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  • Minimizing distance to a weighted grid

    - by Andrew Tomazos - Fathomling
    Lets suppose you have a 1000x1000 grid of positive integer weights W. We want to find the cell that minimizes the average weighted distance.to each cell. The brute force way to do this would be to loop over each candidate cell and calculate the distance: int best_x, best_y, best_dist; for x0 = 1:1000, for y0 = 1:1000, int total_dist = 0; for x1 = 1:1000, for y1 = 1:1000, total_dist += W[x1,y1] * sqrt((x0-x1)^2 + (y0-y1)^2); if (total_dist < best_dist) best_x = x0; best_y = y0; best_dist = total_dist; This takes ~10^12 operations, which is too long. Is there a way to do this in or near ~10^8 or so operations?

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  • PHP: Trying to come up with a "prev" and "next" link

    - by fwaokda
    I'm displaying 10 records per page. The variables I have currently that I'm working with are.. $total = total number of records $page = whats the current page I'm displaying I placed this at the top of my page... if ( $_GET['page'] == '' ) { $page = 1; } //if no page is specified set it to `1` else { $page = ($_GET['page']); } // if page is specified set it Here are my two links... if ( $page != 1 ) { echo '<div style="float:left" ><a href="index.php?page='. ( $page - 1 ) .'" rev="prev" >Prev</a></div>'; } if ( !( ( $total / ( 10 * $page ) ) < $page ) ) { echo '<div style="float:right" ><a href="index.php?page='. ( $page + 1 ) .'" rev="next" >Next</a></div>'; } Now I guess (unless I'm not thinking of something) that I can display the "Prev" link every time except when the page is '1'. How can make it where the "Next" link doesn't show on the last page though?

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  • Print all ways to sum n integers so that they total a given sum.

    - by noghead
    Im trying to come up with an algorithm that will print out all possible ways to sum N integers so that they total a given value. Example. Print all ways to sum 4 integers so that they sum up to be 5. Result should be something like: 5 0 0 0 4 1 0 0 3 2 0 0 3 1 1 0 2 3 0 0 2 2 1 0 2 1 2 0 2 1 1 1 1 4 0 0 1 3 1 0 1 2 2 0 1 2 1 1 1 1 3 0 1 1 2 1 1 1 1 2

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  • Calculating percent "x/y * 100" always results in 0?

    - by Patrick Beninga
    In my assignment i have to make a simple version of Craps, for some reason the percentage assignments always produce 0 even when both variables are non 0, here is the code. import java.util.Random; Header, note the variables public class Craps { private int die1, die2,myRoll ,myBet,point,myWins,myLosses; private double winPercent,lossPercent; private Random r = new Random(); Just rolls two dies and produces their some. public int roll(){ die1 = r.nextInt(6)+1; die2 = r.nextInt(6)+1; return(die1 + die2); } The Play method, this just loops through the game. public void play(){ myRoll = roll(); point = 0; if(myRoll == 2 ||myRoll == 3 || myRoll == 12){ System.out.println("You lose!"); myLosses++; }else if(myRoll == 7 || myRoll == 11){ System.out.println("You win!"); myWins++; }else{ point = myRoll; do { myRoll = roll(); }while(myRoll != 7 && myRoll != point); if(myRoll == point){ System.out.println("You win!"); myWins++; }else{ System.out.println("You lose!"); myLosses++; } } } This is where the bug is, this is the tester method. public void tester(int howMany){ int i = 0; while(i < howMany){ play(); i++; } bug is right here in these assignments statements winPercent = myWins/i * 100; lossPercent = myLosses/i* 100; System.out.println("program ran "+i+" times "+winPercent+"% wins "+ lossPercent+"% losses with "+myWins+" wins and "+myLosses+" losses"); } }

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  • How long would this have to go on for...

    - by Pieman
    I have the Pi formulae -Well one of them... 1 - 1/3 + 1/5 - 1/7 etc. How long would it take to get to like 1000 S.F correct? -Well, not how long, how big would the denominator be? -I have it updating 4 times in one refresh: http://zombiewrath.com/pi.php So the section above would be done in one refresh, then 7 to 13 in another etc. Answer this maths question please :) Also how can I get the 10,002 length variable onto 'seperate lines'? -I want it to fill 100% screen width -no scrolling needed (well downwards only)

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  • Why do we need to estimate the true position in Kalman filters?

    - by Kalla
    I am following a probably well-known tutorial about Kalman filter here From these lines of code: figure; plot(t,pos, t,posmeas, t,poshat); grid; xlabel('Time (sec)'); ylabel('Position (feet)'); title('Figure 1 - Vehicle Position (True, Measured, and Estimated)') I understand that x is the true position, y is measured position, xhat is estimated position. Then, if we can compute x (this code: x = a * x + b * u + ProcessNoise;), why do we need to estimated x anymore?

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  • Scale 2D coordinates and keep their relative euclidean distances intact?

    - by eiaxlid
    I have a set of points like: pointA(3302.34,9392.32), pointB(34322.32,11102.03), etc. I need to scale these so each x- and y-coordinate is in the range (0.0 - 1.0). I tried doing this by first finding the largest x value in the data set (maximum_x_value), and the largest y value in the set (minimum_y_value). I then did the following: pointA.x = (pointA.x - minimum_x_value) / (maximum_x_value - minimum_x_value) pointA.y = (pointA.y - minimum_y_value) / (maximum_y_value - minimum_y_value) This changes the relative distances(?), and therefore makes the data useless for my purposes. Is there a way to scale these coordinates while keeping their relative distances the intact?

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  • How to map a long integer number to a N-dimensional vector of smaller integers (and fast inverse)?

    - by psihodelia
    Given a N-dimensional vector of small integers is there any simple way to map it with one-to-one correspondence to a large integer number? Say, we have N=3 vector space. Can we represent a vector X=[(int32)x1,(int32)x2,(int32)x3] using an integer (int48)y? The obvious answer is "Yes, we can". But the question is: "What is the fastest way to do this and its inverse operation?"

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  • Assigning XY positions to points based on a "weight" between them

    - by sanity
    I have a bunch of points in a graph, and for every pair of these points I have "weight" value indicating what their proximity should be, between -1 and 1. I want to choose XY coordinates for these points such that those that have a proximity of 1 are in the same position, and those with a proximity of -1 are distant from each-other. All points must reside within a bounded area. What algorithms should I investigate to achieve this?

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  • How do I improve this linear regression function?

    - by user558383
    I have the following PHP function that I'm using to draw a trend line. However, it sometimes plots the line below all the points in the scatter graph. Is there an error in my function or is there a better way to do it. I think it might be something to do with that with the line it produces, it treats all the residuals (the distances from the scatter points to the line) as positive regardless of them being above or below the line. function linear_regression($x, $y) { $n = count($x); $x_sum = array_sum($x); $y_sum = array_sum($y); $xx_sum = 0; $xy_sum = 0; for($i = 0; $i < $n; $i++) { $xy_sum+=($x[$i]*$y[$i]); $xx_sum+=($x[$i]*$x[$i]); } $m = (($n * $xy_sum) - ($x_sum * $y_sum)) / (($n * $xx_sum) - ($x_sum * $x_sum)); $b = ($y_sum - ($m * $x_sum)) / $n; return array("m"=>$m, "b"=>$b); }

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  • This codes in Actionscript-2, Can anyone help me translate it into AS-3 please ?.......a newby pulli

    - by Spux
    this.createEmptyMovieClip('mask_mc',0); bg_mc.setMask(mask_mc); var contor:Number=0; // function drawCircle draws a circle on mask_mc MovieClip of radius r and having center to mouse coordinates function drawCircle(mask_mc:MovieClip):Void{ var r:Number = 20; var xcenter:Number = _xmouse; var ycenter:Number = _ymouse; var A:Number = Math.tan(22.5 * Math.PI/180); var endx:Number; var endy:Number; var cx:Number; var cy:Number; mask_mc.beginFill(0x000000, 100); mask_mc.moveTo(xcenter+r, ycenter); for (var angle:Number = Math.PI/4; angle<=2*Math.PI; angle += Math.PI/4) { xend = r*Math.cos(angle); yend = r*Math.sin(angle); xbegin =xend + r* A *Math.cos((angle-Math.PI/2)); ybegin =yend + r* A *Math.sin((angle-Math.PI/2)); mask_mc.curveTo(xbegin+xcenter, ybegin+ycenter, xend+xcenter, yend+ycenter); } mask_mc.endFill(); } // contor variable is used to hold if the mouse is pressed (contor is 1) or not (contor is 0) this.onMouseDown=function(){ drawCircle(mask_mc); contor=1; } // if the mouse is hold and moved then we draw a circle on the mask_mc this.onMouseMove=this.onEnterFrame=function(){ if (contor==1){ drawCircle(mask_mc); } } this.onMouseUp=function(){ contor=0; }

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  • Sync Your Pidgin Profile Across Multiple PCs with Dropbox

    - by Matthew Guay
    Pidgin is definitely our favorite universal chat client, but adding all of your chat accounts to multiple computers can be frustrating.  Here’s how you can easily transfer your Pidgin settings to other computers and keep them in sync using Dropbox. Getting Started Make sure you have both Pidgin and Dropbox installed on any computers you want to sync.  To sync Pidgin, you need to: Move your Pidgin profile folder on your first computer to Dropbox Create a symbolic link from the new folder in Dropbox to your old profile location Delete the default pidgin profile on your other computer, and create a symbolic link from your Dropbox Pidgin profile to the default Pidgin profile location This sounds difficult, but it’s actually easy if you follow these steps.  Here we already had all of our accounts setup in Pidgin in Windows 7, and then synced this profile with an Ubuntu and a XP computer with fresh Pidgin installs.  Our instructions for each OS are based on this, but just swap the sync order if your main Pidgin install is in XP or Ubuntu. Please Note:  Please make sure Pidgin isn’t running on your computer while you are making the changes! Sync Your Pidgin Profile from Windows 7 Here is Pidgin with our accounts already setup.  Our Pidgin profile has a Gtalk, MSN Messenger, and Facebook Chat account, and lots of log files. Let’s move this profile to Dropbox to keep it synced.  Exit Pidgin, and then enter %appdata% in the address bar in Explorer or press Win+R and enter %appdata%.  Select the .purple folder, which is your Pidgin profiles and settings folder, and press Ctrl+X to cut it. Browse to your Dropbox folder, and press Ctrl+V to paste the .purple folder there. Now we need to create the symbolic link.  Enter  “command” in your Start menu search, right-click on the Command Prompt shortcut, and select “Run as administrator”. We can now use the mklink command to create a symbolic link to the .purple folder.  In Command Prompt, enter the following and substitute username for your own username. mklink /D “C:\Users\username\Documents\My Dropbox\.purple” “C:\Users\username\AppData\Roaming\.purple” And that’s it!  You can open Pidgin now to make sure it still works as before, with your files being synced with Dropbox. Please Note:  These instructions work the same for Windows Vista.  Also, if you are syncing settings from another computer to Windows 7, then delete the .purple folder instead of cutting and pasting it, and reverse the order of the file paths when creating the symbolic link. Add your Pidgin Profile to Ubuntu Our Ubuntu computer had a clean install of Pidgin, so we didn’t need any of the information in its settings.  If you’ve run Pidgin, even without creating an account, you will need to first remove its settings folder.  Open your home folder, and click View, and then “Show Hidden Files” to see your settings folders. Select the .purple folder, and delete it. Now, to create the symbolic link, open Terminal and enter the following, substituting username for your username: ln –s /home/username/Dropbox/.purple /home/username/ Open Pidgin, and you will see all of your accounts that were on your other computer.  No usernames or passwords needed; everything is setup and ready to go.  Even your status is synced; we had our status set to Away in Windows 7, and it automatically came up the same in Ubuntu. Please Note: If your primary Pidgin account is in Ubuntu, then cut your .purple folder and paste it into your Dropbox folder instead.  Then, when creating the symbolic link, reverse the order of the folder paths. Add your Pidgin Profile to Windows XP In XP we also had a clean install of Pidgin.  If you’ve run Pidgin, even without creating an account, you will need to first remove its settings folder.  Click Start, the Run, and enter %appdata%. Delete your .purple folder. XP does not include a way to create a symbolic link, so we will use the free Junction tool from Sysinternals.  Download Junction (link below) and unzip the folder. Open Command Prompt (click Start, select All Programs, then Accessories, and select Command Prompt), and enter cd followed by the path of the folder where you saved Junction.   Now, to create the symbolic link, enter the following in Command Prompt, substituting username with your username. junction –d “C:\Documents and Settings\username\Application Data\.purple” “C:\Documents and Settings\username\My Documents\My Dropbox\.purple” Open Pidgin, and you will see all of your settings just as they were on your other computer.  Everything’s ready to go.   Please Note: If your primary Pidgin account is in Windows XP, then cut your .purple folder and paste it into your Dropbox folder instead.  Then, when creating the symbolic link, reverse the order of the folder paths. Conclusion This is a great way to keep all of your chat and IM accounts available from all of your computers.  You can easily access logs from chats you had on your desktop from your laptop, or if you add a chat account on your work computer you can use it seamlessly from your home computer that evening.  Now Pidgin is the universal chat client that is always ready whenever and wherever you need it! Links Downlaod Pidgin Download and signup for Dropbox Download Junction for XP Similar Articles Productive Geek Tips Add "My Dropbox" to Your Windows 7 Start MenuUse Multiple Firefox Profiles at the Same TimeEasily Add Facebook Chat to PidginPut Your Pidgin Buddy List into the Windows Vista SidebarBackup and Restore Firefox Profiles Easily TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Download Free iPad Wallpapers at iPad Decor Get Your Delicious Bookmarks In Firefox’s Awesome Bar Manage Photos Across Different Social Sites With Dropico Test Drive Windows 7 Online Download Wallpapers From National Geographic Site Spyware Blaster v4.3

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  • Local Variables take 7x longer to access than global variables?

    - by ItzWarty
    I was trying to benchmark the gain/loss of "caching" math.floor, in hopes that I could make calls faster. Here was the test: <html> <head> <script> window.onload = function() { var startTime = new Date().getTime(); var k = 0; for(var i = 0; i < 1000000; i++) k += Math.floor(9.99); var mathFloorTime = new Date().getTime() - startTime; startTime = new Date().getTime(); window.mfloor = Math.floor; k = 0; for(var i = 0; i < 1000000; i++) k += window.mfloor(9.99); var globalFloorTime = new Date().getTime() - startTime; startTime = new Date().getTime(); var mfloor = Math.floor; k = 0; for(var i = 0; i < 1000000; i++) k += mfloor(9.99); var localFloorTime = new Date().getTime() - startTime; document.getElementById("MathResult").innerHTML = mathFloorTime; document.getElementById("globalResult").innerHTML = globalFloorTime; document.getElementById("localResult").innerHTML = localFloorTime; }; </script> </head> <body> Math.floor: <span id="MathResult"></span>ms <br /> var mathfloor: <span id="globalResult"></span>ms <br /> window.mathfloor: <span id="localResult"></span>ms <br /> </body> </html> My results from the test: [Chromium 5.0.308.0]: Math.floor: 49ms var mathfloor: 271ms window.mathfloor: 40ms [IE 8.0.6001.18702] Math.floor: 703ms var mathfloor: 9890ms [LOL!] window.mathfloor: 375ms [Firefox [Minefield] 3.7a4pre] Math.floor: 42ms var mathfloor: 2257ms window.mathfloor: 60ms [Safari 4.0.4[531.21.10] ] Math.floor: 92ms var mathfloor: 289ms window.mathfloor: 90ms [Opera 10.10 build 1893] Math.floor: 500ms var mathfloor: 843ms window.mathfloor: 360ms [Konqueror 4.3.90 [KDE 4.3.90 [KDE 4.4 RC1]]] Math.floor: 453ms var mathfloor: 563ms window.mathfloor: 312ms The variance is random, of course, but for the most part In all cases [this shows time taken]: [takes longer] mathfloor Math.floor window.mathfloor [is faster] Why is this? In my projects i've been using var mfloor = Math.floor, and according to my not-so-amazing benchmarks, my efforts to "optimize" actually slowed down the script by ALOT... Is there any other way to make my code more "efficient"...? I'm at the stage where i basically need to optimize, so no, this isn't "premature optimization"...

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  • Help with Collision of spawned object(postion fixed) with objects that there are translating on screen

    - by Amrutha
    Hey guys I am creating a game using Corona SDK and so coding it in Lua. So there are 2 separate functions, To translate the hit objects and change their color when they are tapped The link below is the code I am using to for the first function http://developer.anscamobile.com/sample-code/fishies Spawn objects that will hit the translating objects on collision. Alos on collision the spawned object disappears and the translating object bears a color(indicating the collision). In addition the size of this spawned object is dependent on i/p volume level. The function I have written is as follows: --VOICE INPUT CODE local r = media.newRecording() r:startRecording() r:startTuner() --local function newBar() -- local bar = display.newLine( 0, 0, 1, 0 ) -- bar:setColor( 0, 55, 100, 20 ) -- bar.width = 5 -- bar.y=400 -- bar.x=20 -- return bar --end local c1 = display.newImage("str-minion-small.png") c1.isVisible=false local c2 = display.newImage("str-minion-mid.png") c2.isVisible=false local c3 = display.newImage("str-minion-big.png") c3.isVisible=false --SPAWNING local function spawnDisk( event ) local phase = event.phase local volumeBar = display.newLine( 0, 0, 1, 0 ) volumeBar.y = 400 volumeBar.x = 20 --volumeBar.isVisible=false local v = 20*math.log(r:getTunerVolume()) local MINTHRESH = 30 local LEFTMARGIN = 20 local v2 = MINTHRESH + math.max (v, -MINTHRESH) v2 = (display.contentWidth - 1 * LEFTMARGIN ) * v2 / MINTHRESH volumeBar.xScale = math.max ( 20, v2 ) local l = volumeBar.xScale local cnt1 = 0 local cnt2 = 0 local cnt3 = 0 local ONE =1 local val = event.numTaps --local px=event.x --local py=event.y if "ended" == phase then --audio.play( popSound ) --myLabel.isVisible = false if l > 50 and l <=150 then --c1:setFillColor(10,105,0) --c1.isVisible=false c1.x=math.random( 10, 450 ) c1.y=math.random( 10, 300 ) physics.addBody( c1, { density=1, radius=10.0 } ) c1.isVisible=true cnt1= cnt1+ ONE return c1 elseif l > 100 and l <=250 then --c2:setFillColor(200,10,0) c2.x=math.random( 10, 450 ) c2.y=math.random( 10, 300 ) physics.addBody( c2, { density=2, radius=9000.0 } ) c2.isVisible=true cnt2= cnt2+ ONE return c2 elseif l >=250 then c3.x=math.random( 40, 450 ) c3.y=math.random( 40, 300 ) physics.addBody( c3, { density=2, radius=7000.0 , bounce=0.0 } ) c3.isVisible=true cnt3= cnt3+ ONE return c3 end end end buzzR:addEventListener( "touch", spawnDisk ) -- touch the screen to create disks Now both functions work fine independently but there is no collision happening. Its almost as if the translating object and the spawn object are on different layers. The translating object passes through the spawn object freely. Can anyone please tell me how to resolve this problem. And how can I get them to collide. Its my first attempt at game development, that too for a mobile platform so would appreciate all help. Also if I have not been specific do let me know. I'll try to frame the query better :). Thanks in advance.

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