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  • Get Your Business Going With Website Development and More

    Getting a business of the ground can be tough, and in today's world where most businesses at least have a branch online, website development is going to be key. If you need a website, having a quality one is important to your success and a good professional can take care of this, custom application development, and a whole lot more to help you get your business going.

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  • How to Buy a Site on Flippa

    People buy websites for all sorts of reasons, not least to give them a shortcut (whether it be to fame, finances or whatever). As with most things in life, buying websites is a skill, and is a skill that you can be developed and nourished.

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  • 3 Secret Ways to Improve Your Search Engine Ranking Right Now

    How do I increase my search engine ranking so that I can boost traffic and therefore sales? That is a good question and is one that anyone who is involved with internet marketing has asked at least once in their online career. It is easy to think that SEO is complicated, especially if you are just starting out.

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  • Search Engine Optimization For Beginners

    Learning about the SEO techniques can be scary and a bit overwhelming to say the least. If you do not know anything about SEO now is the time to learn, before trying to do it with just a little bit of knowledge, you will fail if you do it that way.

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  • What Are Search Engine "Spiders" and How Do They Work?

    Although the question "What Are Search Engine "Spiders" and How Do They Work?" may seem a geeky one it is actually an important question for all internet marketers and one that you should have at least a basic understanding of the answer to. Spiders are automated software agents...

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  • How many people should be involved in a game startup? [on hold]

    - by Elemecca
    I want to start an indie game company, and I understand that you need a designer, an engineer, and an artist at the very least. I also know that professional game companies like Bethesda http://bgs.bethsoft.com/ have around 40 people. I want to have a good start, and many of you have been in the industry for a good while. What is a good number of employees to start with for an independent game company? Reasons?

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  • Flex: drawing a connector line between shapes

    - by artemb
    Hi! I am building a diagramming tool using Adobe Flex 3. I am about to implement connector lines and I have a question. Imagine I have 2 squares at random positions on the canvas. I need to draw an arrowed connector line between them. I need it to tend to the target square's center but end on its border. How do I find out the exact points between which to draw the line? Thank you

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  • How to count each digit in a range of integers?

    - by Carlos Gutiérrez
    Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses: 1 to 100 51 to 300 1 to 2,000 with zeros to the left The obvious solution is to do a loop from the first to the last number, convert the counter to a string with or without zeros to the left, extract each digit and use it as an index to increment an array of 10 integers. I wonder if there is a better way to solve this, without having to loop through the entire integers range. Solutions in any language or pseudocode are welcome. Edit: Answers review John at CashCommons and Wayne Conrad comment that my current approach is good and fast enough. Let me use a silly analogy: If you were given the task of counting the squares in a chess board in less than 1 minute, you could finish the task by counting the squares one by one, but a better solution is to count the sides and do a multiplication, because you later may be asked to count the tiles in a building. Alex Reisner points to a very interesting mathematical law that, unfortunately, doesn’t seem to be relevant to this problem. Andres suggests the same algorithm I’m using, but extracting digits with %10 operations instead of substrings. John at CashCommons and phord propose pre-calculating the digits required and storing them in a lookup table or, for raw speed, an array. This could be a good solution if we had an absolute, unmovable, set in stone, maximum integer value. I’ve never seen one of those. High-Performance Mark and strainer computed the needed digits for various ranges. The result for one millon seems to indicate there is a proportion, but the results for other number show different proportions. strainer found some formulas that may be used to count digit for number which are a power of ten. Robert Harvey had a very interesting experience posting the question at MathOverflow. One of the math guys wrote a solution using mathematical notation. Aaronaught developed and tested a solution using mathematics. After posting it he reviewed the formulas originated from Math Overflow and found a flaw in it (point to Stackoverflow :). noahlavine developed an algorithm and presented it in pseudocode. A new solution After reading all the answers, and doing some experiments, I found that for a range of integer from 1 to 10n-1: For digits 1 to 9, n*10(n-1) pieces are needed For digit 0, if not using leading zeros, n*10n-1 - ((10n-1) / 9) are needed For digit 0, if using leading zeros, n*10n-1 - n are needed The first formula was found by strainer (and probably by others), and I found the other two by trial and error (but they may be included in other answers). For example, if n = 6, range is 1 to 999,999: For digits 1 to 9 we need 6*105 = 600,000 of each one For digit 0, without leading zeros, we need 6*105 – (106-1)/9 = 600,000 - 111,111 = 488,889 For digit 0, with leading zeros, we need 6*105 – 6 = 599,994 These numbers can be checked using High-Performance Mark results. Using these formulas, I improved the original algorithm. It still loops from the first to the last number in the range of integers, but, if it finds a number which is a power of ten, it uses the formulas to add to the digits count the quantity for a full range of 1 to 9 or 1 to 99 or 1 to 999 etc. Here's the algorithm in pseudocode: integer First,Last //First and last number in the range integer Number //Current number in the loop integer Power //Power is the n in 10^n in the formulas integer Nines //Nines is the resut of 10^n - 1, 10^5 - 1 = 99999 integer Prefix //First digits in a number. For 14,200, prefix is 142 array 0..9 Digits //Will hold the count for all the digits FOR Number = First TO Last CALL TallyDigitsForOneNumber WITH Number,1 //Tally the count of each digit //in the number, increment by 1 //Start of optimization. Comments are for Number = 1,000 and Last = 8,000. Power = Zeros at the end of number //For 1,000, Power = 3 IF Power 0 //The number ends in 0 00 000 etc Nines = 10^Power-1 //Nines = 10^3 - 1 = 1000 - 1 = 999 IF Number+Nines <= Last //If 1,000+999 < 8,000, add a full set Digits[0-9] += Power*10^(Power-1) //Add 3*10^(3-1) = 300 to digits 0 to 9 Digits[0] -= -Power //Adjust digit 0 (leading zeros formula) Prefix = First digits of Number //For 1000, prefix is 1 CALL TallyDigitsForOneNumber WITH Prefix,Nines //Tally the count of each //digit in prefix, //increment by 999 Number += Nines //Increment the loop counter 999 cycles ENDIF ENDIF //End of optimization ENDFOR SUBROUTINE TallyDigitsForOneNumber PARAMS Number,Count REPEAT Digits [ Number % 10 ] += Count Number = Number / 10 UNTIL Number = 0 For example, for range 786 to 3,021, the counter will be incremented: By 1 from 786 to 790 (5 cycles) By 9 from 790 to 799 (1 cycle) By 1 from 799 to 800 By 99 from 800 to 899 By 1 from 899 to 900 By 99 from 900 to 999 By 1 from 999 to 1000 By 999 from 1000 to 1999 By 1 from 1999 to 2000 By 999 from 2000 to 2999 By 1 from 2999 to 3000 By 1 from 3000 to 3010 (10 cycles) By 9 from 3010 to 3019 (1 cycle) By 1 from 3019 to 3021 (2 cycles) Total: 28 cycles Without optimization: 2,235 cycles Note that this algorithm solves the problem without leading zeros. To use it with leading zeros, I used a hack: If range 700 to 1,000 with leading zeros is needed, use the algorithm for 10,700 to 11,000 and then substract 1,000 - 700 = 300 from the count of digit 1. Benchmark and Source code I tested the original approach, the same approach using %10 and the new solution for some large ranges, with these results: Original 104.78 seconds With %10 83.66 With Powers of Ten 0.07 A screenshot of the benchmark application: If you would like to see the full source code or run the benchmark, use these links: Complete Source code (in Clarion): http://sca.mx/ftp/countdigits.txt Compilable project and win32 exe: http://sca.mx/ftp/countdigits.zip Accepted answer noahlavine solution may be correct, but l just couldn’t follow the pseudo code, I think there are some details missing or not completely explained. Aaronaught solution seems to be correct, but the code is just too complex for my taste. I accepted strainer’s answer, because his line of thought guided me to develop this new solution.

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  • Help with animation

    - by Pennywise83
    I've used the "FeatureList" Jquery Plugin to make my own content slider. The script can be found here: http://pastebin.com/7iyE5ADu Here is an exemplificative image to show what I'm triyng to achieve: http://i41.tinypic.com/6jkeq1.jpg Actually the slider add a "current" class to an item (in the example the squares 1,2 and 3) and for each thumb show a content in the main area. In the example, with an interval of 2 seconds, the script switch from 1 to 2, from 2 to 3, and so on. I'd like to make a continuous animation of the thumbs, anyone can help me?

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  • Android - Camera functionality howto

    - by teepusink
    Hi, I'm using the example from the SDK (CameraPreview) and also the example from this site http://marakana.com/forums/android/android_examples/39.html When I run it, both gives this error "The application AppName(appname) has stopped unexpectedly. Please try again". I can't run the debugger either because it's always "Waiting for debugger. Force close". (I have debuggable=true in the manifest file) The phone I have is a Nexus One. Running on the emulator gives me the black and white squares with a moving square. (So I assume it works on the emulator?). Even on the emulator it gives me the "stopped unexpectedly" error 50% of the time. Does anyone know what caused it? Thanks, Tee

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  • Multiple outliers for two variable linear regression

    - by Dave Jarvis
    Problem Building on my previous question, the "extreme" outliers in the following graph are somewhat obvious: Question Given: T - Set of all temperatures Y - Set of all years ST - Sum of temperatures. SY - Sum of years. N - Number of elements T(n) - Temperature of the nth element in the temperature set How would you implement an efficient MySQL stored procedure or user-defined function (UDF) to determine if T(n) is an outlier? (If such an implementation already exists, that would be good to know as well.) Related Sites I am slowly working through these sites to get a better understanding of the problem: Multiple Outliers Detection Procedures in Linear Regression M-estimator Measure of Surprise for Outlier Detection Ordinary Least Squares Linear Regression Many thanks!

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  • "cannot concatenate 'str' and 'list' objects" keeps coming up :(

    - by ratce003
    I'm writing a python program and passing in a html template but an error keeps coming up, "cannot concatenate 'str' and 'list' objects" here is is the program: #!/usr/bin/env python # -*- coding: UTF-8 -*- # enable debugging import cgi import cgitb cgitb.enable() def template(file, **vars): return open(file, 'r').read() % vars print "Content-type: text/html\n" print form = cgi.FieldStorage() # instantiate only once! num_1 = form.getfirst('num_1') num_2 = form.getfirst('num_2') int1r = str(num_1) int2r = str(num_2) def calc_range(int2r, int1r): start = range(int2r, int1r + 1) end = range(1, int2r) return start+end int1 = int(int1r) int2 = int(int2r) out_str = '' for i in range(0, int1): first_line_num = (int2 + i) % int1 if first_line_num == 0: first_line_num = int1 line = calc_range(first_line_num, int1) out_str += line print template('results.html', output=out_str, title="Latin Squares")

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  • Dynamically pixelate an html image element

    - by Chris Armstrong
    I'm to take an image on a webpage, and then use javascript (or whatever would be best suited) to dynamically 'pixelate' it (e.g. into 20px squares). Then, as the user scrolls down the page, I need the image to gradually increase in resolution, till it is no longer pixelated. Any ideas how I could go about doing this? I realise I could use php to resize an image and several times and just switch out the image, but that would require loading several extra images. Also, I know I could probably do the effect with flash & pixelbender, but I want to achieve it within the limitations of HTML5, CSS & Javascript if possible. Appreciate any thoughts!

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  • Group keywords by site

    - by Skudd
    I am finding a lot of useful help here today, and I really appreciate it. This should be the last one for the day: I have a list of the top 10 keywords per site, sorted by visits, by date. The records need to be sorted as follows (excuse the formatting): 2010-05 2010-04 site1.com keyword1 apples wine keyword1 visits 100 12 keyword2 oranges water keyword2 visits 99 10 site2.com keyword1 blueberry cornbread keyword1 visits 90 100 keyword2 squares biscuits keyword2 visits 80 99 Basically what I need to accomplish involves grouping, but I can't seem to figure it out. Am I heading down the right path, or is there another way to achieve this, or is it just impossible?

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  • JLabel withing another JLabel

    - by out_sider
    I want to do a very simple thing with java swing and I'm not being able to. I added a jLabel as it is the only object using java swing that can hold an image (using the icon property)...at least I wasn't able to set an image using other objects. Now I want to add another jlabel in top of the first one with another image so it can be "clicked" independently from the "background" label. But whenever I try to add it using the graphical editor or by doing jLabel1.add(jLabel2) it doesn't work...it simply sets next to the label1 but not on top of it. This is in order to do a java application like the Tic Tac Toe game...so I can have the background which are squares (first label) and the others the "X" and "O". board This might be the board and I want to put labels on each square so they can be the pieces.

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  • Any ideas on how to implement a 'touchMoveOver' event in Javascript?

    - by gargantaun
    I'm faffing around with SVG, specifically for web content aimed at iPad users. I've created a little dial type thingy that I'm calling a "cheese board" that I'd like to use as an interface element. http://appliedworks.co.uk/files/times/SVGTests/raphael.html Clicking on a piece of cheese (to keep the analogy going) will do "something". That bit's easy. However, I'd like the user to be able to drag their finger around the 'cheese board', firing a new event (touchesMovedOver?) every time they their finger moves over a new piece of cheese. But I can't figure out how to do it since there's no 'mouseOver' equivalent for touch interfaces. If the whole thing was made of squares, I could have created some sort of 'rectContainsPoint' method to be called for every 'touchesMoved', but that approach wouldn't work here. If anyone has any idea about how something like this could be achieved, I'd love to hear it.

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  • How's my pygame code?

    - by Isaiah
    I'm still getting the hang of lots of things and thought I should post some code I made with pygame and get some feedback^^. I posted code here: http://urlvars.com/code/snippet/39272/my-bouncing-program http://urlvars.com/code/snippet/39273/my-bouncing-program-classes There's tome things that I implemented that I'm not using yet I just realized like a timer at the bottom of the main while loop. If my code isn't readable, I'm sorry, I'm self taught and this is the first code I've ever posted anywhere. By the way I made some variables that take the screensize and half it to find a point to spit out the squares, but when I try to use it, it makes a weird effect :/ Try switching the list i have in the newbyte() function with the halfScreen variable and see it freak out o.O thank you

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  • Simple Android OpenGL App Lag

    - by Eugene
    Hi, I have an Android OpenGL application which simply draws 2D squares (using 2 triangles) and animates them moving down the screen. I essentially do this by running: glLoadIdentity(), then glTranslatef, and finally glDrawElements all in a for loop. (The for loop is to draw all 10 blocks on the screen for every frame). In every drawFrame, the y-position of the blocks increments for the animation. The problem I'm having is strange. I run the application and the animation is laggy and not smooth. Then I re-run the application and I get a smooth animation. If I run again, I may get a smooth animation, or possibly not. Is my method correct, or is there a better way of doing this animation? Thanks for the help!

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  • Only Integrating Box2D collision detection in my 2d engine?

    - by Mr.Gando
    I have integrated box2d in my engine, ( Debug Draw, etc. ) and with a world I can throw in some 2d squares/rectangles etc. I saw this post, where the user is basically not using a world for collision detection, however the user doesn't explain anything about how he's using the manifold (b2Manifold), etc. Another post, is in the cocos2d forum, ( scroll down to the user Lam in the third reply ) Could anyone help me a bit with this?, basically want to add collision detection without the need of using b2World ,etc etc. Thanks a lot!

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  • total number of magic square from 9 numbers

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

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  • How to detect touch in triangle area

    - by Romain
    For my application, i need to divide diagonally the screen of my iphone in 4 parts and detect which of the part was touched. I am very confused because my areas are triangles and not squares and I can't find a solution to detect which of the triangle was touched... I get the touched Point with method touchesBegan, and there I'm stuck... :( How to define triangle shapes and test if it was touched? with View? layer? It could be really cool if someone could help me.

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  • stuck on sliders with jquery

    - by chris
    I am building a portfolio page of different work that has been done and im using two different jquery sliding techniques, one nested in amongst the other one. The first slider works great but the second one doesnmt work that great for the first few and then doesnt work at all. If you take a look at this page- You can click the clients up and down the left side to slide the main divs, but when you click on the squares in below the images to slide in and out the content, either it doesnt work as smoothly as it should or it doesnt work right at all. http://justni.com/wip/mdd/martin_duggan_portfolio.html anyone wanna take a look at the code and tell me what am doing wrong?

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