Search Results

Search found 662 results on 27 pages for 'hilbert curve'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 5/27 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Any software transforming broken lines into curves?

    - by user32931
    Hello, do you know of any software that would help me transform a broken line into a curved line? For example, I have an octagon or a heptagon and I want it to be transformed into something resembling a circle. if you know such software, please, let me know. Thank You! Update A: Here is an image from the tutorial given to me by Jamie Keeling (right now it's the first answer below). At least the picture there represents what I want. In that tutorial this process is called "flattening paths". I will try to put that image right here, but if it doesn't get displayed, you can find it by this URL: http://msdn.microsoft.com/en-us/library/ms536364%28v=VS.85%29.aspx The red line in the picture is what I would want to submit, and the blue line is what I would want to get in the end:

    Read the article

  • How to achieve uniform speed of movement on a bezier curve in cocos 2d?

    - by Andrey Chernukha
    I'm an absolute beginner in cocos2 , actually i started dealing with it yesterday. What i'm trying to do is moving an image along Bezier curve. This is how i do it - (void)startFly { [self runAction:[CCSequence actions: [CCBezierBy actionWithDuration:timeFlying bezier:[self getPathWithDirection:currentDirection]], [CCCallFuncN actionWithTarget:self selector:@selector(endFly)], nil]]; } My issue is that the image moves not uniformly. In the beginning it's moving slowly and then it accelerates gradually and at the end it's moving really fast. What should i do to get rid of this acceleration?

    Read the article

  • Drawing a clamped uniform cubic B-spline using Cairo

    - by Tamás
    I have a bunch of coordinates which are the control points of a clamped uniform cubic B-spline on the 2D plane. I would like to draw this curve using Cairo calls (in Python, using Cairo's Python bindings), but as far as I know, Cairo supports Bézier curves only. I also know that the segments of a B-spline between two control points can be drawn using Bézier curves, but I can't find the exact formulae anywhere. Given the coordinates of the control points, how can I derive the control points of the corresponding Bézier curves? Is there any efficient algorithm for that?

    Read the article

  • Kink detection in drawn polylines

    - by David Rutten
    Users can sketch in my app using a very simple tool (move mouse while holding LMB). This results in a series of mousemove events and I record the cursor location at each event. The resulting polyline curve tends to be rather dense, with recorded points almost every other pixel. I'd like to smooth this pixelated polyline, but I don't want to smooth intended kinks. So how do I figure out where the kinks are? The image shows the recorded trail (red pixels) and the 'implied' shape as a human would understand it. People tend to slow down near corners, so there is usually even more noise here than on the straight bits.

    Read the article

  • Python fit polynomial, power law and exponential from data

    - by Nadir
    I have some data (x and y coordinates) coming from a study and I have to plot them and to find the best curve that fits data. My curves are: polynomial up to 6th degree; power law; and exponential. I am able to find the best fit for polynomial with while(i < 6): coefs, val = poly.polyfit(x, y, i, full=True) and I take the degree that minimizes val. When I have to fit a power law (the most probable in my study), I do not know how to do it correctly. This is what I have done. I have applied the log function to all x and y and I have tried to fit it with a linear polynomial. If the error (val) is lower than the others polynomial tried before, I have chosen the power law function. Am I correct? Now how can I reconstruct my power law starting from the line y = mx + q in order to draw it with the original points? I need also to display the function found. I have tried with: def power_law(x, m, q): return q * (x**m) using x_new = np.linspace(x[0], x[-1], num=len(x)*10) y1 = power_law(x_new, coefs[0], coefs[1]) popt, pcov = curve_fit(power_law, x_new, y1) but it seems not to work well.

    Read the article

  • Drawing half of a Bezier path in Raphael

    - by Fibericon
    Let's say I have a cubic Bezier path as follows (formatted for use with the Raphael path function): M55 246S55 247 55 248 Just an example. This was taken from my drawing application, where I use the cursor to draw a line when the user holds the mouse button down, kind of like a pencil or marker. I'm using jquery's mousemove event to draw the line between two points every time the user moves the mouse. There is another (the reference point) that is taken before the line is drawn, so that the Bezier curve can be created. Here's my question: is it possible to make Raphael only draw half of a given path? I'm aware of the getSubpath() function, but if my understanding of Bezier curves is correct, it would be rather difficult to calculate the second argument. The problem with the animate function is that it creates double lines (that is, it creates the curved line that I want, and the boxy line around it which should not be shown, possibly because the mouse is being moved faster than the animation can handle). Of course, if my approach itself is flawed in some way (or my understanding of the possible solutions), I'd like to hear it. Any help would be appreciated.

    Read the article

  • trackroll does not works as i want or expect to do,blackberry

    - by SWATI
    i am working on blackberry curve 8300 i have added some components in the main screen,now i want to move the focus vertically when the trackball moves up or down and move the focus horizontally when track-Wheel moves left or right. ================================================================================== --Title area that contains a focusable field(BACK)-- --Non focusable Label field that indicates the name of the user-- --A horizontal field manager1 that contains 4 buttons-- --A horizontal field manager2 that contains 4 buttons-- --A horizontal field manager2 that contains 4 buttons-- ================================================================================== now suppose currently focus is on BACK button and i scroll the track-wheel downwards then, focus should come on 1st button of manager1 Again when i scroll downwards,then focus should come on the 1st button of manager2 and not the 2nd button of manager1(as its happening on device) my code is ::: protected boolean trackwheelRoll(int amount, int status, int time) { focusIndex = this.getFieldWithFocusIndex(); System.out.println("focus index ::::::::::::::::"+focusIndex); Field f; if(focusIndex!=0) { if(amount==-1) { //move up if(focusIndex>=0) { focusIndex = focusIndex-1; f = getField(focusIndex); f.setFocus(); } } if(amount==1) { //moving down if(focusIndex<=3) { f = getField(++focusIndex); f.setFocus(); } } } return super.trackwheelRoll(amount, status, time); } even after this control moves abruptly on simulator but on device no change took place

    Read the article

  • How to find "y" values of the already estimated monotone function of the non-monotone regression curve corresponding to the original "x" points?

    - by parenthesis
    The title sounds complicated but that is what I am looking for. Focus on the picture. ## data x <- c(1.009648,1.017896,1.021773,1.043659,1.060277,1.074578,1.075495,1.097086,1.106268,1.110550,1.117795,1.143573,1.166305,1.177850,1.188795,1.198032,1.200526,1.223329,1.235814,1.239068,1.243189,1.260003,1.262732,1.266907,1.269932,1.284472,1.307483,1.323714,1.326705,1.328625,1.372419,1.398703,1.404474,1.414360,1.415909,1.418254,1.430865,1.431476,1.437642,1.438682,1.447056,1.456152,1.457934,1.457993,1.465968,1.478041,1.478076,1.485995,1.486357,1.490379,1.490719) y <- c(0.5102649,0.0000000,0.6360097,0.0000000,0.8692671,0.0000000,1.0000000,0.0000000,0.4183691,0.8953987,0.3442624,0.0000000,0.7513169,0.0000000,0.0000000,0.0000000,0.0000000,0.1291901,0.4936121,0.7565551,1.0085108,0.0000000,0.0000000,0.1655482,0.0000000,0.1473168,0.0000000,0.0000000,0.0000000,0.1875293,0.4918018,0.0000000,0.0000000,0.8101771,0.6853480,0.0000000,0.0000000,0.0000000,0.0000000,0.4068802,1.1061434,0.0000000,0.0000000,0.0000000,0.0000000,0.0000000,0.0000000,0.0000000,0.0000000,0.0000000,0.6391678) fit1 <- c(0.5102649100,0.5153380934,0.5177234836,0.5255544980,0.5307668662,0.5068087080,0.5071001179,0.4825657520,0.4832969250,0.4836378194,0.4842147729,0.5004039310,0.4987301366,0.4978800742,0.4978042478,0.4969807064,0.5086987191,0.4989497612,0.4936121200,0.4922210302,0.4904593166,0.4775197108,0.4757040857,0.4729265271,0.4709141776,0.4612406896,0.4459316517,0.4351338346,0.4331439717,0.4318664278,0.3235179189,0.2907908968,0.1665721429,0.1474035158,0.1443999345,0.1398517097,0.1153991839,0.1142140393,0.1022584672,0.1002410843,0.0840033244,0.0663669309,0.0629119398,0.0627979240,0.0473336492,0.0239237481,0.0238556876,0.0084990298,0.0077970954,0.0000000000,-0.0006598571) fit2 <- c(-0.0006598571,0.0153328298,0.0228511733,0.0652889427,0.0975108758,0.1252414661,0.1270195143,0.1922510501,0.2965234797,0.3018551305,0.3108761043,0.3621749370,0.4184150225,0.4359301495,0.4432114081,0.4493565757,0.4510158144,0.4661865431,0.4744926045,0.4766574718,0.4796937554,0.4834718810,0.4836125426,0.4839450098,0.4841092849,0.4877317306,0.4930561638,0.4964939389,0.4970089201,0.4971376528,0.4990394601,0.5005881678,0.5023814257,0.5052125977,0.5056691690,0.5064254338,0.5115481820,0.5117259449,0.5146054557,0.5149729419,0.5184178197,0.5211542908,0.5216215426,0.5216426533,0.5239797875,0.5273573222,0.5273683002,0.5293994824,0.5295130266,0.5306236672,0.5307303109) ## picture plot(x, y) ## red regression curve points(x, fit1, col=2); lines(x, fit1, col=2) ## blue monotonic curve to the regression points(min(x) + cumsum(c(0, rev(diff(x)))), rev(fit2), col="blue"); lines(min(x) + cumsum(c(0, rev(diff(x)))), rev(fit2), col="blue") ## "x" original point matches with the regression estimated point ## but not with the estimated (fit2=estimate) monotonic curve abline(v=1.223329, lty=2, col="grey") Focus on the dashed grey line. The idea is to get y value of the monotonic blue curve corresponding to x original value. The grey line should cross three points (the original one "black", the regression estimate "red", the adjusted regression estimate "blue"). Can we do this? Methodology: The object "fit2" is the output of the function rearrangement(). It is always monotonically increasing. library(Rearrangement) fit2 <- rearrangement(x=as.data.frame(x), y=fit1)

    Read the article

  • Any software transforming broken lines into curves?

    - by brilliant
    Hello, do you know of any software that would help me transform a broken line into a curved line? For example, I have an octagon or a heptagon and I want it to be transformed into something resembling a circle. if you know such software, please, let me know. Thank You! Update A: Here is an image from the tutorial given to me by Jamie Keeling (right now it's the first answer below). At least the picture there represents what I want. In that tutorial this process is called "flattening paths". I will try to put that image right here, but if it doesn't get displayed, you can find it here: http://msdn.microsoft.com/en-us/library/ms536364%28v=VS.85%29.aspx The red line in the picture is what I would want to submit, and the blue line is what I would want to get in the end:

    Read the article

  • Is it worth moving from stored procedures to linq ?

    - by Josef
    I'm looking at standardizing programming in an organisaiton. Half uses stored procedures and the other half Linq. From what i've read there is still some debate going on on this topic. My concern is that MS is trying to slip in it's own proprietry query language 'linq' to make SQL redundant. If a few years back microsoft had tried to win customers from oracle and sybase with their MSSQL database and stated that it didn't use SQL by their own proprietry query langues ie linq. I doubt many would have switched. I believe that is exactly what is happening now by introducting it into the applicaiton business layer. I have used MS for many years but there is one gripe that I have with them and that is that they change their direction a lot. By a lot I mean new releases of .net, silverlight etc are more than 30% different from previous version. So by the time you become productive a new release is on the way. As things stand now a web developer using .net would need to know either vb.net or c#, xml, xaml,javascript,html, sql and now linq. That doesn't make for good productivity in my books. My concern is that once we all start using linq MS will start changing it between releases. and it will become an ever changing landscape. I believe that 'linq to sql' has already been deprecated. At leas with SQL we are dealing with a more stable and standardized language. Are we looking at a programming revolution or a marketing campaign? As far as I know other languages like Cobol have stayed the same for years. A cobol program from 20 years ago could pick up todays code and start working on it. Could a Vb3 person work on a modern .net web app ? Would these large changes need to be made if the underlying original foundation had been sound ? I worry about following MS shaking roadmap with it's deadends and double backs. are there any architects out there who feel the same ? regards Josef

    Read the article

  • Starting work on a Pre-existing Project

    - by Toymakerii
    So this is more of a generic question. I seem to keep finding myself being put on larger and larger projects. Recently I have been assigned to a very large project written in C and VHDL. The goal is for me to become familiar with the code and eventually take the lead on the project. This is by far the largest project I have been assigned to work on that I didn't start. So here is my question: What methods/tools do you use to learn how everything works? Do you just increase and expand on comments? Do you make a UML representation of the project? Any tips would be great! Thanks

    Read the article

  • Use of Curve25519

    - by chmike
    I'm currently investigating the use of curve25519 for signing. Original distribution can be obtained here and a C code implementation here or here. Bernstein suggest to use ECDSA for this but I could not find any code.

    Read the article

  • variable row size list field in blackberry

    - by SWATI
    i have created an application that contains a list field(custom) i want if a certain condition is satisfied then rowheight should be 100 else it should be 50 how can i do that i tried setRowHeight(index,size); but it dnt worked.Moreover its undocumented toooooooooo.. any help will be appreciated

    Read the article

  • An extended Bezier Library or Algorithms of bezier operations

    - by Sorush Rabiee
    Hi, Is there a library of data structures and operations for quadratic bezier curves? I need to implement: bezier to bitmap converting with arbitrary quality optimizing bezier curves common operations like subtraction, extraction, rendering etc. languages: c,c++,.net,python Algorithms without implementation (pseudocode or etc) could be useful too. (especially optimization)

    Read the article

  • Curvey Redraw tabs working fine on Firefox , problem with IE

    - by Rohit
    Hi, I have used curvey redraw library from google code(http://code.google.com/p/curvycorners/) & it has solved my purpose, though now as per new req i am struggling with IE. I want to have two tab rows each containing 2 tabs. <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>curvyCorners - Tab demo</title> <style type="text/css">/*<![CDATA[*/ /* tab styles */ #tabrow { margin:0; padding-left:1ex; min-width:800px; font-size:small; letter-spacing:0.3pt; line-height:1; height:24px; } #tabrow ul { margin:0; padding:0; list-style:none; position:absolute; z-index:2; } #tabrow li { float:left; background-color:#E0DFE3; color:#000; margin-right:5px; padding:5px; -webkit-border-top-left-radius:5px; -webkit-border-top-right-radius:5px; -moz-border-radius-topleft:5px; -moz-border-radius-topright:5px; border-top:solid #9B9B9B 1px; border-left:solid #9B9B9B 1px; border-right:solid #9B9B9B 1px; border-bottom-width:0; border-bottom-color:transparent; cursor:pointer; font-family:verdana;font-weight:bold;font-style:italic } #tabrow li.select { background-color:#ffffff; color:#2470c4; height:14px; } /* page styles */ #midbox { width:220px; height:305px; -webkit-border-radius: 5px; -moz-border-radius: 5px; } #midbox { border: solid #9b9b9b 1px; background-color:#ffffff; } #midbox p { margin:0; padding-bottom:1ex; } h1, #topbox h2 { margin:0 15pt; padding: 5pt 0; } div.subpage { padding:1em; } /*]]>*/ </style> <script type="text/javascript" src="curvs.js"> </script> <script type="text/javascript">//<![CDATA[ var selectedTab = 0; function tabclick(n) { if (n === selectedTab) return; // nothing to do. var li = document.getElementById('tab' + selectedTab); curvyCorners.adjust(li, 'className', ''); // Remove the 'select' style li = document.getElementById('page' + selectedTab); li.style.display = 'none'; // hide the currently selected sub-page li = document.getElementById('page' + n); li.style.display = 'block'; // show the new sub-page li = document.getElementById('tab' + n); // get the new (clicked) tab curvyCorners.adjust(li, 'className', 'select'); // and update its style curvyCorners.redraw(); // Redraw all elements with className curvyRedraw selectedTab = n; // store for future reference } var selectedTab1 = 2; function tabclick1(n) { if (n === selectedTab1) return; // nothing to do. var li = document.getElementById('tab' + selectedTab1); curvyCorners.adjust(li, 'className', ''); // Remove the 'select' style li = document.getElementById('page' + selectedTab1); li.style.display = 'none'; // hide the currently selected sub-page li = document.getElementById('page' + n); li.style.display = 'block'; // show the new sub-page li = document.getElementById('tab' + n); // get the new (clicked) tab curvyCorners.adjust(li, 'className', 'select'); // and update its style curvyCorners.redraw(); // Redraw all elements with className curvyRedraw selectedTab1 = n; // store for future reference } //]]> </script> </head> <body> <div id="tabrow"> <ul> <li id="tab0" onclick="tabclick(0);" class="select curvyRedraw">Categories</li> <li id="tab1" onclick="tabclick(1);" class="curvyRedraw">Services</li> </ul> </div> <div id="midbox" class="curvyRedraw"> <div id="page0" class="subpage"> Category details </div> <div id="page1" class="subpage" style="display:none"> Service details </div> </div> <br/><br/> <div id="tabrow"> <ul> <li id="tab2" onclick="tabclick1(2);" class="select curvyRedraw">Recent Activiites</li> <li id="tab3" onclick="tabclick1(3);" class="curvyRedraw">News</li> </ul> </div> <div id="midbox" class="curvyRedraw"> <div id="page2" class="subpage"> Activities </div> <div id="page3" class="subpage" style="display:none"> News </div> </div> </body> </html> Can you please help me out in this? Thanks, Rohit.

    Read the article

  • How to find control points for a BezierSegment given Start, End, and 2 Intersection Pts in C# - AKA

    - by softwarequestioneer
    Hi, I've been struggling looking for an understandable way to do this. I have four points, a StartPt, EndPoint, and Intersection points to represent the peak and valley in the bezier. The BezierSegment in C# requires start, controlPoint 1, controlPoint 2, endpoint - however I don't have any control points I only have these two points that lie along the bezier curves (i'm calling them intersection points above)... how can I calculate the two control points? Thanks in advance, this has been driving me crazy. There's some kind of explanation here: http://www.tinaja.com/glib/nubz4pts1.pdf but it's written in postscript and that language makes no sense to me at all - it's over my head.

    Read the article

  • "too many threads error" in blackberry OS-4.5

    - by SWATI
    hi in my application i have 20 icons(bitmap fields) on the home screen When i click on any icon an HTTP request is made in a separate thread. I have used invoke later method wherever necessary to take care of multi-threading problems. But still the number of threads goes beyond 16 and an error pops up indicating too many threads error and applications needs to be restarted!! can anybody tell me how to destroy these threads when they are no longer in use. I don't understand why they don't destroy on their own as usually they do.

    Read the article

  • How to create a level with curved lines with cocos2d + Box2d on the iphone?

    - by Steven
    I'd like to create a game that has levels such as this: http://img169.imageshack.us/img169/7294/picdq.png The Player moves "flies" through the level and mustn't collide with the walls. How can I create such levels? I found that piece of software: http://www.sapusmedia.com/levelsvg/ It's not that cheap, so I wonder whether there is another way to create such a level as shown in the picture above...?

    Read the article

  • different brainpoolP521r1 parameters for flexiProvider. Why?

    - by user182846
    I am trying to generate ECC public private key pairs using flexiProvider. I have noticed that values for parameters like p and q are different in brainpoolP521r1 of flexiProvider than those which are specified in many sites. Values specified are Q= AADD.... but what I get is Q=8948... Any idea why flexiProvider does not use specified values and whether having different values is affectes security. I am new to ECC. Any help will be greatly appreciated.

    Read the article

  • Do any well-known CAs issue Elliptic Curve certificates?

    - by erickson
    Background I've seen that Comodo has an elliptic curve root ("COMODO ECC Certification Authority"), but I don't see mention of EC certificates on their web site. Does Certicom have intellectual property rights that prevent other issuers from offering EC certificates? Does a widely-used browser fail to support ECC? Is ECC a bad fit for traditional PKI use like web server authentication? Or is there just no demand for it? I'm interested in switching to elliptic curve because of the NSA Suite B recommendation. But it doesn't seem practical for many applications. Bounty Criteria To claim the bounty, an answer must provide a link to a page or pages at a well-known CA's website that describes the ECC certificate options they offer, prices, and how to purchase one. In this context, "well-known" means that the proper root certificate must be included by default in Firefox 3.5 and IE 8. If multiple qualifying answers are provided (one can hope!), the one with the cheapest certificate from a ubiquitous CA will win the bounty. If that doesn't eliminate any ties (still hoping!), I'll have to choose an answer at my discretion. Remember, someone always claims at least half of the bounty, so please give it a shot even if you don't have all the answers.

    Read the article

  • How do I prevent a curve snapping to a straight line in Flash CS4?

    - by Kelix
    I am trying to do some vector drawing in Flash but am having trouble when "curving" lines. At the moment, unless the curve is significant, it snaps back to a straight line meaning I am finding it impossible to draw shallow curves. Any idea how to turn this snapping off? I have tried turning everything off in View Snapping and it makes no difference.

    Read the article

  • Does the term "Learning Curve" include the knowing of the gotchas?

    - by voroninp
    When you learn new technology you spend time understanding its concepts and tools. But when technology meets real life strange and not pleasant things happen. Reuqirements are often far from ideal and differ from 'classic' scenario. And soon I find myself bending the technology to my real needs. At this point I begin to know bugs of the system or that is is not so flexible as it seemed at the very begining. And this 'fighting' with technology consumes a great part of the time while developing. What is more depressing is that the bunch of such gotchas and workarounds are not concentrated at one place (book, site, etc.) And before you really confront it you cannot really ask the correct question because you do not even suspect the reason for the problem to occur (unknown-unknown). So my question consiststs of three: 1) Do you really manage (and how) to predict possible future problems? 2) How much time do you spend for finding the workaround/fix/solution before you leave it and switch to other problems. 3) What are the criteria for you to think about yourself as experienced in the tecnology. Do you take these gotchas into account?

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >