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• Project Management, Developer being project managers manager

  - by marabutt

  I am in the planning stages of a project and am looking to hire a project manager. I want be doing some coding and keeping an eye on all parts of the project but feel a project manager will get better results than I could. I can project manage the project and not code and hire another coder or code myself and hire a project manager. I am worried that the project manager will fell impeded by having the project owner as part of the development team. If I run the project, the team might fall apart causing the project to fail. To stick within budget, I have to be involved in one capacity or another. Does anyone have experience with this situation or suggestions?

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• Converting a Visual Studio 2003 Web Project to a Visual Studio 2008 Web Application Project

  - by navaneeth

  This walkthrough describes how to convert a Visual Studio .NET 2002 or Visual Studio .NET 2003 Web project to a Visual Studio 2008 Web application project. The Visual Studio 2008 Web application project model is like the Visual Studio 2005 Web application project model. Therefore, the conversion processes are similar. For more information about Web application projects, see ASP.NET Web Application Projects. You can also convert from a Visual Studio .NET Web project to a Visual Studio 2008 Web site project. However, conversion to a Web application project is the approach that is supported, and gives you the convenience of tools to help with the conversion. For example, when you convert to a Visual Studio 2008 Web application project, you can use the Visual Studio Conversion Wizard to automate part of the process. For information about how to convert a Visual Studio .NET Web project to a Visual Studio 2008 Web site, see Common Web Project Conversion Issues and Solutions. There are two parts involved in converting a Visual Studio 2002 or 2003 Web project to a Visual Studio 2008 Web application project. The parts are as follows: Converting the project. You can use the Visual Studio Conversion Wizard for the initial conversion of the project and Web.config files. You can later use the Convert To Web Application command to update the project's files and structure. Upgrading the .NET Framework version of the project. You must upgrade the project's .NET Framework version to either .NET Framework 2.0 SP1 or to .NET Framework 3.5. This .NET Framework version upgrade is required because Visual Studio 2008 cannot target earlier versions of the .NET Framework. You can perform this upgrade during the project conversion, by using the Conversion Wizard. Alternatively, you can upgrade the .NET Framework version after you convert the project.   NoteYou can change a project's .NET Framework version manually. To do so, in Visual Studio open the property pages for the project, click the Application tab, and then select a new version from the Target Framework list. This walkthrough illustrates the following tasks: Opening the Visual Studio .NET project in Visual Studio 2008 and creating a backup of the project files. Upgrading the .NET Framework version that the project targets. Converting the project file and the Web.config file. Converting ASP.NET code files. Testing the converted project. Prerequisites    To complete this walkthrough, you will need: Visual Studio 2008. A Web site project that was created in Visual Studio .NET version 2002 or 2003 that compiles and runs without errors. Converting the Project and Upgrading the .NET Framework Version    To begin, you open the project in Visual Studio 2008, which starts the conversion. It offers you an opportunity to back up the project before converting it. NoteIt is strongly recommended that you back up the project. The conversion works on the original project files, which cannot be recovered if the conversion is not successful.To convert the project and back up the files In Visual Studio 2008, in the File menu, click Open and then click Project. The Open Project dialog box is displayed. Browse to the folder that contains the project or solution file for the Visual Studio .NET project, select the file, and then click Open. NoteMake sure that you open the project by using the Open Project command. If you use the Open Web Site command, the project will be converted to the Web site project format.The Conversion Wizard opens and prompts you to create a backup before converting the project. To create the backup, click Yes. Click Browse, select the folder in which the backup should be created, and then click Next. Click Finish. The backup starts. NoteThere might be significant delays as the Conversion Wizard copies files, with no updates or progress indicated. Wait until the process finishes before you continue.When the conversion finishes, the wizard prompts you to upgrade the targeted version of the .NET Framework for the project. To upgrade to the .NET Framework 3.5, click Yes. To upgrade the project to target the .NET Framework 2.0 SP1, click No. It is recommended that you leave the check box selected that asks whether you want to upgrade all Webs in the solution. If you upgrade to .NET Framework 3.5, the project's Web.config file is modified at the same time as the project file. When the upgrade and conversion have finished, a message is displayed that indicates that you have completed the first step in converting your project. Click OK. The wizard displays status information about the conversion. Click Close. Testing the Converted Project    After the conversion has finished, you can test the project to make sure that it runs. This will also help you identify code in the project that must be updated. To verify that the project runs If you know about changes that are required for the code to run with the new version of the .NET Framework, make those changes. In the Build menu, click Build. Any missing references or other compilation issues in the project are displayed in the Error List window. The most likely issues are missing assembly references or issues with dynamically generated types. In Solution Explorer, right-click the Web page that will be used to launch the application, and then click Set as Start Page. On the Debug menu, click Start Debugging. If debugging is not enabled, the Debugging Not Enabled dialog box is displayed. Select the option to add a Web.config file that has debugging enabled, and then click OK. Verify that the converted project runs as expected. Do not continue with the conversion process until all build and run-time errors are resolved. Converting ASP.NET Code Files    ASP.NET Web page files and user-control files in Visual Studio 2008 that use the code-behind model have an associated designer file. The files that you just converted will have an associated code-behind file, but no designer file. Therefore, the next step is to generate designer files. NoteOnly ASP.NET Web pages and user controls that have their code in a separate code file require a separate designer file. For pages that have inline code and no associated code file, no designer file will be generated.To convert ASP.NET code files In Solution Explorer, right-click the project node, and then click Convert To Web Application. The files are converted. Verify that the converted code files have a code file and a designer file. Build and run the project to verify the results of the conversion.

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• Does it work when a developer is the project manager's boss?

  - by marabutt

  I am in the planning stage of a project and I am looking to hire a project manager. I would like to do some coding and keep eye on all parts of the project. However, i have a feeling that a project manager will get better results. I have the following options: 1) manage the project and not code 2) hire a project manager and code myself I am worried that the project manager will feel impeded by having the project owner in the development team. If I run the project, the team might fall apart causing the project to fail. To stick within budget, I have to be involved in one capacity or another. Does anyone have experience with this situation, any suggestions? more info: 4 in-house developers each responsible for a specific area. The developers can also outsource work if agreed to by the project manager.

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• Project Euler 51: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 51.  I know I started back up with Python this week, but I have three more Ruby solutions in the hopper and I wanted to share. For the record, Project Euler 51 was the second hardest Euler problem for me thus far. Yeah. As always, any feedback is welcome. # Euler 51 # http://projecteuler.net/index.php?section=problems&id=51 # By replacing the 1st digit of *3, it turns out that six # of the nine possible values: 13, 23, 43, 53, 73, and 83, # are all prime. # # By replacing the 3rd and 4th digits of 56**3 with the # same digit, this 5-digit number is the first example # having seven primes among the ten generated numbers, # yielding the family: 56003, 56113, 56333, 56443, # 56663, 56773, and 56993. Consequently 56003, being the # first member of this family, is the smallest prime with # this property. # # Find the smallest prime which, by replacing part of the # number (not necessarily adjacent digits) with the same # digit, is part of an eight prime value family. timer_start = Time.now require 'mathn' def eight_prime_family(prime) 0.upto(9) do |repeating_number| # Assume mask of 3 or more repeating numbers if prime.count(repeating_number.to_s) >= 3 ctr = 1 (repeating_number + 1).upto(9) do |replacement_number| family_candidate = prime.gsub(repeating_number.to_s, replacement_number.to_s) ctr += 1 if (family_candidate.to_i).prime? end return true if ctr >= 8 end end false end # Wanted to loop through primes using Prime.each # but it took too long to get to the starting value. n = 9999 while n += 2 next if !n.prime? break if eight_prime_family(n.to_s) end puts n puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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• Project Euler 10: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 10.  As always, any feedback is welcome. # Euler 10 # http://projecteuler.net/index.php?section=problems&id=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. import time start = time.time() def primes_to_max(max): primes, number = [2], 3 while number < max: 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 primes = primes_to_max(2000000) print sum(primes) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 15: (Iron)Python

  - by Ben Griswold

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

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• Project Euler 9: (Iron)Python

  - by Ben Griswold

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

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• Project Euler 5: (Iron)Python

  - by Ben Griswold

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

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• Project Euler 8: (Iron)Python

  - by Ben Griswold

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

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• 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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• 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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• 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 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 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 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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• 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 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 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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