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Articles indexed in December 2010

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• My Favourite Two Buttons in Denali CTP1 SSIS

  In SSIS for SQL Server 2005 and SQL Server 2008 when you delete something from the design surface it is gone.  The only real way of getting the deleted item(s) back is to revert to a previous version of the package or to redo the deleted items manually.  Neither of these options is particularly great.  I have made this mistake before and cursed not having CTL+Z and CTL+Y.  Denali changes this.  We can now undo and redo.  Very very welcome.  Well done, finally, the SSIS team.

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• Using the Script Component as a Conditional Split

  This is a quick walk through on how you can use the Script Component to perform Conditional Split like behaviour, splitting your data across multiple outputs. We will use C# code to decide what does flows to which output, rather than the expression syntax of the Conditional Split transformation. Start by setting up the source. For my example the source is a list of SQL objects from sys.objects, just a quick way to get some data: SELECT type, name FROM sys.objects type name S syssoftobjrefs F FK_Message_Page U Conference IT queue_messages_23007163 Shown above is a small sample of the data you could expect to see. Once you have setup your source, add the Script Component, selecting Transformation when prompted for the type, and connect it up to the source. Now open the component, but don’t dive into the script just yet. First we need to select some columns. Select the Input Columns page and then select the columns we want to uses as part of our filter logic. You don’t need to choose columns that you may want later, this is just the columns used in the script itself. Next we need to add our outputs. Select the Inputs and Outputs page.You get one by default, but we need to add some more, it wouldn’t be much of a split otherwise. For this example we’ll add just one more. Click the Add Output button, and you’ll see a new output is added. Now we need to set some properties, so make sure our new Output 1 is selected. In the properties grid change the SynchronousInputID property to be our input Input 0, and  change the ExclusionGroup property to 1. Now select Ouput 0 and change the ExclusionGroup property to 2. This value itself isn’t important, provided each output has a different value other than zero. By setting this property on both outputs it allows us to split the data down one or the other, making each exclusive. If we left it to 0, that output would get all the rows. It can be a useful feature allowing you to copy selected rows to one output whilst retraining the full set of data in the other. Now we can go back to the Script page and start writing some code. For the example we will do a very simple test, if the value of the type column is U, for user table, then it goes down the first output, otherwise it ends up in the other. This mimics the exclusive behaviour of the conditional split transformation. public override void Input0_ProcessInputRow(Input0Buffer Row) { // Filter all user tables to the first output, // the remaining objects down the other if (Row.type.Trim() == "U") { Row.DirectRowToOutput0(); } else { Row.DirectRowToOutput1(); } } The code itself is very simple, a basic if clause that determines which of the DirectRowToOutput methods we call, there is one for each output. Of course you could write a lot more code to implement some very complex logic, but the final direction is still just a method call. If we now close the script component, we can hook up the outputs and test the package. Your numbers will vary depending on the sample database but as you can see we have clearly split out input data into two outputs. As a final tip, when adding the outputs I would normally rename them, changing the Name in the Properties grid. This means the generated methods follow the pattern as do the path label shown on the design surface, making everything that much easier to recognise.

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• Import SSIS Project in Denali CTP1

  For years Analysis Services has had the ability to take an existing database from a server and reverse engineer it into a BIDS project.  This is extremely useful when all you have is the running instance of the database and the project that created it has long since disappeared.  Reverse engineering has never been a feature of SSIS until now. Let me walk you through the simple steps. The first step is that you obviously have to have a project deployed to an SSIS Catalog.  I will do a video on this soon but in case you can’t wait then my good buddy Jamie Thomson has written it up here As you can see I have a project called imaginatively “Denali1” with one package “Package.dtsx” The next thing we need to do is fire up BIDS and choose the right project type (Integration Services Import Project) Now we just follow the wizard.  We make sure we specify on which server to find the Catalog and in which folder to look for the project. Next the setting are validated and we are greeted with the familiar review screen before the creation of our new project from the deployed project happens Hit Import and away we go The result is just what we wanted.

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• SSIS Field Notes – SQLBits 7 Presentation

  Here are the slides from my session SSIS Field Notes presented at SQLBits 7 in York earlier this month - SSIS Field Notes – Darren Green.pptx On a similar theme, the video of my session Design patterns for SSIS Performance from is now available. You heard it here first! I know that this because I’ve only just finished updating the SQLBits site with all the videos from SQLBits 6. Hopefully we’ll get them released quicker for SQLBits 7.

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• Issuing Current Time Increments in StreamInsight (A Practical Example)

  The issuing of a Current Time Increment, Cti, in StreamInsight is very definitely one of the most important concepts to learn if you want your Streams to be responsive. A full discussion of how to issue Ctis is beyond the scope of this article but a very good explanation in addition to Books Online can be found in these three articles by a member of the StreamInsight team at Microsoft, Ciprian Gerea. Time in StreamInsight Series http://blogs.msdn.com/b/streaminsight/archive/2010/07/23/time-in-streaminsight-i.aspx http://blogs.msdn.com/b/streaminsight/archive/2010/07/30/time-in-streaminsight-ii.aspx http://blogs.msdn.com/b/streaminsight/archive/2010/08/03/time-in-streaminsight-iii.aspx A lot of the problems I see with unresponsive or stuck streams on the MSDN Forums are to do with how Ctis are enqueued or in a lot of cases not enqueued. If you enqueue events and never enqueue a Cti then StreamInsight will be perfectly happy. You, on the other hand, will never see data on the output as you have not told StreamInsight to flush the stream. This article deals with a specific implementation problem I had recently whilst working on a StreamInsight project. I look at some possible options and discuss why they would not work before showing the way I solved the problem. The stream of data I was dealing with on this project was very bursty that is to say when events were flowing they came through very quickly and in large numbers (1000 events/sec), but when the stream calmed down it could be a few seconds between each event. When enqueuing events into the StreamInsight engne it is best practice to do so with a StartTime that is given to you by the system producing the event . StreamInsight processes events and it doesn't matter whether those events are being pushed into the engine by a source system or the events are being read from something like a flat file in a directory somewhere. You can apply the same logic and temporal algebra to both situations. Reading from a file is an excellent example of where the time of the event on the source itself is very important. We could be reading that file a long time after it was written. Being able to read the StartTime from the events allows us to define windows that will hold the correct sets of events. I was able to do this with my stream but this is where my problems started. Below is a very simple script to create a SQL Server table and populate it with sample data that will show exactly the problem I had. CREATE TABLE [dbo].[t] ( [c1] [int] PRIMARY KEY, [c2] [datetime] NULL ) INSERT t VALUES (1,'20100810'),(2,'20100810'),(3,'20100810') Column c2 defines the StartTime of the event on the source and as you can see the values in all 3 rows of data is the same. If we read Ciprian’s articles we know that we can define how Ctis get injected into the stream in 3 different places The Stream Definition The Input Factory The Input Adapter I personally have always been a fan of enqueing Ctis through the factory. Below is code typical of what I would use to do this On the class itself I do some inheriting public class SimpleInputFactory : ITypedInputAdapterFactory<SimpleInputConfig>, ITypedDeclareAdvanceTimeProperties<SimpleInputConfig> And then I implement the following function public AdapterAdvanceTimeSettings DeclareAdvanceTimeProperties<TPayload>(SimpleInputConfig configInfo, EventShape eventShape) { return new AdapterAdvanceTimeSettings( new AdvanceTimeGenerationSettings(configInfo.CtiFrequency, TimeSpan.FromTicks(-1)), AdvanceTimePolicy.Adjust); } The configInfo .CtiFrequency property is a value I pass through to define after how many events I want a Cti to be injected and this in turn will flush through the stream of data. I usually pass a value of 1 for this setting. The second parameter determines the CTI timestamp in terms of a delay relative to the events. -1 ticks in the past results in 1 tick in the future, i.e., ahead of the event. The problem with this method though is that if consecutive events have the same StartTime then only one of those events will be enqueued. In this example I use the following to define how I assign the StartTime of my events currEvent.StartTime = (DateTimeOffset)dt.c2; If I go ahead and run my StreamInsight process with this configuration i can see on the output adapter that two events have been removed To see this in a little more depth I can use the StreamInsight Debugger and see what happens internally. What is happening here is that the first event arrives and a Cti is injected with a time of 1 tick after the StartTime of that event (Also the EndTime of the event). The second event arrives and it has a StartTime of before the Cti and even though we specified AdvanceTimePolicy.Adjust on the factory we know that a point event can never be adjusted like this and the event is dropped. The same happens for the third event as well (The second and third events get trumped by the Cti). For a more detailed discussion of why this happens look here http://www.sqlis.com/sqlis/post/AdvanceTimePolicy-and-Point-Event-Streams-In-StreamInsight.aspx We end up with a single event being pushed into the output adapter and our result now makes sense. The next way I tried to solve this problem by changing the value of the second parameter to TimeSpan.Zero Here is how my factory code now looks public AdapterAdvanceTimeSettings DeclareAdvanceTimeProperties<TPayload>(SimpleInputConfig configInfo, EventShape eventShape) { return new AdapterAdvanceTimeSettings( new AdvanceTimeGenerationSettings(configInfo.CtiFrequency, TimeSpan.Zero), AdvanceTimePolicy.Adjust); } What I am doing here is declaring a policy that says inject a Cti together with every event and stamp it with a StartTime that is equal to the start time of the event itself (TimeSpan.Zero). This method has plus points as well as a downside. The upside is that no events will be lost by having the same StartTime as previous events. The Downside is that because the Cti is declared with the StartTime of the event itself then it does not actually flush that particular event because in the StreamInsight algebra, a Cti commits only those events that occurred strictly before them. To flush the events we need a Cti to be enqueued with a greater StartTime than the events themselves. Here is what happened when I ran this configuration As you can see all we got through was the Cti and none of the events. The debugger output shows the stamps on the Cti and the events themselves. Because the Cti issued has the same timestamp (StartTime) as the events then none of the events get flushed. I was nearly there but not quite. Because my stream was bursty it was possible that the next event would not come along for a few seconds and this was far too long for an event to be enqueued and not be flushed to the output adapter. I needed another solution. Two possible solutions crossed my mind although only one of them made sense when I explored it some more. Where multiple events have the same StartTime I could add 1 tick to the first event, two to the second, three to third etc thereby giving them unique StartTime values. Add a timer to manually inject Ctis The problem with the first implementation is that I would be giving the events a new StartTime. This would cause me the following problems If I want to define windows over the stream then some events may not be captured in the right windows and therefore any calculations on those windows I did would be wrong What would happen if we had 10,000 events with the same StartTime? I would enqueue them with StartTime + n ticks. Along comes a genuine event with a StartTime of the very first event + 1 tick. It is now too far in the past as far as my stream is concerned and it would be dropped. Not what I would want to do at all. I decided then to look at the Timer based solution I created a timer on my input adapter that elapsed every 200ms. private Timer tmr; public SimpleInputAdapter(SimpleInputConfig configInfo) { ctx = new SimpleTimeExtractDataContext(configInfo.ConnectionString); this.configInfo = configInfo; tmr = new Timer(200); tmr.Elapsed += new ElapsedEventHandler(t_Elapsed); tmr.Enabled = true; } void t_Elapsed(object sender, ElapsedEventArgs e) { ts = DateTime.Now - dtCtiIssued; if (ts.TotalMilliseconds >= 200 && TimerIssuedCti == false) { EnqueueCtiEvent(System.DateTime.Now.AddTicks(-100)); TimerIssuedCti = true; } }   In the t_Elapsed event handler I find out the difference in time between now and when the last event was processed (dtCtiIssued). I then check to see if that is greater than or equal to 200ms and if the last issuing of a Cti was done by the timer or by a genuine event (TimerIssuedCti). If I didn’t do this check then I would enqueue a Cti every time the timer elapsed which is not something I wanted. If the difference between the two times is greater than or equal to 500ms and the last event enqueued was by a real event then I issue a Cti through the timer to flush the event Queue, otherwise I do nothing. When I enqueue the Ctis into my stream in my ProduceEvents method I also set the values of dtCtiIssued and TimerIssuedCti   currEvent = CreateInsertEvent(); currEvent.StartTime = (DateTimeOffset)dt.c2; TimerIssuedCti = false; dtCtiIssued = currEvent.StartTime; If I go ahead and run this configuration I see the following in my output. As we can see the first Cti gets enqueued as before but then another is enqueued by the timer and because this has a later timestamp it flushes the enqueued events through the engine. Conclusion Hopefully this has shown how the enqueuing of Ctis can have a dramatic effect on the responsiveness of your output in StreamInsight. Understanding the temporal nature of the product is for me one of the most important things you can learn. I have attached my solution for the demos. It is all in one project and testing each variation is a simple matter of commenting and un-commenting the parts in the code we have been dealing with here.

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• SQL Profiler Through StreamInsight Sample Solution

  In this postI show how you can use StreamInsight to take events coming from SQL Server Profiler in real-time and do some analytics whilst the data is in flight.  Here is the solution for that post.  The download contains Project that reads events from a previously recorded trace file Project that starts a trace and captures events in real-time from a custom trace definition file (Included) It is a very simple solution and could be extended.  Whilst this example traces against SQL Server it would be trivial to change this so it profiles events in Analysis Services.       Enjoy.

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• Additional Columns in StreamInsight Event Flow Debugger

  This tool is excellent when investigating what is going on in your StreamInsight Streams.  I was looking through the menu items recently and went to Query => Event Fields I found that there were a couple of columns not added by default to the event viewer (Reminds me of the fact that the Variables viewer in SSIS hides columns also) Latency NewEndTime EnqueueTime Here they all are together.     This gives us even more information about what is going on

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• SQL Bits 7 - 30th September - 2nd October 2010 in York

  In case you haven't heard we are planning the next SQL Bits event, and today we have released the agenda for Friday & Saturday, a total of 50 sessions covering all aspects of SQL Server with a great selection of speakers. http://www.sqlbits.com/information/Agenda.aspx From our recent announcement - ...SQLBits 7 will take place over three days from Thursday September 30th to Saturday October 2nd in York. Day one will be a training day, featuring in-depth full day seminars by leading SQL Server professionals such as Chris Testa-O’Neill and Chris Webb (see http://www.sqlbits.com/information/TrainingDay.aspx for more details); day two will be a deep-dive conference day with advanced sessions delivered by the best speakers from the SQL Server community; and day three will be the traditional SQLBits community conference day, with a wide range of sessions covered all aspects of SQL Server at all levels of ability. There will be a charge to attend days one and two, but day three, Saturday October 2nd, will as usual be completely free to attend allowing everyone to attend and experience a great day of training even if they have no training budget. Full details available at http://www.sqlbits.com.

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• A deadlock was detected while trying to lock variables in SSIS

  Error: 0xC001405C at SQL Log Status: A deadlock was detected while trying to lock variables "User::RowCount" for read/write access. A lock cannot be acquired after 16 attempts. The locks timed out. Have you ever considered variable locking when building your SSIS packages? I expect many people haven’t just because most of the time you never see an error like the one above. I’ll try and explain a few key concepts about variable locking and hopefully you never will see that error. First of all, what is all this variable locking all about? Put simply SSIS variables have to be locked before they can be accessed, and then of course unlocked once you have finished with them. This is baked into SSIS, presumably to reduce the risk of race conditions, but with that comes some additional overhead in that you need to be careful to avoid lock conflicts in some scenarios. The most obvious place you will come across any hint of locking (no pun intended) is the Script Task or Script Component with their ReadOnlyVariables and ReadWriteVariables properties. These two properties allow you to enter lists of variables to be used within the task, or to put it another way, these lists of variables to be locked, so that they are available within the task. During the task pre-execute phase the variables and locked, you then use them during the execute phase when you code is run, and then unlocked for you during the post-execute phase. So by entering the variable names in one of the two list, the locking is taken care of for you, and you just read and write to the Dts.Variables collection that is exposed in the task for the purpose. As you can see in the image above, the variable PackageInt is specified, which means when I write the code inside that task I don’t have to worry about locking at all, as shown below. public void Main() { // Set the variable value to something new Dts.Variables["PackageInt"].Value = 199; // Raise an event so we can play in the event handler bool fireAgain = true; Dts.Events.FireInformation(0, "Script Task Code", "This is the script task raising an event.", null, 0, ref fireAgain); Dts.TaskResult = (int)ScriptResults.Success; } As you can see as well as accessing the variable, hassle free, I also raise an event. Now consider a scenario where I have an event hander as well as shown below. Now what if my event handler uses tries to use the same variable as well? Well obviously for the point of this post, it fails with the error quoted previously. The reason why is clearly illustrated if you consider the following sequence of events. Package execution starts Script Task in Control Flow starts Script Task in Control Flow locks the PackageInt variable as specified in the ReadWriteVariables property Script Task in Control Flow executes script, and the On Information event is raised The On Information event handler starts Script Task in On Information event handler starts Script Task in On Information event handler attempts to lock the PackageInt variable (for either read or write it doesn’t matter), but will fail because the variable is already locked. The problem is caused by the event handler task trying to use a variable that is already locked by the task in Control Flow. Events are always raised synchronously, therefore the task in Control Flow that is raising the event will not regain control until the event handler has completed, so we really do have un-resolvable locking conflict, better known as a deadlock. In this scenario we can easily resolve the problem by managing the variable locking explicitly in code, so no need to specify anything for the ReadOnlyVariables and ReadWriteVariables properties. public void Main() { // Set the variable value to something new, with explicit lock control Variables lockedVariables = null; Dts.VariableDispenser.LockOneForWrite("PackageInt", ref lockedVariables); lockedVariables["PackageInt"].Value = 199; lockedVariables.Unlock(); // Raise an event so we can play in the event handler bool fireAgain = true; Dts.Events.FireInformation(0, "Script Task Code", "This is the script task raising an event.", null, 0, ref fireAgain); Dts.TaskResult = (int)ScriptResults.Success; } Now the package will execute successfully because the variable lock has already been released by the time the event is raised, so no conflict occurs. For those of you with a SQL Engine background this should all sound strangely familiar, and boils down to getting in and out as fast as you can to reduce the risk of lock contention, be that SQL pages or SSIS variables. Unfortunately we cannot always manage the locking ourselves. The Execute SQL Task is very often used in conjunction with variables, either to pass in parameter values or get results out. Either way the task will manage the locking for you, and will fail when it cannot lock the variables it requires. The scenario outlined above is clear cut deadlock scenario, both parties are waiting on each other, so it is un-resolvable. The mechanism used within SSIS isn’t actually that clever, and whilst the message says it is a deadlock, it really just means it tried a few times, and then gave up. The last part of the error message is actually the most accurate in terms of the failure, A lock cannot be acquired after 16 attempts. The locks timed out.  Now this may come across as a recommendation to always manage locking manually in the Script Task or Script Component yourself, but I think that would be an overreaction. It is more of a reminder to be aware that in high concurrency scenarios, especially when sharing variables across multiple objects, locking is important design consideration. Update – Make sure you don’t try and use explicit locking as well as leaving the variable names in the ReadOnlyVariables and ReadWriteVariables lock lists otherwise you’ll get the deadlock error, you cannot lock a variable twice!

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• AdvanceTimePolicy and Point Event Streams In StreamInsight.

  There are a number of ways to issues CTIs (Current Time Increments) into your StreamInsight streams but a quite useful way is to do it declaratively on your source factory like this public AdapterAdvanceTimeSettings DeclareAdvanceTimeProperties<TPayload>(InputConfig configInfo, EventShape eventShape) {     return new AdapterAdvanceTimeSettings(         new AdvanceTimeGenerationSettings(configInfo.CtiFrequency, TimeSpan.FromTicks(-1)),         AdvanceTimePolicy.Adjust); } This will issue a CTI after every event and allows no delay (for delayed events) by stamping the CTI with the timestamp of the last event minus 1 tick. The very last statement "AdvanceTimePolicy.Adjust" tells the adapter what to do with events that violate the policy (arrive late).  From BOL "Events that violate the inserted CTI are moved in time if their lifetime overlaps with the CTI timestamp. That is, the start timestamp of the events is set to the most recent CTI timestamp, which renders those events valid. If both start and end time of an event fall before the CTI timestamp, then the event is dropped." This means that if you are using this method of inserting CTIs for a Point event stream and have specified "AdvanceTimePolicy.Adjust" for the violation policy, this setting will be ignored and instead it will use "AdvanceTimePolicy.Drop" because a Point event can never straddle a CTI.

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• Test-Drive ASP.NET MVC Review

  - by Ben Griswold

  A few years back I started dallying with test-driven development, but I never fully committed to the practice. This wasn’t because I didn’t believe in the value of TDD; it was more a matter of not completely understanding how to incorporate “test first” into my everyday development. Back in my web forms days, I could point fingers at the framework for my ignorance and laziness. After all, web forms weren’t exactly designed for testability so who could blame me for not embracing TDD in those conditions, right? But when I switched to ASP.NET MVC and quickly found myself fresh out of excuses and it became instantly clear that it was time to get my head around red-green-refactor once and for all or I would regretfully miss out on one of the biggest selling points the new framework had to offer. I have previously written about how I learned ASP.NET MVC. It was primarily hands on learning but I did read a couple of ASP.NET MVC books along the way. The books I read dedicated a chapter or two to TDD and they certainly addressed the benefits of TDD and how MVC was designed with testability in mind, but TDD was merely an afterthought compared to, well, teaching one how to code the model, view and controller. This approach made some sense, and I learned a bunch about MVC from those books, but when it came to TDD the books were just a teaser and an opportunity missed.  But then I got lucky – Jonathan McCracken contacted me and asked if I’d review his book, Test-Drive ASP.NET MVC, and it was just what I needed to get over the TDD hump. As the title suggests, Test-Drive ASP.NET MVC takes a different approach to learning MVC as it focuses on testing right from the very start. McCracken wastes no time and swiftly familiarizes us with the framework by building out a trivial Quote-O-Matic application and then dedicates the better part of his book to testing first – first by explaining TDD and then coding a full-featured Getting Organized application inspired by David Allen’s popular book, Getting Things Done. If you are a learn-by-example kind of coder (like me), you will instantly appreciate and enjoy McCracken’s style – its fast-moving, pragmatic and focused on only the most relevant information required to get you going with ASP.NET MVC and TDD. The book continues with the test-first theme but McCracken moves away from the sample application and incorporates other practical skills like persisting models with NHibernate, leveraging Inversion of Control with the IControllerFactory and building a RESTful web service. What I most appreciated about this section was McCracken’s use of and praise for open source libraries like Rhino Mocks, SQLite and StructureMap (to name just a few) and productivity tools like ReSharper, Web Platform Installer and ASP.NET SQL Server Setup Wizard.  McCracken’s emphasis on real world, pragmatic development was clearly demonstrated in every tool choice, straight-forward code block and developer tip. Whether one is already familiar with the tools/tips or not, McCracken’s thought process is easily understood and appreciated. The final section of the book walks the reader through security and deployment – everything from error handling and logging with ELMAH, to ASP.NET Health Monitoring, to using MSBuild with automated builds, to the deployment  of ASP.NET MVC to various web environments. These chapters, like those prior, offer enough information and explanation to simply help you get the job done.  Do I believe Test-Drive ASP.NET MVC will turn you into an expert MVC developer overnight?  Well, no.  I don’t think any book can make that claim.  If that were possible, I think book list prices would skyrocket!  That said, Test-Drive ASP.NET MVC provides a solid foundation and a unique (and dare I say necessary) approach to learning ASP.NET MVC.  Along the way McCracken shares loads of very practical software development tips and references numerous tools and libraries. The bottom line is it’s a great ASP.NET MVC primer – if you’re new to ASP.NET MVC it’s just what you need to get started.  Do I believe Test-Drive ASP.NET MVC will give you everything you need to start employing TDD in your everyday development?  Well, I used to think that learning TDD required a lot of practice and, if you’re lucky enough, the guidance of a mentor or coach.  I used to think that one couldn’t learn TDD from a book alone. Well, I’m still no pro, but I’m testing first now and Jonathan McCracken and his book, Test-Drive ASP.NET MVC, played a big part in making this happen.  If you are an MVC developer and a TDD newb, Test-Drive ASP.NET MVC is just the book for you.

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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 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 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 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 15: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 15.  As always, any feedback is welcome. # Euler 15 # http://projecteuler.net/index.php?section=problems&id=15 # Starting in the top left corner of a 2x2 grid, there # are 6 routes (without backtracking) to the bottom right # corner. How many routes are their in a 20x20 grid? import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) rows, cols = 20, 20 print factorial(rows+cols) / (factorial(rows) * factorial(cols)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• 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 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 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 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 53: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 53.  I first attempted to solve this problem using the Ruby combinations libraries. That didn’t work out so well. With a second look at the problem, the provided formula ended up being just the thing to solve the problem effectively. As always, any feedback is welcome. # Euler 53 # http://projecteuler.net/index.php?section=problems&id=53 # There are exactly ten ways of selecting three from five, # 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245, # and 345 # In combinatorics, we use the notation, 5C3 = 10. # In general, # # nCr = n! / r!(n-r)!,where r <= n, # n! = n(n1)...321, and 0! = 1. # # It is not until n = 23, that a value exceeds # one-million: 23C10 = 1144066. # In general: nCr # How many, not necessarily distinct, values of nCr, # for 1 <= n <= 100, are greater than one-million timer_start = Time.now # There's no factorial method in Ruby, I guess. class Integer # http://rosettacode.org/wiki/Factorial#Ruby def factorial (1..self).reduce(1, :*) end end def combinations(n, r) n.factorial / (r.factorial * (n-r).factorial) end answer = 0 100.downto(3) do |c| (2).upto(c-1) { |r| answer += 1 if combinations(c, r) > 1_000_000 } end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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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 51: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 51.  I know I started back up with Python this week, but I have three more Ruby solutions in the hopper and I wanted to share. For the record, Project Euler 51 was the second hardest Euler problem for me thus far. Yeah. As always, any feedback is welcome. # Euler 51 # http://projecteuler.net/index.php?section=problems&id=51 # By replacing the 1st digit of *3, it turns out that six # of the nine possible values: 13, 23, 43, 53, 73, and 83, # are all prime. # # By replacing the 3rd and 4th digits of 56**3 with the # same digit, this 5-digit number is the first example # having seven primes among the ten generated numbers, # yielding the family: 56003, 56113, 56333, 56443, # 56663, 56773, and 56993. Consequently 56003, being the # first member of this family, is the smallest prime with # this property. # # Find the smallest prime which, by replacing part of the # number (not necessarily adjacent digits) with the same # digit, is part of an eight prime value family. timer_start = Time.now require 'mathn' def eight_prime_family(prime) 0.upto(9) do |repeating_number| # Assume mask of 3 or more repeating numbers if prime.count(repeating_number.to_s) >= 3 ctr = 1 (repeating_number + 1).upto(9) do |replacement_number| family_candidate = prime.gsub(repeating_number.to_s, replacement_number.to_s) ctr += 1 if (family_candidate.to_i).prime? end return true if ctr >= 8 end end false end # Wanted to loop through primes using Prime.each # but it took too long to get to the starting value. n = 9999 while n += 2 next if !n.prime? break if eight_prime_family(n.to_s) end puts n puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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• Project Euler 10: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 10.  As always, any feedback is welcome. # Euler 10 # http://projecteuler.net/index.php?section=problems&id=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. import time start = time.time() def primes_to_max(max): primes, number = [2], 3 while number < max: 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 primes = primes_to_max(2000000) print sum(primes) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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