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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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• 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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• Project structure: where to put business logic

  - by Mister Smith

  First of all, I'm not asking where does business logic belong. This has been asked before and most answers I've read agree in that it belongs in the model: Where to put business logic in MVC design? How much business logic should be allowed to exist in the controller layer? How accurate is "Business logic should be in a service, not in a model"? Why put the business logic in the model? What happens when I have multiple types of storage? However people disagree in the way this logic should be distributed across classes. There seem to exist three major currents of thought: Fat model with business logic inside entity classes. Anemic model and business logic in "Service" classes. It depends. I find all of them problematic. The first option is what most Fowlerites stick to. The problem with a fat model is that sometimes a business logic funtion is not only related to a class, and instead uses a bunch of other classes. If, for example, we are developing a web store, there should be a function that calcs an order's total. We could think of putting this function inside the Order class, but what actually happens is that the logic needs to use different classes, not only data contained in the Order class, but also in the User class, the Session class, and maybe the Tax class, Country class, or Giftcard, Payment, etc. Some of these classes could be composed inside the Order class, but some others not. Sorry if the example is not very good, but I hope you understand what I mean. Putting such a function inside the Order class would break the single responsibility principle, adding unnecesary dependences. The business logic would be scattered across entity classes, making it hard to find. The second option is the one I usually follow, but after many projects I'm still in doubt about how to name the class or classes holding the business logic. In my company we usually develop apps with offline capabilities. The user is able to perform entire transactions offline, so all validation and business rules should be implemented in the client, and then there's usually a background thread that syncs with the server. So we usually have the following classes/packages in every project: Data model (DTOs) Data Access Layer (Persistence) Web Services layer (Usually one class per WS, and one method per WS method). Now for the business logic, what is the standard approach? A single class holding all the logic? Multiple classes? (if so, what criteria is used to distribute the logic across them?). And how should we name them? FooManager? FooService? (I know the last one is common, but in our case it is bad naming because the WS layer usually has classes named FooWebService). The third option is probably the right one, but it is also devoid of any useful info. To sum up: I don't like the first approach, but I accept that I might have been unable to fully understand the Zen of it. So if you advocate for fat models as the only and universal solution you are welcome to post links explaining how to do it the right way. I'd like to know what is the standard design and naming conventions for the second approach in OO languages. Class names and package structure, in particular. It would also be helpful too if you could include links to Open Source projects showing how it is done. Thanks in advance.

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• Project Euler 17: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 17.  As always, any feedback is welcome. # Euler 17 # http://projecteuler.net/index.php?section=problems&id=17 # If the numbers 1 to 5 are written out in words: # one, two, three, four, five, then there are # 3 + 3 + 5 + 4 + 4 = 19 letters used in total. # If all the numbers from 1 to 1000 (one thousand) # inclusive were written out in words, how many letters # would be used? # # NOTE: Do not count spaces or hyphens. For example, 342 # (three hundred and forty-two) contains 23 letters and # 115 (one hundred and fifteen) contains 20 letters. The # use of "and" when writing out numbers is in compliance # with British usage. import time start = time.time() def to_word(n): h = { 1 : "one", 2 : "two", 3 : "three", 4 : "four", 5 : "five", 6 : "six", 7 : "seven", 8 : "eight", 9 : "nine", 10 : "ten", 11 : "eleven", 12 : "twelve", 13 : "thirteen", 14 : "fourteen", 15 : "fifteen", 16 : "sixteen", 17 : "seventeen", 18 : "eighteen", 19 : "nineteen", 20 : "twenty", 30 : "thirty", 40 : "forty", 50 : "fifty", 60 : "sixty", 70 : "seventy", 80 : "eighty", 90 : "ninety", 100 : "hundred", 1000 : "thousand" } word = "" # Reverse the numbers so position (ones, tens, # hundreds,...) can be easily determined a = [int(x) for x in str(n)[::-1]] # Thousands position if (len(a) == 4 and a[3] != 0): # This can only be one thousand based # on the problem/method constraints word = h[a[3]] + " thousand " # Hundreds position if (len(a) >= 3 and a[2] != 0): word += h[a[2]] + " hundred" # Add "and" string if the tens or ones # position is occupied with a non-zero value. # Note: routine is broken up this way for [my] clarity. if (len(a) >= 2 and a[1] != 0): # catch 10 - 99 word += " and" elif len(a) >= 1 and a[0] != 0: # catch 1 - 9 word += " and" # Tens and ones position tens_position_value = 99 if (len(a) >= 2 and a[1] != 0): # Calculate the tens position value per the # first and second element in array # e.g. (8 * 10) + 1 = 81 tens_position_value = int(a[1]) * 10 + a[0] if tens_position_value <= 20: # If the tens position value is 20 or less # there's an entry in the hash. Use it and there's # no need to consider the ones position word += " " + h[tens_position_value] else: # Determine the tens position word by # dividing by 10 first. E.g. 8 * 10 = h[80] # We will pick up the ones position word later in # the next part of the routine word += " " + h[(a[1] * 10)] if (len(a) >= 1 and a[0] != 0 and tens_position_value > 20): # Deal with ones position where tens position is # greater than 20 or we have a single digit number word += " " + h[a[0]] # Trim the empty spaces off both ends of the string return word.replace(" ","") def to_word_length(n): return len(to_word(n)) print sum([to_word_length(i) for i in xrange(1,1001)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project euler problem 3 in haskell

  - by shk

  I'm new in Haskell and try to solve 3 problem from http://projecteuler.net/. The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? My solution: import Data.List getD :: Int -> Int getD x = -- find deviders let deriveList = filter (\y -> (x `mod` y) == 0) [1 .. x] filteredList = filter isSimpleNumber deriveList in maximum filteredList -- Check is nmber simple isSimpleNumber :: Int -> Bool isSimpleNumber x = let deriveList = map (\y -> (x `mod` y)) [1 .. x] filterLength = length ( filter (\z -> z == 0) deriveList) in case filterLength of 2 -> True _ -> False I try to run for example: getD 13195 > 29 But when i try: getD 600851475143 I get error Exception: Prelude.maximum: empty list Why? Thank you @Barry Brown, I think i must use: getD :: Integer -> Integer But i get error: Couldn't match expected type `Int' with actual type `Integer' Expected type: [Int] Actual type: [Integer] In the second argument of `filter', namely `deriveList' In the expression: filter isSimpleNumber deriveList Thank you.

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• What to do when opensource project starts to tear apart? (or a manager tries to write code and than shouts at the team)

  - by Kabumbus

  Imagine there is an open source cross-platform project on Google code. It has lots of revisions (1000). It concentrates in itself lots technological stuff - rare stuff - it mixes top tech. It contains server, and more than one client. The project was created by a well-connected team of developers (friends) and a manager that was sponsoring project at its start up during its first few months (project now is more than a year old-sponsoring oss project is a big good deal- also gave the idea of project to developers). The project was growing in complexity and effort reqiered to continue development. Once upon a time a manager - team leader started trying to write code (he was a programmer in some other projects - not the best, but he felt like he was one). He started because one of the developers suggested an idea at the team meeting and he felt he just needed to do it on his own. He failed, and he told the dev team about it. The dev team did what he failed to do in a few days. After that, the manager feels that team codes with out him perfectly and gets the job done in short time. He felt sorry and lost and he started to crash like an old bad PC. Firstly, he started to scream (in forms of messages not in voice) he tried to tell developers that what they were doing was a bad, not-needed thing - developers kindly told him that his "beginnings" were not compilable while dev team product worked as needed. He told the developers that all work they do should be firstly discussed with him. Here is the part where we need to mention that all team members are "project owners" and logically have equal rights. The team leader suggested to the developers these options: change their dev process to go through him, or be moved from project owners to contributers. So what are our options as developers? What arguments we can provide to the team leader/manager for him to calm down? Is it possible to save the project or is it better to fork out now? An important issue is that lately we had no active ticket system, and I personally think that this was the reason the mess appeared. So... any ideas?

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• How to import ejb project classes into web project on eclipse?

  - by Jhonnytunes

  I have 3 projects on eclipse, maven ejb project, maven wicket webapp project and a enterprise application project that includes ejb and web I mentioned before, as modules. But my question is, How from my web project can I use classes on the ejb project. Lets say ClassA is on the ejb project and I want to say from web project: ClassA cl = new ClassA. I wont be allowed since web project doesnt have ClassA but ejb project. I want to use ejb project as a Business Logic, JPA part. Also I want to include here some not beans classes that I want to use from the web. Dont know if its possible. The web project goes without Business logic, just the wicket pages calling ejb, and using ther classes. Im new in the EE world. Never made an ear file including a war and ejb jar for deploying it on Glassfish. Thanks in advanced.

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