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• Implement 3270 protocol in Java

  - by G B

  I've got a big problem with IBM HACL for accessing a server which speaks 3270 protocol. The library keeps crashing, and our JNI wrapper is actually a bug-fixing layer for the poorly-implemented and poorly-documented library (and I suspect we have introduced new bugs with it too). Moreover, in our company, everybody knows Java, and could maintain the software if we didn't have the JNI-Layer and the IBM class library. We have to use the C++ class library, because the IBM Java library is unusable: we get every non-printable character translated, and we lose all control characters along the way. Now the question is: can we ditch this library and implement our solution in Java completely (we'd like to avoid using another library from another vendor)? Is the protocol well documented? Is the implementation of 3270-over-ssl really so complex? Thanks.

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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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• ReSharper File Location

  - by Ben Griswold

  By default, the ReSharper cache is stored in the solution folder.  It’s one extra folder and one extra .user file.  It’s no big deal but it does clutter up your solution a bit – especially since the files provide no real value. I prefer to store the ReSharper cache in the system Temp folder.  This setting is available by visiting ReSharper > Options > Environment > General. Just update where you’d like to store the ReSharper cache and you’re good to go.  Note, the .user file continues to linger around the solution folder but at least the _ReSharper.SolutionName folder is moved out of sight.

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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 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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• Language Club

  - by Ben Griswold

  We started a language club at work this week.  Thus far, we have a collective interest in a number of languages: Python, Ruby, F#, Erlang, Objective-C, Scala, Clojure, Haskell and Go. There are more but these 9 received the most votes. During the first few meetings we are going to determine which language we should tackle first. To help make our selection, each member will provide a quick overview of their favored language by answering the following set of questions: Why are you interested in learning “your” language(s). (There’s lots of work, I’m an MS shill, It’s hip and  fun, etc) What type of language is it?  (OO, dynamic, functional, procedural, declarative, etc) What types of problems is your language best suited to solve?  (Algorithms over big data, rapid application development, modeling, merely academic, etc) Can you provide examples of where/how it is being used?  If it isn’t being used, why not?  (Erlang was invented at Ericsson to provide an extremely fault tolerant, concurrent system.) Quick history – Who created/sponsored the language?  When was it created?  Is it currently active? Does the language have hardware support (an attempt was made at one point to create processor instruction sets specific to Prolog), or can it run as an interpreted language inside another language (like Ruby in the JVM)? Are there facilities for programs written in this language to communicate with other languages?  How does this affect its utility? Does the language have a IDE tool support?  (Think Eclipse or Visual Studio) How well is the language supported in terms of books, community and documentation? What’s the number one things which differentiates the language from others?  (i.e. Why is it cool?) How is the language applicability to us as consultants?  What would the impact be of using the language in terms of cost, maintainability, personnel costs, etc.? What’s the number one things which differentiates the language from others?  (i.e. Why is it cool?) This should provide an decent introduction into nearly a dozen languages and give us enough context to decide which single language deserves our undivided attention for the weeks to come.  Stay tuned for the winner…

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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 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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• Hidden features of VBA

  - by guillermooo

  Which features of the VBA language are either poorly documented, or simply not often used?

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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 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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• client problems - misaligned expectations & not following SDLC protocols

  - by louism

  hi guys, im having some serious problems with a client on a project - i could use some advice please the short version i have been working with this client now for almost 6 months without any problems (a classified website project in the range of 500 hours) over the last few days things have drastically deteriorated to the point where ive had to place the project on-hold whilst i work-out what to do (this has pissed the client off even more) to be simplistic, the root cause of the issue is this: the client doesnt read the specs i make for him, i code the feature, he than wants to change things, i tell him its not to the agreed spec and that that change will have to be postponed and possibly charged for, he gets upset and rants saying 'hes paid for the feature' and im not keeping to the agreement (<- misalignment of expectations) i think the root cause of the root cause is my clients failure to take my SDLC protocols seriously. i have a bug tracking system in place which he practically refuses to use (he still emails me bugs), he doesnt seem to care to much for the protocols i use for dealing with scope creep and change control the whole situation came to a head recently where he 'cracked it' (an aussie term for being fed-up). the more terms like 'postponed for post-launch implementation', 'costed feature addition', and 'not to agreed spec' i kept using, the worse it got finally, he began to bully me - basically insisting i shut-up and do the work im being paid for. i wrote a long-winded email explaining how wrong he was on all these different points, and explaining what all the SDLC protocols do to protect the success of the project. than i deleted that email and wrote a new one in the new email, i suggested as a solution i write up a list of grievances we both had. we than review the list and compromise on different points: he gets some things he wants, i get some things i want. sometimes youve got to give ground to get ground his response to this suggestion was flat-out refusal, and a restatement that i should just get on with the work ive been paid to do so there you have the very subjective short version. if you have the time and inclination, the long version may be a little less bias as it has the email communiques between me and my client the long version (with background) the long version works by me showing you the email communiques which lead to the situation coming to a head. so here it is, judge for yourself where the trouble started... 1. client asked me why something was missing from a feature i just uploaded, my response was to show him what was in the spec: it basically said the item he was looking for was never going to be included 2. [clients response...] Memo Louis, We are following your own title fields and keeping a consistent layout. Why the big fuss about not adding "Part". It simply replaces "model" and is consistent with your current title fields. 3. [my response...] hi [client], the 'part' field appeared to me as a redundancy / mistake. i requested clarification but never received any in a timely manner (about 2 weeks ago) the specification for this feature also indicated it wasnt going to be included: RE: "Why the big fuss about not adding "Part" " it may not appear so, but it would actually be a lot of work for me to now add a 'Part' field it could take me up to 15-20 minutes to properly explain why its such a big undertaking to do this, but i would prefer to use that time instead to work on completing your v1.1 features as a simplistic explanation - it connects to the change in paradigm from a 'generic classified ad' model to a 'specific attributes for specific categories' model basically, i am saying it is a big fuss, but i understand that it doesnt look that way - after all, it is just one ity-bitty field :) if you require a fuller explanation, please let me know and i will commit the time needed to write that out also, if you recall when we first started on the project, i said that with the effort/time required for features, you would likely not know off the top of your head. you may think something is really complex, but in reality its quite simple, you might think something is easy - but it could actually be a massive trauma to code (which is the case here with the 'Part' field). if you also recalled, i said the best course of action is to just ask, and i would let you know on a case-by-case basis 4. [email from me to client...] hi [client], the online catalogue page is now up live (see my email from a few days ago for information on how it works) note: the window of opportunity for input/revisions on what data the catalogue stores has now closed (as i have put the code up live now) RE: the UI/layout of the online catalogue page you may still do visual/ui tweaks to the page at the moment (this window for input/revisions will close in a couple of days time) 5. [email from client to me...] *(note: i had put up the feature & asked the client to review it, never heard back from them for a few days)* Memo Louis, Here you go again. CLOSED without a word of input from the customer. I don't think so. I will reply tomorrow regarding the content and functionality we require from this feature. 5. [from me to client...] hi [client]: RE: from my understanding, you are saying that the mini-sale yard control would change itself based on the fact someone was viewing for parts & accessories <- is that correct? this change is outside the scope of the v1.1 mini-spec and therefore will need to wait 'til post launch for costing/implementation 6. [email from client to me...] Memo Louis, Following your v1.1 mini-spec and all your time paid in full for the work selected. We need to make the situation clear. There will be no further items held for post-launch. Do not expect us to pay for any further items other than those we have agreed upon. You have undertaken to complete the Parts and accessories feature as follows. Obviously, as part of this process the "mini search" will be effected, and will require "adaption to make sense". 7. [email from me to client...] hi [client], RE: "There will be no further items held for post-launch. Do not expect us to pay for any further items other than those we have agreed upon." a few points to consider: 1) the specification for the 'parts & accessories' feature was as follows: (i.e. [what] "...we have agreed upon.") 2) you have received the 'parts & accessories' feature free of charge (you have paid $0 for it). ive spent two days coding that feature as a gesture of good will i would request that you please consider these two facts carefully and sincerely 8. [email from client to me...] Memo Louis, I don't see how you are giving us anything for free. From your original fee proposal you have deleted more than 30 hours of included features. Your title "shelved features". Further you have charged us twice by adding back into the site, at an addition cost, some of those "shelved features" features. See v1.1 mini-spec. Did include in your original fee proposal a change request budget but then charge without discussion items included in v1.1 mini-spec. Included a further Features test plan for a regression test, a fee of 10 hours that would not have been required if the "shelved features" were not left out of the agreed fee proposal. I have made every attempt to satisfy your your uneven business sense by offering you everything your heart desired, in the v1.1 mini-spec, to be left once again with your attitude of "its too hard, lets leave it for post launch". I am no longer accepting anything less than what we have contracted you to do. That is clearly defined in v1.1 mini-spec, and you are paid in advance for delivering those items as an acceptable function. a few notes about the above email... i had to cull features from the original spec because it didnt fit into the budget. i explained this to the client at the start of the project (he wanted more features than he had budget hours to do them all) nothing has been charged for twice, i didnt charge the client for culled features. im charging him to now do those culled features the draft version of the project schedule included a change request budget of 10 hours, but i had to remove that to meet the budget (the client may not have been aware of this to be fair to them) what the client refers to as my attitude of 'too hard/leave it for post-launch', i called a change request protocol and a method for keeping scope creep under control 9. [email from me to client...] hi [client], RE: "...all your grievances..." i had originally written out a long email response; it was fantastic, it had all these great points of how 'you were wrong' and 'i was right', you would of loved it (and by 'loved it', i mean it would of just infuriated you more) so, i decided to deleted it start over, for two reasons: 1) a long email is being disrespectful of your time (youre a busy businessman with things to do) 2) whos wrong or right gets us no closer to fixing the problems we are experiencing what i propose is this... i prepare a bullet point list of your grievances and my grievances (yes, im unhappy too about how things are going - and it has little to do with money) i submit this list to you for you to add to as necessary we then both take a good hard look at this list, and we decide which areas we are willing to give ground on as an example, the list may look something like this: "louis, you keep taking away features you said you would do" [your grievance 2] [your grievance 3] [your grievance ...] "[client], i feel you dont properly read the specs i prepare for you..." [my grievance 2] [my grievance 3] [my grievance ...] if you are willing to give this a try, let me know will it work? who knows. but if it doesnt, we can always go back to arguing some more :) obviously, this will only work if you are willing to give it a genuine try, and you can accept that you may have to 'give some ground to get some ground' what do you think? 10. [email from client to me ...] Memo Louis, Instead of wasting your time listing grievances, I would prefer you complete the items in v1.1 mini-spec, to a satisfactory conclusion. We almost had the website ready for launch until you brought the v1.1 mini-spec into the frame. Obviously I expected you could complete the v1.1 mini-spec in a two-week time frame as you indicated and give the site a more profession presentation. Most of the problems have been caused by you not following our instructions, but deciding to do what you feel like at the time. And then arguing with us how the missing information is not necessary. For instance "Parts and Accessories". Why on earth would you leave out the parts heading, when it ties-in with the fields you have already developed. It replaces "model" and is just as important in the context of information that appears in the "Details" panel. We are at a stage where the the v1.1 mini-spec needs to be completed without further time wasting and the site is complete (subject to all features working). We are on standby at this end to do just that. Let me know when you are back, working on the site and we will process and complete each v1.1 mini-spec, item by item, until the job is complete. 11. [last email from me to client...] hi [client], based on this reply, and your demonstrated unwillingness to compromise/give any ground on issues at hand, i have decided to place your project on-hold for the moment i will be considering further options on how to over-come our challenges over the next few days i will contact you by monday 17/may to discuss any new options i have come up with, and if i believe it is appropriate to restart work on your project at that point or not told you it was long... what do you think?

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