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• Dell Docking Station Doesn’t Detect USB Mouse and Keyboard

  - by Ben Griswold

  I’ve found myself in this situation with multiple Dell docking stations and multiple Dell laptops running various Windows operating systems.  I don’t know why the docking station stops recognizing my USB mouse and keyboard – it just does.  It’s black magic.  The last time around I just starting plugging the mouse and keyboard into the docked laptop directly and went about my business (as if I wasn’t completing missing out on a couple of the core benefits of using a docking station.)  I guess that’s what happens when you forget how you got yourself out of the mess the last time around.  I had been in this half-assed state for a couple of weeks now, but a coworker fortunately got themselves in and out of the same pickle this morning.  Procrastinate long enough and the solution will just come to you, right? Here’s how to get yourself out of this mess: Undock your computer Unplug your docking station Count to an arbitrary number greater than 12.  (Not sure this is really required, but…) Plug your docking station back in Redock your machine I put my machine to sleep before taking the aforementioned actions.  My coworker completely shutdown his laptop instead.  The steps worked on both of our Win 7 machines this morning and, who knows, it might just work for you too. 

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• Jquery hide show list

  - by faxtion

  Having few issues with a jquery hide show item. Got a list of images, want to show the first one but hide the rest and when the user clicks next the current one is hidden and then the next list item is shown. Been banging head against wall for a while, and there is probably a very easy solution but for some reason I am unable to see it. Would be grateful of any advice. $(function() { $('#feat>li:gt(0)').hide(); var $links = $('#feat>li'); var $item = $('#feat>li'); $links.click(function() { var $par = $(this); $par.slideUp(700, function() { var index = $links.index( $par.get(0) ) $item.eq( index ).slideDown(700); }); }); }); <div id="feature"> <ul id="feat"> <li><img src="images/sample.png" /><a href="#">Next</a></li> <li><img src="images/sample2.jpg" /><a href="#">Next</a></li> <li><img src="images/sample.png" /><a href="#">Next</a></li> </ul> </div>

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• Project Euler 18: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 18.  As always, any feedback is welcome. # Euler 18 # http://projecteuler.net/index.php?section=problems&id=18 # By starting at the top of the triangle below and moving # to adjacent numbers on the row below, the maximum total # from top to bottom is 23. # # 3 # 7 4 # 2 4 6 # 8 5 9 3 # # That is, 3 + 7 + 4 + 9 = 23. # Find the maximum total from top to bottom of the triangle below: # 75 # 95 64 # 17 47 82 # 18 35 87 10 # 20 04 82 47 65 # 19 01 23 75 03 34 # 88 02 77 73 07 63 67 # 99 65 04 28 06 16 70 92 # 41 41 26 56 83 40 80 70 33 # 41 48 72 33 47 32 37 16 94 29 # 53 71 44 65 25 43 91 52 97 51 14 # 70 11 33 28 77 73 17 78 39 68 17 57 # 91 71 52 38 17 14 91 43 58 50 27 29 48 # 63 66 04 68 89 53 67 30 73 16 69 87 40 31 # 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 # NOTE: As there are only 16384 routes, it is possible to solve # this problem by trying every route. However, Problem 67, is the # same challenge with a triangle containing one-hundred rows; it # cannot be solved by brute force, and requires a clever method! ;o) import time start = time.time() triangle = [ [75], [95, 64], [17, 47, 82], [18, 35, 87, 10], [20, 04, 82, 47, 65], [19, 01, 23, 75, 03, 34], [88, 02, 77, 73, 07, 63, 67], [99, 65, 04, 28, 06, 16, 70, 92], [41, 41, 26, 56, 83, 40, 80, 70, 33], [41, 48, 72, 33, 47, 32, 37, 16, 94, 29], [53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14], [70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57], [91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48], [63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31], [04, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 04, 23]] # Loop through each row of the triangle starting at the base. for a in range(len(triangle) - 1, -1, -1): for b in range(0, a): # Get the maximum value for adjacent cells in current row. # Update the cell which would be one step prior in the path # with the new total. For example, compare the first two # elements in row 15. Add the max of 04 and 62 to the first # position of row 14.This provides the max total from row 14 # to 15 starting at the first position. Continue to work up # the triangle until the maximum total emerges at the # triangle's apex. triangle [a-1][b] += max(triangle [a][b], triangle [a][b+1]) print triangle [0][0] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• print a linear linked list into a table

  - by user1796970

  I am attempting to print some values i have stored into a LLL into a readable table. The data i have stored is the following : DEBBIE STARR F 3 W 1000.00 JOAN JACOBUS F 9 W 925.00 DAVID RENN M 3 H 4.75 ALBERT CAHANA M 3 H 18.75 DOUGLAS SHEER M 5 W 250.00 SHARI BUCHMAN F 9 W 325.00 SARA JONES F 1 H 7.50 RICKY MOFSEN M 6 H 12.50 JEAN BRENNAN F 6 H 5.40 JAMIE MICHAELS F 8 W 150.00 i have stored each firstname, lastname, gender, tenure, payrate, and salary into their own List. And would like to be able to print them out in the same format that they are viewed on the text file i read them in from. i have messed around with a few methods that allow me to traverse and print the Lists, but i end up with ugly output. . . here is my code for the storage of the text file and the format i would like to print out: public class Payroll { private LineWriter lw; private ObjectList output; ListNode input; private ObjectList firstname, lastname, gender, tenure, rate, salary; public Payroll(LineWriter lw) { this.lw = lw; this.firstname = new ObjectList(); this.lastname = new ObjectList(); this.gender = new ObjectList(); this.tenure = new ObjectList(); this.rate = new ObjectList(); this.salary = new ObjectList(); this.output = new ObjectList(); this.input = new ListNode(); } public void readfile() { File file = new File("payfile.txt"); try{ Scanner scanner = new Scanner(file); while(scanner.hasNextLine()) { String line = scanner.nextLine(); Scanner lineScanner = new Scanner(line); lineScanner.useDelimiter("\\s+"); while(lineScanner.hasNext()) { firstname.insert1(lineScanner.next()); lastname.insert1(lineScanner.next()); gender.insert1(lineScanner.next()); tenure.insert1(lineScanner.next()); rate.insert1(lineScanner.next()); salary.insert1(lineScanner.next()); } } }catch(FileNotFoundException e) {e.printStackTrace();} } public void printer(LineWriter lw) { String msg = " FirstName " + " LastName " + " Gender " + " Tenure " + " Pay Rate " + " Salary "; output.insert1(msg); System.out.println(output.getFirst()); System.out.println(" " + firstname.getFirst() + " " + lastname.getFirst() + "\t" + gender.getFirst() + "\t" + tenure.getFirst() + "\t" + rate.getFirst() + "\t" + salary.getFirst()); } }

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• jQuery animation queues

  - by OneNerd

  I am at a dead end, so hoping you jQuery gurus can help. I have a total of 10 elements (actually small images) on a page. I need to animate them like this: first 2 show up then the next 2 show up then the next 3 show up then the next 1 shows up then the last 2 show up So, I have added attributes to each one (sequence_num = "1" (or 2 or 3 etc) so I can easily choose via the $() which ones to animate using the animate() function.) My goal is to write a function that does the animation (I can do that - i think i have grasped the animate() function). What I am getting stuck on is how to delay the animation so the proper groups of objects are animated in before the next group starts. I have tried the queue parameter of the animate() function, but that doesn't seem to work for what I am trying to do. Does anyone have any experience with this?

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• T4Toolbox and Visual Studio 2010

  - by Ben Griswold

  I’ve been using the T4Toolbox to help generate my ASP.NET MVC models and scaffolding for a while now.  Another developer tried using my generator project last week and ran into troubles due to a breaking change around the RenderCore() and TransformText() methods in support for VS 2010.  If you upgraded to the latest version of T4Toolbox and receive a build error similar to the following, you are probably in the same boat: GeneratedTextTransformation.[Template].RenderCore(): no suitable method found to override We took the easy way out.  I had him uninstall the latest version of T4Toolbox and install version 9.7.25.1 which my templates were initially coded against.  For now, that worked great, but it sounds like I’ll be doing some rework of the 20+ templates in my project to support Visual Studio 2010 when we migrate later this month.

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• SVN Export or Recursively Remove .SVN Folders

  - by Ben Griswold

  I shared this script with a coworker yesterday. It doesn’t do much; it recursively deletes .svn folders from a source tree.  It comes in handy if you want to share your codebase or you get in a terrible spot with SVN and you just want to start all over. Just blow away all svn artifacts and use your mulligan. It’s true. You can nearly get the same result using the SVN export command which copies your source sans the .svn folders to an alternate location.  The catch is an export only includes those files/folders which exist under version control.  If you want a clean copy of your source – versioned or not – export just might not do. The contents of the .cmd file include the following: for /f "tokens=* delims=" %%i in (’dir /s /b /a:d *.svn’) do ( rd /s /q "%%i" ) Just download and drop the unzipped “SVN Cleanup.cmd” file into the root of the project, execute and away you go.  If you search around enough, I know you can find similar scripts and approaches elsewhere, but I’m still uploading my script for completeness and future reference. Download SVN Cleanup

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• Loops inside loops

  - by cozzy

  Hi, I'd like to find easier way to write loops inside loops. Here is example code of 3 levels of 'for' loops: int level = 0; int value = 0; bool next = false; for (int i0 = 0; i0 < 6; i0++) { level = 0; value = i0; method(); if (next) for (int i1 = 0; i1 < 6; i1++) { level = 1; value = i1; method(); if (next) for (int i2 = 0; i2 < 6; i2++) { level = 2; value = i2; method(); if (next) { //Do somethnig } } } } private void method() { //use int 'level' and 'value' //determine bool 'next' } I wonder if it's possible to write the same thing different way. To set number of levels(number of loops) and loop repeats. In this case levels = 3; repeats = 6;. I need it because I am using more than 20 loops inside themselves and than the code is not comprehensible. I hope my explanation was ok and thanks for help.

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• Forced Learning

  - by Ben Griswold

  If you ask me, it can be a little intimidating to stand in front of a group and walkthrough anything remotely technical. Even if you know “Technical Thingy #52” inside and out, public speaking can be unsettling.  And if you don’t have your stuff together, well, it can be downright horrifying. With that said, if given the choice, I still like to schedule myself to present on unfamiliar topics. Over the past few months, I’ve talked about Aspect-Oriented Programming, Functional Programming, Lean Software Development and Kanban Systems, Domain-Driven Design and Behavior Driven Development.  What do these topics have in common? You guessed it: I was truly interested in them. I had only a superficial understanding of each. Huh?  Why in the world would I ever want to to put myself in that intimidating situation? Actually, I rarely want to put myself into that situation but I often do as I like the results.  There’s nothing remotely clever going on here.  All I’m doing is putting myself into a compromising situation knowing that I’ll likely work myself out of it by learning the topic prior to the presentation.  I’m simply time-boxing myself to learn something new while knowing there are negative repercussions if I fall short. So, I end up doing tons of research and I learn bunches to ensure I have my head firmly wrap around the material before my talk. I’m not saying I become an expert overnight (or over a couple of weeks) but I’ll definitely know enough to be confident and comfortable and I’ll know more than enough to ensure the audience will learn a thing or two from me. It’s forced learning and though it might sound a little scary to some, it works for me. Now I could very easily rename this post to something like Fear Is My Motivator because, in a sense, fear of failure and embarrassment is what’s driving my learning. However, I’m the guy signing up for the presentation and since the entire process is self-imposed I’m not sure Fear deserves too much credit.  Anyway…

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• Jquery JCarousel, itemLastInCallback how to change out the controls

  - by TJ Sherrill

  I am new to js and Jquery other than plug and play scripts and plugins. I have Jcarousel running great on a Ruby on Rails site and I need to use the itemLastInCallback method to change out the next and prev buttons. Basically if the end user clicks through to the last item in a carousel, the next button would become a link to the next carousel. <%= link_to (image_tag 'scroll_right.png'), @next_group % is the code I need to render in the end. So something like: if this is not the last item in the carousel button is normal jcarousel next button else <%= link_to (image_tag 'scroll_right.png'), @next_group % end is there a best way to do this? am I on the right track?

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• Project Euler 15: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 15.  As always, any feedback is welcome. # Euler 15 # http://projecteuler.net/index.php?section=problems&id=15 # Starting in the top left corner of a 2x2 grid, there # are 6 routes (without backtracking) to the bottom right # corner. How many routes are their in a 20x20 grid? import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) rows, cols = 20, 20 print factorial(rows+cols) / (factorial(rows) * factorial(cols)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• ReSharper File Location

  - by Ben Griswold

  By default, the ReSharper cache is stored in the solution folder.  It’s one extra folder and one extra .user file.  It’s no big deal but it does clutter up your solution a bit – especially since the files provide no real value. I prefer to store the ReSharper cache in the system Temp folder.  This setting is available by visiting ReSharper > Options > Environment > General. Just update where you’d like to store the ReSharper cache and you’re good to go.  Note, the .user file continues to linger around the solution folder but at least the _ReSharper.SolutionName folder is moved out of sight.

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• Project Euler 9: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 9.  As always, any feedback is welcome. # Euler 9 # http://projecteuler.net/index.php?section=problems&id=9 # A Pythagorean triplet is a set of three natural numbers, # a b c, for which, # a2 + b2 = c2 # For example, 32 + 42 = 9 + 16 = 25 = 52. # There exists exactly one Pythagorean triplet for which # a + b + c = 1000. Find the product abc. import time start = time.time() product = 0 def pythagorean_triplet(): for a in range(1,501): for b in xrange(a+1,501): c = 1000 - a - b if (a*a + b*b == c*c): return a*b*c print pythagorean_triplet() print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 5: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 5.  As always, any feedback is welcome. # Euler 5 # http://projecteuler.net/index.php?section=problems&id=5 # 2520 is the smallest number that can be divided by each # of the numbers from 1 to 10 without any remainder. # What is the smallest positive number that is evenly # divisible by all of the numbers from 1 to 20? import time start = time.time() def gcd(a, b): while b: a, b = b, a % b return a def lcm(a, b): return a * b // gcd(a, b) print reduce(lcm, range(1, 20)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 14: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 14.  As always, any feedback is welcome. # Euler 14 # http://projecteuler.net/index.php?section=problems&id=14 # The following iterative sequence is defined for the set # of positive integers: # n -> n/2 (n is even) # n -> 3n + 1 (n is odd) # Using the rule above and starting with 13, we generate # the following sequence: # 13 40 20 10 5 16 8 4 2 1 # It can be seen that this sequence (starting at 13 and # finishing at 1) contains 10 terms. Although it has not # been proved yet (Collatz Problem), it is thought that all # starting numbers finish at 1. Which starting number, # under one million, produces the longest chain? # NOTE: Once the chain starts the terms are allowed to go # above one million. import time start = time.time() def collatz_length(n): # 0 and 1 return self as length if n <= 1: return n length = 1 while (n != 1): if (n % 2 == 0): n /= 2 else: n = 3*n + 1 length += 1 return length starting_number, longest_chain = 1, 0 for x in xrange(1, 1000001): l = collatz_length(x) if l > longest_chain: starting_number, longest_chain = x, l print starting_number print longest_chain # Slow 31 seconds print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Language Club

  - by Ben Griswold

  We started a language club at work this week.  Thus far, we have a collective interest in a number of languages: Python, Ruby, F#, Erlang, Objective-C, Scala, Clojure, Haskell and Go. There are more but these 9 received the most votes. During the first few meetings we are going to determine which language we should tackle first. To help make our selection, each member will provide a quick overview of their favored language by answering the following set of questions: Why are you interested in learning “your” language(s). (There’s lots of work, I’m an MS shill, It’s hip and  fun, etc) What type of language is it?  (OO, dynamic, functional, procedural, declarative, etc) What types of problems is your language best suited to solve?  (Algorithms over big data, rapid application development, modeling, merely academic, etc) Can you provide examples of where/how it is being used?  If it isn’t being used, why not?  (Erlang was invented at Ericsson to provide an extremely fault tolerant, concurrent system.) Quick history – Who created/sponsored the language?  When was it created?  Is it currently active? Does the language have hardware support (an attempt was made at one point to create processor instruction sets specific to Prolog), or can it run as an interpreted language inside another language (like Ruby in the JVM)? Are there facilities for programs written in this language to communicate with other languages?  How does this affect its utility? Does the language have a IDE tool support?  (Think Eclipse or Visual Studio) How well is the language supported in terms of books, community and documentation? What’s the number one things which differentiates the language from others?  (i.e. Why is it cool?) How is the language applicability to us as consultants?  What would the impact be of using the language in terms of cost, maintainability, personnel costs, etc.? What’s the number one things which differentiates the language from others?  (i.e. Why is it cool?) This should provide an decent introduction into nearly a dozen languages and give us enough context to decide which single language deserves our undivided attention for the weeks to come.  Stay tuned for the winner…

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• using python 'with' statement with iterators?

  - by Stephen

  Hi, I'm using Python 2.5. I'm trying to use this 'with' statement. from __future__ import with_statement a = [] with open('exampletxt.txt','r') as f: while True: a.append(f.next().strip().split()) print a The contents of 'exampletxt.txt' are simple: a b In this case, I get the error: Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/tmp/python-7036sVf.py", line 5, in <module> a.append(f.next().strip().split()) StopIteration And if I replace f.next() with f.read(), it seems to be caught in an infinite loop. I wonder if I have to write a decorator class that accepts the iterator object as an argument, and define an __exit__ method for it? I know it's more pythonic to use a for-loop for iterators, but I wanted to implement a while loop within a generator that's called by a for-loop... something like def g(f): while True: x = f.next() if test1(x): a = x elif test2(x): b = f.next() yield [a,x,b] a = [] with open(filename) as f: for x in g(f): a.append(x)

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• create graph using adjacency list

  - by sum1needhelp

  #include<iostream> using namespace std; class TCSGraph{ public: void addVertex(int vertex); void display(); TCSGraph(){ head = NULL; } ~TCSGraph(); private: struct ListNode { string name; struct ListNode *next; }; ListNode *head; } void TCSGraph::addVertex(int vertex){ ListNode *newNode; ListNode *nodePtr; string vName; for(int i = 0; i < vertex ; i++ ){ cout << "what is the name of the vertex"<< endl; cin >> vName; newNode = new ListNode; newNode->name = vName; if (!head) head = newNode; else nodePtr = head; while(nodePtr->next) nodePtr = nodePtr->next; nodePtr->next = newNode; } } void TCSGraph::display(){ ListNode *nodePtr; nodePtr = head; while(nodePtr){ cout << nodePtr->name<< endl; nodePtr = nodePtr->next; } } int main(){ int vertex; cout << " how many vertex u wan to add" << endl; cin >> vertex; TCSGraph g; g.addVertex(vertex); g.display(); return 0; }

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• how to set tab order in jquery

  - by Martin Ongtangco

  Hello, I'm using Telerik controls, specifically the numerical textbox where you can set a up-down arrow to increment/decrement values in a textbox. I am required to set the tab order to move to the next field but since there's a button on the up-down arrow, the browser will go through those buttons first then move to the next textbox field. How do you set the jquery to detect the next visible textbox/dropdown/etc input field and move to that on pressing the tab button instead of running through the buttons near it?

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• Copy a linked list

  - by emkrish

  typedef struct Node { int data; Node *next; Node *other; }; Node *pHead; pHead is a singly linked list. The next field points to the next element in the list. The other field may point to any other element (could be one of the previous nodes or one of the nodes ahead) in the list or NULL. How does one write a copy function that duplicates the linked list and its connectivity? None of the elements (next and other) in the new list should point to any element in the old list.

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• Project Euler 8: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Managing .NET External Dependencies

  - by Ben Griswold

  Noah and I continue our screencast series by sharing our approach to managing external dependencies referenced within a .NET solution.  This is another introductory episode but you might find a hidden gem in the short 4-minute clip.  ELMAH (Error Logging Modules and Handlers) is the external dependencies we are focusing on in the presentation.  If you are not familiar with ELMAH, this episode may be worth your time.   YouTube - Managing .NET External Dependencies This is one of our first screencasts.  If you have feedback, I’d love to hear it.

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• Android Application is unexpectedly stopped error when button is clicked

  - by user1794499

  Hi there I'm totally new to Android development and I'm working in my android application my application includes a forum where users can post, comment and have their discussion there.... So I'm working in the interface but I get error when I click on the button I directs me to the signup page can somebody please help me with this error this is the code of the mainuserinterface.java for the mainuserinterface.xml file where the button resides. and the signupform.class is the java file for the next activity triggered when the button is clicked the error I receive is the application is unexpectedly stopped.. Hope I make it clear for you guys package com.mohammed.watzIslam; import android.app.Activity; import android.content.Intent; import android.os.Bundle; import android.view.View; import android.widget.Button; public class mainuserinterface extends Activity { @Override protected void onCreate(Bundle savedInstanceState) { // TODO Auto-generated method stub super.onCreate(savedInstanceState); setContentView(R.layout.mainuserinterface); // this is the button where I receive errors when I click Button forum = (Button) findViewById(R.id.next); forum.setOnClickListener(new View.OnClickListener() { public void onClick(View view){ Intent myIntent = new Intent (view.getContext(), signupform.class); startActivityForResult (myIntent, 0); } }); //these two button still not directing to any next activity yet Button button1 = (Button) findViewById(R.id.next); forum.setOnClickListener(new View.OnClickListener() { public void onClick(View view){ Intent myIntent1 = new Intent (view.getContext(), signupform.class); startActivityForResult (myIntent1, 0); } }); Button button2 = (Button) findViewById(R.id.next); forum.setOnClickListener(new View.OnClickListener() { public void onClick(View view){ Intent myIntent2 = new Intent (view.getContext(), signupform.class); startActivityForResult (myIntent2, 0); } }); } }

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• Project Euler 52: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 52.  Compared to Problem 51, this problem was a snap. Brute force and pretty quick… As always, any feedback is welcome. # Euler 52 # http://projecteuler.net/index.php?section=problems&id=52 # It can be seen that the number, 125874, and its double, # 251748, contain exactly the same digits, but in a # different order. # # Find the smallest positive integer, x, such that 2x, 3x, # 4x, 5x, and 6x, contain the same digits. timer_start = Time.now def contains_same_digits?(n) value = (n*2).to_s.split(//).uniq.sort.join 3.upto(6) do |i| return false if (n*i).to_s.split(//).uniq.sort.join != value end true end i = 100_000 answer = 0 while answer == 0 answer = i if contains_same_digits?(i) i+=1 end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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• Simple Java calculator

  - by Kevin Duke

  Firstly this is not a homework question. I am practicing my knowledge on java. I figured a good way to do this is to write a simple program without help. Unfortunately, my compiler is telling me errors I don't know how to fix. Without changing much logic and code, could someone kindly point out where some of my errors are? Thanks import java.lang.*; import java.util.*; public class Calculator { private int solution; private int x; private int y; private char operators; public Calculator() { solution = 0; Scanner operators = new Scanner(System.in); Scanner operands = new Scanner(System.in); } public int addition(int x, int y) { return x + y; } public int subtraction(int x, int y) { return x - y; } public int multiplication(int x, int y) { return x * y; } public int division(int x, int y) { solution = x / y; return solution; } public void main (String[] args) { System.out.println("What operation? ('+', '-', '*', '/')"); System.out.println("Insert 2 numbers to be subtracted"); System.out.println("operand 1: "); x = operands; System.out.println("operand 2: "); y = operands.next(); switch(operators) { case('+'): addition(operands); operands.next(); break; case('-'): subtraction(operands); operands.next(); break; case('*'): multiplication(operands); operands.next(); break; case('/'): division(operands); operands.next(); break; } } }

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