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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 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 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 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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• MS-DOSdivide overflow

  - by repozitor

  when i want to install my dear application on MS-DOS os, i see error "Divide Overflow" what is the meaning of this error and how to fix it? the procedure of installing is: 1-partitioing my HDD disk 2-format C drive 3-installing MS-DOS 4-add the flowing lines to config.sys "DEVICE=C:\DOS\HIMEM.SYS DEVICE=C:\DOS\EMM386.EXE RAM DEVICE=C:\DOS\RAMDRIVE.SYS 6000 512 64 /e" 5-insert my floppy application and then restart 6-when boot process completed, all of the thing appear correctly and i now how to do the next steps note:for installing my application i need to boot from floppy app when i have ms-dos on my hdd disk i do it successfully on the Virtual machine, Q emulator but i can't do it on the real machine, Vectra HP PC, because of divide overflow

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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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• Connect Access 2007 to SQL Server 2008 Database

  - by Peter

  Hi, I've seen numerous answers to similar questions like this one. I haven't seen on the web many people have asked the seemingly simple question "How do I connect Access 2007 to an SQL server 2008 database" - but all of the answers describe how you can migrate from access 2007 to an sql server 2008 database, or they describe how to connect access 2007 to an sql server 2005 database. I can't find any simple solution to my problem (and probably this is a problem for many others). Here is the question (sorry for the over emphasis): How do I connect to an sql server 2008 database (and I mean 2008, not 2005 :) ) from access 2007? Apologies again for the over emphasis, but this very simple question, and what I thought should be a very simple task seems, yes, ... impossible! I tried running sql server browser, enabling pipes, TCP etc, but it seems that with 2008 SQLEXPRESS just isn't recognised! Please can someone help with this. Peter

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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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• Project Server 2007 Task Updates hangs on 'Loading Grid...'

  - by entens

  A strange problem began occurring after applying MOSS2 (KB953334) and the August 2009 cumulative update to our Project server. When a user enters the 'Task Update' screen they are prompted to download a new ActiveX control. Upon refresh, and subsequent access attempts, the user is presented with a blank grid with the caption 'Loading Grid...' We have attempted to fix this issue by updating the 'Trusted Sites' list and changing the security settings according to KB818046. However, nothing seems to definitely fix the problem. Also, when the problem randomly fixes itself, it still occurs when viewing specific projects. Any ideas on a fix?

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