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• C or assembly code to find current cpu core speed

  - by honestann

  How can my application efficiently determine the following information peroidically while it executes: 1: current speed of each of the 8 CPU cores. 2: which core the code is currently executing on. My application is C and assembly-language, so any solution in either C or assembly-language is fine. This code needs to execute quickly, so creating, reading and processing a file generated by "cat /proc/cpuinfo" is much too slow. The cores slow-down and speed-up automatically, probably to keep CPU temperature under control. Therefore, a one-time measure is not sufficient for my purposes. My application already reads and subtracts the cpu cycle counter in assembly language to determine number of clock cycles, but my program cannot compute elapsed time in nanoseconds unless it knows the current clock frequency of the cpu cores (and which core the code is executing on). Thanks!

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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 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 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 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 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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• is gnome 3 is slow or my machine?

  - by user103054

  I have Dell XPS 15 i7 - 2360 with nvidia 525M (1GB) with optimus running Ubuntu 12.10 64 bit and have bumblebee installed for optimus. Recently I installed Gnome Shell to give it a try and found it sort of lagging under few scenarios though rest all is butter smooth. When I click on Date gnome 3 it appears as if a second has elapsed before the window is drawn. But others panel items like network or volume looks fine. In gnome 3 when I'm on dash with (window preview) if I remove any of the windows before the remaining windows reshuffle it takes about 2 seconds (noticeable clearly). Everything else is fine in both unity and Gnome Shell. What is causing this slowness? Is Gnome Shell is really slower?

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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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• Material, Pass, Technique and shaders

  - by Papi75

  I'm trying to make a clean and advanced Material class for the rendering of my game, here is my architecture: class Material { void sendToShader() { program->sendUniform( nameInShader, valueInMaterialOrOther ); } private: Blend blendmode; ///< Alpha, Add, Multiply, … Color ambient; Color diffuse; Color specular; DrawingMode drawingMode; // Line Triangles, … Program* program; std::map<string, TexturePacket> textures; // List of textures with TexturePacket = { Texture*, vec2 offset, vec2 scale} }; How can I handle the link between the Shader and the Material? (sendToShader method) If the user want to send additionals informations to the shader (like time elapsed), how can I allow that? (User can't edit Material class) Thanks!

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• Auto-Invoke Update Manager to update everything and shutdown after system idle for x minutes?

  - by unknownthreat

  I have Ubuntu 10.10 installed on a machine for my parents. The thing is they never request updates from Update Manager even the manager itself prompted them so. Moreover, when they are done with whatever they are doing on Ubuntu, they always leave the computer on. And I always have to come back and shut the machine down. Sometimes, the computer even sit idle for hours. So I want to know whether this is possible in Ubuntu. I am thinking of a script that will be activated after the machine is idle for x minutes. When x minutes have elapsed, Update Manager will automatically update everything listed. (I recall that you need the admin password for this, so is there a workaround?) After all the updates are done, the machine will automatically shutdown. Is this possible?

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• Bug Tracking Etiquette - Necromancy or Duplicate?

  - by Shauna

  I came across a really old (2+ years) feature request issue in a bug tracker for an open source project that was marked as "resolved (won't fix)" due to the lack of tools required to make the requested enhancement. In the time elapsed since that determination was made, new tools have been developed that would allow it to be resolved, and I'd like to bring that to the attention of the community for that application. However, I'm not sure as to what the generally accepted etiquette is for bug tracking in cases like this. Obviously, if the system explicitly states to not duplicate and will actively mark new items as duplicates (much in the way the SE sites do), then the answer would be to follow what the system says. But what about when the system doesn't explicitly say that, or a new user can't easily find a place that says with the system's preference is? Is it generally considered better to err on the side of duplication or necromancy? Does this differ depending on whether it's a bug or a feature request?

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• An "Invoke Update Manager to update everything and shutdown" script after idle for x minutes?

  - by unknownthreat

  I have Ubuntu 10.10 installed on a machine for my parents. The thing is they never request updates from Update Manager even the manager itself prompted them so. Moreover, when they are done with whatever they are doing on Ubuntu, they always leave the computer on. And I always have to come back and shut the machine down. Sometimes, the computer even sit idle for hours. So I want to know whether this is possible in Ubuntu. I am thinking of a script that will be activated after the machine is idle for x minutes. When x minutes have elapsed, Update Manager will automatically update everything listed. (I recall that you need the admin password for this, so is there a workaround?) After all the updates are done, the machine will automatically shutdown. Is this possible?

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• Why don't smart phones have an auto-forget password feature? [closed]

  - by Kelvin

  Storing passwords to external services (e.g. corporate email servers) on smart phones is very insecure, since phones are more easily stolen. Has any vendor implemented a feature to only cache a password in memory for a limited amount of time? After the time period has elapsed, the app would ask for the password again. EDIT: I should've clarified - I'm aware that many (most?) users are lazy and want to just "set it and forget it". The always-remember feature will probably always be present. I was curious about an option to enable auto-forget for the security-conscious.

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• Equation / formula to determine an objects position on an ellipitcal path

  - by David Murphy

  I'm making a space game, as such I need objects to follow an elliptical path (orbit). I've worked out how to calculate all the important aspects of my orbits, the only remaining thing is how to have an object follow it. My Orbit class contains the major, minor (and by extension semi-major,semi-minor) lengths. The focii radius, area and circumference even. What is the equation to determine an objects x/y position (only need 2D) on an ellipse with a certain speed after a period of time. Basically, every frame I want to update the position based on the amount of elapsed time. I would like to have the speed along the path speed up and slow down according to the distance from the object it's orbiting, but not sure how to factor this in to the above given that at any point in time the speed has changed from it's previous speed. EDIT I can't answer my own question. But I found the question and answer is already on stackexchange: Kepler orbit : get position on the orbit over time

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• XNA GameTime TotalGameTime slower than real time

  - by robasaurus

  I have set-up an empty test project consisting of a System.Diagnostics.Stopwatch and this in the draw method: spriteBatch.DrawString(font, gameTime.TotalGameTime.TotalSeconds.ToString(), new Vector2(100, 100), Color.White); spriteBatch.DrawString(font, stopwatch.Elapsed.TotalSeconds.ToString(), new Vector2(100, 200), Color.White); The GameTime.TotalGameTime displayed is slower than the stop watch (by about 5 seconds per minute) even though GameTime.IsRunningSlowly is always false, why is this? The reason this is an issue is because I have a server which uses stopwatch and it is faster than my client game. For instance my client notifies the server it has dropped a mine which explodes in one minute. Because the stopwatch is faster the server state explodes the mine before the client and they are out of sync. I don't want to have to notify the client when the server explodes it as this would use unnecessary bandwidth.

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• Should I use a workflow engine?

  - by Fernando

  I need to add some new features to a PHP application. It is to follow the steps of a order. A process create some orders, the order goes to confirmation, then if approved is sent to a provider, later the provider confirm that can deliver the order, a request is made to the provider and so on... I need to register when every step is made and send notifications. Also, some steps have a estimate time, and if that time is elapsed I need to send notifications so everybody know about the delay. When a process starts, it have a predefined set of steps, but in a middle the user should be able to create new sub-steps, and delete or skip future steps.. Should I use a workflow engine? Which one do you suggests (free-opensource only)?

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• Variable-step update() in game loop is falling behind, how can I get around this?

  - by ThatsGobbles

  I'm working on a minimal game engine for my next game. I'm using the delta update method like shown: void update(double delta) { // Update code that uses `delta` goes here } I have a deep hierarchy of updatable objects, with a root updatable that contains several updatables, each of which contains more updatables, etc. Normally I'd just iterate through each of the root's children and update each one, which would then do the same for its children, and so on. However, passing a fixed value of delta to the root means that by the time the leaf updatables are reached, it's been longer since delta seconds that have elapsed. This is causing noticable desyncing in my game, and time synchronization is very important in my case (I'm working on a rhythm game). Any ideas on how I should tackle this? I've considered using StopWatches and a global readable timer, but any advice would be helpful. I'm also open to moving to fixed timesteps as opposed to variable.

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• Variable-step update() in game loop is falling behind, how can I get around this?

  - by ThatsGobbles

  I'm working on a minimal game engine for my next game. I'm using the delta update method like shown: void update(double delta) { // Update code that uses `delta` goes here } I have a deep hierarchy of updatable objects, with a root updatable that contains several updatables, each of which contains more updatables, etc. Normally I'd just iterate through each of the root's children and update each one, which would then do the same for its children, and so on. However, passing a fixed value of delta to the root means that by the time the leaf updatables are reached, it's been longer since delta seconds that have elapsed. This is causing noticable desyncing in my game, and time synchronization is very important in my case (I'm working on a rhythm game). Any ideas on how I should tackle this? I've considered using StopWatches and a global readable timer, but any advice would be helpful. I'm also open to moving to fixed timesteps as opposed to variable.

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• Masked Time control in WPF

  - by Ashish Ashu

  We are in need of a control in which we can show the elapsed time in "Hour:Min:Sec" format. This control should have a spin control attached with it. Whenever we spin the spin window, it should increment the selected option ( Either hour, or min , or sec). Also the Hour may have values between 0 to 99 , Min many have values between 0 to 59 and Sec may have values between 0 to 59. Is anybody know this type of control in WPF?

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• Calculating timespan with t-sql

  - by jdiaz

  Given two date/times: @start_date = '2009-04-15 10:24:00.000' @end_date = '2009-04-16 19:43:01.000' Is it possible to calculate the time elapsed between the two dates in the following format 1d 9h 19m

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• How to play multiple audio sources simultaneously in Silverlight

  - by Shurup

  I want to play simultaneous multiply audio sources in Silverlight. So I've created a prototype in Silverlight 4 that should play a two mp3 files containing the same ticks sound with an intervall 1 second. So these files must be sounded as one sound if they will be played together with any whole second offsets (0 and 1, 0 and 2, 1 and 1 seconds, etc.) I my prototype I use two MediaElement (me and me2) objects. DateTime startTime; private void Play_Clicked(object sender, RoutedEventArgs e) { me.SetSource(new FileStream(file1), FileMode.Open))); me2.SetSource(new FileStream(file2), FileMode.Open))); var timer = new DispatcherTimer { Interval = TimeSpan.FromMilliseconds(1) }; timer.Tick += RefreshData; timer.Start(); } First file should be played at 00:00 sec. and the second in 00:02 second. void RefreshData(object sender, EventArgs e) { if(me.CurrentState != MediaElementState.Playing) { startTime = DateTime.Now; me.Play(); return; } var elapsed = DateTime.Now - startTime; if(me2.CurrentState != MediaElementState.Playing && elapsed >= TimeSpan.FromSeconds(2)) { me2.Play(); ((DispatcherTimer)sender).Stop(); } } The tracks played every time different and not simultaneous as they should (as one sound). Addition: I've tested a code from the Bobby's answer. private void Play_Clicked(object sender, RoutedEventArgs e) { me.SetSource(new FileStream(file1), FileMode.Open))); me2.SetSource(new FileStream(file2), FileMode.Open))); // This code plays well enough. // me.Play(); // me2.Play(); // But adding the 2 second offset using the timer, // they play no simultaneous. var timer = new DispatcherTimer { Interval = TimeSpan.FromSeconds(2) }; timer.Tick += (source, arg) => { me2.Play(); ((DispatcherTimer)source).Stop(); }; timer.Start(); } Is it possible to play them together using only one MediaElement or any implementation of MediaStreamSource that can play multiply sources?

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• What is the most useful script you've written for everyday life?

  - by Kyle Walsh

  Just wondering what little scripts/programs people here have written that helps one with his or her everyday life (aka not work related). Anything goes, groundbreaking or not. For me right now, it's a small python script to calculate running pace given distance and time elapsed.

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