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  • Project Euler 52: Ruby

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

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  • How to determine number of resources to be allocated in a software project

    - by aditi
    Last day I have been interviewed and the interviwer asked me as given the outline of a project, how can we determine the number of resources to be needed for the same? I donot know to do do so? Is there any standard way of doing so? or is it based on the experience? or how.... I am pretty new in this activity and my knowledge is zero at present .... so any clear explanation with some example(simple) will help me(and people like me) to understand this. Thanks

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  • Project Euler 2: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 2.  As always, any feedback is welcome. # Euler 2 # http://projecteuler.net/index.php?section=problems&id=2 # Find the sum of all the even-valued terms in the # Fibonacci sequence which do not exceed four million. # Each new term in the Fibonacci sequence is generated # by adding the previous two terms. By starting with 1 # and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # Find the sum of all the even-valued terms in the # sequence which do not exceed four million. import time start = time.time() total = 0 previous = 0 i = 1 while i <= 4000000: if i % 2 == 0: total +=i # variable swapping removes the need for a temp variable i, previous = previous, previous + i print total print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 16: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 16.  As always, any feedback is welcome. # Euler 16 # http://projecteuler.net/index.php?section=problems&id=16 # 2^15 = 32768 and the sum of its digits is # 3 + 2 + 7 + 6 + 8 = 26. # What is the sum of the digits of the number 2^1000? import time start = time.time() print sum([int(i) for i in str(2**1000)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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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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  • Can't open DDMS

    - by Emerald214
    When I open a standalone DDMS besides Eclipse, it generates the below error. hieund@hieund:~$ ddms 01:51:58 E/ddms: Could not open Selected VM debug port (8700). Make sure you do not have another instance of DDMS or of the eclipse plugin running. If it's being used by something else, choose a new port number in the preferences. (DDMS:6904): Gtk-WARNING **: gtk_widget_size_allocate(): attempt to allocate widget with width -5 and height 17 (DDMS:6904): Gtk-WARNING **: gtk_widget_size_allocate(): attempt to allocate widget with width -5 and height 17 01:52:18 E/DDMS: device offline com.android.ddmlib.AdbCommandRejectedException: device offline at com.android.ddmlib.AdbHelper.setDevice(AdbHelper.java:736) at com.android.ddmlib.AdbHelper.executeRemoteCommand(AdbHelper.java:373) at com.android.ddmlib.Device.executeShellCommand(Device.java:364) at com.android.ddmuilib.SysinfoPanel.loadFromDevice(SysinfoPanel.java:159) at com.android.ddmuilib.SysinfoPanel.deviceSelected(SysinfoPanel.java:126) at com.android.ddmuilib.SelectionDependentPanel.deviceSelected(SelectionDependentPanel.java:52) at com.android.ddms.UIThread.selectionChanged(UIThread.java:1721) at com.android.ddmuilib.DevicePanel.notifyListeners(DevicePanel.java:752) at com.android.ddmuilib.DevicePanel.notifyListeners(DevicePanel.java:740) at com.android.ddmuilib.DevicePanel.access$1100(DevicePanel.java:56) at com.android.ddmuilib.DevicePanel$1.widgetSelected(DevicePanel.java:357) at org.eclipse.swt.widgets.TypedListener.handleEvent(Unknown Source) at org.eclipse.swt.widgets.EventTable.sendEvent(Unknown Source) at org.eclipse.swt.widgets.Widget.sendEvent(Unknown Source) at org.eclipse.swt.widgets.Display.runDeferredEvents(Unknown Source) at org.eclipse.swt.widgets.Display.readAndDispatch(Unknown Source) at com.android.ddms.UIThread.runUI(UIThread.java:517) at com.android.ddms.Main.main(Main.java:116) 01:52:32 E/ddms: shutting down due to uncaught exception 01:52:32 E/ddms: Failed to execute runnable (java.lang.ArrayIndexOutOfBoundsException: -1) org.eclipse.swt.SWTException: Failed to execute runnable (java.lang.ArrayIndexOutOfBoundsException: -1) at org.eclipse.swt.SWT.error(Unknown Source) at org.eclipse.swt.SWT.error(Unknown Source) at org.eclipse.swt.widgets.Synchronizer.runAsyncMessages(Unknown Source) at org.eclipse.swt.widgets.Display.runAsyncMessages(Unknown Source) at org.eclipse.swt.widgets.Display.readAndDispatch(Unknown Source) at com.android.ddms.UIThread.runUI(UIThread.java:517) at com.android.ddms.Main.main(Main.java:116) Caused by: java.lang.ArrayIndexOutOfBoundsException: -1 at org.eclipse.jface.viewers.AbstractTableViewer$VirtualManager.resolveElement(AbstractTableViewer.java:100) at org.eclipse.jface.viewers.AbstractTableViewer$1.handleEvent(AbstractTableViewer.java:70) at org.eclipse.swt.widgets.EventTable.sendEvent(Unknown Source) at org.eclipse.swt.widgets.Widget.sendEvent(Unknown Source) at org.eclipse.swt.widgets.Widget.sendEvent(Unknown Source) at org.eclipse.swt.widgets.Widget.sendEvent(Unknown Source) at org.eclipse.swt.widgets.Table.checkData(Unknown Source) at org.eclipse.swt.widgets.Table.cellDataProc(Unknown Source) at org.eclipse.swt.widgets.Display.cellDataProc(Unknown Source) at org.eclipse.swt.internal.gtk.OS._gtk_list_store_append(Native Method) at org.eclipse.swt.internal.gtk.OS.gtk_list_store_append(Unknown Source) at org.eclipse.swt.widgets.Table.setItemCount(Unknown Source) at org.eclipse.jface.viewers.TableViewer.doSetItemCount(TableViewer.java:217) at org.eclipse.jface.viewers.AbstractTableViewer.internalVirtualRefreshAll(AbstractTableViewer.java:661) at org.eclipse.jface.viewers.AbstractTableViewer.internalRefresh(AbstractTableViewer.java:635) at org.eclipse.jface.viewers.AbstractTableViewer.internalRefresh(AbstractTableViewer.java:620) at org.eclipse.jface.viewers.StructuredViewer$7.run(StructuredViewer.java:1430) at org.eclipse.jface.viewers.StructuredViewer.preservingSelection(StructuredViewer.java:1365) at org.eclipse.jface.viewers.StructuredViewer.preservingSelection(StructuredViewer.java:1328) at org.eclipse.jface.viewers.StructuredViewer.refresh(StructuredViewer.java:1428) at org.eclipse.jface.viewers.ColumnViewer.refresh(ColumnViewer.java:537) at org.eclipse.jface.viewers.StructuredViewer.refresh(StructuredViewer.java:1387) at com.android.ddmuilib.logcat.LogCatPanel$LogCatTableRefresherTask.run(LogCatPanel.java:1000) at org.eclipse.swt.widgets.RunnableLock.run(Unknown Source) ... 5 more I tried to change port for DDMS in Eclipse but it still doesn't work.

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  • Open-sourcing a proprietary library without certain features

    - by nha
    I hope I'm in the right place to ask that. I have a question regarding the practice of open-sourcing a proprietary library that we built and use at work. The licence will probably be MIT. I like the idea, but here comes the unusual part : I have been tasked to remove some of the most advanced features. Those will remain on our servers, available as a service. We will open-source the (JavaScript in case it is of interest) library, along with a minimal associated server code. I am not asking a question about the technical problems (I imagine we will have to maintain and synchronize somehow different repositories, maybe with incompatible pull requests, but this for stack overflow). What I would like to know is: How that would be perceived by the community at large ? Does it risk killing the eventual interest in this library? I don't personally know of any library that works like that. I'm pretty sure it is possible however, but any evidence of such a library is welcome (successful if possible). That's also because I'd like to see how they present it. More importantly, what could be the rationale for/against it? I'm not sure I understand the consequences of doing it so.

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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 do you go about checking your open source libraries for keystroke loggers?

    - by asd
    A random person on the internet told me that a technology was secure(1), safe to use and didn't contain keyloggers because it is open source. While I can trivially detect the key stroke logger in this open source application, what can developers(2) do to protect themselves against rouge committers to open source projects? Doing a back of the envelope threat analysis, if I were a rogue developer, I'd fork a branch on git and promote it's download since it would have twitter support (and a secret key stroke logger). If it was an SVN repo, I'd create just create a new project. Even better would be to put the malicious code in the automatic update routines. (1) I won't mention which because I can only deal with one kind of zealot at a time. (2) Ordinary users are at the mercy of their virus and malware detection software-- it's absurd to expect grandma to read the source of code of their open source word processor's source code to find the keystroke logger.

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  • Can I use the test suite from an open source project to verify that my own 'compatible library' is compatible?

    - by Mark Booth
    The question Is it illegal to rewrite every line of an open source project in a slightly different way, and use it in a closed source project? makes me wonder what would be considered a clean-room implementation in the era of open source projects. Hypothetically, if I were to develop a library which duplicates the publicly documented interface of an open-source library, without ever looking at the source code for that library, could that code ever be considered a derivative work? Obviously it would need the same class hierarchy and method signatures, so that it could be a drop-in replacement - could that in itself, be enough to provoke a copyright claim? What about if I used the test suite of the open source project to verify whether my clean implementation behaved in the same way as the original library? Would using the test suite be enough to dirty my clean code? As should be expected from a question like this, I am not looking for specific legal advice, but looking to document experiences people may have had with this sort of issue.

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