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• Why do large IT projects tend to fail or have big cost/schedule overruns?

  - by Pratik

  I always read about large scale transformation or integration project that are total or almost total disaster. Even if they somehow manage to succeed the cost and schedule blow out is enormous. What is the real reason behind large projects being more prone to failure. Can agile be used in these sort of projects or traditional approach is still the best. One example from Australia is the Queensland Payroll project where they changed test success criteria to deliver the project. See some more failed projects in this SO question Have you got any personal experience to share?

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• Project Euler 7: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 4: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 4.  As always, any feedback is welcome. # Euler 4 # http://projecteuler.net/index.php?section=problems&id=4 # Find the largest palindrome made from the product of # two 3-digit numbers. A palindromic number reads the # same both ways. The largest palindrome made from the # product of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of # two 3-digit numbers. import time start = time.time() def isPalindrome(s): return s == s[::-1] max = 0 for i in xrange(100, 999): for j in xrange(i, 999): n = i * j; if (isPalindrome(str(n))): if (n > max): max = n print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 13: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 6: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 6.  As always, any feedback is welcome. # Euler 6 # http://projecteuler.net/index.php?section=problems&id=6 # Find the difference between the sum of the squares of # the first one hundred natural numbers and the square # of the sum. import time start = time.time() square_of_sums = sum(range(1,101)) ** 2 sum_of_squares = reduce(lambda agg, i: agg+i**2, range(1,101)) print square_of_sums - sum_of_squares print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 20: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 20.  As always, any feedback is welcome. # Euler 20 # http://projecteuler.net/index.php?section=problems&id=20 # n! means n x (n - 1) x ... x 3 x 2 x 1 # Find the sum of digits in 100! import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) print sum([int(i) for i in str(factorial(100))]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 1: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 1.  As always, any feedback is welcome. # Euler 1 # http://projecteuler.net/index.php?section=problems&amp;id=1 # If we list all the natural numbers below 10 that are # multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of # these multiples is 23. Find the sum of all the multiples # of 3 or 5 below 1000. import time start = time.time() print sum([x for x in range(1000) if x % 3== 0 or x % 5== 0]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue') # Also cool def constraint(x): return x % 3 == 0 or x % 5 == 0 print sum(filter(constraint, range(1000)))

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• Project Euler 3: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 3.  As always, any feedback is welcome. # Euler 3 # http://projecteuler.net/index.php?section=problems&id=3 # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number # 600851475143? import time start = time.time() def largest_prime_factor(n): max = n divisor = 2 while (n >= divisor ** 2): if n % divisor == 0: max, n = n, n / divisor else: divisor += 1 return max print largest_prime_factor(600851475143) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Motivating developers in a project perceived as **dull** ?

  - by Fanatic23

  As a manager, I can't always end up generating work that'd be cutting edge. Some of the projects do run on maintenance mode, and generate a healthy free cash flow for the company. As a developer what would it take for you to stick around in this project? I have been thinking of re-branding the work, but I could do with a lot of help here. Appreciate a single response per post. Please don't suggest an increased pay-packet, this creates more problems than it solves.

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• Updating and organizing class diagrams in a growing C++ project

  - by vanna

  I am working on a C++ project that is getting bigger and bigger. I do a lot of UML so it is not really hard to explain my work to co-workers. Lately though I implemented a lot of new features and I gave up updating by hand my Dia UML diagrams. I once used the class diagram of Visual Studio, which is my IDE but didn't get clear results. I need to show my work on a regular basis and I would like to be as clear as possible. Is there any tool that could generate a sort of organized map of my work (namespaces, classes, interactions, etc.) ?

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• Web Application Project Deployment VS2010 - Precompile Views

  - by Malcolm Frexner

  Using Visual Studio 2010 Ultimate I created a ASP.NET MVC 2.0 Web Application. I read http://msdn.microsoft.com/query/dev10.query?appId=Dev10IDEF1&l=EN-US&k=k%28WEBAPPLICATIONPROJECTS.PACKAGEPUBLISHOVERVIEW%29;k%28TargetFrameworkMoniker-%22.NETFRAMEWORK%2cVERSION%3dV4.0%22%29&rd=true. Its about the new features for Web Application Deployment. I dont see an option to precompile Views. Also I dont see other options that where available in previous version of Web Deployment Project: ie if the web is compiled into a single assembly or into one assembly per page. I had the impression that Application Project Deployment is the scuccessor of Web Deployment Project.. maybe I am wrong about it. How should I precompile views now?

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• Curing the Database-Application mismatch

  - by Phil Factor

  If an application requires access to a database, then you have to be able to deploy it so as to be version-compatible with the database, in phase. If you can deploy both together, then the application and database must normally be deployed at the same version in which they, together, passed integration and functional testing.  When a single database supports more than one application, then the problem gets more interesting. I’ll need to be more precise here. It is actually the application-interface definition of the database that needs to be in a compatible ‘version’.  Most databases that get into production have no separate application-interface; in other words they are ‘close-coupled’.  For this vast majority, the whole database is the application-interface, and applications are free to wander through the bowels of the database scot-free.  If you’ve spurned the perceived wisdom of application architects to have a defined application-interface within the database that is based on views and stored procedures, any version-mismatch will be as sensitive as a kitten.  A team that creates an application that makes direct access to base tables in a database will have to put a lot of energy into keeping Database and Application in sync, to say nothing of having to tackle issues such as security and audit. It is not the obvious route to development nirvana. I’ve been in countless tense meetings with application developers who initially bridle instinctively at the apparent restrictions of being ‘banned’ from the base tables or routines of a database.  There is no good technical reason for needing that sort of access that I’ve ever come across.  Everything that the application wants can be delivered via a set of views and procedures, and with far less pain for all concerned: This is the application-interface.  If more than zero developers are creating a database-driven application, then the project will benefit from the loose-coupling that an application interface brings. What is important here is that the database development role is separated from the application development role, even if it is the same developer performing both roles. The idea of an application-interface with a database is as old as I can remember. The big corporate or government databases generally supported several applications, and there was little option. When a new application wanted access to an existing corporate database, the developers, and myself as technical architect, would have to meet with hatchet-faced DBAs and production staff to work out an interface. Sure, they would talk up the effort involved for budgetary reasons, but it was routine work, because it decoupled the database from its supporting applications. We’d be given our own stored procedures. One of them, I still remember, had ninety-two parameters. All database access was encapsulated in one application-module. If you have a stable defined application-interface with the database (Yes, one for each application usually) you need to keep the external definitions of the components of this interface in version control, linked with the application source,  and carefully track and negotiate any changes between database developers and application developers.  Essentially, the application development team owns the interface definition, and the onus is on the Database developers to implement it and maintain it, in conformance.  Internally, the database can then make all sorts of changes and refactoring, as long as source control is maintained.  If the application interface passes all the comprehensive integration and functional tests for the particular version they were designed for, nothing is broken. Your performance-testing can ‘hang’ on the same interface, since databases are judged on the performance of the application, not an ‘internal’ database process. The database developers have responsibility for maintaining the application-interface, but not its definition,  as they refactor the database. This is easily tested on a daily basis since the tests are normally automated. In this setting, the deployment can proceed if the more stable application-interface, rather than the continuously-changing database, passes all tests for the version of the application. Normally, if all goes well, a database with a well-designed application interface can evolve gracefully without changing the external appearance of the interface, and this is confirmed by integration tests that check the interface, and which hopefully don’t need to be altered at all often.  If the application is rapidly changing its ‘domain model’  in the light of an increased understanding of the application domain, then it can change the interface definitions and the database developers need only implement the interface rather than refactor the underlying database.  The test team will also have to redo the functional and integration tests which are, of course ‘written to’ the definition.  The Database developers will find it easier if these tests are done before their re-wiring  job to implement the new interface. If, at the other extreme, an application receives no further development work but survives unchanged, the database can continue to change and develop to keep pace with the requirements of the other applications it supports, and needs only to take care that the application interface is never broken. Testing is easy since your automated scripts to test the interface do not need to change. The database developers will, of course, maintain their own source control for the database, and will be likely to maintain versions for all major releases. However, this will not need to be shared with the applications that the database servers. On the other hand, the definition of the application interfaces should be within the application source. Changes in it have to be subject to change-control procedures, as they will require a chain of tests. Once you allow, instead of an application-interface, an intimate relationship between application and database, we are in the realms of impedance mismatch, over and above the obvious security problems.  Part of this impedance problem is a difference in development practices. Whereas the application has to be regularly built and integrated, this isn’t necessarily the case with the database.  An RDBMS is inherently multi-user and self-integrating. If the developers work together on the database, then a subsequent integration of the database on a staging server doesn’t often bring nasty surprises. A separate database-integration process is only needed if the database is deliberately built in a way that mimics the application development process, but which hampers the normal database-development techniques.  This process is like demanding a official walking with a red flag in front of a motor car.  In order to closely coordinate databases with applications, entire databases have to be ‘versioned’, so that an application version can be matched with a database version to produce a working build without errors.  There is no natural process to ‘version’ databases.  Each development project will have to define a system for maintaining the version level. A curious paradox occurs in development when there is no formal application-interface. When the strains and cracks happen, the extra meetings, bureaucracy, and activity required to maintain accurate deployments looks to IT management like work. They see activity, and it looks good. Work means progress.  Management then smile on the design choices made. In IT, good design work doesn’t necessarily look good, and vice versa.

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• Ant: project dependencies in a flat project layout with ivy

  - by MH

  Hello, I have two (Eclipse-) projects. Project A depends on project B, but the projects aren't nested i.e. project A is not a subproject of project B. Apache Ivy is responsible for the dependency management. When I run the compile task in Project A, is there any way to trigger the compile task (in project B) automatically (for example if the jar file of project B doesn't exist)? Thanks a million in advance.

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• Project Euler 53: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 53.  I first attempted to solve this problem using the Ruby combinations libraries. That didn’t work out so well. With a second look at the problem, the provided formula ended up being just the thing to solve the problem effectively. As always, any feedback is welcome. # Euler 53 # http://projecteuler.net/index.php?section=problems&id=53 # There are exactly ten ways of selecting three from five, # 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245, # and 345 # In combinatorics, we use the notation, 5C3 = 10. # In general, # # nCr = n! / r!(n-r)!,where r <= n, # n! = n(n1)...321, and 0! = 1. # # It is not until n = 23, that a value exceeds # one-million: 23C10 = 1144066. # In general: nCr # How many, not necessarily distinct, values of nCr, # for 1 <= n <= 100, are greater than one-million timer_start = Time.now # There's no factorial method in Ruby, I guess. class Integer # http://rosettacode.org/wiki/Factorial#Ruby def factorial (1..self).reduce(1, :*) end end def combinations(n, r) n.factorial / (r.factorial * (n-r).factorial) end answer = 0 100.downto(3) do |c| (2).upto(c-1) { |r| answer += 1 if combinations(c, r) > 1_000_000 } end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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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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• Project Euler 12: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 12.  As always, any feedback is welcome. # Euler 12 # http://projecteuler.net/index.php?section=problems&id=12 # The sequence of triangle numbers is generated by adding # the natural numbers. So the 7th triangle number would be # 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms # would be: # 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... # Let us list the factors of the first seven triangle # numbers: # 1: 1 # 3: 1,3 # 6: 1,2,3,6 # 10: 1,2,5,10 # 15: 1,3,5,15 # 21: 1,3,7,21 # 28: 1,2,4,7,14,28 # We can see that 28 is the first triangle number to have # over five divisors. What is the value of the first # triangle number to have over five hundred divisors? import time start = time.time() from math import sqrt def divisor_count(x): count = 2 # itself and 1 for i in xrange(2, int(sqrt(x)) + 1): if ((x % i) == 0): if (i != sqrt(x)): count += 2 else: count += 1 return count def triangle_generator(): i = 1 while True: yield int(0.5 * i * (i + 1)) i += 1 triangles = triangle_generator() answer = 0 while True: num = triangles.next() if (divisor_count(num) >= 501): answer = num break; print answer print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 19: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 19.  As always, any feedback is welcome. # Euler 19 # http://projecteuler.net/index.php?section=problems&id=19 # You are given the following information, but you may # prefer to do some research for yourself. # # - 1 Jan 1900 was a Monday. # - Thirty days has September, # April, June and November. # All the rest have thirty-one, # Saving February alone, # Which has twenty-eight, rain or shine. # And on leap years, twenty-nine. # - A leap year occurs on any year evenly divisible by 4, # but not on a century unless it is divisible by 400. # # How many Sundays fell on the first of the month during # the twentieth century (1 Jan 1901 to 31 Dec 2000)? import time start = time.time() import datetime sundays = 0 for y in range(1901,2001): for m in range(1,13): # monday == 0, sunday == 6 if datetime.datetime(y,m,1).weekday() == 6: sundays += 1 print sundays print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Codenames - Yea or Nay?

  - by rmx

  Where I work, most of our projects have (or at least attempt) descriptive, useful names. However we have a few with names that make no sense: I found that an assembly named WiFi which actually has nothing whatsoever to do with wi-fi, but is a codename. When I asked why, I was told that it's to protect company secrets incase some intern has few too many at the pub on Friday and starts chatting about the brand new 'WiFi' project he's been working on. Its clear that some people find enjoyment in finding silly / amusing codenames for their projects (like in this question). My question is: is it really a good idea to use codenames for your projects or are you better off spending the time to decide upon a descriptive name? My opinion is that in the long-run its better to give your projects relevant names. My reasoning is that if you can't think of a decent name, perhaps you don't really know the requirements well enough. I think there are better ways to 'protect company secrets' and I find it quite confusing when the name does not correlate at all with the content. It's just common sense, surely?! So do you use codenames and what the your reasons for or against this seemingly common, yet annoying (to me at least) practice?

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• A project idea I've got...

  - by Mr Teeth

  Hi, Next year I will be doing a final year project at Uni. I've already thought of one and was wondering what you guys think of it. I want to create a University Information Search for prospective students who are trying to look for an affordable University to attend. It will depend on the student's family income and the grades they get. They enter in those two parameters (and some more) and it comes up with a list of suitable Unis based on their criteria. This is not about the price of tution fee. It's mostly to do with the cost of living. Stuff like: Rent (if living in a private flat). Student Accomadation. Cost of traveling to your home and back (for holidays). ...and some other stuff stuff I haven't thought about yet. It'll mostly be GUI driven with some textual information. I'm also thinking of using it as a website interface. What do you guys think? Can I program something like this Java? If there's any holes you see in my idea please tell me.

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• How can you force the start and end date of a task in Microsoft Project to be on the same day?

  - by Hauke P.

  I have a task called "Interview person A about topic X". The task's duration is set to 2 hours. The start date of the task should automatically be calculated taking dependencies and resource availabilities into account. My question boils down to: How can I force this task to start and end on the same date? Background: In my case, Microsoft Project sets the start date to a Friday at 5pm. As my working hours are set to 8am to 12am and 1pm to 6pm (Mon-Fri), Microsoft Project "splits up" the task at 6pm on Friday and plans to continue it at 8am on the following Monday. However, it does not make any sense to stop the interview on a Friday and restart it on Monday. Therefore the automatic suggestion is not helpful in this case. That's why I'm looking for a way way to force the task to start and end on the very same day. (In my example, I'd like Microsoft Project to delay the start date of the task until Monday 8am as this is the first time slot in which the task "fits in completely".) By the way: I have lots of such cases... for that reason it would be really great if there was a solution that doesn't just deal with this single special case.)

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• How to force the start and end date of a task in Microsoft Project to be on the same day?

  - by Hauke P.

  I have a task called "Interview person A about topic X". The task's duration is set to 2 hours. The start date of the task should automatically be calculated taking dependencies and resource availabilities into account. My question boils down to: How can I force this task to start and end on the same date? Background: In my case, Microsoft Project sets the start date to a Friday at 5pm. As my working hours are set to 8am to 12am and 1pm to 6pm (Mon-Fri), Microsoft Project "splits up" the task at 6pm on Friday and plans to continue it at 8am on the following Monday. However, it does not make any sense to stop the interview on a Friday and restart it on Monday. Therefore the automatic suggestion is not helpful in this case. That's why I'm looking for a way way to force the task to start and end on the very same day. (In my example, I'd like Microsoft Project to delay the start date of the task until Monday 8am as this is the first time slot in which the task "fits in completely".) By the way: I have lots of such cases... for that reason it would be really great if there was a solution that doesn't just deal with this single special case.

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• New project created with Flex Mojo's archetype throws Cannot Find Parent Project-Maven Exception

  - by ignorant

  This is probably a silly question but I just cant seem to figure out. I'm completely new to flex and maven. Maven 2.2.1: Maven 2.2.1 unzipped,M2_HOME set and repository altered to point to different drive location in settings.xml Flex 4.0: Installed Created a multi-modular webapp project using flexmojo: mvn archetype:generate -DarchetypeRepository=http://repository.sonatype.org/content/groups/flexgroup -DarchetypeGroupId=org.sonatype.flexmojos -DarchetypeArtifactId=flexmojos-archetypes-modular-webapp -DarchetypeVersion=RELEASE with following options groupId=com.test artifactId=test version=1.0-snapshot package=com.tests * Creates * test |-- pom.xml |--swc -pom.xml |--swf -pom.xml `--war -pom.xml Parent pom has swc, swf, war as modules. Dependency is war-swf-swc. With parent artifactId of swf, swc, war set to swf, swc, test respectively. On executing mvn on test folder(for that matter clean or anything) I get this following error. G:\Projects\testmvn -e + Error stacktraces are turned on. [INFO] Scanning for projects... Downloading: http://repo1.maven.org/maven2/com/test/swc/1.0-snapshot/swc-1.0-snapshot.pom [INFO] Unable to find resource 'com.test:swc:pom:1.0-snapshot' in repository central (http://repo1.maven.org/maven2) [INFO] ------------------------------------------------------------------------ [ERROR] FATAL ERROR [INFO] ------------------------------------------------------------------------ [INFO] Failed to resolve artifact. GroupId: com.test ArtifactId: swc Version: 1.0-snapshot Reason: Unable to download the artifact from any repository com.test:swc:pom:1.0-snapshot from the specified remote repositories: central (http://repo1.maven.org/maven2) [INFO] ------------------------------------------------------------------------ [INFO] Trace org.apache.maven.reactor.MavenExecutionException: Cannot find parent: com.test:swc for project: com.test:swc-swc:swc:1.0-snapshot for project com.test:swc-swc:swc:1.0-snapshot at org.apache.maven.DefaultMaven.getProjects(DefaultMaven.java:404) at org.apache.maven.DefaultMaven.doExecute(DefaultMaven.java:272) at org.apache.maven.DefaultMaven.execute(DefaultMaven.java:138) at org.apache.maven.cli.MavenCli.main(MavenCli.java:362) at org.apache.maven.cli.compat.CompatibleMain.main(CompatibleMain.java:60) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25) at java.lang.reflect.Method.invoke(Method.java:585) at org.codehaus.classworlds.Launcher.launchEnhanced(Launcher.java:315) at org.codehaus.classworlds.Launcher.launch(Launcher.java:255) at org.codehaus.classworlds.Launcher.mainWithExitCode(Launcher.java:430) at org.codehaus.classworlds.Launcher.main(Launcher.java:375) Caused by: org.apache.maven.project.ProjectBuildingException: Cannot find parent: com.test:swc for project: com.test:swc-swc:swc:1.0-snapshot for project com.test:swc-swc:swc:1.0-snapshot at org.apache.maven.project.DefaultMavenProjectBuilder.assembleLineage(DefaultMavenProjectBuilder.java:1396) at org.apache.maven.project.DefaultMavenProjectBuilder.buildInternal(DefaultMavenProjectBuilder.java:823) at org.apache.maven.project.DefaultMavenProjectBuilder.buildFromSourceFileInternal(DefaultMavenProjectBuilder.java:508) at org.apache.maven.project.DefaultMavenProjectBuilder.build(DefaultMavenProjectBuilder.java:200) at org.apache.maven.DefaultMaven.getProject(DefaultMaven.java:604) at org.apache.maven.DefaultMaven.collectProjects(DefaultMaven.java:487) at org.apache.maven.DefaultMaven.collectProjects(DefaultMaven.java:560) at org.apache.maven.DefaultMaven.getProjects(DefaultMaven.java:391) ... 12 more Caused by: org.apache.maven.project.ProjectBuildingException: POM 'com.test:swc' not found in repository: Unable to download the artifact from any repository com.test:swc:pom:1.0-snapshot from the specified remote repositories: central (http://repo1.maven.org/maven2) for project com.test:swc at org.apache.maven.project.DefaultMavenProjectBuilder.findModelFromRepository(DefaultMavenProjectBuilder.java:605) at org.apache.maven.project.DefaultMavenProjectBuilder.assembleLineage(DefaultMavenProjectBuilder.java:1392) ... 19 more Caused by: org.apache.maven.artifact.resolver.ArtifactNotFoundException: Unable to download the artifact from any repository com.test:swc:pom:1.0-snapshot from the specified remote repositories: central (http://repo1.maven.org/maven2) at org.apache.maven.artifact.resolver.DefaultArtifactResolver.resolve(DefaultArtifactResolver.java:228) at org.apache.maven.artifact.resolver.DefaultArtifactResolver.resolve(DefaultArtifactResolver.java:90) at org.apache.maven.project.DefaultMavenProjectBuilder.findModelFromRepository(DefaultMavenProjectBuilder.java:558) ... 20 more Caused by: org.apache.maven.wagon.ResourceDoesNotExistException: Unable to download the artifact from any repository at org.apache.maven.artifact.manager.DefaultWagonManager.getArtifact(DefaultWagonManager.java:404) at org.apache.maven.artifact.resolver.DefaultArtifactResolver.resolve(DefaultArtifactResolver.java:216) ... 22 more [INFO] ------------------------------------------------------------------------ [INFO] Total time: 1 second [INFO] Finished at: Tue Jun 15 19:22:15 GMT+02:00 2010 [INFO] Final Memory: 1M/2M [INFO] ------------------------------------------------------------------------ Looks like its trying to download the project from maven's central repository instead of building it. What am I missing?

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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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• Project Euler 11: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 11.  As always, any feedback is welcome. # Euler 11 # http://projecteuler.net/index.php?section=problems&id=11 # What is the greatest product # of four adjacent numbers in any direction (up, down, left, # right, or diagonally) in the 20 x 20 grid? import time start = time.time() grid = [\ [8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8],\ [49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00],\ [81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65],\ [52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91],\ [22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],\ [24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50],\ [32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],\ [67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],\ [24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],\ [21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95],\ [78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92],\ [16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57],\ [86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58],\ [19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40],\ [04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66],\ [88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69],\ [04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],\ [20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16],\ [20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54],\ [01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48]] # left and right max, product = 0, 0 for x in range(0,17): for y in xrange(0,20): product = grid[y][x] * grid[y][x+1] * \ grid[y][x+2] * grid[y][x+3] if product > max : max = product # up and down for x in range(0,20): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x] * \ grid[y+2][x] * grid[y+3][x] if product > max : max = product # diagonal right for x in range(0,17): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x+1] * \ grid[y+2][x+2] * grid[y+3][x+3] if product > max: max = product # diagonal left for x in range(0,17): for y in xrange(0,17): product = grid[y][x+3] * grid[y+1][x+2] * \ grid[y+2][x+1] * grid[y+3][x] if product > max : max = product print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Help in (re)designing my Swing application

  - by Harihar Das

  I have developed a Swing application that controls execution of several script like jobs. I need to display the interim output of the jobs concurrently. I have followed MVC while writing the application. The application is working as expected. But off late I have the following requirements in hand: A few of the script jobs need special user privileges to execute so as to access specialized resources. There seems to be now way in Java to impersonate as a different user while running an application.[examined in this question]. Also trying to run the Swing application as a scheduled task in windows is not helping. Once started the jobs should be running even if the user logs off after starting the jobs. I am thinking of separating the execution logic from the UI and run that as a service; and introduce JMS in between the two layers so as to store/retrieve the interim the output. Note: I need to run this application on windows Any ideas on meeting my requirements will be highly appreciated.

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