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  • Project Euler 20: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 20.  As always, any feedback is welcome. # Euler 20 # http://projecteuler.net/index.php?section=problems&id=20 # n! means n x (n - 1) x ... x 3 x 2 x 1 # Find the sum of digits in 100! import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) print sum([int(i) for i in str(factorial(100))]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 3: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 3.  As always, any feedback is welcome. # Euler 3 # http://projecteuler.net/index.php?section=problems&id=3 # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number # 600851475143? import time start = time.time() def largest_prime_factor(n): max = n divisor = 2 while (n >= divisor ** 2): if n % divisor == 0: max, n = n, n / divisor else: divisor += 1 return max print largest_prime_factor(600851475143) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 1: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 1.  As always, any feedback is welcome. # Euler 1 # http://projecteuler.net/index.php?section=problems&amp;id=1 # If we list all the natural numbers below 10 that are # multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of # these multiples is 23. Find the sum of all the multiples # of 3 or 5 below 1000. import time start = time.time() print sum([x for x in range(1000) if x % 3== 0 or x % 5== 0]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue') # Also cool def constraint(x): return x % 3 == 0 or x % 5 == 0 print sum(filter(constraint, range(1000)))

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  • Project Euler 14: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 14.  As always, any feedback is welcome. # Euler 14 # http://projecteuler.net/index.php?section=problems&id=14 # The following iterative sequence is defined for the set # of positive integers: # n -> n/2 (n is even) # n -> 3n + 1 (n is odd) # Using the rule above and starting with 13, we generate # the following sequence: # 13 40 20 10 5 16 8 4 2 1 # It can be seen that this sequence (starting at 13 and # finishing at 1) contains 10 terms. Although it has not # been proved yet (Collatz Problem), it is thought that all # starting numbers finish at 1. Which starting number, # under one million, produces the longest chain? # NOTE: Once the chain starts the terms are allowed to go # above one million. import time start = time.time() def collatz_length(n): # 0 and 1 return self as length if n <= 1: return n length = 1 while (n != 1): if (n % 2 == 0): n /= 2 else: n = 3*n + 1 length += 1 return length starting_number, longest_chain = 1, 0 for x in xrange(1, 1000001): l = collatz_length(x) if l > longest_chain: starting_number, longest_chain = x, l print starting_number print longest_chain # Slow 31 seconds print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 12: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 12.  As always, any feedback is welcome. # Euler 12 # http://projecteuler.net/index.php?section=problems&id=12 # The sequence of triangle numbers is generated by adding # the natural numbers. So the 7th triangle number would be # 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms # would be: # 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... # Let us list the factors of the first seven triangle # numbers: # 1: 1 # 3: 1,3 # 6: 1,2,3,6 # 10: 1,2,5,10 # 15: 1,3,5,15 # 21: 1,3,7,21 # 28: 1,2,4,7,14,28 # We can see that 28 is the first triangle number to have # over five divisors. What is the value of the first # triangle number to have over five hundred divisors? import time start = time.time() from math import sqrt def divisor_count(x): count = 2 # itself and 1 for i in xrange(2, int(sqrt(x)) + 1): if ((x % i) == 0): if (i != sqrt(x)): count += 2 else: count += 1 return count def triangle_generator(): i = 1 while True: yield int(0.5 * i * (i + 1)) i += 1 triangles = triangle_generator() answer = 0 while True: num = triangles.next() if (divisor_count(num) >= 501): answer = num break; print answer print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 19: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 19.  As always, any feedback is welcome. # Euler 19 # http://projecteuler.net/index.php?section=problems&id=19 # You are given the following information, but you may # prefer to do some research for yourself. # # - 1 Jan 1900 was a Monday. # - Thirty days has September, # April, June and November. # All the rest have thirty-one, # Saving February alone, # Which has twenty-eight, rain or shine. # And on leap years, twenty-nine. # - A leap year occurs on any year evenly divisible by 4, # but not on a century unless it is divisible by 400. # # How many Sundays fell on the first of the month during # the twentieth century (1 Jan 1901 to 31 Dec 2000)? import time start = time.time() import datetime sundays = 0 for y in range(1901,2001): for m in range(1,13): # monday == 0, sunday == 6 if datetime.datetime(y,m,1).weekday() == 6: sundays += 1 print sundays print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 18: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 18.  As always, any feedback is welcome. # Euler 18 # http://projecteuler.net/index.php?section=problems&id=18 # By starting at the top of the triangle below and moving # to adjacent numbers on the row below, the maximum total # from top to bottom is 23. # # 3 # 7 4 # 2 4 6 # 8 5 9 3 # # That is, 3 + 7 + 4 + 9 = 23. # Find the maximum total from top to bottom of the triangle below: # 75 # 95 64 # 17 47 82 # 18 35 87 10 # 20 04 82 47 65 # 19 01 23 75 03 34 # 88 02 77 73 07 63 67 # 99 65 04 28 06 16 70 92 # 41 41 26 56 83 40 80 70 33 # 41 48 72 33 47 32 37 16 94 29 # 53 71 44 65 25 43 91 52 97 51 14 # 70 11 33 28 77 73 17 78 39 68 17 57 # 91 71 52 38 17 14 91 43 58 50 27 29 48 # 63 66 04 68 89 53 67 30 73 16 69 87 40 31 # 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 # NOTE: As there are only 16384 routes, it is possible to solve # this problem by trying every route. However, Problem 67, is the # same challenge with a triangle containing one-hundred rows; it # cannot be solved by brute force, and requires a clever method! ;o) import time start = time.time() triangle = [ [75], [95, 64], [17, 47, 82], [18, 35, 87, 10], [20, 04, 82, 47, 65], [19, 01, 23, 75, 03, 34], [88, 02, 77, 73, 07, 63, 67], [99, 65, 04, 28, 06, 16, 70, 92], [41, 41, 26, 56, 83, 40, 80, 70, 33], [41, 48, 72, 33, 47, 32, 37, 16, 94, 29], [53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14], [70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57], [91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48], [63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31], [04, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 04, 23]] # Loop through each row of the triangle starting at the base. for a in range(len(triangle) - 1, -1, -1): for b in range(0, a): # Get the maximum value for adjacent cells in current row. # Update the cell which would be one step prior in the path # with the new total. For example, compare the first two # elements in row 15. Add the max of 04 and 62 to the first # position of row 14.This provides the max total from row 14 # to 15 starting at the first position. Continue to work up # the triangle until the maximum total emerges at the # triangle's apex. triangle [a-1][b] += max(triangle [a][b], triangle [a][b+1]) print triangle [0][0] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 11: (Iron)Python

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

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  • 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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  • Stir Trek 2: Iron Man Edition

    Next month (7 May 2010) Ill be presenting at the second annual Stir Trek event in Columbus, Ohio. Stir Trek (so named because last year its themes mixed MIX and the opening of the Star Trek movie) is a very cool local event.  Its a lot of fun to present at and to attend, because of its unique venue: a movie theater.  And whats more, the cost of admission includes a private showing of a new movie (this year: Iron Man 2).  The sessions cover a variety of topics (not just Microsoft), similar to CodeMash.  The event recently sold out, so Im not telling you all of this so that you can go and sign up (though I believe you can get on the waitlist still).  Rather, this is pretty much just an excuse for me to talk about my session as a way to organize my thoughts. Im actually speaking on the same topic as I did last year, but the key difference is that last year the subject of my session was nowhere close to being released, and this year, its RTM (as of last week).  Thats right, the topic is Whats New in ASP.NET 4 how did you guess? Whats New in ASP.NET 4 So, just what *is* new in ASP.NET 4?  Hasnt Microsoft been spending all of their time on Silverlight and MVC the last few years?  Well, actually, no.  There are some pretty cool things that are now available out of the box in ASP.NET 4.  Theres a nice summary of the new features on MSDN.  Here is my super-brief summary: Extensible Output Caching use providers like distributed cache or file system cache Preload Web Applications IIS 7.5 only; avoid the startup tax for your site by preloading it. Permanent (301) Redirects are finally supported by the framework in one line of code, not two. Session State Compression Can speed up session access in a web farm environment.  Test it to see. Web Forms Features several of which mirror ASP.NET MVC advantages (viewstate, control ids) Set Meta Keywords and Description easily Granular and inheritable control over ViewState Support for more recent browsers and devices Routing (introduced in 3.5 SP1) some new features and zero web.config changes required Client ID control makes client manipulation of DOM elements much simpler. Row Selection in Data Controls fixed (id based, not row index based) FormView and ListView enhancements (less markup, more CSS compliant) New QueryExtender control makes filtering data from other Data Source Controls easy More CSS and Accessibility support Reduction of Tables and more control over output for other template controls Dynamic Data enhancements More control templates Support for inheritance in the Data Model New Attributes ASP.NET Chart Control (learn more) Lots of IDE enhancements Web Deploy tool My session will cover many but not all of these features.  Theres only an hour (3pm-4pm), and its right before the prize giveaway and movie showing, so Ill be moving quickly and most likely answering questions off-line via email after the talk. Hope to see you there! Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Stark Expo Needs You

    - by [email protected]
    Train to Become a Master Cloud Operative Can't wait until September to get your Oracle fix? Then come visit us at the Stark Expo now. Marvel Entertainment has turned itself into one of the hottest media companies of the digital age, and at the heart of Marvel's growth and transformation is Oracle technology. Now, this successful collaboration finds its way to the big screen, as Oracle joins forces with Marvel to launch a special showcase Website and movie trailer for the upcoming Iron Man 2. In Iron Man 2, Oracle is a proud sponsor of Stark Expo, a world-class tradeshow that depends on a cloud computing architecture to ensure that systems are free from overload. Starting today, visitors to the showcase Website are invited to become Master Cloud Operatives and keep Stark Expo up and running. Complete your training, test your troubleshooting skills in the Oracle Pavilion, and qualify to receive a free movie poster.

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  • Add Firefox’s Awesome Bar Bookmark Search Function to Chrome and Iron

    - by Asian Angel
    Do you have a large number of bookmarks saved in your Chromium-based browser and need a quick way to search through them? Then see how easy it is to search through those bookmarks just like Firefox users do with the AwesomeBar extension. To engage the bookmark search function type “ab” in the Address Bar as seen above and press either Tab or the Space Bar. That will display the AwesomeBar prefix-bar as seen below. Enter the desired text to begin your search. For our example we decided to conduct a search for bookmarks related to the Ubuntu Twitter client Hotot. The results will continue to narrow down nicely as you type… Typing just a bit more finishes narrowing our search down the rest of the way for Hotot related items. Install the AwesomeBar Extension [Google Chrome Extensions] How to Enable Google Chrome’s Secret Gold IconHow to Create an Easy Pixel Art Avatar in Photoshop or GIMPInternet Explorer 9 Released: Here’s What You Need To Know

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  • how to open multiple projects into the CAST IRON integration tool?

    - by Mishal
    Hi, I am learning the cast iron tool which is widely used now a days for integration purpose,but i can only open 1 project and if i want to open the other project at the same time than i have to close the 1st project and open the 2nd project. So many times i need to have to open the 2 projects at the same time but i dont know in which way i can open the projects ? can any body give me any urgent solution for the same to open the multiple projects at the same time and to switch between them ? Thanks, Mishal Shah

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  • An Interview with Wim Coekaerts

    - by [email protected]
    It isn't everyday you get to hear an interview with an SVP at Oracle, nor do you often get glimpses into the future of Oracle products. However - in this interview you get both. listen to Wim talk about Sun Rays, VDI and what Virtual Iron might mean to the mix of products coming...Enjoy

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  • Oracle ACE????????

    - by Kazuhiro.Yamaguchi
    ?????????Oracle ACE?????????????????????????????2????????????????,?????????????????????????????????????????????????????????????????????????? ·?????? ·11g R2 for Windows ?????? ?????????????????????????????????????????????????????????????????????????????????50?????????????????????????????????????????????????????????????????????????! ???????????????????????????????????? wmo6hash::blog 2010/05/29(Sat) Iron Man 2 ????????????????(?)?????????????????????????????????????????????????????? ????: ????????????????????????????????????????????????????????????????????????????????????????????????????????2????????????????????????????????????????????????????????????????????CEO ????????????????????????????????????????????????????????????? ????: ?(??????)1?????????????!?????Marvel????????????????????????????·??????????????·???????????????????????????2????????·?????????????????????????????????????????????????(?)???????????????????????????????·???????????????????AI???????????IT??????????CEO????·????????????!?????????????????????????????(?)?????????????????? Oracle ACE???????????????????!(???:??????:????) ??????????????2???????????????????????????????????????????????? ???????2???????? - 3????=3????????! ????????????????????????????????·??????????????????????????????????????????????????????????·????????????SF????·?????????????????????????????????????

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  • Failed dependency while installing browser Iron(A google chrome clone)

    - by Krishnadas PC
    Installation failed while trying to install Iron browser. [root@localhost softwares]# rpm -ivh iron64.rpm error: Failed dependencies: libc.so.6(GLIBC_2.15)(64bit) is needed by iron64-29.0.1600-2.x86_64 libudev.so.1()(64bit) is needed by iron64-29.0.1600-2.x86_64 libudev.so.1(LIBUDEV_183)(64bit) is needed by iron64-29.0.1600-2.x86_64 and when tried to install using yum it failed also.

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  • What do I need to develop an Iron Python web app in Visual Studio 2010

    - by Greg
    Hi, I've got Visual Studio 2010. To develop a web app in Iron Python (i.e. to use a Ruby like language not C#) what downloads to I need? e.g. is the DLR already in VS2010, Iron Python itself Once setup would I actually be still developing an ASP.net MVC web app but just using Ruby for the language, or is the model something different to this? thanks

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  • Give Chromium-Based Browser Desktop Notifications a Native System Look in Ubuntu

    - by Asian Angel
    Desktop notifications from Chromium-based browsers are an awesome feature, but they do not blend in well at all with the native system theming in Ubuntu. Now you can fix that small problem using the wonderful Chromify-OSD extension created by Marco Ceppi. Once you get the extension installed you can give it a quick test run using the link and information we have listed below. As you can see in the image above the new notification style looks absolutely wonderful. Chromify-OSD (Chrome Web Store) [via OMG! Ubuntu!] You can test the new look of the notifications for yourself using the following webpage. Keep in mind that the extension needs to be installed first before this will work though. Note: Enter the following image URL into the Icon Blank (http://www.rgraph.net/images/logo.png) or the URL for an appropriate image, otherwise the notification may not work properly during your test. Chromify Sample HTML5 Notification Test Page The wallpaper shown in the screenshot above can be downloaded here: anime sport [DesktopNexus] Latest Features How-To Geek ETC How to Enable User-Specific Wireless Networks in Windows 7 How to Use Google Chrome as Your Default PDF Reader (the Easy Way) How To Remove People and Objects From Photographs In Photoshop Ask How-To Geek: How Can I Monitor My Bandwidth Usage? Internet Explorer 9 RC Now Available: Here’s the Most Interesting New Stuff Here’s a Super Simple Trick to Defeating Fake Anti-Virus Malware The Citroen GT – An Awesome Video Game Car Brought to Life [Video] Final Man vs. Machine Round of Jeopardy Unfolds; Watson Dominates Give Chromium-Based Browser Desktop Notifications a Native System Look in Ubuntu Chrome Time Track Is a Simple Task Time Tracker Google Sky Map Turns Your Android Phone into a Digital Telescope Walking Through a Seaside Village Wallpaper

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  • Cisco Spam Blocker, Iron Port, Lotus Domino, Integration Help

    - by NickToyota
    Hi serverfault universe, I work for a medium sized (roughly 200 user) company. We are attempting to intagrate our new Cisco Spam Video Blocker (ironport) device into our network so that it acts as an incoming filter then passes it off to our Lotus domino mail server. And also vise versa. The way our network is setup currently has an mx record pointing to our Domino mail SMTP incoming server which is currently setup to be an inbound gateway and filter (using symantec domino mail software). We want to replace the inbound gateway with the ironport. Our company has also invested in a pool of external IP addresses which I believe has been currently assigned to our web, email, servers. What would the proper course of action be to successfully integrate the device be? Mx record change? Replace the domino gateway completely with the ironport? We attempted to set the ironport device to the external IP of what our mx record is pointing to without much success. Any help on proper setup would be greatly appreciated.

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  • Cisco Spam Blocker, Iron Port, Lotus Domino, Integration Help

    - by NickToyota
    Hi serverfault universe, I work for a medium sized (roughly 200 user) company. We are attempting to intagrate our new Cisco Spam Video Blocker (ironport) device into our network so that it acts as an incoming filter then passes it off to our Lotus domino mail server. And also vise versa. The way our network is setup currently has an mx record pointing to our Domino mail SMTP incoming server which is currently setup to be an inbound gateway and filter (using symantec domino mail software). We want to replace the inbound gateway with the ironport. Our company has also invested in a pool of external IP addresses which I believe has been currently assigned to our web, email, servers. What would the proper course of action be to successfully integrate the device be? Mx record change? Replace the domino gateway completely with the ironport? We attempted to set the ironport device to the external IP of what our mx record is pointing to without much success. Any help on proper setup would be greatly appreciated.

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