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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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  • Managing .NET External Dependencies

    - by Ben Griswold
    Noah and I continue our screencast series by sharing our approach to managing external dependencies referenced within a .NET solution.  This is another introductory episode but you might find a hidden gem in the short 4-minute clip.  ELMAH (Error Logging Modules and Handlers) is the external dependencies we are focusing on in the presentation.  If you are not familiar with ELMAH, this episode may be worth your time.   YouTube - Managing .NET External Dependencies This is one of our first screencasts.  If you have feedback, I’d love to hear it.

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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 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 17: (Iron)Python

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

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  • Project Euler 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 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 11: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 11.  As always, any feedback is welcome. # Euler 11 # http://projecteuler.net/index.php?section=problems&id=11 # What is the greatest product # of four adjacent numbers in any direction (up, down, left, # right, or diagonally) in the 20 x 20 grid? import time start = time.time() grid = [\ [8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8],\ [49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00],\ [81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65],\ [52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91],\ [22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],\ [24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50],\ [32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],\ [67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],\ [24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],\ [21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95],\ [78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92],\ [16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57],\ [86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58],\ [19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40],\ [04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66],\ [88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69],\ [04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],\ [20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16],\ [20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54],\ [01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48]] # left and right max, product = 0, 0 for x in range(0,17): for y in xrange(0,20): product = grid[y][x] * grid[y][x+1] * \ grid[y][x+2] * grid[y][x+3] if product > max : max = product # up and down for x in range(0,20): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x] * \ grid[y+2][x] * grid[y+3][x] if product > max : max = product # diagonal right for x in range(0,17): for y in xrange(0,17): product = grid[y][x] * grid[y+1][x+1] * \ grid[y+2][x+2] * grid[y+3][x+3] if product > max: max = product # diagonal left for x in range(0,17): for y in xrange(0,17): product = grid[y][x+3] * grid[y+1][x+2] * \ grid[y+2][x+1] * grid[y+3][x] if product > max : max = product print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project 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 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 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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  • InSync12 and Australia Visits: UX is Global, Regional, Everywhere!

    - by ultan o'broin
    I attended the Australian Oracle User Group (AUSOUG) and Quest International User Group's InSync12 event in Melbourne, Australia: the user group conference for Oracle products in the ANZ region. I demoed Oracle Fusion Applications and then presented how Oracle crafted the world class Fusion Apps user experience (UX). I explained about the Oracle user experience design pattern strategy of uptake for all apps, not just Fusion, and what our UX pattern externalization strategy means for customers, partners, and ADF developers. A great conference, lots of energy, the InSync12 highlights for me were Oracle's Senior Vice President Cliff Godwin’s fast-moving Oracle E-Business Suite (EBS) roadshow with the killer Oracle Endeca user experience uptake, and Oracle ADF product outreachmeister Chris Muir’s (@chriscmuir) session on Oracle ADF Mobile solution and his hands-on mobile app development showing how existing ADF/JDev skills can build a secure, code once-deploy-to-many-device hybrid app solution in minutes. Cliff Godwin shows off the Oracle Endeca integration with Oracle E-Business Suite. Chris Muir talked the talk and then walked the walked with Oracle ADF Mobile. Applications UX was mixing it up with the crowd at InSync12 too, showing off cool mobile UX solutions, gathering data for future innovations, and engaging with EBS, JD Edwards, and PeopleSoft apps customers and partners. User conferences such as InSync12 are an important part of our Oracle Applications UX user-centered design process, giving real apps users the opportunity to make real inputs and a way for us to watch and to listen to their needs and wants and get views on current and emerging UX too. Eric Stilan (@icondaddy) of Applications UX uses an iPad to gather feedback on the latest UX designs from conference attendees. While in Melbourne, I also visited impressive Oracle partner, Callista for a major ADF and UX pow-wow, and was the er, star of a very proactive event hosted by another partner Park Lane Information Technology (coordinated by Bambi Price (@bambiprice) of ODTUG) where I explained what UX is about, and how partner and customers can engage, participate and deploy that Applications UX scientific insight to advantage for their entire business. I also paired up with Oracle Australia in Sydney to visit key customers while there, and back at Oracle in Melbourne I spoke with sales consultants and account managers about regional opportunities and UX strategy, and came away with an understanding of what makes the Oracle market tick in Australia. Mobile worker solution development and user experience is hot news in Australia, and this was a great opportunity to team up with Chris Muir and show how the alignment of the twin stars of UX design patterns and ADF technology enables developers to make great-looking, usable apps that really sparkle. Our UX design patterns--or functional (UI) patterns, to use the developer world language--means that developers now have not only a great tool set to build apps on Oracle ADF/FMW but proven, tested usability solutions to solve common problems they can apply in the IDE too. In all, a whirlwind UX visit, packed with events and delivery opportunities, and all too short a time in the wonderful city of Melbourne. I need to get back there soon! For those who need a reminder, there's a website explaining how to get involved with, and participate in, Applications User Experience (including the Oracle Usability Advisory Board) events and programs. Thank you to AUSOUG, Quest, InSync, Callista, Park Lane IT, everyone at Oracle Australia, Chris Muir, and all the other people who came together to make this a productive visit. Stay tuned for more UX developments and engagements in the region on the Oracle VoX blog and Usable Apps website too!

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  • Simple dynamic memory allocation bug.

    - by M4design
    I'm sure you (pros) can identify the bug's' in my code, I also would appreciate any other comments on my code. BTW, the code crashes after I run it. #include <stdlib.h> #include <stdio.h> #include <stdbool.h> typedef struct { int x; int y; } Location; typedef struct { bool walkable; unsigned char walked; // number of times walked upon } Cell; typedef struct { char name[40]; // Name of maze Cell **grid; // 2D array of cells int rows; // Number of rows int cols; // Number of columns Location entrance; } Maze; Maze *maz_new() { int i = 0; Maze *mazPtr = (Maze *)malloc(sizeof (Maze)); if(!mazPtr) { puts("The memory couldn't be initilised, Press ENTER to exit"); getchar(); exit(-1); } else { // allocating memory for the grid mazPtr->grid = (Cell **) malloc((sizeof (Cell)) * (mazPtr->rows)); for(i = 0; i < mazPtr->rows; i++) mazPtr->grid[i] = (Cell *) malloc((sizeof (Cell)) * (mazPtr->cols)); } return mazPtr; } void maz_delete(Maze *maz) { int i = 0; if (maz != NULL) { for(i = 0; i < maz->rows; i++) free(maz->grid[i]); free(maz->grid); } } int main() { Maze *ptr = maz_new(); maz_delete(ptr); getchar(); return 0; } Thanks in advance.

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  • What's the best Open Php newsletter manager ?

    - by Bilel
    Hi :) I'm looking for a nice newsletter management solution. I tried CCmail a good script but whaere I can't imort usernames !!! I would like to find a system that is able to import Opt-in lists in the following format : John Smith;[email protected];other paramaeters...;[like] ;Male;Age... I will develop my own module if I could find another emailing manager Are you already satisfied with a similar application with a trusted (spam-prevention) emailer ? Thank you :)

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  • Algorithm to compare people names to detect identicalness

    - by Pentium10
    I am working on address book synchronization algorithm. I would like to reuse some code if there exists, but couldn't find one yet. Does someone know about an algorithm that will tell me in numbers/float/procent how much two names are identical. Levenstein distance is not good in this approach, as names and our adddress books are matching the begining of each of the name sections. John Smith should match Smith Jon, Jonathan Smith, Johnny Smith

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  • SQL & Linq To SQL Help

    - by cre-johnny07
    I have a table which is some thing like below.. Date ID 2009-07-01 1 2009-07-01 2 2009-07-01 3 2009-08-01 4 2009-08-01 5 2009-08-01 6 2009-09-01 7 2009-09-01 8 2009-10-01 9 2009-10-01 10 2009-11-01 11 .... Now I need to write a query which will show a output like below. Date Start End 2009-07 1 3 2009-08 4 6 2009-09 7 8 ... How can I do this.. Any help would be highly appreciated Thanking In Advance Johnny

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  • Efficient way to store tuples in the datastore

    - by Drew Sears
    If I have a pair of floats, is it any more efficient (computationally or storage-wise) to store them as a GeoPtProperty than it would be pickle the tuple and store it as a BlobProperty? If GeoPt is doing something more clever to keep multiple values in a single property, can it be leveraged for arbitrary data? Can I store the tuple ("Johnny", 5) in a single entity property in a similarly efficient manner?

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  • Tykie

    - by Brian
    Here’s the obituary my mother wrote for Tykie, I still miss the little guy quite a bit. Anyone who’s interested in further information on hearing dogs should check out the IHDI website. I cannot begin to express how helpful a hearing dog can be for the hearing impaired. If you feel so inclined, please make a donation. In Memoriam, Tykie 1993-2010 The American Legion Post 401, South Wichita, KS, supported one of its members and commander by sponsoring a service dog for him. Unlike most service dogs this one was for the hearing impaired. Both Ocie and Betty Sims had hearing loss – Ocie more than Betty. The Post and Auxilliary had garage sales, auctions and other fund-raising endeavors to get donations for the dog. Betty made Teddy bears with growlers that were auctioned for donations to bring a hearing dog from International Hearing Dog, Henderson, Colorado. Tykie, a small wiry, salt and pepper terrier, arrived September 1, 1994 to begin his work that included attending Post 401 meetings and celebrations as well as raising more money to be donated to IHD to help others have hearing dogs. Tykie was a young dog less than a year old when he came to Wichita. He was always anxious to please and seldom barked, though he did put out a kind of cry when he was giving his urgent announcement that someone was at the door or the telephone was ringing. He also enjoyed chasing squirrels in the backyard garden that Ocie prized. In 1995, Betty almost died of a lung infection. Tykie was at the hospital with Ocie when he could visit. Several weeks after she was able to come home after a miraculous recovery, Tykie and Ocie went to a car show in downtown Wichita. Ocie’s retina tore loose in the only eye he could see out of and he almost blind was in great pain. How Ocie and Tykie got home is still a mystery, but the family legend goes that Tykie added seeing eye dog to his repertoire and helped drive him home. Health problems continued for Ocie and when he was placed in a nursing home, Tykie was moved to be Betty’s hearing dog. No problem for Tykie, he still saw his friends at the post and continued to help with visitors at the door. The night of May 3, 1999, Betty and Tykie were in the bedroom watching TV when Tykie began hitting her with both front paws as he would if something were urgent. She said later she thought he wanted to go out. As she and the dog walked down the hall towards the back of the house, Tykie hit her again with his front paws with such urgency that she fell into a small coat closet. That small 2-by-2 closet became their refuge as that very second the roof of her house went off as the f4 tornado raced through the city. Betty acquired one small wound on her hand from a piece of flying glass as she pulled Tykie into the closet with her. Tykie was a hero that day and a lot of days after. He kept Betty going as she rebuilt her home and after her husband died April 15, 2000. Tykie had to be cared for so she had to take him outside and bring him inside. He attended weddings of grandchildren and funerals of Post friends. When Betty died February 17, 2002 Tykie’s life changed again. IHD gave approval for his transfer and retirement to Betty and Ocie’s grandson, Brian Laird, who has a similar hearing loss to his grandfather. A few days after the funeral Tykie flew to his new home in Rutherford, NJ where he was able to take long walks for a couple of years before moving back to the Kansas City area. He was still full of adventure. He was written up in a book about service dogs and his story of the tornado and his picture appeared. He spent weekends at Brian’s mother’s farm to get muddy and be afraid of cats and chickens. He also took on an odyssey as he slipped from his fenced yard in Lenexa one day and walked more than seven miles in Overland Park traffic before being found by a good Samaritan who called IHD to find out where he belonged. Tykie was deaf for about the last two years of his long life and became blind as well, but he continued to strive to please. Tykie was 16 years and 4 months when he was cremated. His ashes were scattered on the graves of Betty and Ocie Sims at Greenwood Cemetery west of Wichita on the afternoon of March 21, 2010, with about a dozen family and Post 401 members. It is still the rule. Service dogs are the only dogs allowed inside the Post home. Submitted by Linda Laird, daughter of Betty and Ocie and mother of Brian Laird.

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