + bounce + pan + paging code.

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  • Finding edge and corner values of an image in matlab

    - by James
    Hi, this problem links to two other questions i've asked on here. I am tracing the outline of an image and plotting this to a dxf file. I would like to use the bwboundaries function to find the coordinates of the edges of the image, find the corner coordinates using the cornermetric function and then remove any edge coordinates that are not a corner. The important thing I need to be able to do is keep the order of the corner elements obtained from bwboundaries, so that the section traces properly. The dxf function I have that draws from the coordinates draws lines between coordinates that are next to each other, so the line has to be drawn "around" the section rather than straight between the corner points. The reason I am doing this is because there are less coordinates obtained this way, so it is easier to amend the dxf file (as there are less points to manipulate). The code I have so far is: %# Shape to be traced bw = zeros(200); bw(20:40,20:180) = 1; bw(20:180,90:110) = 1; bw(140:180,20:185) = 1; %# Boundary Finding Section [Boundary] = bwboundaries(bw); %Traces the boundary of each section figure, imshow(bw); hold on; colors=['b' 'g' 'r' 'c' 'm' 'y']; for k=1:length(Boundary) perim = Boundary{k}; %Obtains perimeter coordinates (as a 2D matrix) from the cell array cidx = mod(k,length(colors))+1;% Obtains colours for the plot plot(perim(:,2), perim(:,1),... colors(cidx),'LineWidth',2); end Coordmat = cell2mat(Boundary) %Converts the traced regions to a matrix X = Coordmat(:,1) Y = Coordmat(:,2) % This gives the edge coordinates in matrix form %% Corner Finding Section (from Jonas' answer to a previous question %# get corners cornerProbability = cornermetric(bw); cornerIdx = find(cornerProbability==max(cornerProbability(:))); %# Label the image. bwlabel puts 1 for the first feature, 2 for the second, etc. %# Since concave corners are placed just outside the feature, grow the features %# a little before labeling bw2 = imdilate(bw,ones(3)); labeledImage = bwlabel(bw2); %# read the feature number associated with the corner cornerLabels = labeledImage(cornerIdx); %# find all corners that are associated with feature 1 corners_1 = cornerIdx(cornerLabels==1) [Xcorners, Ycorners] = ind2sub(200,corners_1) % Convert subscripts The code I have is, to give a matrix Xfin for the final x coordinates (which are on the edge AND at a corner. Xfin = zeros(length(X),1) for i = Xcorners XFin(i) = Xcorners if i~= Xcorners XFin(i) = [] end end However, this does not work correctly, because the values in the solution are sorted into order, and only one of each value remains. As I said, I would like the corner elements to be in the same order as obtained from bwboundaries, to allow the image to trace properly. Thanks

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  • Convert a PDF to a Transparent PNG with GhostScript

    - by Jonathon Wolfe
    Hi all. I am attempting, unsuccessfully, to use Ghostscript to rasterize PDF files with a transparent background to PNG files with a transparent background. I've searched high and low for questions from others attempting the same thing and none of the posted solutions, which as far as I can tell come down to specifying -sDEVICE=pngalpha, have worked with my test files. At this point I would really appreciate any advice or tips a more experienced hand could provide. My test PDF is located here: http://www.kolossus.com/files/test.pdf It could be that the issue is with this file, but I doubt it. As far as I can tell, it has no specified background, and when I open the file with a transparency-aware app like Photoshop or Illustrator, sure enough it displays with a transparent background. However, when opened with an application like Adobe Reader the file is rendered with a white background. I believe that this has more to do with the application rendering the PDF than with the PDF itself -- apps like Adobe Reader assume you want to see what a printed document will look like and therefore always show a white canvas behind the artwork -- but I can't be sure. The gs command I'm using is: gs -dNOPAUSE -dBATCH -sDEVICE=pngalpha -r72 -sOutputFile=test.png test.pdf This produces a PNG that has transparent pixels outside of the bounding box of the artwork in the file, but all pixels that are inside the artwork's bounding box are rasterized against a white background. This is a problem for me, as my artwork has drop shadows and antialiased edges that need to be preserved in the final output, and can't just be postprocessed out with ImageMagick. A sample of my PNG output is at the same location as the pdf above, with .png at the end (stackoverflow won't let me include more than one url in my post). Interestingly, I see no effects from using the -dBackgroundColor flag, even if I set it to something non-white like -dBackgroundColor=16#ff0000. Perhaps my understanding of the syntax of this flag is wrong. Also interestingly, I see no effects from using the -dTextAlphaBits=4 -dGraphicsAlphaBits=4 flags to try to enable subpixel antialiasing. I would also appreciate any advice on how to enable subpixel antialiasing, especially on text. Finally, I'm using GPL Ghostscript 8.64 on Mac OS 10.5.7, and the rendering workflow I'm trying to get set up is to generate transparent PNG images from PDFs output by PrinceXML. I'm calling Ghostscript directly for the rasterization instead of using ImageMagick because ImageMagick delegates to Ghostscript for PDF rasterization and I should be able to control the rasterization better by calling GS directly. Thanks for your help. -Jon Wolfe

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  • What's the best way to do base36 arithmetic in perl?

    - by DVK
    What's the best way to do base36 arithmetic in Perl? To be more specific, I need to be able to do the following: Operate on positive N-digit numbers in base 36 (e.g. digits are 0-9 A-Z) N is finite, say 9 Provide basic arithmetic, at the very least the following 3: Addition (A+B) Subtraction (A-B) Whole division, e.g. floor(A/B). Strictly speaking, I don't really need a base10 conversion ability - the numbers will 100% of time be in base36. So I'm quite OK if the solution does NOT implement conversion from base36 back to base10 and vice versa. I don't much care whether the solution is brute-force "convert to base 10 and back" or converting to binary, or some more elegant approach "natively" performing baseN operations (as stated above, to/from base10 conversion is not a requirement). My only 3 considerations are: It fits the minimum specifications above It's "standard". Currently we're using and old homegrown module based on base10 conversion done by hand that is buggy and sucks. I'd much rather replace that with some commonly used CPAN solution instead of re-writing my own bicycle from scratch, but I'm perfectly capable of building it if no better standard possibility exists. It must be fast-ish (though not lightning fast). Something that takes 1 second to sum up 2 9-digit base36 numbers is worse than anything I can roll on my own :) P.S. Just to provide some context in case people decide to solve my XY problem for me in addition to answering the technical question above :) We have a fairly large tree (stored in DB as a bunch of edges), and we need to superimpose order on a subset of that tree. The tree dimentions are big both depth- and breadth- wise. The tree is VERY actively updated (inserts and deletes and branch moves). This is currently done by having a second table with 3 columns: parent_vertex, child_vertex, local_order, where local_order is an 9-character string built of A-Z0-9 (e.g. base 36 number). Additional considerations: It is required that the local order is unique per child (and obviously unique per parent), Any complete re-ordering of a parent is somewhat expensive, and thus the implementation is to try and assign - for a parent with X children - the orders which are somewhat evenly distributed between 0 and 36**10-1, so that almost no tree inserts result in a full re-ordering.

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  • WPF TreeView: How to style selected items with rounded corners like in Explorer

    - by Helge Klein
    The selected item in a WPF TreeView has a dark blue background with "sharp" corners. That looks a bit dated today: I would like to change the background to look like in Explorer of Windows 7 (with/without focus): What I tried so far does not remove the original dark blue background but paints a rounded border on top of it so that you see the dark blue color at the edges and at the left side - ugly. Interestingly, when my version does not have the focus, it looks pretty OK: I would like to refrain from redefining the control template as shown here or here. I want to set the minimum required properties to make the selected item look like in Explorer. Alternative: I would also be happy to have the focused selected item look like mine does now when it does not have the focus. When losing the focus, the color should change from blue to grey. Here is my code: <TreeView x:Name="TreeView" ItemsSource="{Binding TopLevelNodes}" VirtualizingStackPanel.IsVirtualizing="True" VirtualizingStackPanel.VirtualizationMode="Recycling"> <TreeView.ItemContainerStyle> <Style TargetType="{x:Type TreeViewItem}"> <Setter Property="IsExpanded" Value="{Binding IsExpanded, Mode=TwoWay}" /> <Setter Property="IsSelected" Value="{Binding IsSelected, Mode=TwoWay}" /> <Style.Triggers> <Trigger Property="IsSelected" Value="True"> <Setter Property="BorderBrush" Value="#FF7DA2CE" /> <Setter Property="Background" Value="#FFCCE2FC" /> </Trigger> </Style.Triggers> </Style> </TreeView.ItemContainerStyle> <TreeView.Resources> <HierarchicalDataTemplate DataType="{x:Type viewmodels:ObjectBaseViewModel}" ItemsSource="{Binding Children}"> <Border Name="ItemBorder" CornerRadius="2" Background="{Binding Background, RelativeSource={RelativeSource AncestorType=TreeViewItem}}" BorderBrush="{Binding BorderBrush, RelativeSource={RelativeSource AncestorType=TreeViewItem}}" BorderThickness="1"> <StackPanel Orientation="Horizontal" Margin="2"> <Image Name="icon" Source="/ExplorerTreeView/Images/folder.png"/> <TextBlock Text="{Binding Name}"/> </StackPanel> </Border> </HierarchicalDataTemplate> </TreeView.Resources> </TreeView>

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  • <li> titles and descriptions align across top

    - by AMC
    Not quite sure how to go about describing what it is I'm trying to achieve, so I have some screenshots.. Basically, I'd like the <h2>s all have their top edges align, as well as the <p>'s. What I currently have: What I would like to have: index.html <div id="services"> <p><span>Services</span></p> <ul> <li><h2>Identity and Logo Design</h2><p>Nulla et diam risus. Praesent vestibulum augue non purus tincidunt placerat. Sed in orci leo. Duis dignissim nibh vitae lacus placerat et posuere</p></li> <li><h2>Branding</h2><p>Nulla et diam risus. Praesent vestibulum augue non purus tincidunt placerat. Sed in orci leo. Duis dignissim nibh vitae lacus placerat et posuere</p></li> <li><h2>Web Design and Development</h2><p>Nulla et diam risus. Praesent vestibulum augue non purus tincidunt placerat. Sed in orci leo. Duis dignissim nibh vitae lacus placerat et posuere</p></li> <li><h2>Social Media</h2><p>Nulla et diam risus. Praesent vestibulum augue non purus tincidunt placerat. Sed in orci leo. Duis dignissim nibh vitae lacus placerat et posuere</p></li> </ul> <div class="next"> <a href="#"><img src="img/next.png" alt="Click for more information" /></a> </div><!-- end next --> </div><!-- end services --> style.css #services { border-top: 1px solid #202020; padding-bottom: 60px; padding-top: 40px; } #services p span { font-family: nevis-webfont; font-size: 112.5%; font-weight: normal; letter-spacing: 1px; text-transform: uppercase; } #services ul li { display: inline-block; padding-left: 100px; width: 160px; } #services ul li:first-child { padding-left: 0px; } #services ul li h2 { padding-top: 20px; text-align: center; } #services .next { float: right }

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  • Algorithm for finding symmetries of a tree

    - by Paxinum
    I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them). The sectors live in the complex plane, and for n even, sector 0 and n/2 are bisected by the real axis, and the sectors are evenly spaced. These branches meet at certain points, called junctions. Each junction is adjacent to a subset of the sectors (at least 3 of them). Specifying the junctions, (in pre-fix order, lets say, starting from junction adjacent to sector 0 and 1), and the distance between the junctions, uniquely describes the tree. Now, given such a representation, how can I see if it is symmetric wrt the real axis? For example, n=6, the tree (0,1,5)(1,2,4,5)(2,3,4) have three junctions on the real line, so it is symmetric wrt the real axis. If the distances between (015) and (1245) is equal to distance from (1245) to (234), this is also symmetric wrt the imaginary axis. The tree (0,1,5)(1,2,5)(2,4,5)(2,3,4) have 4 junctions, and this is never symmetric wrt either imaginary or real axis, but it has 180 degrees rotation symmetry if the distance between the first two and the last two junctions in the representation are equal. Edit: This is actually for my research. I have posted the question at mathoverflow as well, but my days in competition programming tells me that this is more like an IOI task. Code in mathematica would be excellent, but java, python, or any other language readable by a human suffices. Here are some examples (pretend the double edges are single and we have a tree) http://www2.math.su.se/~per/files.php?file=contr_ex_1.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_2.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_5.pdf Example 1 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,3). Example 2 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,1). Example 5 is described as (0,1,4,5)(1,2,3,4) with distances (2). So, given the description/representation, I want to find some algorithm to decide if it is symmetric wrt real, imaginary, and rotation 180 degrees. The last example have 180 degree symmetry. (These symmetries corresponds to special kinds of potential in the Schroedinger equation, which has nice properties in quantum mechanics.)

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  • Closure Tables - Is this enough data to display a tree view?

    - by James Pitt
    Here is the table I have created by testing the closure table method. | id | parentId | childId | hops | | | | | 270 | 6 | 6 | 0 | 271 | 7 | 7 | 0 | 272 | 8 | 8 | 0 | 273 | 9 | 9 | 0 | 276 | 10 | 10 | 0 | 281 | 9 | 10 | 1 | 282 | 7 | 9 | 1 | 283 | 7 | 10 | 2 | 285 | 7 | 8 | 1 | 286 | 6 | 7 | 1 | 287 | 6 | 9 | 2 | 288 | 6 | 10 | 3 | 289 | 6 | 8 | 2 | 293 | 6 | 9 | 1 | 294 | 6 | 10 | 2 I am trying to create a simple tree of this using PHP. There does not seem to be enough data to create the table. For example, when I look purely at parentId = 6: -Part 6 -Part 7 - ? - ? -Part 9 - ? - ? We know that parts 8 and 10 exists below Part 7 or 9, but not which. We know that part 10 exists at both 3 and 4 nodes deep but where? If I look at other data in the table it is possible to tell it should be: - Part 6 - Part 7 - Part 9 - Part 10 - Part 9 - Part 10 I thought one of the benefits of closure tables was there was no need for recursive queries? Could you help explain what I am doing wrong? EDIT: For clarification, this is a mapping table. There is another table called "parts" which has a column called part_id that correlates to both the parentId and childId columns in the "closure" table. The "id" column in the table above (closure) is just for the purposes of maintaining a primary key. It is not really necessary. The methods I have used to create this closure table is described in the following article: http://dirtsimple.org/2010/11/simplest-way-to-do-tree-based-queries.html EDIT2: It can have two and three hops. I will explain easier by assigning names to the items. Part 6 = Bicycle Part 7 = Gears Part 8 = Chain Part 9 = Bolt Part 10 = Nut Nut is part of Bolt. The Bolt and Nut combo exists directly within Bicycle and within Gears which is part of Bicycle. In relation to what method to use I have looked at Adjacency, Edges, Enum Paths, Closures, DAGS(networks) and the Nested Set Model. I am still trying to work out what is what, but this is an extremely complex component database where there are multiple parents and any modification to a sub-tree must propogate through the other trees. More importantly there will be insertions, deletions and tree views that I wish to avoid recursion during general use, even at the cost of database space and query time during entry.

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  • Why do I get rows of zeros in my 2D fft?

    - by Nicholas Pringle
    I am trying to replicate the results from a paper. "Two-dimensional Fourier Transform (2D-FT) in space and time along sections of constant latitude (east-west) and longitude (north-south) were used to characterize the spectrum of the simulated flux variability south of 40degS." - Lenton et al(2006) The figures published show "the log of the variance of the 2D-FT". I have tried to create an array consisting of the seasonal cycle of similar data as well as the noise. I have defined the noise as the original array minus the signal array. Here is the code that I used to plot the 2D-FT of the signal array averaged in latitude: import numpy as np from numpy import ma from matplotlib import pyplot as plt from Scientific.IO.NetCDF import NetCDFFile ### input directory indir = '/home/nicholas/data/' ### get the flux data which is in ### [time(5day ave for 10 years),latitude,longitude] nc = NetCDFFile(indir + 'CFLX_2000_2009.nc','r') cflux_southern_ocean = nc.variables['Cflx'][:,10:50,:] cflux_southern_ocean = ma.masked_values(cflux_southern_ocean,1e+20) # mask land nc.close() cflux = cflux_southern_ocean*1e08 # change units of data from mmol/m^2/s ### create an array that consists of the seasonal signal fro each pixel year_stack = np.split(cflux, 10, axis=0) year_stack = np.array(year_stack) signal_array = np.tile(np.mean(year_stack, axis=0), (10, 1, 1)) signal_array = ma.masked_where(signal_array > 1e20, signal_array) # need to mask ### average the array over latitude(or longitude) signal_time_lon = ma.mean(signal_array, axis=1) ### do a 2D Fourier Transform of the time/space image ft = np.fft.fft2(signal_time_lon) mgft = np.abs(ft) ps = mgft**2 log_ps = np.log(mgft) log_mgft= np.log(mgft) Every second row of the ft consists completely of zeros. Why is this? Would it be acceptable to add a randomly small number to the signal to avoid this. signal_time_lon = signal_time_lon + np.random.randint(0,9,size=(730, 182))*1e-05 EDIT: Adding images and clarify meaning The output of rfft2 still appears to be a complex array. Using fftshift shifts the edges of the image to the centre; I still have a power spectrum regardless. I expect that the reason that I get rows of zeros is that I have re-created the timeseries for each pixel. The ft[0, 0] pixel contains the mean of the signal. So the ft[1, 0] corresponds to a sinusoid with one cycle over the entire signal in the rows of the starting image. Here are is the starting image using following code: plt.pcolormesh(signal_time_lon); plt.colorbar(); plt.axis('tight') Here is result using following code: ft = np.fft.rfft2(signal_time_lon) mgft = np.abs(ft) ps = mgft**2 log_ps = np.log1p(mgft) plt.pcolormesh(log_ps); plt.colorbar(); plt.axis('tight') It may not be clear in the image but it is only every second row that contains completely zeros. Every tenth pixel (log_ps[10, 0]) is a high value. The other pixels (log_ps[2, 0], log_ps[4, 0] etc) have very low values.

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  • Scrolling RelativeLayout- white border over part of the content

    - by Tanis.7x
    I have a fairly simply Fragment that adds a handful of colored ImageViews to a RelativeLayout. There are more images than can fit on screen, so I implemented some custom scrolling. However, When I scroll around, I see that there is an approximately 90dp white border overlapping part of the content right where the edges of the screen are before I scroll. It is obvious that the ImageViews are still being created and drawn properly, but they are being covered up. How do I get rid of this? I have tried: Changing both the RelativeLayout and FrameLayout to WRAP_CONTENT, FILL_PARENT, MATCH_PARENT, and a few combinations of those. Setting the padding and margins of both layouts to 0dp. Example: Fragment: public class MyFrag extends Fragment implements OnTouchListener { int currentX; int currentY; RelativeLayout container; final int[] colors = {Color.BLACK, Color.RED, Color.BLUE}; @Override public View onCreateView(LayoutInflater inflater, ViewGroup fragContainer, Bundle savedInstanceState) { return inflater.inflate(R.layout.fragment_myfrag, null); } @Override public void onActivityCreated(Bundle savedInstanceState) { super.onActivityCreated(savedInstanceState); container = (RelativeLayout) getView().findViewById(R.id.container); container.setOnTouchListener(this); // Temp- Add a bunch of images to test scrolling for(int i=0; i<1500; i+=100) { for (int j=0; j<1500; j+=100) { int color = colors[(i+j)%3]; ImageView image = new ImageView(getActivity()); image.setScaleType(ImageView.ScaleType.CENTER); image.setBackgroundColor(color); LayoutParams lp = new RelativeLayout.LayoutParams(100, 100); lp.setMargins(i, j, 0, 0); image.setLayoutParams(lp); container.addView(image); } } } @Override public boolean onTouch(View v, MotionEvent event) { switch (event.getAction()) { case MotionEvent.ACTION_DOWN: { currentX = (int) event.getRawX(); currentY = (int) event.getRawY(); break; } case MotionEvent.ACTION_MOVE: { int x2 = (int) event.getRawX(); int y2 = (int) event.getRawY(); container.scrollBy(currentX - x2 , currentY - y2); currentX = x2; currentY = y2; break; } case MotionEvent.ACTION_UP: { break; } } return true; } } XML: <FrameLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:tools="http://schemas.android.com/tools" android:layout_width="fill_parent" android:layout_height="fill_parent" tools:context=".FloorPlanFrag"> <RelativeLayout android:id="@+id/container" android:layout_width="fill_parent" android:layout_height="fill_parent" /> </FrameLayout>

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  • Editing Neo Slider to move in one direction

    - by user2106416
    I have a website with brandspace theme from pixelentity. Currently, the slider move from right to left ( auto transition) . But, when the slider reaches the last image, it will go back to the first image by making all previous images sliding back from left to right. I want to make the slider to have a normal transition,i.e. going back to first image but still with the image sliding from right to left. Below is the slider.php code.: <?php $t =& peTheme(); ?> <?php list($pid,$conf,$loop) = $t->template->data(); ?> <?php $w = $t->media->width(940); ?> <?php $h = $t->media->height(300); ?> <div class="peSlider peVolo" data-autopause="disabled" data-plugin="<?php echo apply_filters("pe_theme_slider_plugin","peVolo"); ?>" data-controls-arrows="edges-full" data-controls-bullets="disabled" data-icon-font="enabled"> <?php while ($slide =& $loop->next()): ?> <?php $link = empty($slide->link) ? false: $slide->link; ?> <?php $img = $t->image->resizedImg($slide->img,$w,$h); ?> <div data-delay="4" <?php echo $slide->idx == 0 ? ' class="visible"' : ''; ?>> <?php if (isset($slide->caption)) $t->slider->caption($slide->caption); ?> <?php if ($link): ?> <a href="<?php echo $link ?>" data-flare-gallery="fsGallery<?php echo $pid ?>"> <?php echo $img; ?> </a> <?php else: ?> <?php echo $img; ?> <?php endif; ?> </div> <?php endwhile; ?> </div>

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  • How can you prevent both jumpiness, and interrupting tweens with animated Flash buttons?

    - by Kevin Suttle
    This is something I've never been able to figure out. You've got a button offscreen you want to animate in. We'll call it 'btn.' You've got a hit area that serves as the proximity sensor to trigger btn's animation. We'll call it 'hitZone' (as to not cause confusion with the hitArea property of display objects). Both btn and hitZone are MovieClips. The listeners go something like this. import com.greensock.*; import com.greensock.easing.*; import flash.events.MouseEvent; var endPoint:Number = 31; hitZone.addEventListener(MouseEvent.ROLL_OVER, onHitZoneOver); hitZone.addEventListener(MouseEvent.ROLL_OUT, onHitZoneOut); hitZone.addEventListener(MouseEvent.CLICK, onHitZoneClick); btn.addEventListener(MouseEvent.ROLL_OVER, onBtnOver); btn.addEventListener(MouseEvent.ROLL_OUT, onBtnOut); btn.addEventListener(MouseEvent.CLICK, onBtnClick); btn.mouseChildren = false; function onHitZoneOver(e:MouseEvent):void { TweenLite.to(btn, 0.75, {x:endPoint, ease:Expo.easeOut}); trace("over hitZone"); } function onHitZoneOut(e:MouseEvent):void { TweenLite.to(btn, 0.75, {x:-1, ease:Expo.easeOut}); trace("out hitZone"); } function onBtnOver(e:MouseEvent):void { hitZone.mouseEnabled = false; hitZone.removeEventListener(MouseEvent.ROLL_OVER, onHitZoneOver); hitZone.removeEventListener(MouseEvent.ROLL_OUT, onHitZoneOut); trace("over BTN"); // This line is the only thing keeping the btn animation from being fired continuously // causing jumpiness. However, calling this allows the animation to be interrupted // at any point. TweenLite.killTweensOf(btn); } function onBtnOut(e:MouseEvent):void { hitZone.mouseEnabled = true; hitZone.addEventListener(MouseEvent.ROLL_OVER, onHitZoneOver); hitZone.addEventListener(MouseEvent.ROLL_OUT, onHitZoneOut); trace("out BTN"); } function onBtnClick(e:MouseEvent):void { trace("click BTN"); } function onHitZoneClick(e:MouseEvent):void { trace("click hitZone"); } The issue is when your mouse is over both the hitZone and btn. The button continuously jumps unless you call TweenLite.killAllTweensOf(). This fixes the jumpiness, but it introduces a new problem. Now, it's very easy to interrupt the animation of the btn at any point, stopping it before it's totally visible on the stage. I've seen similar posts, but even they suffer from the same issue. Perhaps it's a problem with how Flash detects edges, because I've never once seen a workaround for this.

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  • Finding contained bordered regions from Excel imports.

    - by dmaruca
    I am importing massive amounts of data from Excel that have various table layouts. I have good enough table detection routines and merge cell handling, but I am running into a problem when it comes to dealing with borders. Namely performance. The bordered regions in some of these files have meaning. Data Setup: I am importing directly from Office Open XML using VB6 and MSXML. The data is parsed from the XML into a dictionary of cell data. This wonks wonderfully and is just as fast as using docmd.transferspreadsheet in Access, but returns much better results. Each cell contains a pointer to a style element which contains a pointer to a border element that defines the visibility and weight of each border (this is how the data is structured inside OpenXML, also). Challenge: What I'm trying to do is find every region that is enclosed inside borders, and create a list of cells that are inside that region. What I have done: I initially created a BFS(breadth first search) fill routine to find these areas. This works wonderfully and fast for "normal" sized spreadsheets, but gets way too slow for imports into the thousands of rows. One problem is that a border in Excel could be stored in the cell you are checking or the opposing border in the adjacent cell. That's ok, I can consolidate that data on import to reduce the number of checks needed. One thing I thought about doing is to create a separate graph that outlines the cells using the borders as my edges and using a graph algorithm to find regions that way, but I'm having trouble figuring out how to implement the algorithm. I've used Dijkstra in the past and thought I could do similar with this. So I can span out using no endpoint to search the entire graph, and if I encounter a closed node I know that I just found an enclosed region, but how can I know if the route I've found is the optimal one? I guess I could flag that to run a separate check for the found closed node to the previous node ignoring that one edge. This could work, but wouldn't be much better performance wise on dense graphs. Can anyone else suggest a better method? Thanks for taking the time to read this.

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  • PHP Parse Error unexpected '{'

    - by Laxmidi
    Hi, I'm getting a "Parse error: syntax error, unexpected '{' in line 2". And I don't see the problem. <?php class pointLocation {     var $pointOnVertex = true; // Check if the point sits exactly on one of the vertices     function pointLocation() {     }                   function pointInPolygon($point, $polygon, $pointOnVertex = true) {         $this->pointOnVertex = $pointOnVertex;                  // Transform string coordinates into arrays with x and y values         $point = $this->pointStringToCoordinates($point);         $vertices = array();          foreach ($polygon as $vertex) {             $vertices[] = $this->pointStringToCoordinates($vertex);          }                  // Check if the point sits exactly on a vertex         if ($this->pointOnVertex == true and $this->pointOnVertex($point, $vertices) == true) {             return "vertex";         }                  // Check if the point is inside the polygon or on the boundary         $intersections = 0;          $vertices_count = count($vertices);              for ($i=1; $i < $vertices_count; $i++) {             $vertex1 = $vertices[$i-1];              $vertex2 = $vertices[$i];             if ($vertex1['y'] == $vertex2['y'] and $vertex1['y'] == $point['y'] and $point['x'] > min($vertex1['x'], $vertex2['x']) and $point['x'] < max($vertex1['x'], $vertex2['x'])) { // Check if point is on an horizontal polygon boundary                 return "boundary";             }             if ($point['y'] > min($vertex1['y'], $vertex2['y']) and $point['y'] <= max($vertex1['y'], $vertex2['y']) and $point['x'] <= max($vertex1['x'], $vertex2['x']) and $vertex1['y'] != $vertex2['y']) {                  $xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x'];                  if ($xinters == $point['x']) { // Check if point is on the polygon boundary (other than horizontal)                     return "boundary";                 }                 if ($vertex1['x'] == $vertex2['x'] || $point['x'] <= $xinters) {                     $intersections++;                  }             }          }          // If the number of edges we passed through is even, then it's in the polygon.          if ($intersections % 2 != 0) {             return "inside";         } else {             return "outside";         }     }               function pointOnVertex($point, $vertices) {         foreach($vertices as $vertex) {             if ($point == $vertex) {                 return true;             }         }          }                   function pointStringToCoordinates($pointString) {         $coordinates = explode(" ", $pointString);         return array("x" => $coordinates[0], "y" => $coordinates[1]);     }           } $pointLocation = new pointLocation(); $points = array("30 19", "0 0", "10 0", "30 20", "11 0", "0 11", "0 10", "30 22", "20 20"); $polygon = array("10 0", "20 0", "30 10", "30 20", "20 30", "10 30", "0 20", "0 10", "10 0"); foreach($points as $key => $point) { echo "$key ($point) is " . $pointLocation->pointInPolygon($point, $polygon) . "<br>"; } ?> Does anyone see the problem? Thanks, -Laxmidi

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  • Algorithm: Determine shape of two sectors delineated by an arbitrary path, and then fill one.

    - by Arseniy Banayev
    NOTE: This is a challenging problem for anybody who likes logic problems, etc. Consider a rectangular two-dimensional grid of height H and width W. Every space on the grid has a value, either 0 1 or 2. Initially, every space on the grid is a 0, except for the spaces along each of the four edges, which are initially a 2. Then consider an arbitrary path of adjacent (horizontally or vertically) grid spaces. The path begins on a 2 and ends on a different 2. Every space along the path is a 1. The path divides the grid into two "sectors" of 0 spaces. There is an object that rests on an unspecified 0 space. The "sector" that does NOT contain the object must be filled completely with 2. Define an algorithm that determines the spaces that must become 2 from 0, given an array (list) of values (0, 1, or 2) that correspond to the values in the grid, going from top to bottom and then from left to right. In other words, the element at index 0 in the array contains the value of the top-left space in the grid (initially a 2). The element at index 1 contains the value of the space in the grid that is in the left column, second from the top, and so forth. The element at index H contains the value of the space in the grid that is in the top row but second from the left, and so forth. Once the algorithm finishes and the empty "sector" is filled completely with 2s, the SAME algorithm must be sufficient to do the same process again. The second (and on) time, the path is still drawn from a 2 to a different 2, across spaces of 0, but the "grid" is smaller because the 2s that are surrounded by other 2s cannot be touched by the path (since the path is along spaces of 0). I thank whomever is able to figure this out for me, very very much. This does not have to be in a particular programming language; in fact, pseudo-code or just English is sufficient. Thanks again! If you have any questions, just leave a comment and I'll specify what needs to be specified.

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  • Is there a way to make PHP's SplHeap recalculate? (aka: add up-heap to SplHeap?)

    - by md2k7
    I am using an SplHeap to hold graph nodes of a tree with directed edges that will be traversed from the leaves to the root. For this, I precalculate the "fan-in" of nodes and put them into the heap so that I can always retrieve the node with the smallest fan-in (0) from it. After visiting a node, I reduce the fan-in of its successor by 1. Then obviously, the heap needs to be recalculated because the successor is now in the wrong place there. I have tried recoverFromCorruption(), but it doesn't do anything and keeps the heap in the wrong order (node with larger fanIn stays in front of smaller fanIn). As a workaround, I'm now creating a new heap after each visit, amounting to a full O(N*log(N)) sort each time. It should be possible, however, to make up-heap operations on the changed heap entry until it's in the right position in O(log(N)). The API for SplHeap doesn't mention an up-heap (or deletion of an arbitrary element - it could then be re-added). Can I somehow derive a class from SplHeap to do this or do I have to create a pure PHP heap from scratch? EDIT: Code example: class VoteGraph { private $nodes = array(); private function calculateFanIn() { /* ... */ } // ... private function calculateWeights() { $this->calculateFanIn(); $fnodes = new GraphNodeHeap(); // heap by fan-in ascending (leaves are first) foreach($this->nodes as $n) { // omitted: filter loops $fnodes->insert($n); } // traversal from leaves to root while($fnodes->valid()) { $node = $fnodes->extract(); // fetch a leaf from the heap $successor = $this->nodes[$node->successor]; // omitted: actual job of traversal $successor->fanIn--; // will need to fix heap (sift up successor) because of this //$fnodes->recoverFromCorruption(); // doesn't work for what I want // workaround: rebuild $fnodes from scratch $fixedHeap = new GraphNodeHeap(); foreach($fnodes as $e) $fixedHeap->insert($e); $fnodes = $fixedHeap; } } } class GraphNodeHeap extends SplHeap { public function compare($a, $b) { if($a->fanIn === $b->fanIn) return 0; else return $a->fanIn < $b->fanIn ? 1 : -1; } }

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  • XSL transformation of SVG adds namespace attribute to new tag

    - by Steve
    I have a SVG file that I want to extend by adding onclick handlers to edges and nodes. I also want to add a script tag referring to a JavaScript. The problem is that the script tag gets an empty namespace attribute added to it. I haven't found any information regarding this that I understand. Why does XSLT add an empty namespace? XSL file: <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:svg="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"> <xsl:output method="xml" encoding="utf-8" /> <xsl:template match="/svg:svg"> <xsl:copy> <script type="text/ecmascript" xlink:href="base.js" /> <!-- this tag gets a namespace attr --> <xsl:apply-templates /> </xsl:copy> </xsl:template> <!-- Identity transform http://www.w3.org/TR/xslt#copying --> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <!-- Check groups and add functions --> <xsl:template match="svg:g"> <xsl:copy> <xsl:if test="@class = 'node'"> <xsl:attribute name="onclick">node_clicked()</xsl:attribute> </xsl:if> <xsl:if test="@class = 'edge'"> <xsl:attribute name="onclick">edge_clicked()</xsl:attribute> </xsl:if> <xsl:apply-templates select="@*|node()" /> </xsl:copy> </xsl:template> </xsl:stylesheet>

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  • refactoring this function in Java

    - by Joel
    Hi folks, I'm learning Java, and I know one of the big complaints about newbie programmers is that we make really long and involved methods that should be broken into several. Well here is one I wrote and is a perfect example. :-D. public void buildBall(){ /* sets the x and y value for the center of the canvas */ double i = ((getWidth() / 2)); double j = ((getHeight() / 2)); /* randomizes the start speed of the ball */ vy = 3.0; vx = rgen.nextDouble(1.0, 3.0); if (rgen.nextBoolean(.05)) vx = -vx; /* creates the ball */ GOval ball = new GOval(i,j,(2 *BALL_RADIUS),(2 * BALL_RADIUS)); ball.setFilled(true); ball.setFillColor(Color.RED); add(ball); /* animates the ball */ while(true){ i = (i + (vx* 2)); j = (j + (vy* 2)); if (i > APPLICATION_WIDTH-(2 * BALL_RADIUS)){ vx = -vx; } if (j > APPLICATION_HEIGHT-(2 * BALL_RADIUS)){ vy = -vy; } if (i < 0){ vx = -vx; } if (j < 0){ vy = -vy; } ball.move(vx + vx, vy + vy); pause(10); /* checks the edges of the ball to see if it hits an object */ colider = getElementAt(i, j); if (colider == null){ colider = getElementAt(i + (2*BALL_RADIUS), j); } if (colider == null){ colider = getElementAt(i + (2*BALL_RADIUS), j + (2*BALL_RADIUS)); } if (colider == null){ colider = getElementAt(i, j + (2*BALL_RADIUS)); } /* If the ball hits an object it reverses direction */ if (colider != null){ vy = -vy; /* removes bricks when hit but not the paddle */ if (j < (getHeight() -(PADDLE_Y_OFFSET + PADDLE_HEIGHT))){ remove(colider); } } } You can see from the title of the method that I started with good intentions of "building the ball". There are a few issues I ran up against: The problem is that then I needed to move the ball, so I created that while loop. I don't see any other way to do that other than just keep it "true", so that means any other code I create below this loop won't happen. I didn't make the while loop a different function because I was using those variables i and j. So I don't see how I can refactor beyond this loop. So my main question is: How would I pass the values of i and j to a new method: "animateBall" and how would I use ball.move(vx + vx, vy + vy); in that new method if ball has been declared in the buildBall method? I understand this is probably a simple thing of better understanding variable scope and passing arguments, but I'm not quite there yet...

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  • XNA 4.0 draw a cube with DrawUserIndexedPrimitives method [on hold]

    - by Leggy7
    EDIT Since I read what Mark H suggested (thanks a lot, I found it very useful) I think my question can become clearer structured this way: Using XNA 4.0, I'm trying to draw a cube. Im using this method: GraphicsDevice.DrawUserIndexedPrimitives<VertexPositionColor>( PrimitiveType.LineList, primitiveList, 0, // vertex buffer offset to add to each element of the index buffer 8, // number of vertices in pointList lineListIndices, // the index buffer 0, // first index element to read 7 // number of primitives to draw ); I got the code sample from this page which simply draw a serie of triangles. I want to modify this code in order to draw a cube. I was able to slitghly move the camera so I can have the perception of solidity, I set the vertex array to contain the 8 points defining a cube. But I can't fully understand how many primitives I have to draw (last parameter) for each of PrimitiveType. So, I wasn't able to draw the cube (just some of the edges in a non-defined order). More in detail: to build the vertex index list, the sample used // Initialize an array of indices of type short. lineListIndices = new short[(points * 2) - 2]; // Populate the array with references to indices in the vertex buffer for (int i = 0; i < points - 1; i++) { lineListIndices[i * 2] = (short)(i); lineListIndices[(i * 2) + 1] = (short)(i + 1); } I'm ashamed to say I cannot do the same in the case of a cube. what has to be the size of the lineListIndices? how should I populate it? In which order? And how do these things change when I use a different PrimitiveType? In the code sample there are also another couple of calls which I cannot fully understand, which are: // Initialize the vertex buffer, allocating memory for each vertex. vertexBuffer = new VertexBuffer(graphics.GraphicsDevice, vertexDeclaration, points, BufferUsage.None); // Set the vertex buffer data to the array of vertices. vertexBuffer.SetData<VertexPositionColor>(pointList); and vertexDeclaration = new VertexDeclaration(new VertexElement[] { new VertexElement(0, VertexElementFormat.Vector3, VertexElementUsage.Position, 0), new VertexElement(12, VertexElementFormat.Color, VertexElementUsage.Color, 0) } ); that is, for VertexBuffer and VertexDeclaration I could not find and significant (monkey-like) guide. I reported them too because I think they could be involded in understanding things. I think I also have to understand something related to the order the vertexes are stored in the array. But actually I have no clue of what I should learn to have this function drawing a cube. So, if anybody could point me to the right direction, it wil be appreciated. Hope to have made myself clear this time

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  • Custom InputIterator for Boost graph (BGL)

    - by Shadow
    Hi, I have a graph with custom properties to the vertices and edges. I now want to create a copy of this graph, but I don't want the vertices to be as complex as in the original. By this I mean that it would suffice that the vertices have the same indices (vertex_index_t) as they do in the original graph. Instead of doing the copying by hand I wanted to use the copy-functionality of boost::adjacency_list (s. http://www.boost.org/doc/libs/1_37_0/libs/graph/doc/adjacency_list.html): template <class EdgeIterator> adjacency_list(EdgeIterator first, EdgeIterator last, vertices_size_type n, edges_size_type m = 0, const GraphProperty& p = GraphProperty()) The description there says: The EdgeIterator must be a model of InputIterator. The value type of the EdgeIterator must be a std::pair, where the type in the pair is an integer type. The integers will correspond to vertices, and they must all fall in the range of [0, n). Unfortunately I have to admit that I don't quite get it how to define an EdgeIterator that is a model of InputIterator. Here's what I've succeded so far: template< class EdgeIterator, class Edge > class MyEdgeIterator// : public input_iterator< std::pair<int, int> > { public: MyEdgeIterator() {}; MyEdgeIterator(EdgeIterator& rhs) : actual_edge_it_(rhs) {}; MyEdgeIterator(const MyEdgeIterator& to_copy) {}; bool operator==(const MyEdgeIterator& to_compare) { return actual_edge_it_ == to_compare.actual_edge_it_; } bool operator!=(const MyEdgeIterator& to_compare) { return !(*this == to_compare); } Edge operator*() const { return *actual_edge_it_; } const MyEdgeIterator* operator->() const; MyEdgeIterator& operator ++() { ++actual_edge_it_; return *this; } MyEdgeIterator operator ++(int) { MyEdgeIterator<EdgeIterator, Edge> tmp = *this; ++*this; return tmp; } private: EdgeIterator& actual_edge_it_; } However, this doesn't work as it is supposed to and I ran out of clues. So, how do I define the appropriate InputIterator?

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  • Fix overscan in Linux with Intel graphics Vizio HDTV

    - by Padenton
    I am connecting my server to my HDTV so that I can conveniently display it there. My VIZIO HDTV cuts off all 4 edges. I already realize it is not optimal to be running a GUI on a server; this server will not have much external traffic so I prefer it for convenience. I have already spent countless hours searching for a fix, but all I could find required an ATI or NVIDIA graphics card, or didn’t work. In Windows, the Intel driver has a setting for underscan, though it seems only to be available by a glitch. Here’s my specs: Ubuntu Linux (Quantal 12.10) (Likely to switch to Arch) This is a home server computer, with KDE for managing(for now, at least) Graphics: Intel HD Graphics 4000 from Ivy Bridge Motherboard: ASRock Z77 Extreme4 CPU: Intel Core i5-3450 My monitors: Dell LCD monitor Vizio VX37L_HDTV10A 37" on HDMI input I have tried all of the following from both HDMI?HDMI and DVI?HDMI cables connected to the ports on my motherboard: Setting properties in xrandr Making sure drivers are all up to date Trying several different modes The TV was “cheap”; max resolution 1080i. I am able to get a 1920x1080 modeline, in both GNU/Linux and Windows, without difficulty. There is no setting in the menu to fix the overscan (I have tried all of them, I realize it’s not always called overscan). I have been in the service menu for the TV, which still does not contain an option to fix it. No aspect ratio settings, etc. The TV has a VGA connector but I am unsure if it would fix it, as I don’t have a VGA cable long enough, and am not sure it would get me the 1920x1080 resolution which I want. Using another resolution does not fix the problem. I tried custom modelines with the dimensions of my screen’s viewable area, but it wouldn’t let me use them. Ubuntu apparently doesn’t automatically generate an xorg.conf file for use. I read somewhere that modifying it may help solve it. I tried X -configure several times(with reboots, etc.) but it consistently gave the following error messages: In log file: … (WW) Falling back to old probe method for vesa Number of created screens does not match number of detected devices. Configuration failed. In output: … (++) Using config file: "/root/xorg.conf.new" (==) Using system config directory "/usr/share/X11/xorg.conf.d" Number of created screens does not match number of detected devices. Configuration failed. Server terminated with error (2). Closing log file. Tried using 'overscan' prop in xrandr: root@xxx:/home/xxx# xrandr --output HDMI1 --set overscan off X Error of failed request: BadName (named color or font does not exist) Major opcode of failed request: 140 (RANDR) Minor opcode of failed request: 11 (RRQueryOutputProperty) Serial number of failed request: 42 Current serial number in output stream: 42 'overscan on', 'underscan off', 'underscan on' were all also tried. Originally tried with Ubuntu 12.04, but failed and so updated to 12.10 when it was released. All software is up to date. I am not opposed to reinstalling my OS, likely will anyways (my preference being Arch).

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  • Multiple data centers and HTTP traffic: DNS Round Robin is the ONLY way to assure instant fail-over?

    - by vmiazzo
    Hi, Multiple A records pointing to the same domain seem to be used almost exclusively to implement DNS Round Robin as a cheap load balancing technique. The usual warning against DNS RR is that it is not good for high availability. When 1 IP goes down clients will continue to use it for minutes. A load balancer is often suggested as a better choice. Both claims are not completely true: When the traffic is HTTP then, most of the HTML browsers are able to automatically try the next A record if the previous is down, without a new DNS look-up. Read here chapter 3.1 and here. When multiple data centers are involved then, DNS RR is the only option to distribute traffic across them. So, is it true that, with multiple data centers and HTTP traffic, the use of DNS RR is the ONLY way to assure instant fail-over when one data center goes down? Thanks, Valentino Edit: Off course each data center has a local Load Balancer with hot spare. It's OK to sacrifice session affinity for an instant fail-over. AFAIK the only way for a DNS to suggest a data center instead of another is to reply with just the IP (or IPs) associated to that data center. If the data center becomes unreachable then all those IP are also unreachables. This means that, even if smart HTML browsers are able to instantly try another A record , all the attempts will fail until the local cache entry expires and a new DNS lookup is done, fetching the new working IPs (I assume DNS automatically suggests to a new data center when one fail). So, "smart DNS" cannot assure instant fail-over. Conversely a DNS round-robin permits it. When one data center fail, the smart HTML browsers (most of them) instantly try the other cached A records jumping to another (working) data center. So, DNS round-robin doesn't assure session affinity or the lowest RTT but seems to be the only way to assure instant fail-over when the clients are "smart" HTML browsers. Edit 2: Some people suggest TCP Anycast as a definitive solution. In this paper (chapter 6) is explained that Anycast fail-over is related to BGP convergence. For this reason Anycast can employ from 15 minutes to 20 seconds to complete. 20 seconds are possible on networks where the topology was optimized for this. Probably just CDN operators can grant such fast fail-overs. Edit 3:* I did some DNS look-ups and traceroutes (maybe some expert can double check) and: The only CDN using TCP Anycast seems to be CacheFly, other operators like CDN networks and BitGravity use CacheFly. Seems that their edges cannot be used as reverse proxies. Therefore, they cannot be used to grant instant failover. Akamai and LimeLight seems to use geo-aware DNS. But! They return multiple A records. From traceroutes seems that the returned IPs are on the same data center. So, I'm puzzled on how they can offer a 100% SLA when one data center goes down.

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  • Another Marketing Conference, part two – the afternoon

    - by Roger Hart
    In my previous post, I’ve covered the morning sessions at AMC2012. Here’s the rest of the write-up. I’ve skipped Charles Nixon’s session which was a blend of funky futurism and professional development advice, but you can see his slides here. I’ve also skipped the Google presentation, as it was a little thin on insight. 6 – Brand ambassadors: Getting universal buy in across the organisation, Vanessa Northam Slides are here This was the strongest enforcement of the idea that brand and campaign values need to be delivered throughout the organization if they’re going to work. Vanessa runs internal communications at e-on, and shared her experience of using internal comms to align an organization and thereby get the most out of a campaign. She views the purpose of internal comms as: “…to help leaders, to communicate the purpose and future of an organization, and support change.” This (and culture) primes front line staff, which creates customer experience and spreads brand. You ensure a whole organization knows what’s going on with both internal and external comms. If everybody is aligned and informed, if everybody can clearly articulate your brand and campaign goals, then you can turn everybody into an advocate. Alignment is a powerful tool for delivering a consistent experience and message. The pathological counter example is the one in which a marketing message goes out, which creates inbound customer contacts that front line contact staff haven’t been briefed to handle. The NatWest campaign was again mentioned in this context. The good example was e-on’s cheaper tariff campaign. Building a groundswell of internal excitement, and even running an internal launch meant everyone could contribute to a good customer experience. They found that meter readers were excited – not a group they’d considered as obvious in providing customer experience. But they were a group that has a lot of face-to-face contact with customers, and often were asked questions they may not have been briefed to answer. Being able to communicate a simple new message made it easier for them, and also let them become a sales and marketing asset to the organization. 7 – Goodbye Internet, Hello Outernet: the rise and rise of augmented reality, Matt Mills I wasn’t going to write this up, because it was essentially a sales demo for Aurasma. But the technology does merit some discussion. Basically, it replaces QR codes with visual recognition, and provides a simple-looking back end for attaching content. It’s quite sexy. But here’s my beef with it: QR codes had a clear visual language – when you saw one you knew what it was and what to do with it. They were clunky, but they had the “getting started” problem solved out of the box once you knew what you were looking at. However, they fail because QR code reading isn’t native to the platform. You needed an app, which meant you needed to know to download one. Consequentially, you can’t use QR codes with and ubiquity, or depend on them. This means marketers, content providers, etc, never pushed them, and they remained and awkward oddity, a minority sport. Aurasma half solves problem two, and re-introduces problem one, making it potentially half as useful as a QR code. It’s free, and you can apparently build it into your own apps. Add to that the likelihood of it becoming native to the platform if it takes off, and it may have legs. I guess we’ll see. 8 – We all need to code, Helen Mayor Great title – good point. If there was anybody in the room who didn’t at least know basic HTML, and if Helen’s presentation inspired them to learn, that’s fantastic. However, this was a half hour sales pitch for a basic coding training course. Beyond advocating coding skills it contained no useful content. Marketers may also like to consider some of these resources if they’re looking to learn code: Code Academy – free interactive tutorials Treehouse – learn web design, web dev, or app dev WebPlatform.org – tutorials and documentation for web tech  11 – Understanding our inner creativity, Margaret Boden This session was the most theoretical and probably least actionable of the day. It also held my attention utterly. Margaret spoke fluently, fascinatingly, without slides, on the subject of types of creativity and how they work. It was splendid. Yes, it raised a wry smile whenever she spoke of “the content of advertisements” and gave an example from 1970s TV ads, but even without the attempt to meet the conference’s theme this would have been thoroughly engaging. There are, Margaret suggested, three types of creativity: Combinatorial creativity The most common form, and consisting of synthesising ideas from existing and familiar concepts and tropes. Exploratory creativity Less common, this involves exploring the limits and quirks of a particular constraint or style. Transformational creativity This is uncommon, and arises from finding a way to do something that the existing rules would hold to be impossible. In essence, this involves breaking one of the constraints that exploratory creativity is composed from. Combinatorial creativity, she suggested, is particularly important for attaching favourable ideas to existing things. As such is it probably worth developing for marketing. Exploratory creativity may then come into play in something like developing and optimising an idea or campaign that now has momentum. Transformational creativity exists at the edges of this exploration. She suggested that products may often be transformational, but that marketing seemed unlikely to in her experience. This made me wonder about Listerine. Crucially, transformational creativity is characterised by there being some element of continuity with the strictures of previous thinking. Once it has happened, there may be  move from a revolutionary instance into an explored style. Again, from a marketing perspective, this seems to chime well with the thinking in Youngme Moon’s book: Different Talking about the birth of Modernism is visual art, Margaret pointed out that transformational creativity has historically risked a backlash, demanding what is essentially an education of the market. This is best accomplished by referring back to the continuities with the past in order to make the new familiar. Thoughts The afternoon is harder to sum up than the morning. It felt less concrete, and was troubled by a short run of poor presentations in the middle. Mainly, I found myself wrestling with the internal comms issue. It’s one of those things that seems astonishingly obvious in hindsight, but any campaign – particularly any large one – is doomed if the people involved can’t believe in it. We’ve run things here that haven’t gone so well, of course we have; who hasn’t? I’m not going to air any laundry, but people not being informed (much less aligned) feels like a common factor. It’s tough though. Managing and anticipating information needs across an organization of any size can’t be easy. Even the simple things like ensuring sales and support departments know what’s in a product release, and what messages go with it are easy to botch. The thing I like about framing this as a brand and campaign advocacy problem is that it makes it likely to get addressed. Better is always sexier than less-worse. Any technical communicator who’s ever felt crowded out by a content strategist or marketing copywriter  knows this – increasing revenue gets a seat at the table far more readily than reducing support costs, even if the financial impact is identical. So that’s it from AMC. The big thought-provokers were social buying behaviour and eliciting behaviour change, and the value of internal communications in ensuring successful campaigns and continuity of customer experience. I’ll be chewing over that for a while, and I’d definitely return next year.      

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  • Move penetrating OBB out of another OBB to resolve collision

    - by Milo
    I'm working on collision resolution for my game. I just need a good way to get an object out of another object if it gets stuck. In this case a car. Here is a typical scenario. The red car is in the green object. How do I correctly get it out so the car can slide along the edge of the object as it should. I tried: if(buildings.size() > 0) { Entity e = buildings.get(0); Vector2D vel = new Vector2D(); vel.x = vehicle.getVelocity().x; vel.y = vehicle.getVelocity().y; vel.normalize(); while(vehicle.getRect().overlaps(e.getRect())) { vehicle.setCenter(vehicle.getCenterX() - vel.x * 0.1f, vehicle.getCenterY() - vel.y * 0.1f); } colided = true; } But that does not work too well. Is there some sort of vector I could calculate to use as the vector to move the car away from the object? Thanks Here is my OBB2D class: public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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