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• Create Custom Sized Thumbnail Images with Simple Image Resizer [Cross-Platform]

  - by Asian Angel

  Are you looking for an easy way to create custom sized thumbnail images for use in blog posts, photo albums, and more? Whether is it a single image or a CD full, Simple Image Resizer is the right app to get the job done for you. To add the new PPA for Simple Image Resizer open the Ubuntu Software Center, go to the Edit Menu, and select Software Sources. Access the Other Software Tab in the Software Sources Window and add the first of the PPAs shown below (outlined in red). The second PPA will be automatically added to your system. Once you have the new PPAs set up, go back to the Ubuntu Software Center and click on the PPA listing for Rafael Sachetto on the left (highlighted with red in the image). The listing for Simple Image Resizer will be right at the top…click Install to add the program to your system. After the installation is complete you can find Simple Image Resizer listed as Sir in the Graphics sub-menu. When you open Simple Image Resizer you will need to browse for the directory containing the images you want to work with, select a destination folder, choose a target format and prefix, enter the desired pixel size for converted images, and set the quality level. Convert your image(s) when ready… Note: You will need to determine the image size that best suits your needs before-hand. For our example we chose to convert a single image. A quick check shows our new “thumbnailed” image looking very nice. Simple Image Resizer can convert “into and from” the following image formats: .jpeg, .png, .bmp, .gif, .xpm, .pgm, .pbm, and .ppm Command Line Installation Note: For older Ubuntu systems (9.04 and previous) see the link provided below. sudo add-apt-repository ppa:rsachetto/ppa sudo apt-get update && sudo apt-get install sir Links Note: Simple Image Resizer is available for Ubuntu, Slackware Linux, and Windows. Simple Image Resizer PPA at Launchpad Simple Image Resizer Homepage Command Line Installation for Older Ubuntu Systems Bonus The anime wallpaper shown in the screenshots above can be found here: The end where it begins [DesktopNexus] Latest Features How-To Geek ETC Macs Don’t Make You Creative! So Why Do Artists Really Love Apple? MacX DVD Ripper Pro is Free for How-To Geek Readers (Time Limited!) HTG Explains: What’s a Solid State Drive and What Do I Need to Know? How to Get Amazing Color from Photos in Photoshop, GIMP, and Paint.NET Learn To Adjust Contrast Like a Pro in Photoshop, GIMP, and Paint.NET Have You Ever Wondered How Your Operating System Got Its Name? Create Shortcuts for Your Favorite or Most Used Folders in Ubuntu Create Custom Sized Thumbnail Images with Simple Image Resizer [Cross-Platform] Etch a Circuit Board using a Simple Homemade Mixture Sync Blocker Stops iTunes from Automatically Syncing The Journey to the Mystical Forest [Wallpaper] Trace Your Browser’s Roots on the Browser Family Tree [Infographic]

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• Cientos de Directores Financieros se congregaron en el evento “Innovación y Excelencia en la Función Financiera”

  - by Noelia Gomez

  v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} El pasado 24 de Octubre tuvo lugar el evento “Innovación y Excelencia en la Función Financiera” en la Fundación Rafael de Pino, Madrid (que ya anunciamos aquí). APD, en colaboración con Oracle, organizaron esta jornada con el objetivo de analizar el proceso de transformación del Director Financiero en las compañías (aquí puedes ver un estudio sobre ello). Enrique Sanchez de Leon, Director de APD, fue el encargado de abrir la jornada con una calurosa bienvenida a los invitados. Tras él, Fernando Rumbero, Iberia Applications Cluster Leader de Oracle , comenzó dando unas pinceladas sobre los cambios a los que los Directores Financieros deben estar preparados para convertirse en parte de la estrategia de la compañía. Después de que todos los ponentes fueran presentados y se acomodaran en su lugar del escenario de aquella gran sala, Oriol Farré, Presales Director de Oracle, tomó la palabra para profundizar sobre el nuevo rol estratégico del Director Financiero y cómo éste se está convirtiendo cada vez más en el catalizador del cambio dentro de las empresas (¿tú lo eres? aquí hablamos de cómo puedes evaluarlo) Por su parte, Maria Jesús Carrato, Profesora de Dirección Financiera Internacional en el IE y Directora Financiera del Grupo SM mostró su visión sobre cómo serán los Departamentos Financieros del futuro. Después llego el turno de Ramón Arguelaguet, Financial Controller & Reporting Senior Manager de Vodafone, que profundizo en la innovación y la transformación lideradas por los Directores Financieros dentro de las organizaciones. Por último, pero no menos importante, Juan Jesús Donoso, Director Económico de Cruz Roja Española, nos mostro el punto de vista de la gestión de una organización sin ánimo de lucro. Finalmente, en la mesa redonda, cada uno de los integrantes dio su punto de vista sobre el nuevo rol de Director Financiero y los nuevos retos a los que se enfrentan. El broche final de la jornada la puso el coctel para abrir paso a un espacio de networking que sin duda los cientos de Directores Financieros aprovecharon para intercambiar puntos de vista, conocer a nuevos compañeros y reencontrarse con muchos otros. Si estuviste en el evento… ¿qué te pareció? Tal vez no encontraste el momento de plantear alguna cuestión. Ahora puedes hacerlo en los comentarios y se lo trasladaremos a los ponentes. Contact 12.00 Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Calibri","sans-serif";}

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• Project Euler #15

  - by Aistina

  Hey everyone, Last night I was trying to solve challenge #15 from Project Euler: Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner. How many routes are there through a 20×20 grid? I figured this shouldn't be so hard, so I wrote a basic recursive function: const int gridSize = 20; // call with progress(0, 0) static int progress(int x, int y) { int i = 0; if (x < gridSize) i += progress(x + 1, y); if (y < gridSize) i += progress(x, y + 1); if (x == gridSize && y == gridSize) return 1; return i; } I verified that it worked for a smaller grids such as 2×2 or 3×3, and then set it to run for a 20×20 grid. Imagine my surprise when, 5 hours later, the program was still happily crunching the numbers, and only about 80% done (based on examining its current position/route in the grid). Clearly I'm going about this the wrong way. How would you solve this problem? I'm thinking it should be solved using an equation rather than a method like mine, but that's unfortunately not a strong side of mine. Update: I now have a working version. Basically it caches results obtained before when a n×m block still remains to be traversed. Here is the code along with some comments: // the size of our grid static int gridSize = 20; // the amount of paths available for a "NxM" block, e.g. "2x2" => 4 static Dictionary<string, long> pathsByBlock = new Dictionary<string, long>(); // calculate the surface of the block to the finish line static long calcsurface(long x, long y) { return (gridSize - x) * (gridSize - y); } // call using progress (0, 0) static long progress(long x, long y) { // first calculate the surface of the block remaining long surface = calcsurface(x, y); long i = 0; // zero surface means only 1 path remains // (we either go only right, or only down) if (surface == 0) return 1; // create a textual representation of the remaining // block, for use in the dictionary string block = (gridSize - x) + "x" + (gridSize - y); // if a same block has not been processed before if (!pathsByBlock.ContainsKey(block)) { // calculate it in the right direction if (x < gridSize) i += progress(x + 1, y); // and in the down direction if (y < gridSize) i += progress(x, y + 1); // and cache the result! pathsByBlock[block] = i; } // self-explanatory :) return pathsByBlock[block]; } Calling it 20 times, for grids with size 1×1 through 20×20 produces the following output: There are 2 paths in a 1 sized grid 0,0110006 seconds There are 6 paths in a 2 sized grid 0,0030002 seconds There are 20 paths in a 3 sized grid 0 seconds There are 70 paths in a 4 sized grid 0 seconds There are 252 paths in a 5 sized grid 0 seconds There are 924 paths in a 6 sized grid 0 seconds There are 3432 paths in a 7 sized grid 0 seconds There are 12870 paths in a 8 sized grid 0,001 seconds There are 48620 paths in a 9 sized grid 0,0010001 seconds There are 184756 paths in a 10 sized grid 0,001 seconds There are 705432 paths in a 11 sized grid 0 seconds There are 2704156 paths in a 12 sized grid 0 seconds There are 10400600 paths in a 13 sized grid 0,001 seconds There are 40116600 paths in a 14 sized grid 0 seconds There are 155117520 paths in a 15 sized grid 0 seconds There are 601080390 paths in a 16 sized grid 0,0010001 seconds There are 2333606220 paths in a 17 sized grid 0,001 seconds There are 9075135300 paths in a 18 sized grid 0,001 seconds There are 35345263800 paths in a 19 sized grid 0,001 seconds There are 137846528820 paths in a 20 sized grid 0,0010001 seconds 0,0390022 seconds in total I'm accepting danben's answer, because his helped me find this solution the most. But upvotes also to Tim Goodman and Agos :) Bonus update: After reading Eric Lippert's answer, I took another look and rewrote it somewhat. The basic idea is still the same but the caching part has been taken out and put in a separate function, like in Eric's example. The result is some much more elegant looking code. // the size of our grid const int gridSize = 20; // magic. static Func<A1, A2, R> Memoize<A1, A2, R>(this Func<A1, A2, R> f) { // Return a function which is f with caching. var dictionary = new Dictionary<string, R>(); return (A1 a1, A2 a2) => { R r; string key = a1 + "x" + a2; if (!dictionary.TryGetValue(key, out r)) { // not in cache yet r = f(a1, a2); dictionary.Add(key, r); } return r; }; } // calculate the surface of the block to the finish line static long calcsurface(long x, long y) { return (gridSize - x) * (gridSize - y); } // call using progress (0, 0) static Func<long, long, long> progress = ((Func<long, long, long>)((long x, long y) => { // first calculate the surface of the block remaining long surface = calcsurface(x, y); long i = 0; // zero surface means only 1 path remains // (we either go only right, or only down) if (surface == 0) return 1; // calculate it in the right direction if (x < gridSize) i += progress(x + 1, y); // and in the down direction if (y < gridSize) i += progress(x, y + 1); // self-explanatory :) return i; })).Memoize(); By the way, I couldn't think of a better way to use the two arguments as a key for the dictionary. I googled around a bit, and it seems this is a common solution. Oh well.

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• Remove/squash entries in a vertical hash

  - by Forkrul Assail

  I have a grid that represents an X, Y matrix, stored as a hash here. Some points on the X Y matrix may have values (as type string), and some may not. A typical grid could look like this: {[9, 5]=>"Alaina", [10, 3]=>"Courtney", [11, 1]=>"Gladys", [8, 7]=>"Alford", [14, 11]=>"Lesley", [17, 2]=>"Lawson", [0, 5]=>"Katrine", [2, 1]=>"Tyra", [3, 3]=>"Fredy", [1, 7]=>"Magnus", [6, 9]=>"Nels", [7, 11]=>"Kylie", [11, 0]=>"Kellen", [10, 2]=>"Johan", [14, 10]=>"Justice", [0, 4]=>"Barton", [2, 0]=>"Charley", [3, 2]=>"Magnolia", [1, 6]=>"Maximo", [7, 10]=>"Olga", [19, 5]=>"Isadore", [16, 3]=>"Delfina", [17, 1]=>"Noe", [20, 11]=>"Francis", [10, 5]=>"Creola", [9, 3]=>"Bulah", [8, 1]=>"Lempi", [11, 7]=>"Raquel", [13, 11]=>"Jace", [1, 5]=>"Garth", [3, 1]=>"Ernest", [2, 3]=>"Malcolm", [0, 7]=>"Alejandrin", [7, 9]=>"Marina", [6, 11]=>"Otilia", [16, 2]=>"Hailey", [20, 10]=>"Brandt", [8, 0]=>"Madeline", [9, 2]=>"Leanne", [13, 10]=>"Jenifer", [1, 4]=>"Humberto", [3, 0]=>"Nicholaus", [2, 2]=>"Nadia", [0, 6]=>"Abigail", [6, 10]=>"Zola", [20, 5]=>"Clementina", [23, 3]=>"Alvah", [19, 11]=>"Wallace", [11, 5]=>"Tracey", [8, 3]=>"Hulda", [9, 1]=>"Jedidiah", [10, 7]=>"Annetta", [12, 11]=>"Nicole", [2, 5]=>"Alison", [0, 1]=>"Wilma", [1, 3]=>"Shana", [3, 7]=>"Judd", [4, 9]=>"Lucio", [5, 11]=>"Hardy", [19, 10]=>"Immanuel", [9, 0]=>"Uriel", [8, 2]=>"Milton", [12, 10]=>"Elody", [5, 10]=>"Alexanne", [1, 2]=>"Lauretta", [0, 0]=>"Louvenia", [2, 4]=>"Adelia", [21, 5]=>"Erling", [18, 11]=>"Corene", [22, 3]=>"Haskell", [11, 11]=>"Leta", [10, 9]=>"Terrence", [14, 1]=>"Giuseppe", [15, 3]=>"Silas", [12, 5]=>"Johnnie", [4, 11]=>"Aurelie", [5, 9]=>"Meggie", [2, 7]=>"Phoebe", [0, 3]=>"Sister", [1, 1]=>"Violet", [3, 5]=>"Lilian", [18, 10]=>"Eusebio", [11, 10]=>"Emma", [15, 2]=>"Theodore", [14, 0]=>"Cassidy", [4, 10]=>"Edmund", [2, 6]=>"Claire", [0, 2]=>"Madisen", [1, 0]=>"Kasey", [3, 4]=>"Elijah", [17, 11]=>"Susana", [20, 1]=>"Nicklaus", [21, 3]=>"Kelsie", [10, 11]=>"Garnett", [11, 9]=>"Emanuel", [15, 1]=>"Louvenia", [14, 3]=>"Otho", [13, 5]=>"Vincenza", [3, 11]=>"Tate", [2, 9]=>"Beau", [5, 7]=>"Jason", [6, 1]=>"Jayde", [7, 3]=>"Lamont", [4, 5]=>"Curt", [17, 10]=>"Mack", [21, 2]=>"Lilyan", [10, 10]=>"Ruthe", [14, 2]=>"Georgianna", [4, 4]=>"Nyasia", [6, 0]=>"Sadie", [16, 11]=>"Emil", [21, 1]=>"Melba", [20, 3]=>"Delia", [3, 10]=>"Rosalee", [2, 8]=>"Myrtle", [7, 2]=>"Rigoberto", [14, 5]=>"Jedidiah", [13, 3]=>"Flavie", [12, 1]=>"Evie", [8, 9]=>"Olaf", [9, 11]=>"Stan", [20, 2]=>"Judge", [5, 5]=>"Cassie", [7, 1]=>"Gracie", [6, 3]=>"Armando", [4, 7]=>"Delia", [3, 9]=>"Marley", [16, 10]=>"Robyn", [2, 11]=>"Richie", [12, 0]=>"Gilberto", [13, 2]=>"Dedrick", [9, 10]=>"Liam", [5, 4]=>"Jabari", [7, 0]=>"Enola", [6, 2]=>"Lela", [3, 8]=>"Jade", [2, 10]=>"Johnson", [15, 5]=>"Willow", [12, 3]=>"Fredrick", [13, 1]=>"Beau", [9, 9]=>"Carlie", [8, 11]=>"Daisha", [6, 5]=>"Declan", [4, 1]=>"Carolina", [5, 3]=>"Cruz", [7, 7]=>"Jaime", [0, 9]=>"Anthony", [1, 11]=>"Esta", [13, 0]=>"Shaina", [12, 2]=>"Alec", [8, 10]=>"Lora", [6, 4]=>"Emely", [4, 0]=>"Rodger", [5, 2]=>"Cedrick", [0, 8]=>"Collin", [1, 10]=>"Armani", [16, 5]=>"Brooks", [19, 3]=>"Eleanora", [18, 1]=>"Alva", [7, 5]=>"Melissa", [5, 1]=>"Tabitha", [4, 3]=>"Aniya", [6, 7]=>"Marc", [1, 9]=>"Marjorie", [0, 11]=>"Arvilla", [19, 2]=>"Adela", [7, 4]=>"Zakary", [5, 0]=>"Emely", [4, 2]=>"Alison", [1, 8]=>"Lorenz", [0, 10]=>"Lisandro", [17, 5]=>"Aylin", [18, 3]=>"Giles", [19, 1]=>"Kyleigh", [8, 5]=>"Mary", [11, 3]=>"Claire", [10, 1]=>"Avis", [9, 7]=>"Manuela", [15, 11]=>"Chesley", [18, 2]=>"Kristopher", [24, 3]=>"Zola", [8, 4]=>"Pietro", [10, 0]=>"Delores", [11, 2]=>"Timmy", [15, 10]=>"Khalil", [18, 5]=>"Trudie", [17, 3]=>"Rafael", [16, 1]=>"Anthony"} What I need to do though, is basically remove all the empty entries. Let's say [17,3] = Raphael does not have an element in front of if (let's say - no [16,3] exists) then [17,3] should become [16,3] etc. So basically all empty items will be popped off the vertical (row) structure of the hash. Are there functions I should have a look at or is there an easy squash-like method that would just remove blanks and adjust and move other items? Thanks in advance for your help.

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