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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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• 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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• Can you access registers from python functions in vim

  - by Michael Anderson

  It seems vims python sripting is designed to edit buffer and files rather than work nicely with vims registers. You can use some of the vim packages commands to get access to the registers but its not pretty. My solution for creating a vim function using python that uses a register is something like this. function printUnnamedRegister() python >> EOF print vim.eval('@@') EOF Setting registers may also be possible using something like function setUnnamedRegsiter() python >> EOF s = "Some \"crazy\" string\nwith interesting characters" vim.command('let @@="%s"' % myescapefn(s) ) EOF However this feels a bit cumbersome and I'm not sure exactly what myescapefn should be. So I've never been able to get the setting version to work properly. So if there's a way to do something more like function printUnnamedRegister() python >> EOF print vim.getRegister('@') EOF function setUnnamedRegsiter() python >> EOF s = "Some \"crazy\" string\nwith interesting characters" vim.setRegister('@',s) EOF Or even a nice version of myescapefn I could use then that would be very handy.

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• What kind of knowledge do you need to invent a new programming language?

  - by systempuntoout

  I just finished to read "Coders at works", a brilliant book by Peter Seibel with 15 interviews to some of the most interesting computer programmers alive today. Well, many of the interviewees have (co)invented\implemented a new programming language. Some examples: Joe Armstrong: Inventor of Erlang L. Peter Deutsch: implementer of Smalltalk-80 Brendan Eich: Inventor of JavaScript Dan Ingalls: Smalltalk implementor and designer Simon Peyton Jones: Coinventor of Haskell Guy Steele: Coinventor of Scheme Is out of any doubt that their minds have something special and unreachable, and i'm not crazy to think i will ever able to create a new language; i'm just interested in this topic. So, imagine a funny\grotesque scenario where your crazy boss one day will come to your desk to say "i want a new programming language with my name on it..take the time you need and do it", which is the right approach to studying this fascinating\intimidating\magic topic? What kind of knowledge do you need to model, design and implement a brand new programming language?

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• What kind of knowledge you need to invent a new programming language?

  - by systempuntoout

  I just finished to read "coders at works", a brilliant book by Peter Seibel with 15 interviews to some of the most interesting computer programmers alive today. Well, many of the interviewees have (co)invented\implemented a new programming language. For example: * Joe Armstrong: Inventor of Erlang * L. Peter Deutsch: implementer of Smalltalk-80 * Brendan Eich: Inventor of JavaScript * Dan Ingalls: Smalltalk implementor and designer * Simon Peyton Jones: Coinventor of Haskell * Guy Steele: Coinventor of Scheme Is out of any doubt that their minds have something special and unreachable, and i'm not crazy to think i will ever able to create a new language; i'm just interested in this topic. So, imagine a funny\grotesque scenario where your crazy boss one day will come to your desk to say "i want a new programming language with my name on it..take the time you need and do it", what will you start to study? What kind of knowledge do you need to model, design and implement a brand new programming language?

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• Is x a reserved keyword in Javascript FF/Safari not in IE?

  - by Marco Demaio

  A web page of a web application was showing a strange error. I regressively removed all the HTML/CSS/JS code and arrived to the basic and simple code below. <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html><head> <title>test</title> <script type="text/javascript"> var TestObj = { foo: function() {} } alert(x); //ok displays "undefined" var x = TestObj.foo; var z = TestObj.foo; </script> </head><body> <p onclick='alert(x);'>Click shows function foo!</p> <img onclick='alert(x);' alt='CRAZY click displays a number in FF/Safari not function foo' src='' style='display: block; width: 100px; height: 100px; border: 1px solid #00ff00;'> <p onclick='alert(x);'>Click shows function foo!</p> </body></html> It's crazy: when clicking on P elements the string "function(){}" is displaied as expected. But when clicking on IMG element it shows a number as if x function got in some way removed from memory or deinstantiated (it does not even show x as "undefined" but as a number). To let you test it quickly I placed the working test above also here. This can be reproduced on both Firefox 3.6 and Safari 4.0.4. Everything works properly only on IE7+. I'm really clueless, I was wondering if x is maybe a reserved keyword in JS Firefox/Safari. Thanks to anyone who could help! FYI: if you repalce x() with z() everything work prefectly in all browsers (this is even more crazy to me) adding a real image in src attribute does not fix the problem removing style in img does not fix the problem (i gave style to image only to help you clicking on image thus you can see the imnage border)

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