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  • Diamond-square terrain generation problem

    - by kafka
    I've implemented a diamond-square algorithm according to this article: http://www.lighthouse3d.com/opengl/terrain/index.php?mpd2 The problem is that I get these steep cliffs all over the map. It happens on the edges, when the terrain is recursively subdivided: Here is the source: void DiamondSquare(unsigned x1,unsigned y1,unsigned x2,unsigned y2,float range) { int c1 = (int)x2 - (int)x1; int c2 = (int)y2 - (int)y1; unsigned hx = (x2 - x1)/2; unsigned hy = (y2 - y1)/2; if((c1 <= 1) || (c2 <= 1)) return; // Diamond stage float a = m_heightmap[x1][y1]; float b = m_heightmap[x2][y1]; float c = m_heightmap[x1][y2]; float d = m_heightmap[x2][y2]; float e = (a+b+c+d) / 4 + GetRnd() * range; m_heightmap[x1 + hx][y1 + hy] = e; // Square stage float f = (a + c + e + e) / 4 + GetRnd() * range; m_heightmap[x1][y1+hy] = f; float g = (a + b + e + e) / 4 + GetRnd() * range; m_heightmap[x1+hx][y1] = g; float h = (b + d + e + e) / 4 + GetRnd() * range; m_heightmap[x2][y1+hy] = h; float i = (c + d + e + e) / 4 + GetRnd() * range; m_heightmap[x1+hx][y2] = i; DiamondSquare(x1, y1, x1+hx, y1+hy, range / 2.0); // Upper left DiamondSquare(x1+hx, y1, x2, y1+hy, range / 2.0); // Upper right DiamondSquare(x1, y1+hy, x1+hx, y2, range / 2.0); // Lower left DiamondSquare(x1+hx, y1+hy, x2, y2, range / 2.0); // Lower right } Parameters: (x1,y1),(x2,y2) - coordinates that define a region on a heightmap (default (0,0)(128,128)). range - basically max. height. (default 32) Help would be greatly appreciated.

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  • Use TV (hdmi) with non-square pixels

    - by labsin
    I am having a problem when I connect my LG plasma tv (with a native resolution of 1024x768 pixels) to my 12.04 laptop. The pixels (actual pixels, not the signal) of my TV are stretched so it gets his 16:9 ratio. The pixels are rectangular (1.3333x1). Everything I display from my laptop oviously get stretched (4:3 stretched to 16:9). There is a different dpi in X and Y needed for it to display properly (some kind of anamorphic mode). Default Ubuntu uses a dpi of 96x96. I can change it using xrandr, but only square eg 100x100 or 70x70. Already looked here, but it seems Ubuntu totally ignore the displaySize in xorg.conf When I use the code below to see the dpi and nothing I do changes it. The displaySize also stays the same (calculated using 96 dpi and the resolution) xdpyinfo | grep -B2 resolution I use the propretary ATI drivers for my ATI Mobility Radeon HD 50xx but it is the same with the Radeon drivers. My temporary solution is to use: xrandr --output DFP1 --mode 1024x768 --scale 1.333333333333x1 --output LVDS --off But with this the right side of the screen is nog accesable. This is a known problem with xrandr --scale and ubuntu. This is because of a patch for the mouse/windows not going outside the screen. I search a way to change the DisplaySize or the dpi(to something not square like 128x96) when I connect the display.

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  • Filling array with numbers from given range so that sum of adjacent numbers is square number

    - by REACHUS
    Problem: Fill all the cells using distinct numbers from <1,25 set, so that sum of two adjacent cells is a square number. (source: http://grymat.im.pwr.wroc.pl/etap1/zad1etp1213.pdf; numbers 20 and 13 have been given) I've already solved this problem analytically and now I would like to approach it using an algorithm. I would like to know how should I approach these kind of problems in general (not a solution, just a point for me to start).

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  • Terrain square loading

    - by AndroidXTr3meN
    Games like Skyrim, Morrowind, and more are using quads or square to divide the terrain if im correct. The player is always at #5 1 | 2 | 3 4 | 5 | 6 7 | 8 | 9 So whenever you cross the border you unload and load the new "areas" But if the user goes just over the edge and then the second after goes back previous area a lot of unnecessary loading and unloading is done. Is there a general approach to this because I dont think games like skyrim have this issue? Cheers!

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  • total number of magic square from 9 numbers

    - by Peeyush
    9 numbers need to be arranged in a magic number square. A magic number square is a square of numbers that is arranged such that every row and column has the same sum.(condition for diagonal has been relaxed) For example: 1 2 3 3 2 1 2 2 2 How do we calculate total number of distinct magic square from 9 numbers. Two magic number squares are distinct if they differ in value at one or more positions. For example, there is only one magic number square that can be made of 9 instances of the same number. e.g. for these 9 numbers { 4, 4, 4, 4, 4, 4, 4, 4, 4 }, answer should be 1. Also the complexity should be optimal. Do we need to iterate through all the permutations , discarding if a[0]+a[1]+a[2] %3!=0 such combinations ? moreover how do we remove duplicate magic square?

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  • Sliding Response after a Point-Square Collision

    - by mars
    In general terms and pseudo-code, what would be the best way to have a collision response of sliding along a wall if the wall is actually just a part of an entire square that a point is colliding into? The collision test method used is a test to see if the point lies in the square. Should I divide the square into four lines and just calculate the shortest distance to the line and then move the point back that distance?If so, then how can I determine which edge of the square the point is closest to after collision?

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  • Notepad++ Move Caret Outside Bracket

    - by marknadal
    I've searched everywhere and can't find an answer to something incredibly simple and useful for Notepad++. When using autoclose, the cursor gets stuck inside, which is good until I've finished typing the parameters. How do I get the caret to jump outside of the currently nested autoclosed element? Regardless of it being (), {}, [], "", '', ? CTRL+B does does something similar, but it goes to the inside of the left bracket first, and then to the outside of the right bracket, which is where I want to be. This only works for brackets though, not tags and qoutations Hitting CTRL+B twice is cumbersome and manually hitting "right arrow" requires too much hand motion while coding. And finally, is there anyway I can map this to "SHIFT+Space"?

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  • Detect a white square in a black and white image

    - by gcc
    I saw that question i one web site (cite name like that programm.) then i tried to solve but icannot (not my and myfriend homework ) how can i approach to that one (in program.net no solution there is ) Read black & white image data from standard input, and detect a white square in the image. Output the coordinates of the upper left corner of the square, and the width of the square. In the preliminary work, you can print the output and terminate your program after you detect your first square. If you can't find any square on the image, you will print the string: "NO DETECTION". Input (which represents a 2 by 2 square in the center of a 5 by 4 image): 2 2 5 4 0 0 0 0 0 0 0 255 255 0 0 0 255 255 0 0 0 0 0 0 Output: 3 2 2 Input (more comprehensible format of the image, with the same output): 2 6 4 000 000 000 000 000 000 000 000 255 255 255 000 000 000 000 255 255 000 000 000 000 000 000 000 Output: no detection Input can be: 000 255 255 000 000 000 000 255 255 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 255 255 255 000 000 000 255 255 255 000 000 000 255 255 255 000 000 000 000 000 000 000 If there are two squares detected, we should use the biggest one

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  • Attempting to find a formula for tessellating rectangles onto a board, where middle square can't be

    - by timemirror
    I'm working on a spatial stacking problem... at the moment I'm trying to solve in 2D but will eventually have to make this work in 3D. I divide up space into n x n squares around a central block, therefore n is always odd... and I'm trying to find the number of locations that a rectangle of any dimension less than n x n (eg 1x1, 1x2, 2x2 etc) can be placed, where the middle square is not available. So far I've got this.. total number of rectangles = ((n^2 + n)^2 ) / 4 ..also the total number of squares = (n (n+1) (2n+1)) / 6 However I'm stuck in working out a formula to find how many of those locations are impossible as the middle square would be occupied. So for example: [] [] [] [] [x] [] [] [] [] 3 x 3 board... with 8 possible locations for storing stuff as mid square is in use. I can use 1x1 shapes, 1x2 shapes, 2x1, 3x1, etc... Formula gives me the number of rectangles as: (9+3)^2 / 4 = 144/4 = 36 stacking locations However as the middle square is unoccupiable these can not all be realized. By hand I can see that these are impossible options: 1x1 shapes = 1 impossible (mid square) 2x1 shapes = 4 impossible (anything which uses mid square) 3x1 = 2 impossible 2x2 = 4 impossible etc Total impossible combinations = 16 Therefore the solution I'm after is 36-16 = 20 possible rectangular stacking locations on a 3x3 board. I've coded this in C# to solve it through trial and error, but I'm really after a formula as I want to solve for massive values of n, and also to eventually make this 3D. Can anyone point me to any formulas for these kind of spatial / tessellation problem? Also any idea on how to take the total rectangle formula into 3D very welcome! Thanks!

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  • Javascript Square bracket notation for global variables

    - by Yousuf Haider
    I ran into an interesting issue the other day and was wondering if someone could shed light on why this is happening. Here is what I am doing (for the purposes of this example I have dumbed down the example somewhat): I am creating a globally scoped variable using the square bracket notation and assigning it a value. Later I declare a var with the same name as the one I just created above. Note I am not assigning a value. Since this is a redeclaration of the same variable the old value should not be overriden as described here: http://www.w3schools.com/js/js_variables.asp //create global variable with square bracket notation window['y'] = 'old'; //redeclaration of the same variable var y; if (!y) y = 'new'; alert(y); //shows New instead of Old The problem is that the old value actually does get overriden and in the above eg. the alert shows 'new' instead of 'old'. Why ? I guess another way to state my question is how is the above code different in terms of semantics from the code below: //create global variable var y = 'old'; //redeclaration of the same variable var y; if (!y) y = 'new'; alert(y); //shows New instead of Old

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  • Square One to Game Development

    - by Ian Quach
    How does someone even get into developing a game. What would they need to know, how would someone find the knowledge to program a game? I've always looked at game development as a future career. Now that I'm getting closer to university I was hoping to find a way to head start this future in game development. What would be the best place to start? I would love any help or tips from anyone. Thanks for reading this. :)

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  • Fastest way to check if two square 2D arrays are rotationally and reflectively distinct

    - by kustrle
    The best idea I have so far is to rotate first array by {0, 90, 180, 270} degrees and reflect it horizontally or/and vertically. We basically get 16 variations [1] of first array and compare them with second array. if none of them matches the two arrays are rotationally and reflectively distinct. I am wondering if there is more optimal solution than this brute-force approach? [1] 0deg, no reflection 0deg, reflect over x 0deg, reflect over y 0deg, reflect over x and y 90deg, no reflection ...

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  • How to replace round bracket tag in javascript string

    - by tomaszs
    I have trouble with changing round bracket tag in Javascript. I try to do this: var K = 1; var Text = "This a value for letter K: {ValueOfLetterK}"; Text = Text.replace("{ValueOfLetterK}", K); and after that I get: Text = "This a value for letter K: {ValueOfLetterK}" What can be done to make this work? When I remove round brackets it works fine.

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  • Tournament bracket

    - by Luke
    Not sure of the best way to go about this? I want to create a tournament bracket of 2,4,8,16,32, etc teams. The winner of the first two will play winner of the next 2 etc. All the way until there is a winner. Like this Can anyone help me?

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  • How to display two-rows bracket in Latex?

    - by niko
    Hi, Does anyone know how to modify the following string in order to display the two-lines bracket? str = '$$c_i =\{\begin{array}{l l} 1 \quad L\left(Q_i\right) < 0 \\ 0 \quad L\left(Q_i\right) \geq 0 \\ \end{array}$$'; The current output is the following: The sign '{' has to embrace both rows (1 and 0).

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  • bracket style border around elements

    - by user1255049
    I'm looking for a way to implement a bracket style border around my <h2> headings; I've attached an image showing exactly what I'm trying to accomplish. The only way I can think of to achieve this effect is by using images, but I'm unsure of exactly how to do so(all of my <h2>s are of varying length/height, or if maybe there is a better way. Any tips & insight are greatly appreciated.

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  • what does square bracket syntax mean above a method in C#, ASP.NET

    - by Alexander
    I am just looking a bunch of codes that I am trying to learn from an open source project and sometimes I see a square brackets above a function such as: [EdmFunction("NerdDinnerModel.Store", "DistanceBetween")] public static double DistanceBetween(double lat1, double long1, double lat2, double long2) or [Bind(Include = "Title,Description,EventDate,Address,Country,ContactPhone,Latitude,Longitude")] [MetadataType(typeof(Dinner_Validation))] public partial class Dinner

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  • Square Brackets in XSL-FO

    - by Igman
    I am attempting to create a list in XSL-FO using a square bracket. I have been able to get it working using the standard unicode bullet character (&#8226;) but I just can't seem to get it working for square brackets. I have tried using &#9632;, but that does not seem to work. It is important that i can get the square bullets working because I am matching an existing file format.Any help in getting this working would be greatly appreciated.

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  • Javascript Regex to convert dot notation to bracket notation

    - by Tauren
    Consider this javascript: var joe = { name: "Joe Smith", location: { city: "Los Angeles", state: "California" } } var string = "{name} is currently in {location.city}, {location.state}"; var out = string.replace(/{([\w\.]+)}/g, function(wholematch,firstmatch) { return typeof values[firstmatch] !== 'undefined' ? values[firstmatch] : wholematch; }); This will output the following: Joe Smith is currently in {location.city}, {location.state} But I want to output the following: Joe Smith is currently in Los Angeles, California I'm looking for a good way to convert multiple levels of dot notation found between braces in the string into multiple parameters to be used with bracket notation, like this: values[first][second][third][etc] Essentially, for this example, I'm trying to figure out what regex string and function I would need to end up with the equivalent of: out = values[name] + " is currently in " + values["location"]["city"] + values["location"]["state"];

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  • Scite Lua - escaping right bracket in regex?

    - by ~sd-imi
    Hi all, Bumped into a somewhat weird problem... I want to turn the string: a\left(b_{d}\right) into a \left( b_{d} \right) in Scite using a Lua script. So, I made the following Lua script for Scite: function SpaceTexEquations() editor:BeginUndoAction() local sel = editor:GetSelText() local cln3 = string.gsub(sel, "\\left(", " \\left( ") local cln4 = string.gsub(cln3, "\\right)", " \\right) ") editor:ReplaceSel(cln4) editor:EndUndoAction() end The cln3 line works fine, however, cln4 crashes with: /home/user/sciteLuaFunctions.lua:49: invalid pattern capture >Lua: error occurred while processing command I think this is because bracket characters () are reserved characters in Lua; but then, how come the cln3 line works without escaping? By the way I also tried: -- using backslash \ as escape char: local cln4 = string.gsub(cln3, "\\right\)", " \\right) ") -- crashes all the same -- using percentage sign % as escape chare local cln4 = string.gsub(cln3, "\\right%)", " \\right) ") -- does not crash, but does not match either Could anyone tell me what would be the correct way to do this? Thanks, Cheers!

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  • createChildren Called Before Component's MXML Bracket Logic Is Evaluated

    - by Nalandial
    I have the following MXML: <mx:Script> var someBoolean:Boolean = determineSomeCondition(); </mx:Script> .... <foo:MyComponent somePropertyExpectingIDataRenderer="{ someBoolean ? new Component1ThatImplementsIDataRenderer() : new Component2ThatImplementsIDataRenderer() }"> </foo:MyComponent> I have also overridden the createChildren() function: override protected function createChildren():void { super.createChildren(); //do something with somePropertyExpectingIDataRenderer } My problem is: createChildren() is being called before the squiggly bracket logic is being evaluated, so in createChildren(), somePropertyExpectingIDataRenderer is null. However if I pass the component via MXML like this: <foo:MyComponent> <bar:somePropertyExpectingIDataRenderer> <baz:Component1ThatImplementsIDataRenderer/> </bar:somePropertyExpectingIDataRenderer> </foo:MyComponent> Then when createChildren() is called, that same property isn't null. Is this supposed to happen and if so, what other workarounds should I consider?

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