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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 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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  • nfs mountpoint named ``share'' breaks ls and man

    - by freddyb
    I mounted a nfs server to ~/share. This works fine as long as I'm at home, where the nfs share is in reach. Whenver I'm not, this seems to break access to all manpages. Using man (or ls in my homedir) waits forever. Checking with strace reveals that they try to access the folder called share. Unmounting fails too. Even with -l (lazy) and -f (force). I am asking for three things here: Is ``share'' a magic name? Does something like MANPATH exist, which I should avoid? How do I unmount without rebooting? (I already commented the share out in fstab) What would you suggest me to do, to have network/position based mounting of NFS shares?

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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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  • Secret Agent Man

    - by Bil Simser
    Just a quick one this morning as we all get started in the week. Something that comes into play (sometimes in a big way) is the user agent string your browser gives off. So for example using the User-Agent field in the request header, you can determine what browser the user is running and act accordingly.Internet Explorer 9 modified the UA string slightly so just in case you're looking for it here are the user agent strings for IE9 (in various modes):Internet Explorer 9 Mode: Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.1; WOW64; Trident/5.0)Internet Explorer 8 Mode: Mozilla/4.0 (compatible; MSIE 8.0; Windows NT 6.1; WOW64; Trident/4.0; SLCC2; .NET CLR 2.0.50727; .NET CLR 3.5.30729; .NET CLR 3.0.30729; MS-RTC LM 8; InfoPath.3; .NET4.0C; .NET4.0E; Zune 4.7)Internet Explorer 7 Mode: Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 6.1; WOW64; SLCC2; .NET CLR 2.0.50727; .NET CLR 3.5.30729; .NET CLR 3.0.30729; MS-RTC LM 8; InfoPath.3; .NET4.0C; .NET4.0E; Zune 4.7)Internet Explorer 9 (Compatibility Mode): Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 6.1; WOW64; Trident/5.0; SLCC2; .NET CLR 2.0.50727; .NET CLR 3.5.30729; .NET CLR 3.0.30729; MS-RTC LM 8; InfoPath.3; .NET4.0C; .NET4.0E; Zune 4.7)A couple of things to note here:This was from a 64-bit Windows 7 client so that might account for the WOW64 in the agent string (I don't have a 32-bit client to test from)Various applications and platforms add to the UA string just like they do in previous IE releases. So for example you can see I have various .NET versions installed as well as Zune. You can take advantage of this by querying the UA string for compatibilities and present options accordingly to the end user.As applications will continue to add and modify this string you'll want to query the string for parts not the entire string. For example if you want to detect if you're coming from IE running  on a Windows Phone 7 just look for "iemobile" in the user agent stringHappy hacking!

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  • Poor Man’s PowerShell TFS SSMS Integration

    - by merrillaldrich
    This is lame. Still, here goes: I need, increasingly, to author both PowerShell and SQL Server scripts, bundle them into a solution and store that in TFS. Usually the PowerShell scripts are very closely related to SQL Server, and have a lot of SQL in them. I am hopeful that 2012 SSDT, or the tighter integration of SSMS and Visual Studio in 2012, might help put all of this in one place, but for now I am stuck in SSMS 2008 R2. So here are my blunt attempts to marry these activities. (This post is rather...(read more)

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  • What are the advantages of Maven when it comes to single man, educational projects

    - by Leron
    I've spend a few hours playing around with Maven + reading some stuff on the apache official site and also a few random googled articles. By this I mean that I really tried to find the answers myself - both by reading and by doing things on my own. Also maybe worth to mention that I installed the m2e plugin so most of the time I've tried things out from Eclipse and not using the command line too much. However aside from the generated project that for example prevent me from using the default package I didn't see that much of a difference with the standard way I've created my projects before try Maven. In fact I've almost decided to skip Maven for now and move on to the other technology I wanted to learn more in-depth - Hibernate, but when I start with opening the official page the first thing I've read was the recommendation to use Hibernate with Maven. That get me confused and made me taking a step back and trying once more to find what I'm obviously missing right now. As it's said in the maven.apache.. site, the true strength of Maven is shown when you work on large projects with other people, but I lack the option to see how Maven is really used in this scenario, still i think that there are maybe advantages even when it comes to working with small projects alone, but I really have difficulties to point them out. So what do you think are the advantages of Maven when it's used for small projects writing from a single person. What are the things that I should be aware of and try to exploit (I mean features offered by Maven) that can come in handy in this situations?

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  • To Serve Man?

    - by Dave Convery
    Since the announcement of Windows 8 and its 'Metro' interface, the .NET community has wondered if the skills they've spent so long developing might be swept aside,in favour of HTML5 and JavaScript. Mercifully, that only seems to be true of SilverLight (as Simon Cooper points out), but it did leave me thinking how easy it is to impose a technology upon people without directly serving their needs. Case in point: QR codes. Once, probably, benign in purpose, they seem to have become a marketer's tool for determining when someone has engaged with an advert in the real world, with the same certainty as is possible online. Nobody really wants to use QR codes - it's far too much hassle. But advertisers want that data - they want to know that someone actually read their billboard / poster / cereal box, and so this flawed technology is suddenly everywhere, providing little to no value to the people who are actually meant to use it. What about 3D cinema? Profits from the film industry have been steadily increasing throughout the period that digital piracy and mass sharing has been possible, yet the industry cinema chains have forced 3D films upon a broadly uninterested audience, as a way of providing more purpose to going to a cinema, rather than watching it at home. Despite advances in digital projection, 3D cinema is scarcely more immersive to us than were William Castle's hoary old tricks of skeletons on wires and buzzing chairs were to our grandparents. iTunes - originally just a piece of software that catalogued and ripped music for you, but which is now multi-purpose bloatware; a massive, system-hogging behemoth. If it was being built for the people that used it, it would have been split into three or more separate pieces of software long ago. But as bloatware, it serves Apple primarily rather than us, stuffed with Music, Video, Various stores and phone / iPad management all bolted into one. Why? It's because, that way, you're more likely to bump into something you want to buy. You can't even buy a new laptop without finding that a significant chunk of your hard drive has been sold to 'select partners' - advertisers, suppliers of virus-busting software, and endless bloatware-flogging pop-ups that make using a new laptop without reformatting the hard drive like stepping back in time. The product you want is not the one you paid for. This is without even looking at services like Facebook and Klout, who provide a notional service with the intention of slurping up as much data about you as possible (in Klout's case, whether you create an account with them or not). What technologies do you find annoying or intrusive, and who benefits from keeping them around?

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  • How to make transition from one-man to small team successfull

    - by si2w
    I have started a big project 1 1/2 ago. It's time to have some help to face the future challenges. Actually I'm the engineer, the architect, the dba, the sysadmin etc... The transistion is not so simple and it's difficult to find the good person. When we will find this person, what's the best ways to manage him/her to give enough freedom to be happy and productive but have a clean, fast, reuseable, working, ... code ? Is there good books you advice ? Thanks !

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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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