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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 Management - Asana / activeCollab / basecamp / alternative / none

    - by rickyduck
    I don't know whether this should be on programmers - I've been looking at the above three apps over the past few weeks just for myself and I'm in two minds. All three look good, are easy to use, and I came to this conclusion; Asana is the easiest to use ActiveCollab is the feature rich and easiest flow BaseCamp is the best UX / design But I didn't really find my workflow was any more quicker / efficient, in fact it was a bit slower and organized. Is there a realistic place for them in workflow - should programmers use them for themselves, or only when a project manager can take control of it?

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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 Kapros: A Custom-Built Workstation Featuring an In-Desk Computer

    - by Jason Fitzpatrick
    While we’ve seen our fair share of case mods, it’s infrequent we see one as polished and built-in as this custom built work station. What started as an IKEA Galant desk, ended as a stunningly executed desk-as-computer build. High gloss paint, sand-blasted plexiglass windows, custom lighting, and some quality hardware all come together in this build to yield a gorgeous setup with plenty of power and style to go around. Hit up the link below for a massive photo album build guide detailing the process from start to finish. Project Kapros: IKEA Galant PC Desk Mod [via Kotaku] How to Stress Test the Hard Drives in Your PC or Server How To Customize Your Android Lock Screen with WidgetLocker The Best Free Portable Apps for Your Flash Drive Toolkit

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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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  • 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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