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  • Announcing: Oracle Database 11g R2 Certification on Oracle Linux 6

    - by Monica Kumar
    Oracle Announces the Certification of the Oracle Database on Oracle Linux 6 and Red Hat Enterprise Linux 6 Yesterday we announced the certification of Oracle Database 11g R2 with Oracle Linux 6 and Unbreakable Enterprise Kernel. Here are the key highlights: Oracle Database 11g Release 2 (R2) and Oracle Fusion Middleware 11g Release 1 (R1) are immediately available on Oracle Linux 6 with the Unbreakable Enterprise Kernel. Oracle Database 11g R2 and Oracle Fusion Middleware 11g R1 will be available on Red Hat Enterprise Linux 6 (RHEL6) and Oracle Linux 6 with the Red Hat Compatible Kernel in 90 days. Oracle offers direct Linux support to customers running RHEL6, Oracle Linux 6, or a combination of both. Oracle Linux will continue to maintain compatibility with Red Hat Linux. Read the full press release. 

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  • Conky to Monitor WLS Managed Servers

    - by John Graves
    I've been using a little utility on my linux-based machines for years called conky.  It can be used to monitor system resources, but I wanted to modify it to monitor my WebLogic managed servers too. Once installing conky, you'll need to update the .conkyrc file.  Here is a simple example. Basically, the important lines are these: - Admin (7001) ${if_empty ${exec /usr/sbin/lsof -i :7001 | grep LISTEN}}${color red}DOWN${color} ${else}${color green} UP ${color}(${tcp_portmon 7001 7001 count}) ${endif} - OSB (8011) ${if_empty ${exec /usr/sbin/lsof -i :8011 | grep LISTEN}}${color red}DOWN${color} ${else}${color green} UP ${color}(${tcp_portmon 8011 8011 count}) ${endif} - BAM (9001) ${if_empty ${exec /usr/sbin/lsof -i :9001 | grep LISTEN}}${color red}DOWN${color} ${else}${color green} UP ${color}(${tcp_portmon 9001 9001 count}) ${endif} - DB (1521) ${if_empty ${exec /usr/sbin/lsof -i :1521 | grep LISTEN}}${color red}DOWN${color} ${else}${color green} UP ${color}(${tcp_portmon 1521 1521 count}) ${endif} It uses lsof to find out if ports are in use. Here is a video showing it in action.

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  • How to change the sprite colors

    - by Mr_Qqn
    In my rhythm game, I have a note object which can be of different colors depending on the note chart. I could use a sprite sheet with all the different color variations I use, but I would prefer to parametrize this. (For information, a note sprite is compound with one main color, for example a red note has only red, light red and dark red.) So, how to change the colors of a sprite basing on a new color ? I'm working with opengl, but any algorithm or math explanation will do. :) Thanks

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  • Boundary fill problem

    - by Taaseen
    hi...Im stuck in this bunch of codes...i cant get the pixel to fill up the circle??...any help #include<iostream> #include<glut.h> struct Color{ float red, green, blue; }; Color getPixel(int x, int y){ // gets the color of the pixel at (x,y) Color c; float color[4]; glReadPixels(x,y,1,1,GL_RGBA, GL_FLOAT, color); c.red = color[0]; c.green = color[1]; c.blue = color[2]; return c; } void setPixel(int x, int y, Color c){ glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); glPushAttrib(GL_ALL_ATTRIB_BITS); glColor3f(c.red, c.green, c.blue); glBegin(GL_POINTS); glVertex2i(x,y); glEnd(); glPopAttrib(); glFlush(); } void init() { glClearColor(1.0,1.0,1.0,0.0); gluOrtho2D(0.0,300.0,0.0,300.0); } void drawPixel(int x,int y) { glBegin(GL_POINTS); glVertex2i(x,y); glEnd(); glFlush(); } void Boundary_fill(int x,int y,Color thisColor){ Color boundary_color; boundary_color.red=0.0; boundary_color.green=1.0; boundary_color.blue=0.0; Color nextpixel=getPixel(x,y); if((nextpixel.red!=boundary_color.red)&&(nextpixel.blue!=boundary_color.blue)&&(nextpixel.green!=boundary_color.green) && (nextpixel.red!=thisColor.red)&& (nextpixel.blue!=thisColor.blue)&& (nextpixel.green!=thisColor.green)){ setPixel(x,y,thisColor); Boundary_fill((x+1),y,thisColor); Boundary_fill((x-1),y,thisColor); Boundary_fill(x,(y+1),thisColor); Boundary_fill(x,(y-1),thisColor); } } void draw(int x1,int y1, int x, int y){ drawPixel(x1+x,y1+y);//quadrant1 drawPixel(x1+x,y1-y);//quadrant2 drawPixel(x1-x,y1+y);//quadrant3 drawPixel(x1-x,y1-y);//quadrant4 drawPixel(x1+y,y1+x);//quadrant5 drawPixel(x1+y,y1-x);//quadrant6 drawPixel(x1-y,y1+x);//quadrant7 drawPixel(x1-y,y1-x);//quadrant8 } void circle(int px,int py,int r){ int a,b; float p; a=0; b=r; p=(5/4)-r; while(a<=b){ draw(px,py,a,b); if(p<0){ p=p+(2*a)+1; } else{ b=b-1; p=p+(2*a)+1-(2*b); } a=a+1; } } void Circle(void) { Color thisColor; thisColor.red=1.0; thisColor.blue=0.0; thisColor.green=0.0; glClear(GL_COLOR_BUFFER_BIT); glColor3f(0.0,1.0,0.0); glPointSize(2.0); int x0 = 100; int y0 = 150; circle(x0,y0,50); glColor3f(thisColor.red,thisColor.blue,thisColor.green); Boundary_fill(x0,y0,thisColor); } void main(int argc, char**argv) { glutInit(&argc,argv); glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB); glutInitWindowSize(400,400); glutInitWindowPosition(1,1); glutCreateWindow("Boundary fill in a circle:Taaseen And Abhinav"); init(); glutDisplayFunc(Circle); glutMainLoop(); }

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  • What Linux platform is Ubuntu Server

    - by Graeme
    I'm just trying to get an old HP ML150 up and running to learn on. I've installed ubuntu server on it and have found the drivers are all not installed. HP has drivers but I have no clue what platform to download.This is what I have to choose from. » Novell NetWare 6.5 » Red Hat Enterprise Linux 4 (AMD64/EM64T) » Red Hat Enterprise Linux 4 (x86) » Red Hat Enterprise Linux 5 Server (x86) » Red Hat Enterprise Linux 5 Server (x86-64) » SCO OpenServer 5 » SCO OpenServer 6 » SCO UnixWare 7 » SUSE Linux Enterprise Server 10 (AMD64/EM64T) » SUSE Linux Enterprise Server 10 (x86) » SUSE Linux Enterprise Server 9 (AMD64/EM64T) » SUSE Linux Enterprise Server 9 (x86) Any suggestions? Thank you in advance guys for your help.

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  • Image 1 becomes image 2 with sliding effect from left to right?

    - by Paul
    I would like to show a second image appearing while a "door" is closing on my character. I've got my character in the middle of the screen and a door coming from the left. When the door passes my character, I would like to have this second image appearing little by little. So far, I've gotten by with fadingOut the character and then fadingIn my second image of the character at the same position when the door is completely closed, but I would like to have both of them at the same time. (the effect that image 1 becomes image 2 when the door is sliding from left to right). Would you know how to do this with Cocos2d? Here are the images : at first, the character is blue, and the door is coming from the left : Then, behind the black door, the character becomes red, but only behind this door, so it stays blue when the door is not on him, and will become completely red when the door passes the character : EDIT : with this code, the black door hides the red and blue rectangles : (And if i add each of my layers at a different depth, and only use GL_LESS, same thing) blue.position = ccp( size.width*0.5 , size.height/2 ); red.position = ccp( size.width*0.46 , size.height/2 ); black.position = ccp( size.width*0.1 , size.height/2 ); glEnable(GL_DEPTH_TEST); [batch addChild:red z:0]; [batch addChild:black z:2]; glDepthFunc(GL_GREATER); [batch addChild:blue z:1]; glDepthFunc(GL_LESS); id action1 = [CCMoveTo actionWithDuration:3 position:ccp(size.width,size.height/2)]; [black runAction: [CCSequence actions:action1, nil]];

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  • Bouncing off a circular Boundary with multiple balls?

    - by Anarkie
    I am making a game like this : Yellow Smiley has to escape from red smileys, when yellow smiley hits the boundary game is over, when red smileys hit the boundary they should bounce back with the same angle they came, like shown below: Every 10 seconds a new red smiley comes in the big circle, when red smiley hits yellow, game is over, speed and starting angle of red smileys should be random. I control the yellow smiley with arrow keys. The biggest problem I have reflecting the red smileys from the boundary with the angle they came. I don't know how I can give a starting angle to a red smiley and bouncing it with the angle it came. I would be glad for any tips! My js source code : var canvas = document.getElementById("mycanvas"); var ctx = canvas.getContext("2d"); // Object containing some global Smiley properties. var SmileyApp = { radius: 15, xspeed: 0, yspeed: 0, xpos:200, // x-position of smiley ypos: 200 // y-position of smiley }; var SmileyRed = { radius: 15, xspeed: 0, yspeed: 0, xpos:350, // x-position of smiley ypos: 65 // y-position of smiley }; var SmileyReds = new Array(); for (var i=0; i<5; i++){ SmileyReds[i] = { radius: 15, xspeed: 0, yspeed: 0, xpos:350, // x-position of smiley ypos: 67 // y-position of smiley }; SmileyReds[i].xspeed = Math.floor((Math.random()*50)+1); SmileyReds[i].yspeed = Math.floor((Math.random()*50)+1); } function drawBigCircle() { var centerX = canvas.width / 2; var centerY = canvas.height / 2; var radiusBig = 300; ctx.beginPath(); ctx.arc(centerX, centerY, radiusBig, 0, 2 * Math.PI, false); // context.fillStyle = 'green'; // context.fill(); ctx.lineWidth = 5; // context.strokeStyle = '#003300'; // green ctx.stroke(); } function lineDistance( positionx, positiony ) { var xs = 0; var ys = 0; xs = positionx - 350; xs = xs * xs; ys = positiony - 350; ys = ys * ys; return Math.sqrt( xs + ys ); } function drawSmiley(x,y,r) { // outer border ctx.lineWidth = 3; ctx.beginPath(); ctx.arc(x,y,r, 0, 2*Math.PI); //red ctx.fillStyle="rgba(255,0,0, 0.5)"; ctx.fillStyle="rgba(255,255,0, 0.5)"; ctx.fill(); ctx.stroke(); // mouth ctx.beginPath(); ctx.moveTo(x+0.7*r, y); ctx.arc(x,y,0.7*r, 0, Math.PI, false); // eyes var reye = r/10; var f = 0.4; ctx.moveTo(x+f*r, y-f*r); ctx.arc(x+f*r-reye, y-f*r, reye, 0, 2*Math.PI); ctx.moveTo(x-f*r, y-f*r); ctx.arc(x-f*r+reye, y-f*r, reye, -Math.PI, Math.PI); // nose ctx.moveTo(x,y); ctx.lineTo(x, y-r/2); ctx.lineWidth = 1; ctx.stroke(); } function drawSmileyRed(x,y,r) { // outer border ctx.lineWidth = 3; ctx.beginPath(); ctx.arc(x,y,r, 0, 2*Math.PI); //red ctx.fillStyle="rgba(255,0,0, 0.5)"; //yellow ctx.fillStyle="rgba(255,255,0, 0.5)"; ctx.fill(); ctx.stroke(); // mouth ctx.beginPath(); ctx.moveTo(x+0.4*r, y+10); ctx.arc(x,y+10,0.4*r, 0, Math.PI, true); // eyes var reye = r/10; var f = 0.4; ctx.moveTo(x+f*r, y-f*r); ctx.arc(x+f*r-reye, y-f*r, reye, 0, 2*Math.PI); ctx.moveTo(x-f*r, y-f*r); ctx.arc(x-f*r+reye, y-f*r, reye, -Math.PI, Math.PI); // nose ctx.moveTo(x,y); ctx.lineTo(x, y-r/2); ctx.lineWidth = 1; ctx.stroke(); } // --- Animation of smiley moving with constant speed and bounce back at edges of canvas --- var tprev = 0; // this is used to calculate the time step between two successive calls of run function run(t) { requestAnimationFrame(run); if (t === undefined) { t=0; } var h = t - tprev; // time step tprev = t; SmileyApp.xpos += SmileyApp.xspeed * h/1000; // update position according to constant speed SmileyApp.ypos += SmileyApp.yspeed * h/1000; // update position according to constant speed for (var i=0; i<SmileyReds.length; i++){ SmileyReds[i].xpos += SmileyReds[i].xspeed * h/1000; // update position according to constant speed SmileyReds[i].ypos += SmileyReds[i].yspeed * h/1000; // update position according to constant speed } // change speed direction if smiley hits canvas edges if (lineDistance(SmileyApp.xpos, SmileyApp.ypos) + SmileyApp.radius > 300) { alert("Game Over"); } // redraw smiley at new position ctx.clearRect(0,0,canvas.height, canvas.width); drawBigCircle(); drawSmiley(SmileyApp.xpos, SmileyApp.ypos, SmileyApp.radius); for (var i=0; i<SmileyReds.length; i++){ drawSmileyRed(SmileyReds[i].xpos, SmileyReds[i].ypos, SmileyReds[i].radius); } } // uncomment these two lines to get every going // SmileyApp.speed = 100; run(); // --- Control smiley motion with left/right arrow keys function arrowkeyCB(event) { event.preventDefault(); if (event.keyCode === 37) { // left arrow SmileyApp.xspeed = -100; SmileyApp.yspeed = 0; } else if (event.keyCode === 39) { // right arrow SmileyApp.xspeed = 100; SmileyApp.yspeed = 0; } else if (event.keyCode === 38) { // up arrow SmileyApp.yspeed = -100; SmileyApp.xspeed = 0; } else if (event.keyCode === 40) { // right arrow SmileyApp.yspeed = 100; SmileyApp.xspeed = 0; } } document.addEventListener('keydown', arrowkeyCB, true); JSFiddle : http://jsfiddle.net/gj4Q7/

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  • Best solution for getting referral information in PHP

    - by absentx
    I am currently redoing some link structuring on a website. In the past we have used specific php files on the last step to direct the user to the proper place. Example: www.mysite.com/action/go-to-blue.php or www.mysite.com/action/short/go-to-red.php www.mysite.com/action/tall/go-to-red.php We are now restructuring to eliminate the /short/ or /tall/ directory. What this means is now "go-to-blue.php" will be doing some extra processing to make sure it sends the visitor to the proper place. The static method of the past was quite effective, because, well, if they left from that page we knew we had it right. Now since we are 301 redirecting action/short/go-to-red.php to just action/go-to-red.php it is quite important on "go-to-red.php" that we realize a user may have been redirected from /short/ or /tall/. So right now I am using HTTP_REFERRER and of course in my testing that works fine, but after a lot of reading it is clear that this is not a solid solution, so I was starting to brainstorm on other ways to check and make sure we get the proper referral information. If we could check HTTP_REFERRER plus some other test, I would feel confident we have a pretty good system in place to send the visitor to the right place. Some questions/comments: Could I use a session variable or a cookie to accomplish this goal? If so, would that be maintained through the 301 redirect? I don't see why it wouldn't be.. Passing the url in the url is not an option in this case.

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  • Debugging Minimum Translation Vector

    - by SyntheCypher
    I implemented the minimum translation vector from codezealot's tutorial on SAT (Separating Axis Theorem) but I'm having an issue I can't quite figure out. Here's the example I have: As you can see in top and bottom left images regardless of the side the of the green car which red car is penetrating the MTV for the red car still remains as a negative number also here is the same example when the front of the red car is facing the opposite direction the number will always be positive. When the red car is past the half way through the green car it should switch polarity. I thought I'd compensated for this in my code, but apparently not either that or it's a bug I can find. Here is my function for finding and returning the MTV, any help would be much appreciated: Code

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  • What's the best way to compare blocks in a matching game that can be multiple colors?

    - by Ryan Detzel
    I have a match 3-4 game and the blocks can be one of 7 colors. There are an addition 7 blocks that are a mix of the original 7 colors so for example there is a red and blue block and there is also a red/blue block which can be matched with either the red or the blue. My original thought is just to use binary operations so. int red = 0x000000001; int blue = 0x000000010; int redblue = 0x000000011; Then just do an & operation so see if they match. Does this sound like a decent plan or am I over complicating it? edit: Better yet so it's more readable. int red = 1; int blue = 2; int red_blue = 3; int yellow = 4; int red_yellow = 5; maybe as defines or static vars?

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  • BoundingSpheres move when they should not

    - by NDraskovic
    I have a XNA 4.0 project in which I load a file that contains type and coordinates of items I need to draw to the screen. Also I need to check if one particular type (the only movable one) is passing in front or trough other items. This is the code I use to load the configuration: if (ks.IsKeyDown(Microsoft.Xna.Framework.Input.Keys.L)) { this.GraphicsDevice.Clear(Color.CornflowerBlue); Otvaranje.ShowDialog(); try { using (StreamReader sr = new StreamReader(Otvaranje.FileName)) { String linija; while ((linija = sr.ReadLine()) != null) { red = linija.Split(','); model = red[0]; x = red[1]; y = red[2]; z = red[3]; elementi.Add(Convert.ToInt32(model)); podatci.Add(new Vector3(Convert.ToSingle(x), Convert.ToSingle(y), Convert.ToSingle(z))); sfere.Add(new BoundingSphere(new Vector3(Convert.ToSingle(x), Convert.ToSingle(y), Convert.ToSingle(z)), 1f)); } } } catch (Exception ex) { Window.Title = ex.ToString(); } } The "Otvaranje" is an OpenFileDialog object, "elementi" is a List (determines the type of item that would be drawn), podatci is a List (determines the location where the items will be drawn) and sfere is a List. Now I solved the picking algorithm (checking for ray and bounding sphere intersection) and it works fine, but the collision detection does not. I noticed, while using picking, that BoundingSphere's move even though the objects that they correspond to do not. The movable object is drawn to the world1 Matrix, and the static objects are drawn into the world2 Matrix (world1 and world2 have the same values, I just separated them so that the static elements would not move when the movable one does). The problem is that when I move the item I want, all boundingSpheres move accordingly. How can I move only the boundingSphere that corresponds to that particular item, and leave the rest where they are?

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  • Conditional formatting & vlookup

    - by zorama
    Please help me with the formula: Main Sheet is Sheet2 B COLUMN I want to look up sheet1 A & B columns with Sheet2 A & B columns from 1 workbook that if sheet2 A are same/equal as Sheet1 A column, also if Sheet2 B column are same/equal as Sheet1 B column , how will I highlight the Sheet2 B column that if Sheet1 A & B + Sheet2 A & B are exactly equal . EXAMPLE: SHEET 1 SHEET 2 SHEET 2 Result A B A B A B CODE NO CODE NO CODE NO A 12 B 205 B 205 (highlight to red) B 105 B 20 B 20 (highlight to red) A 45 B 100 B 100 A 56 A 56 A 56 (highlight to red) A 78 B 25 B 25 A 100 A 12 A 12 (highlight to red) B 77 A 45 A 45 (highlight to red) B 108 A 20000 A 20000 B 20 B 205

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  • simple sql group by custom groups question [migrated]

    - by alex
    imagine a mysql table that only has 2 columns, an id and a name of a color. with this query I know how many id's do I have for each color. SELECT color_name, count(id) FROM color_table GROUP BY (color_name); red:10 blue:5 yellow:3 green:1 my question is, is there a way I can specify to the "group by" some custom groups?? i mean, is there a query that results in this??: red:10 colors different than red: 9

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  • Windows Form color changing

    - by Sef
    Hello, So i am attempting to make a MasterMind program as sort of exercise. Field of 40 picture boxes (line of 4, 10 rows) 6 buttons (red, green, orange, yellow, blue, purple) When i press one of these buttons (lets assume the red one) then a picture box turns red. My question is how do i iterate trough all these picture boxes? I can get it to work but only if i write : And this is offcourse no way to write this, would take me countless of lines that contain basicly the same. private void picRood_Click(object sender, EventArgs e) { UpdateDisplay(); pb1.BackColor = System.Drawing.Color.Red; } Press the red button - first picture box turns red Press the blue button - second picture box turns blue Press the orange button - third picture box turns orange And so on... Ive had a previous similar program that simulates a traffic light, there i could assign a value to each color (red 0, orange 1, green 2). Is something similar needed or how exactly do i adress all those picture boxes and make them correspond to the proper button. Best Regards.

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  • Building Reducisaurus URLs

    - by Alix Axel
    I'm trying to use Reducisaurus Web Service to minify CSS and Javascript but I've run into a problem... Suppose I've two unminified CSS at: http:/domain.com/dynamic/styles/theme.php?color=red http:/domain.com/dynamic/styles/typography.php?font=Arial According to the docs I should call the web service like this: http:/reducisaurus.appspot.com/css?url=http:/domain.com/dynamic/styles/theme.php?color=red And if I want to minify both CSS files at once: http:/reducisaurus.appspot.com/css?url1=http:/domain.com/dynamic/styles/theme.php?color=red&url2=http:/domain.com/dynamic/styles/theme.php?color=red If I wanted to specify a different number of seconds for the cache (3600 for instance) I would use: http:/reducisaurus.appspot.com/css?url=http:/domain.com/dynamic/styles/theme.php?color=red&expire_urls=3600 And again for both CSS files at once: http:/reducisaurus.appspot.com/css?url1=http:/domain.com/dynamic/styles/theme.php?color=red&url2=http:/domain.com/dynamic/styles/theme.php?color=red&expire_urls=3600 Now my question is, how does Reducisaurus knows how to separate the URLs I want? How does it know that &expire_urls=3600 is not part of my URL? And how does it know that &url2=... is not a GET argument of url1? I'm I doing this right? Do I need to urlencode my URLs? I took a peek into the source code and although my Java is very poor it seems that the methods acquireFromRemoteUrl() and getSortedParameterNames() from the BaseServlet.java file hold the answers to my question - if a GET argument name contains - or _ they should be ignored?! What about multiple &url(n)s?

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  • python global variable not working in apache

    - by Suhail
    I am facing issue with the global variable, when i run in the django development server it works fine, but in apache it doesn't work here is the code below: red= "/foodfolio3/test/" def showAddRecipe(request): #global objc if "userid" in request.session: objc["ErrorMsgURL"]= "" try: urlList= request.POST URL= str(urlList['url']) URL= URL.strip('http://') URL= "http://" + URL recipe= __addRecipeUrl__(URL) if (recipe == 'FailToOpenURL') or (recipe == 'Invalid-website-URL'): #request.session["ErrorMsgURL"]= "Kindly check URL, Please enter a valid URL" objc["ErrorMsgURL"]= "Kindly check URL, Please enter a valid URL" print "here global_context =", objc arurl= HttpResponseRedirect("/foodfolio3/add/import/") arurl['ErrorMsgURL']= objc["ErrorMsgURL"] #return HttpResponseRedirect("/foodfolio3/add/import/") #return render_to_response('addRecipeUrl.html', objc, context_instance = RequestContext(request)) return (arurl) else: objc["recipe"] = recipe return render_to_response('addRecipe.html', objc, context_instance = RequestContext(request)) except: objc["recipe"] = "" return render_to_response('addRecipe.html', objc, context_instance = RequestContext(request)) else: global red red= "/foodfolio3/add/" return HttpResponseRedirect("/foodfolio3/login") def showAddRecipeUrl(request): if "userid" in request.session: return render_to_response('addRecipeUrl.html', objc, context_instance = RequestContext(request)) else: global red red= "/foodfolio3/add/import/" return HttpResponseRedirect("/foodfolio3/login") def showLogin(request): obj = {} obj["error_message"] = "" obj["registered"] = "" if request.method == "POST": if (red == "/foodfolio3/test"): next= '/foodfolio3/recipes' else: next= red try: username = request.POST['username'] password = request.POST['password'] user = authenticate(username=username, password=password) except: user = authenticate(request=request) if user is not None: if user.is_active: login(request, user) request.session["userid"] = user.id # Redirect to a success page. return HttpResponseRedirect(next) this code works fine in django development server, but in apache, the url is getting redirected to '/foodfolio3/recipes'

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  • Four-color theorem in Prolog (using a dynamic predicate)

    - by outa
    Hi, I'm working on coloring a map according to the four-color theorem (http://en.wikipedia.org/wiki/Four_color_theorem) with SWI-Prolog. So far my program looks like this: colour(red). colour(blue). map_color(A,B,C) :- colour(A), colour(B), colour(C), C \= B, C \= A. (the actual progam would be more complex, with 4 colors and more fields, but I thought I'd start out with a simple case) Now, I want to avoid double solutions that have the same structure. E.g. for a map with three fields, the solution "red, red, blue" would have the same structure as "blue, blue, red", just with different color names, and I don't want both of them displayed. So I thought I would have a dynamic predicate solution/3, and call assert(solution(A,B,C)) at the end of my map_color predicate. And then, for each solution, check if they already exist as a solution/3 fact. The problem is that I would have to assert something like solution(Color1,Color1,Color2), i.e. with variables in order to make a unification check. And I can't think of a way to achieve this. So, the question is, what is the best way to assert a found solution and then make a unification test so that "red, red, blue" would unify with "blue, blue, red"?

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  • Convert a Door Peephole Viewer into a Fisheye Camera Lens

    - by Jason Fitzpatrick
    Commercial fish eye lenses are a niche product and carry a hefty price tag; if you’re looking to goof around with fish eye photography on the cheap, this $6 tutorial is for you. Courtesy of Dave from Knobtop–a thrifty DIY photography video blog–this hack uses dirt cheap parts (the whole build is composed of a PVC pipe reducer and a door peephole lens) to bring you fun fish eye photography on a budget. Check out the video above to see the build and the results, then hit up the link below to check out the notes on the video for more information. Fisheye Lens for $6 [via DIY Photography] HTG Explains: What Is Two-Factor Authentication and Should I Be Using It? HTG Explains: What Is Windows RT and What Does It Mean To Me? HTG Explains: How Windows 8′s Secure Boot Feature Works & What It Means for Linux

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  • View matrix in opengl

    - by user5584
    Hi! Sorry for my clumsy question. But I don't know where I am wrong at creating view matrix. I have the following code: createMatrix(vec4f(xAxis.x, xAxis.y, xAxis.z, dot(xAxis,eye)), vec4f( yAxis.x_, yAxis.y_, yAxis.z_, dot(yAxis,eye)), vec4f(-zAxis.x_, -zAxis.y_, -zAxis.z_, -dot(zAxis,eye)), vec4f(0, 0, 0, 1)); //column1, column2,... I have tried to transpose it, but with no success. I have also tried to use gluLookAt(...) with success. At the reference page, I watched the remarks about the to-be-created matrix, and it seems the same as mine. Where I am wrong?

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  • There's no Sound Mixer menu, missing menu option in Sound Recorder

    - by AlexN
    I am using: -Ubuntu 11.10 -Skype -PS3 Eye Toy camera to input video and sound This setup has been properly working in former Ubuntu releases. To use the mic already built in on the PS3 Eye Toy camera I open de Sound Recorder app (notice: not inside Skype, from inside Skype it is not possible to do this) that is included in Gnome and then I go to FileSound Mixer, from this menu I can choose Gnome to get the input audio from the PS3 Eye Toy, instead of from the Audio-In of the computer. Now in Ubuntu 11.10 this Sound Mixer menu inside Sound Recorder is missing, Gnome says something like this: gnome-volume-control is not installed in the proper directory Note: I have tried this on Unity, Unity 2D, Gnome Classic, Gnome Classic 2D and Gnome Shell. In all of them the problem is the same. What can I do? Basically what I want to do is to be able to tell the computer to get the audio in from the PS3 Camera. Thanks in advance.

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  • Java image conversion to RGB565

    - by Vladimir
    I try to convert image to RGB565 format. I read this image: BufferedImage bufImg = ImageIO.read(imagePathFile); sendImg = new BufferedImage(CONTROLLER_LCD_WIDTH/*320*/, CONTROLLER_LCD_HEIGHT/*240*/, BufferedImage.TYPE_USHORT_565_RGB); sendImg .getGraphics().drawImage(bufImg, 0, 0, CONTROLLER_LCD_WIDTH/*320*/, CONTROLLER_LCD_HEIGHT/*240*/, null); Here is it: Then I convert it to RGB565: int numByte=0; byte[] OutputImageArray = new byte[CONTROLLER_LCD_WIDTH*CONTROLLER_LCD_HEIGHT*2]; int i=0; int j=0; int len = OutputImageArray.length; for (i=0;i<CONTROLLER_LCD_WIDTH;i++) { for (j=0;j<CONTROLLER_LCD_HEIGHT;j++) { Color c = new Color(sendImg.getRGB(i, j)); int aRGBpix = sendImg.getRGB(i, j); int alpha; int red = c.getRed(); int green = c.getGreen(); int blue = c.getBlue(); //RGB888 red = (aRGBpix >> 16) & 0x0FF; green = (aRGBpix >> 8) & 0x0FF; blue = (aRGBpix >> 0) & 0x0FF; alpha = (aRGBpix >> 24) & 0x0FF; //RGB565 red = red >> 3; green = green >> 2; blue = blue >> 3; //A pixel is represented by a 4-byte (32 bit) integer, like so: //00000000 00000000 00000000 11111111 //^ Alpha ^Red ^Green ^Blue //Converting to RGB565 short pixel_to_send = 0; int pixel_to_send_int = 0; pixel_to_send_int = (red << 11) | (green << 5) | (blue); pixel_to_send = (short) pixel_to_send_int; //dividing into bytes byte byteH=(byte)((pixel_to_send >> 8) & 0x0FF); byte byteL=(byte)(pixel_to_send & 0x0FF); //Writing it to array - High-byte is second OutputImageArray[numByte]=byteH; OutputImageArray[numByte+1]=byteL; numByte+=2; } } Then I try to restore this from resulting array OutputImageArray: i=0; j=0; numByte=0; BufferedImage NewImg = new BufferedImage(CONTROLLER_LCD_WIDTH, CONTROLLER_LCD_HEIGHT, BufferedImage.TYPE_USHORT_565_RGB); for (i=0;i<CONTROLLER_LCD_WIDTH;i++) { for (j=0;j<CONTROLLER_LCD_HEIGHT;j++) { int curPixel=0; int alpha=0x0FF; int red; int green; int blue; byte byteL=0; byte byteH=0; byteH = OutputImageArray[numByte]; byteL = OutputImageArray[numByte+1]; curPixel= (byteH << 8) | (byteL); //RGB565 red = (curPixel >> (6+5)) & 0x01F; green = (curPixel >> 5) & 0x03F; blue = (curPixel) & 0x01F; //RGB888 red = red << 3; green = green << 2; blue = blue << 3; //aRGB curPixel = 0; curPixel = (alpha << 24) | (red << 16) | (green << 8) | (blue); NewImg.setRGB(i, j, curPixel); numByte+=2; } } I output this restored image. But I see that it looks very poor. I expected the lost of pictures quality. But as I thought, this picture has to have almost the same quality as the previous picture. - Is it right? Is my code right?

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  • First order logic formula

    - by user177883
    R(x) is a red block B(x) is a blue block T(x,y) block x is on top of block y Question: Write a formula asserting that if no red block is on top of a red block then no red block is on top of itself. My answer: (Ax)(Ay)(R(x) and R(y) - ~T(x,y))-(Ax)(R(x)- ~T(x,x)) A = For all

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  • Game show game engine [closed]

    - by Red
    So, I am pretty new to the world of game development, so I am a bit fuzzy on what I require. Could someone suggest a game engine that I could use? I need it to be light weight (my game won't require that much power) and have networking functionality for multiplay or even an MMO aspect. The game I am making is like a game show, so it is your basic choose and answer hit the buzzer kind of game. Any suggestions? I would also like it to be open source or at the least free. I would like to support open source projects.

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  • ¿Qué es Social Cloud o computación en la nube social?

    - by RED League Heroes-Oracle
    La computación en nube es la creación de nuevas posibilidades para las empresas en sus negocios, como acercarse a los clientes a través de herramientas digitales. Es cruzar informaciones del registro de los clientes almacenadas en los servidores de la empresa, con informaciones sociales, o sea, con informaciones disponibles en la internet (redes sociales, blogs, geolocalización). Este cruce, seguramente, puede ayudar a entender mejor el comportamiento de sus consumidores y, a través de estos análisis, tomar diferentes acciones para estar cada vez más cerca de ellos o entender nuevas necesidades. El comportamiento de consumo se está alterando con el avance de la internet y de las nuevas tecnologías. Integrar estas nuevas tecnologías al negocio de la empresa es una gran oportunidad para acompañar los consumidores y observar nuevos patrones de comportamiento. Estos nuevos patrones pueden presentar nuevas oportunidades. Utilizar la computación en la nube para agregar conocimiento adicional a los que ya lo poseen puede ser una de las claves para la transformación de la realidad de la empresa. Actualmente, ¿cómo se comportan tus consumidores? ¿Qué suelen hacer? ¿Viajan mensualmente? ¿Tienen hijos? ¿Están buscando nuevos productos? ¿Qué productos buscan? ¿Dónde la mayor parte de tus consumidores está en el momento de ocio? ¡Saber dónde están en el momento de ocio puede ser una excelente oportunidad para que vean tu marca! ¿Cómo tratas hoy en día esos temas? ¡Las soluciones de Social Cloud de Oracle pueden ayudarte! Aprovecha y descarga GRATUITAMENTE el e-book – Simplifica tu movilidad empresarial y conoce el poder de transformación de la movilidad en tu negocio. LINK PARA DESCARGA: http://bit.ly/e-bookmobilidad

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