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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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  • Why hill climbing is called anytime algorithm?

    - by crucified soul
    From wikipedia, Anytime algorithm In computer science an anytime algorithm is an algorithm that can return a valid solution to a problem even if it's interrupted at any time before it ends. The algorithm is expected to find better and better solutions the more time it keeps running. Hill climbing Hill climbing can often produce a better result than other algorithms when the amount of time available to perform a search is limited, such as with real-time systems. It is an anytime algorithm: it can return a valid solution even if it's interrupted at any time before it ends. Hill climbing algorithm can stuck into local optima or ridge, after that even if it runs infinite time, the result won't be any better. Then, why hill climbing is called anytime algorithm?

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  • Performance Optimization &ndash; It Is Faster When You Can Measure It

    - by Alois Kraus
    Performance optimization in bigger systems is hard because the measured numbers can vary greatly depending on the measurement method of your choice. To measure execution timing of specific methods in your application you usually use Time Measurement Method Potential Pitfalls Stopwatch Most accurate method on recent processors. Internally it uses the RDTSC instruction. Since the counter is processor specific you can get greatly different values when your thread is scheduled to another core or the core goes into a power saving mode. But things do change luckily: Intel's Designer's vol3b, section 16.11.1 "16.11.1 Invariant TSC The time stamp counter in newer processors may support an enhancement, referred to as invariant TSC. Processor's support for invariant TSC is indicated by CPUID.80000007H:EDX[8]. The invariant TSC will run at a constant rate in all ACPI P-, C-. and T-states. This is the architectural behavior moving forward. On processors with invariant TSC support, the OS may use the TSC for wall clock timer services (instead of ACPI or HPET timers). TSC reads are much more efficient and do not incur the overhead associated with a ring transition or access to a platform resource." DateTime.Now Good but it has only a resolution of 16ms which can be not enough if you want more accuracy.   Reporting Method Potential Pitfalls Console.WriteLine Ok if not called too often. Debug.Print Are you really measuring performance with Debug Builds? Shame on you. Trace.WriteLine Better but you need to plug in some good output listener like a trace file. But be aware that the first time you call this method it will read your app.config and deserialize your system.diagnostics section which does also take time.   In general it is a good idea to use some tracing library which does measure the timing for you and you only need to decorate some methods with tracing so you can later verify if something has changed for the better or worse. In my previous article I did compare measuring performance with quantum mechanics. This analogy does work surprising well. When you measure a quantum system there is a lower limit how accurately you can measure something. The Heisenberg uncertainty relation does tell us that you cannot measure of a quantum system the impulse and location of a particle at the same time with infinite accuracy. For programmers the two variables are execution time and memory allocations. If you try to measure the timings of all methods in your application you will need to store them somewhere. The fastest storage space besides the CPU cache is the memory. But if your timing values do consume all available memory there is no memory left for the actual application to run. On the other hand if you try to record all memory allocations of your application you will also need to store the data somewhere. This will cost you memory and execution time. These constraints are always there and regardless how good the marketing of tool vendors for performance and memory profilers are: Any measurement will disturb the system in a non predictable way. Commercial tool vendors will tell you they do calculate this overhead and subtract it from the measured values to give you the most accurate values but in reality it is not entirely true. After falling into the trap to trust the profiler timings several times I have got into the habit to Measure with a profiler to get an idea where potential bottlenecks are. Measure again with tracing only the specific methods to check if this method is really worth optimizing. Optimize it Measure again. Be surprised that your optimization has made things worse. Think harder Implement something that really works. Measure again Finished! - Or look for the next bottleneck. Recently I have looked into issues with serialization performance. For serialization DataContractSerializer was used and I was not sure if XML is really the most optimal wire format. After looking around I have found protobuf-net which uses Googles Protocol Buffer format which is a compact binary serialization format. What is good for Google should be good for us. A small sample app to check out performance was a matter of minutes: using ProtoBuf; using System; using System.Diagnostics; using System.IO; using System.Reflection; using System.Runtime.Serialization; [DataContract, Serializable] class Data { [DataMember(Order=1)] public int IntValue { get; set; } [DataMember(Order = 2)] public string StringValue { get; set; } [DataMember(Order = 3)] public bool IsActivated { get; set; } [DataMember(Order = 4)] public BindingFlags Flags { get; set; } } class Program { static MemoryStream _Stream = new MemoryStream(); static MemoryStream Stream { get { _Stream.Position = 0; _Stream.SetLength(0); return _Stream; } } static void Main(string[] args) { DataContractSerializer ser = new DataContractSerializer(typeof(Data)); Data data = new Data { IntValue = 100, IsActivated = true, StringValue = "Hi this is a small string value to check if serialization does work as expected" }; var sw = Stopwatch.StartNew(); int Runs = 1000 * 1000; for (int i = 0; i < Runs; i++) { //ser.WriteObject(Stream, data); Serializer.Serialize<Data>(Stream, data); } sw.Stop(); Console.WriteLine("Did take {0:N0}ms for {1:N0} objects", sw.Elapsed.TotalMilliseconds, Runs); Console.ReadLine(); } } The results are indeed promising: Serializer Time in ms N objects protobuf-net   807 1000000 DataContract 4402 1000000 Nearly a factor 5 faster and a much more compact wire format. Lets use it! After switching over to protbuf-net the transfered wire data has dropped by a factor two (good) and the performance has worsened by nearly a factor two. How is that possible? We have measured it? Protobuf-net is much faster! As it turns out protobuf-net is faster but it has a cost: For the first time a type is de/serialized it does use some very smart code-gen which does not come for free. Lets try to measure this one by setting of our performance test app the Runs value not to one million but to 1. Serializer Time in ms N objects protobuf-net 85 1 DataContract 24 1 The code-gen overhead is significant and can take up to 200ms for more complex types. The break even point where the code-gen cost is amortized by its faster serialization performance is (assuming small objects) somewhere between 20.000-40.000 serialized objects. As it turned out my specific scenario involved about 100 types and 1000 serializations in total. That explains why the good old DataContractSerializer is not so easy to take out of business. The final approach I ended up was to reduce the number of types and to serialize primitive types via BinaryWriter directly which turned out to be a pretty good alternative. It sounded good until I measured again and found that my optimizations so far do not help much. After looking more deeper at the profiling data I did found that one of the 1000 calls did take 50% of the time. So how do I find out which call it was? Normal profilers do fail short at this discipline. A (totally undeserved) relatively unknown profiler is SpeedTrace which does unlike normal profilers create traces of your applications by instrumenting your IL code at runtime. This way you can look at the full call stack of the one slow serializer call to find out if this stack was something special. Unfortunately the call stack showed nothing special. But luckily I have my own tracing as well and I could see that the slow serializer call did happen during the serialization of a bool value. When you encounter after much analysis something unreasonable you cannot explain it then the chances are good that your thread was suspended by the garbage collector. If there is a problem with excessive GCs remains to be investigated but so far the serialization performance seems to be mostly ok.  When you do profile a complex system with many interconnected processes you can never be sure that the timings you just did measure are accurate at all. Some process might be hitting the disc slowing things down for all other processes for some seconds as well. There is a big difference between warm and cold startup. If you restart all processes you can basically forget the first run because of the OS disc cache, JIT and GCs make the measured timings very flexible. When you are in need of a random number generator you should measure cold startup times of a sufficiently complex system. After the first run you can try again getting different and much lower numbers. Now try again at least two times to get some feeling how stable the numbers are. Oh and try to do the same thing the next day. It might be that the bottleneck you found yesterday is gone today. Thanks to GC and other random stuff it can become pretty hard to find stuff worth optimizing if no big bottlenecks except bloatloads of code are left anymore. When I have found a spot worth optimizing I do make the code changes and do measure again to check if something has changed. If it has got slower and I am certain that my change should have made it faster I can blame the GC again. The thing is that if you optimize stuff and you allocate less objects the GC times will shift to some other location. If you are unlucky it will make your faster working code slower because you see now GCs at times where none were before. This is where the stuff does get really tricky. A safe escape hatch is to create a repro of the slow code in an isolated application so you can change things fast in a reliable manner. Then the normal profilers do also start working again. As Vance Morrison does point out it is much more complex to profile a system against the wall clock compared to optimize for CPU time. The reason is that for wall clock time analysis you need to understand how your system does work and which threads (if you have not one but perhaps 20) are causing a visible delay to the end user and which threads can wait a long time without affecting the user experience at all. Next time: Commercial profiler shootout.

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  • Understanding G1 GC Logs

    - by poonam
    The purpose of this post is to explain the meaning of GC logs generated with some tracing and diagnostic options for G1 GC. We will take a look at the output generated with PrintGCDetails which is a product flag and provides the most detailed level of information. Along with that, we will also look at the output of two diagnostic flags that get enabled with -XX:+UnlockDiagnosticVMOptions option - G1PrintRegionLivenessInfo that prints the occupancy and the amount of space used by live objects in each region at the end of the marking cycle and G1PrintHeapRegions that provides detailed information on the heap regions being allocated and reclaimed. We will be looking at the logs generated with JDK 1.7.0_04 using these options. Option -XX:+PrintGCDetails Here's a sample log of G1 collection generated with PrintGCDetails. 0.522: [GC pause (young), 0.15877971 secs] [Parallel Time: 157.1 ms] [GC Worker Start (ms): 522.1 522.2 522.2 522.2 Avg: 522.2, Min: 522.1, Max: 522.2, Diff: 0.1] [Ext Root Scanning (ms): 1.6 1.5 1.6 1.9 Avg: 1.7, Min: 1.5, Max: 1.9, Diff: 0.4] [Update RS (ms): 38.7 38.8 50.6 37.3 Avg: 41.3, Min: 37.3, Max: 50.6, Diff: 13.3] [Processed Buffers : 2 2 3 2 Sum: 9, Avg: 2, Min: 2, Max: 3, Diff: 1] [Scan RS (ms): 9.9 9.7 0.0 9.7 Avg: 7.3, Min: 0.0, Max: 9.9, Diff: 9.9] [Object Copy (ms): 106.7 106.8 104.6 107.9 Avg: 106.5, Min: 104.6, Max: 107.9, Diff: 3.3] [Termination (ms): 0.0 0.0 0.0 0.0 Avg: 0.0, Min: 0.0, Max: 0.0, Diff: 0.0] [Termination Attempts : 1 4 4 6 Sum: 15, Avg: 3, Min: 1, Max: 6, Diff: 5] [GC Worker End (ms): 679.1 679.1 679.1 679.1 Avg: 679.1, Min: 679.1, Max: 679.1, Diff: 0.1] [GC Worker (ms): 156.9 157.0 156.9 156.9 Avg: 156.9, Min: 156.9, Max: 157.0, Diff: 0.1] [GC Worker Other (ms): 0.3 0.3 0.3 0.3 Avg: 0.3, Min: 0.3, Max: 0.3, Diff: 0.0] [Clear CT: 0.1 ms] [Other: 1.5 ms] [Choose CSet: 0.0 ms] [Ref Proc: 0.3 ms] [Ref Enq: 0.0 ms] [Free CSet: 0.3 ms] [Eden: 12M(12M)->0B(10M) Survivors: 0B->2048K Heap: 13M(64M)->9739K(64M)] [Times: user=0.59 sys=0.02, real=0.16 secs] This is the typical log of an Evacuation Pause (G1 collection) in which live objects are copied from one set of regions (young OR young+old) to another set. It is a stop-the-world activity and all the application threads are stopped at a safepoint during this time. This pause is made up of several sub-tasks indicated by the indentation in the log entries. Here's is the top most line that gets printed for the Evacuation Pause. 0.522: [GC pause (young), 0.15877971 secs] This is the highest level information telling us that it is an Evacuation Pause that started at 0.522 secs from the start of the process, in which all the regions being evacuated are Young i.e. Eden and Survivor regions. This collection took 0.15877971 secs to finish. Evacuation Pauses can be mixed as well. In which case the set of regions selected include all of the young regions as well as some old regions. 1.730: [GC pause (mixed), 0.32714353 secs] Let's take a look at all the sub-tasks performed in this Evacuation Pause. [Parallel Time: 157.1 ms] Parallel Time is the total elapsed time spent by all the parallel GC worker threads. The following lines correspond to the parallel tasks performed by these worker threads in this total parallel time, which in this case is 157.1 ms. [GC Worker Start (ms): 522.1 522.2 522.2 522.2Avg: 522.2, Min: 522.1, Max: 522.2, Diff: 0.1] The first line tells us the start time of each of the worker thread in milliseconds. The start times are ordered with respect to the worker thread ids – thread 0 started at 522.1ms and thread 1 started at 522.2ms from the start of the process. The second line tells the Avg, Min, Max and Diff of the start times of all of the worker threads. [Ext Root Scanning (ms): 1.6 1.5 1.6 1.9 Avg: 1.7, Min: 1.5, Max: 1.9, Diff: 0.4] This gives us the time spent by each worker thread scanning the roots (globals, registers, thread stacks and VM data structures). Here, thread 0 took 1.6ms to perform the root scanning task and thread 1 took 1.5 ms. The second line clearly shows the Avg, Min, Max and Diff of the times spent by all the worker threads. [Update RS (ms): 38.7 38.8 50.6 37.3 Avg: 41.3, Min: 37.3, Max: 50.6, Diff: 13.3] Update RS gives us the time each thread spent in updating the Remembered Sets. Remembered Sets are the data structures that keep track of the references that point into a heap region. Mutator threads keep changing the object graph and thus the references that point into a particular region. We keep track of these changes in buffers called Update Buffers. The Update RS sub-task processes the update buffers that were not able to be processed concurrently, and updates the corresponding remembered sets of all regions. [Processed Buffers : 2 2 3 2Sum: 9, Avg: 2, Min: 2, Max: 3, Diff: 1] This tells us the number of Update Buffers (mentioned above) processed by each worker thread. [Scan RS (ms): 9.9 9.7 0.0 9.7 Avg: 7.3, Min: 0.0, Max: 9.9, Diff: 9.9] These are the times each worker thread had spent in scanning the Remembered Sets. Remembered Set of a region contains cards that correspond to the references pointing into that region. This phase scans those cards looking for the references pointing into all the regions of the collection set. [Object Copy (ms): 106.7 106.8 104.6 107.9 Avg: 106.5, Min: 104.6, Max: 107.9, Diff: 3.3] These are the times spent by each worker thread copying live objects from the regions in the Collection Set to the other regions. [Termination (ms): 0.0 0.0 0.0 0.0 Avg: 0.0, Min: 0.0, Max: 0.0, Diff: 0.0] Termination time is the time spent by the worker thread offering to terminate. But before terminating, it checks the work queues of other threads and if there are still object references in other work queues, it tries to steal object references, and if it succeeds in stealing a reference, it processes that and offers to terminate again. [Termination Attempts : 1 4 4 6 Sum: 15, Avg: 3, Min: 1, Max: 6, Diff: 5] This gives the number of times each thread has offered to terminate. [GC Worker End (ms): 679.1 679.1 679.1 679.1 Avg: 679.1, Min: 679.1, Max: 679.1, Diff: 0.1] These are the times in milliseconds at which each worker thread stopped. [GC Worker (ms): 156.9 157.0 156.9 156.9 Avg: 156.9, Min: 156.9, Max: 157.0, Diff: 0.1] These are the total lifetimes of each worker thread. [GC Worker Other (ms): 0.3 0.3 0.3 0.3Avg: 0.3, Min: 0.3, Max: 0.3, Diff: 0.0] These are the times that each worker thread spent in performing some other tasks that we have not accounted above for the total Parallel Time. [Clear CT: 0.1 ms] This is the time spent in clearing the Card Table. This task is performed in serial mode. [Other: 1.5 ms] Time spent in the some other tasks listed below. The following sub-tasks (which individually may be parallelized) are performed serially. [Choose CSet: 0.0 ms] Time spent in selecting the regions for the Collection Set. [Ref Proc: 0.3 ms] Total time spent in processing Reference objects. [Ref Enq: 0.0 ms] Time spent in enqueuing references to the ReferenceQueues. [Free CSet: 0.3 ms] Time spent in freeing the collection set data structure. [Eden: 12M(12M)->0B(13M) Survivors: 0B->2048K Heap: 14M(64M)->9739K(64M)] This line gives the details on the heap size changes with the Evacuation Pause. This shows that Eden had the occupancy of 12M and its capacity was also 12M before the collection. After the collection, its occupancy got reduced to 0 since everything is evacuated/promoted from Eden during a collection, and its target size grew to 13M. The new Eden capacity of 13M is not reserved at this point. This value is the target size of the Eden. Regions are added to Eden as the demand is made and when the added regions reach to the target size, we start the next collection. Similarly, Survivors had the occupancy of 0 bytes and it grew to 2048K after the collection. The total heap occupancy and capacity was 14M and 64M receptively before the collection and it became 9739K and 64M after the collection. Apart from the evacuation pauses, G1 also performs concurrent-marking to build the live data information of regions. 1.416: [GC pause (young) (initial-mark), 0.62417980 secs] ….... 2.042: [GC concurrent-root-region-scan-start] 2.067: [GC concurrent-root-region-scan-end, 0.0251507] 2.068: [GC concurrent-mark-start] 3.198: [GC concurrent-mark-reset-for-overflow] 4.053: [GC concurrent-mark-end, 1.9849672 sec] 4.055: [GC remark 4.055: [GC ref-proc, 0.0000254 secs], 0.0030184 secs] [Times: user=0.00 sys=0.00, real=0.00 secs] 4.088: [GC cleanup 117M->106M(138M), 0.0015198 secs] [Times: user=0.00 sys=0.00, real=0.00 secs] 4.090: [GC concurrent-cleanup-start] 4.091: [GC concurrent-cleanup-end, 0.0002721] The first phase of a marking cycle is Initial Marking where all the objects directly reachable from the roots are marked and this phase is piggy-backed on a fully young Evacuation Pause. 2.042: [GC concurrent-root-region-scan-start] This marks the start of a concurrent phase that scans the set of root-regions which are directly reachable from the survivors of the initial marking phase. 2.067: [GC concurrent-root-region-scan-end, 0.0251507] End of the concurrent root region scan phase and it lasted for 0.0251507 seconds. 2.068: [GC concurrent-mark-start] Start of the concurrent marking at 2.068 secs from the start of the process. 3.198: [GC concurrent-mark-reset-for-overflow] This indicates that the global marking stack had became full and there was an overflow of the stack. Concurrent marking detected this overflow and had to reset the data structures to start the marking again. 4.053: [GC concurrent-mark-end, 1.9849672 sec] End of the concurrent marking phase and it lasted for 1.9849672 seconds. 4.055: [GC remark 4.055: [GC ref-proc, 0.0000254 secs], 0.0030184 secs] This corresponds to the remark phase which is a stop-the-world phase. It completes the left over marking work (SATB buffers processing) from the previous phase. In this case, this phase took 0.0030184 secs and out of which 0.0000254 secs were spent on Reference processing. 4.088: [GC cleanup 117M->106M(138M), 0.0015198 secs] Cleanup phase which is again a stop-the-world phase. It goes through the marking information of all the regions, computes the live data information of each region, resets the marking data structures and sorts the regions according to their gc-efficiency. In this example, the total heap size is 138M and after the live data counting it was found that the total live data size dropped down from 117M to 106M. 4.090: [GC concurrent-cleanup-start] This concurrent cleanup phase frees up the regions that were found to be empty (didn't contain any live data) during the previous stop-the-world phase. 4.091: [GC concurrent-cleanup-end, 0.0002721] Concurrent cleanup phase took 0.0002721 secs to free up the empty regions. Option -XX:G1PrintRegionLivenessInfo Now, let's look at the output generated with the flag G1PrintRegionLivenessInfo. This is a diagnostic option and gets enabled with -XX:+UnlockDiagnosticVMOptions. G1PrintRegionLivenessInfo prints the live data information of each region during the Cleanup phase of the concurrent-marking cycle. 26.896: [GC cleanup ### PHASE Post-Marking @ 26.896### HEAP committed: 0x02e00000-0x0fe00000 reserved: 0x02e00000-0x12e00000 region-size: 1048576 Cleanup phase of the concurrent-marking cycle started at 26.896 secs from the start of the process and this live data information is being printed after the marking phase. Committed G1 heap ranges from 0x02e00000 to 0x0fe00000 and the total G1 heap reserved by JVM is from 0x02e00000 to 0x12e00000. Each region in the G1 heap is of size 1048576 bytes. ### type address-range used prev-live next-live gc-eff### (bytes) (bytes) (bytes) (bytes/ms) This is the header of the output that tells us about the type of the region, address-range of the region, used space in the region, live bytes in the region with respect to the previous marking cycle, live bytes in the region with respect to the current marking cycle and the GC efficiency of that region. ### FREE 0x02e00000-0x02f00000 0 0 0 0.0 This is a Free region. ### OLD 0x02f00000-0x03000000 1048576 1038592 1038592 0.0 Old region with address-range from 0x02f00000 to 0x03000000. Total used space in the region is 1048576 bytes, live bytes as per the previous marking cycle are 1038592 and live bytes with respect to the current marking cycle are also 1038592. The GC efficiency has been computed as 0. ### EDEN 0x03400000-0x03500000 20992 20992 20992 0.0 This is an Eden region. ### HUMS 0x0ae00000-0x0af00000 1048576 1048576 1048576 0.0### HUMC 0x0af00000-0x0b000000 1048576 1048576 1048576 0.0### HUMC 0x0b000000-0x0b100000 1048576 1048576 1048576 0.0### HUMC 0x0b100000-0x0b200000 1048576 1048576 1048576 0.0### HUMC 0x0b200000-0x0b300000 1048576 1048576 1048576 0.0### HUMC 0x0b300000-0x0b400000 1048576 1048576 1048576 0.0### HUMC 0x0b400000-0x0b500000 1001480 1001480 1001480 0.0 These are the continuous set of regions called Humongous regions for storing a large object. HUMS (Humongous starts) marks the start of the set of humongous regions and HUMC (Humongous continues) tags the subsequent regions of the humongous regions set. ### SURV 0x09300000-0x09400000 16384 16384 16384 0.0 This is a Survivor region. ### SUMMARY capacity: 208.00 MB used: 150.16 MB / 72.19 % prev-live: 149.78 MB / 72.01 % next-live: 142.82 MB / 68.66 % At the end, a summary is printed listing the capacity, the used space and the change in the liveness after the completion of concurrent marking. In this case, G1 heap capacity is 208MB, total used space is 150.16MB which is 72.19% of the total heap size, live data in the previous marking was 149.78MB which was 72.01% of the total heap size and the live data as per the current marking is 142.82MB which is 68.66% of the total heap size. Option -XX:+G1PrintHeapRegions G1PrintHeapRegions option logs the regions related events when regions are committed, allocated into or are reclaimed. COMMIT/UNCOMMIT events G1HR COMMIT [0x6e900000,0x6ea00000]G1HR COMMIT [0x6ea00000,0x6eb00000] Here, the heap is being initialized or expanded and the region (with bottom: 0x6eb00000 and end: 0x6ec00000) is being freshly committed. COMMIT events are always generated in order i.e. the next COMMIT event will always be for the uncommitted region with the lowest address. G1HR UNCOMMIT [0x72700000,0x72800000]G1HR UNCOMMIT [0x72600000,0x72700000] Opposite to COMMIT. The heap got shrunk at the end of a Full GC and the regions are being uncommitted. Like COMMIT, UNCOMMIT events are also generated in order i.e. the next UNCOMMIT event will always be for the committed region with the highest address. GC Cycle events G1HR #StartGC 7G1HR CSET 0x6e900000G1HR REUSE 0x70500000G1HR ALLOC(Old) 0x6f800000G1HR RETIRE 0x6f800000 0x6f821b20G1HR #EndGC 7 This shows start and end of an Evacuation pause. This event is followed by a GC counter tracking both evacuation pauses and Full GCs. Here, this is the 7th GC since the start of the process. G1HR #StartFullGC 17G1HR UNCOMMIT [0x6ed00000,0x6ee00000]G1HR POST-COMPACTION(Old) 0x6e800000 0x6e854f58G1HR #EndFullGC 17 Shows start and end of a Full GC. This event is also followed by the same GC counter as above. This is the 17th GC since the start of the process. ALLOC events G1HR ALLOC(Eden) 0x6e800000 The region with bottom 0x6e800000 just started being used for allocation. In this case it is an Eden region and allocated into by a mutator thread. G1HR ALLOC(StartsH) 0x6ec00000 0x6ed00000G1HR ALLOC(ContinuesH) 0x6ed00000 0x6e000000 Regions being used for the allocation of Humongous object. The object spans over two regions. G1HR ALLOC(SingleH) 0x6f900000 0x6f9eb010 Single region being used for the allocation of Humongous object. G1HR COMMIT [0x6ee00000,0x6ef00000]G1HR COMMIT [0x6ef00000,0x6f000000]G1HR COMMIT [0x6f000000,0x6f100000]G1HR COMMIT [0x6f100000,0x6f200000]G1HR ALLOC(StartsH) 0x6ee00000 0x6ef00000G1HR ALLOC(ContinuesH) 0x6ef00000 0x6f000000G1HR ALLOC(ContinuesH) 0x6f000000 0x6f100000G1HR ALLOC(ContinuesH) 0x6f100000 0x6f102010 Here, Humongous object allocation request could not be satisfied by the free committed regions that existed in the heap, so the heap needed to be expanded. Thus new regions are committed and then allocated into for the Humongous object. G1HR ALLOC(Old) 0x6f800000 Old region started being used for allocation during GC. G1HR ALLOC(Survivor) 0x6fa00000 Region being used for copying old objects into during a GC. Note that Eden and Humongous ALLOC events are generated outside the GC boundaries and Old and Survivor ALLOC events are generated inside the GC boundaries. Other Events G1HR RETIRE 0x6e800000 0x6e87bd98 Retire and stop using the region having bottom 0x6e800000 and top 0x6e87bd98 for allocation. Note that most regions are full when they are retired and we omit those events to reduce the output volume. A region is retired when another region of the same type is allocated or we reach the start or end of a GC(depending on the region). So for Eden regions: For example: 1. ALLOC(Eden) Foo2. ALLOC(Eden) Bar3. StartGC At point 2, Foo has just been retired and it was full. At point 3, Bar was retired and it was full. If they were not full when they were retired, we will have a RETIRE event: 1. ALLOC(Eden) Foo2. RETIRE Foo top3. ALLOC(Eden) Bar4. StartGC G1HR CSET 0x6e900000 Region (bottom: 0x6e900000) is selected for the Collection Set. The region might have been selected for the collection set earlier (i.e. when it was allocated). However, we generate the CSET events for all regions in the CSet at the start of a GC to make sure there's no confusion about which regions are part of the CSet. G1HR POST-COMPACTION(Old) 0x6e800000 0x6e839858 POST-COMPACTION event is generated for each non-empty region in the heap after a full compaction. A full compaction moves objects around, so we don't know what the resulting shape of the heap is (which regions were written to, which were emptied, etc.). To deal with this, we generate a POST-COMPACTION event for each non-empty region with its type (old/humongous) and the heap boundaries. At this point we should only have Old and Humongous regions, as we have collapsed the young generation, so we should not have eden and survivors. POST-COMPACTION events are generated within the Full GC boundary. G1HR CLEANUP 0x6f400000G1HR CLEANUP 0x6f300000G1HR CLEANUP 0x6f200000 These regions were found empty after remark phase of Concurrent Marking and are reclaimed shortly afterwards. G1HR #StartGC 5G1HR CSET 0x6f400000G1HR CSET 0x6e900000G1HR REUSE 0x6f800000 At the end of a GC we retire the old region we are allocating into. Given that its not full, we will carry on allocating into it during the next GC. This is what REUSE means. In the above case 0x6f800000 should have been the last region with an ALLOC(Old) event during the previous GC and should have been retired before the end of the previous GC. G1HR ALLOC-FORCE(Eden) 0x6f800000 A specialization of ALLOC which indicates that we have reached the max desired number of the particular region type (in this case: Eden), but we decided to allocate one more. Currently it's only used for Eden regions when we extend the young generation because we cannot do a GC as the GC-Locker is active. G1HR EVAC-FAILURE 0x6f800000 During a GC, we have failed to evacuate an object from the given region as the heap is full and there is no space left to copy the object. This event is generated within GC boundaries and exactly once for each region from which we failed to evacuate objects. When Heap Regions are reclaimed ? It is also worth mentioning when the heap regions in the G1 heap are reclaimed. All regions that are in the CSet (the ones that appear in CSET events) are reclaimed at the end of a GC. The exception to that are regions with EVAC-FAILURE events. All regions with CLEANUP events are reclaimed. After a Full GC some regions get reclaimed (the ones from which we moved the objects out). But that is not shown explicitly, instead the non-empty regions that are left in the heap are printed out with the POST-COMPACTION events.

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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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  • OS Analytics with Oracle Enterprise Manager (by Eran Steiner)

    - by Zeynep Koch
    Oracle Enterprise Manager Ops Center provides a feature called "OS Analytics". This feature allows you to get a better understanding of how the Operating System is being utilized. You can research the historical usage as well as real time data. This post will show how you can benefit from OS Analytics and how it works behind the scenes. The recording of our call to discuss this blog is available here: https://oracleconferencing.webex.com/oracleconferencing/ldr.php?AT=pb&SP=MC&rID=71517797&rKey=4ec9d4a3508564b3Download the presentation here See also: Blog about Alert Monitoring and Problem Notification Blog about Using Operational Profiles to Install Packages and other content Here is quick summary of what you can do with OS Analytics in Ops Center: View historical charts and real time value of CPU, memory, network and disk utilization Find the top CPU and Memory processes in real time or at a certain historical day Determine proper monitoring thresholds based on historical data Drill down into a process details Where to start To start with OS Analytics, choose the OS asset in the tree and click the Analytics tab. You can see the CPU utilization, Memory utilization and Network utilization, along with the current real time top 5 processes in each category (click the image to see a larger version):  In the above screen, you can click each of the top 5 processes to see a more detailed view of that process. Here is an example of one of the processes: One of the cool things is that you can see the process tree for this process along with some port binding and open file descriptors. Next, click the "Processes" tab to see real time information of all the processes on the machine: An interesting column is the "Target" column. If you configured Ops Center to work with Enterprise Manager Cloud Control, then the two products will talk to each other and Ops Center will display the correlated target from Cloud Control in this table. If you are only using Ops Center - this column will remain empty. The "Threshold" tab is particularly helpful - you can view historical trends of different monitored values and based on the graph - determine what the monitoring values should be: You can ask Ops Center to suggest monitoring levels based on the historical values or you can set your own. The different colors in the graph represent the current set levels: Red for critical, Yellow for warning and Blue for Information, allowing you to quickly see how they're positioned against real data. It's important to note that when looking at longer periods, Ops Center smooths out the data and uses averages. So when looking at values such as CPU Usage, try shorter time frames which are more detailed, such as one hour or one day. Applying new monitoring values When first applying new values to monitored attributes - a popup will come up asking if it's OK to get you out of the current Monitoring Policy. This is OK if you want to either have custom monitoring for a specific machine, or if you want to use this current machine as a "Gold image" and extract a Monitoring Policy from it. You can later apply the new Monitoring Policy to other machines and also set it as a default Monitoring Profile. Once you're done with applying the different monitoring values, you can review and change them in the "Monitoring" tab. You can also click the "Extract a Monitoring Policy" in the actions pane on the right to save all the new values to a new Monitoring Policy, which can then be found under "Plan Management" -> "Monitoring Policies". Visiting the past Under the "History" tab you can "go back in time". This is very helpful when you know that a machine was busy a few hours ago (perhaps in the middle of the night?), but you were not around to take a look at it in real time. Here's a view into yesterday's data on one of the machines: You can see an interesting CPU spike happening at around 3:30 am along with some memory use. In the bottom table you can see the top 5 CPU and Memory consumers at the requested time. Very quickly you can see that this spike is related to the Solaris 11 IPS repository synchronization process using the "pkgrecv" command. The "time machine" doesn't stop here - you can also view historical data to determine which of the zones was the busiest at a given time: Under the hood The data collected is stored on each of the agents under /var/opt/sun/xvm/analytics/historical/ An "os.zip" file exists for the main OS. Inside you will find many small text files, named after the Epoch time stamp in which they were taken If you have any zones, there will be a file called "guests.zip" containing the same small files for all the zones, as well as a folder with the name of the zone along with "os.zip" in it If this is the Enterprise Controller or the Proxy Controller, you will have folders called "proxy" and "sat" in which you will find the "os.zip" for that controller The actual script collecting the data can be viewed for debugging purposes as well: On Linux, the location is: /opt/sun/xvmoc/private/os_analytics/collect If you would like to redirect all the standard error into a file for debugging, touch the following file and the output will go into it: # touch /tmp/.collect.stderr   The temporary data is collected under /var/opt/sun/xvm/analytics/.collectdb until it is zipped. If you would like to review the properties for the Analytics, you can view those per each agent in /opt/sun/n1gc/lib/XVM.properties. Find the section "Analytics configurable properties for OS and VSC" to view the Analytics specific values. I hope you find this helpful! Please post questions in the comments below. Eran Steiner

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  • Project Euler 9: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 9.  As always, any feedback is welcome. # Euler 9 # http://projecteuler.net/index.php?section=problems&id=9 # A Pythagorean triplet is a set of three natural numbers, # a b c, for which, # a2 + b2 = c2 # For example, 32 + 42 = 9 + 16 = 25 = 52. # There exists exactly one Pythagorean triplet for which # a + b + c = 1000. Find the product abc. import time start = time.time() product = 0 def pythagorean_triplet(): for a in range(1,501): for b in xrange(a+1,501): c = 1000 - a - b if (a*a + b*b == c*c): return a*b*c print pythagorean_triplet() print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 5: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 5.  As always, any feedback is welcome. # Euler 5 # http://projecteuler.net/index.php?section=problems&id=5 # 2520 is the smallest number that can be divided by each # of the numbers from 1 to 10 without any remainder. # What is the smallest positive number that is evenly # divisible by all of the numbers from 1 to 20? import time start = time.time() def gcd(a, b): while b: a, b = b, a % b return a def lcm(a, b): return a * b // gcd(a, b) print reduce(lcm, range(1, 20)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • New OTL Top Error Documents

    - by Oracle_EBS
    We would like to take this opportunity to announce new documents that are aimed at easing your experience when faced with troubleshooting Oracle Time and Labor issues. To this end we would like to highlight related and updated documentation regarding the top most reported OTL issues. Similar to the iRecruitment top error document updates announced in our EBS HCM Newsletter for December 2011, we proactively analyzed the issues reported on Oracle Time and Labor, identifying and consolidating knowledge content for the top 3 - 4 error messages in My Oracle Support documents. These new documents are as follows: Document Content Type Note ID: Oracle Time and Labor (OTL) Timekeeper issues Functional 1380612.1 Oracle Time and Labor (OTL) Approval issues Functional 1383990.1 Oracle Time and Labor (OTL) Retrieval issues Functional 1385426.1 These documents are now available via our Oracle Time and Labor Information Center Doc ID 1293475.1. As always, we very much welcome your feedback should you use these documents. Please add your views by using the "Rate This Document" feature should you wish to share your experience and any further improvement suggestions.

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  • Project Euler 8: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Unity3D: How To Smoothly Switch From One Camera To Another

    - by www.Sillitoy.com
    The Question is basically self explanatory. I have a scene with many cameras and I'd like to smoothly switch from one to another. I am not looking for a cross fade effect but more to a camera moving and rotating the view in order to reach the next camera point of view and so on. To this end I have tried the following code: firstCamera.transform.position.x = Mathf.Lerp(firstCamera.transform.position.x, nextCamer.transform.position.x,Time.deltaTime*smooth); firstCamera.transform.position.y = Mathf.Lerp(firstCamera.transform.position.y, nextCamera.transform.position.y,Time.deltaTime*smooth); firstCamera.transform.position.z = Mathf.Lerp(firstCamera.transform.position.z, nextCamera.transform.position.z,Time.deltaTime*smooth); firstCamera.transform.rotation.x = Mathf.Lerp(firstCamera.transform.rotation.x, nextCamera.transform.rotation.x,Time.deltaTime*smooth); firstCamera.transform.rotation.z = Mathf.Lerp(firstCamera.transform.rotation.z, nextCamera.transform.rotation.z,Time.deltaTime*smooth); firstCamera.transform.rotation.y = Mathf.Lerp(firstCamera.transform.rotation.y, nextCamera.transform.rotation.y,Time.deltaTime*smooth); But the result is actually not that good.

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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 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 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 13: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 6: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 6.  As always, any feedback is welcome. # Euler 6 # http://projecteuler.net/index.php?section=problems&id=6 # Find the difference between the sum of the squares of # the first one hundred natural numbers and the square # of the sum. import time start = time.time() square_of_sums = sum(range(1,101)) ** 2 sum_of_squares = reduce(lambda agg, i: agg+i**2, range(1,101)) print square_of_sums - sum_of_squares print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Project Euler 20: (Iron)Python

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

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  • Project Euler 1: (Iron)Python

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

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  • Project Euler 3: (Iron)Python

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

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  • How To Smoothly Animate From One Camera Position To Another

    - by www.Sillitoy.com
    The Question is basically self explanatory. I have a scene with many cameras and I'd like to smoothly switch from one to another. I am not looking for a cross fade effect but more to a camera moving and rotating the view in order to reach the next camera point of view and so on. To this end I have tried the following code: firstCamera.transform.position.x = Mathf.Lerp(firstCamera.transform.position.x, nextCamer.transform.position.x,Time.deltaTime*smooth); firstCamera.transform.position.y = Mathf.Lerp(firstCamera.transform.position.y, nextCamera.transform.position.y,Time.deltaTime*smooth); firstCamera.transform.position.z = Mathf.Lerp(firstCamera.transform.position.z, nextCamera.transform.position.z,Time.deltaTime*smooth); firstCamera.transform.rotation.x = Mathf.Lerp(firstCamera.transform.rotation.x, nextCamera.transform.rotation.x,Time.deltaTime*smooth); firstCamera.transform.rotation.z = Mathf.Lerp(firstCamera.transform.rotation.z, nextCamera.transform.rotation.z,Time.deltaTime*smooth); firstCamera.transform.rotation.y = Mathf.Lerp(firstCamera.transform.rotation.y, nextCamera.transform.rotation.y,Time.deltaTime*smooth); But the result is actually not that good.

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  • Setting Up and Using a WebLogic Cluster –Webcast October 17th 2012

    - by JuergenKress
    Date and time: Wednesday, October 17, 2012 8:00 am Pacific Daylight Time (San Francisco, GMT-07:00) Change time zone Wednesday, October 17, 2012 4:00 pm GMT Summer Time (London, GMT+01:00) Wednesday, October 17, 2012 11:00 am Eastern Daylight Time (New York, GMT-04:00) Wednesday, October 17, 2012 8:00 am Pacific Daylight Time (San Francisco, GMT-07:00) Duration: 1 hour Description: This one-hour session is recommended for administrators and developpers who work with Oracle Weblogic Server. The focus in this presentation and demos is to go through entire cycle of cluster configuration, best practices and troubleshooting capabilities. * Configuration * Best practices * Troubleshooting and Debugging capabilities Details and registration WebLogic Partner Community For regular information become a member in the WebLogic Partner Community please visit: http://www.oracle.com/partners/goto/wls-emea ( OPN account required). If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Mix Forum Wiki Technorati Tags: WebLogic Cluster,education,ExaLogic,Exalogic training,training,Exalogic roadmap,exalogic installation,WebLogic Community,Oracle,OPN,Jürgen Kress

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  • Help to solve "Robbery Problem"

    - by peiska
    Hello, Can anybody help me with this problem in C or Java? The problem is taken from here: http://acm.pku.edu.cn/JudgeOnline/problem?id=1104 Inspector Robstop is very angry. Last night, a bank has been robbed and the robber has not been caught. And this happened already for the third time this year, even though he did everything in his power to stop the robber: as quickly as possible, all roads leading out of the city were blocked, making it impossible for the robber to escape. Then, the inspector asked all the people in the city to watch out for the robber, but the only messages he got were of the form "We don't see him." But this time, he has had enough! Inspector Robstop decides to analyze how the robber could have escaped. To do that, he asks you to write a program which takes all the information the inspector could get about the robber in order to find out where the robber has been at which time. Coincidentally, the city in which the bank was robbed has a rectangular shape. The roads leaving the city are blocked for a certain period of time t, and during that time, several observations of the form "The robber isn't in the rectangle Ri at time ti" are reported. Assuming that the robber can move at most one unit per time step, your program must try to find the exact position of the robber at each time step. Input The input contains the description of several robberies. The first line of each description consists of three numbers W, H, t (1 <= W,H,t <= 100) where W is the width, H the height of the city and t is the time during which the city is locked. The next contains a single integer n (0 <= n <= 100), the number of messages the inspector received. The next n lines (one for each of the messages) consist of five integers ti, Li, Ti, Ri, Bi each. The integer ti is the time at which the observation has been made (1 <= ti <= t), and Li, Ti, Ri, Bi are the left, top, right and bottom respectively of the (rectangular) area which has been observed. (1 <= Li <= Ri <= W, 1 <= Ti <= Bi <= H; the point (1, 1) is the upper left hand corner, and (W, H) is the lower right hand corner of the city.) The messages mean that the robber was not in the given rectangle at time ti. The input is terminated by a test case starting with W = H = t = 0. This case should not be processed. Output For each robbery, first output the line "Robbery #k:", where k is the number of the robbery. Then, there are three possibilities: If it is impossible that the robber is still in the city considering the messages, output the line "The robber has escaped." In all other cases, assume that the robber really is in the city. Output one line of the form "Time step : The robber has been at x,y." for each time step, in which the exact location can be deduced. (x and y are the column resp. row of the robber in time step .) Output these lines ordered by time . If nothing can be deduced, output the line "Nothing known." and hope that the inspector will not get even more angry. Output a blank line after each processed case.

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  • WMI permissions: Select CommandLine, ProcessId FROM Win32_Process returns no data for CommandLine

    - by user57935
    Hi all, I am gathering performance data via WMI and would like to avoid having to use an account in the Administrators group for this purpose. The target machine is running Windows Server 2003 with the latest SP/updates. I've done what I believe to be the appropriate configuration to allow our user access to WMI (similar to what is described here: http://msdn.microsoft.com/en-us/library/aa393266.aspx). Here are the specific steps that were followed: Open Administrative Tools - Computer Management: Under Computer Management (Local) Expand Services and Applications, right click WMI Control and select properties. In the Security tab, expand Root, highlight CIMV2, click Security (near bottom of window); add Performance Monitor Users and enable the options : Enable Account and Remote Enable. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - Right click My Computer and select properties, select the COM security tab, in “Access Permissions” click "Edit Default" select(or add then select) “Performance Monitor Users” group and allow local access and remote access and click ok. In “Launch and Activation Permissions” click “Edit Default” select(or add then select) “Performance Monitor Users” group and allow Local and Remote Launch and Activation Permissions. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - My Computer - DCOM Config - highlight “Windows Management and Instrumentation” right click and select properties, Select the Security tab, Under “Launch and Activation Permissions” select Customize, then click edit, add the “Performance Users Group” and allow local and remote Remote Launch and Remote Activation privileges. I am able to connect remotely via WMI Explorer but when I perform this query: Select CommandLine, ProcessId FROM Win32_Process I get a valid result but every row has an empty CommandLine. If I add the user to the Administrators group and re-run the query, the CommandLine column contains the expected data. It seems there is a permission I am missing somewhere but I am not having much luck tracking it down. Many thanks in advance.

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  • WMI permissions: Select CommandLine, ProcessId FROM Win32_Process returns no data for CommandLine

    - by user57935
    I am gathering performance data via WMI and would like to avoid having to use an account in the Administrators group for this purpose. The target machine is running Windows Server 2003 with the latest SP/updates. I've done what I believe to be the appropriate configuration to allow our user access to WMI (similar to what is described here: http://msdn.microsoft.com/en-us/library/aa393266.aspx). Here are the specific steps that were followed: Open Administrative Tools - Computer Management: Under Computer Management (Local) Expand Services and Applications, right click WMI Control and select properties. In the Security tab, expand Root, highlight CIMV2, click Security (near bottom of window); add Performance Monitor Users and enable the options : Enable Account and Remote Enable. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - Right click My Computer and select properties, select the COM security tab, in “Access Permissions” click "Edit Default" select(or add then select) “Performance Monitor Users” group and allow local access and remote access and click ok. In “Launch and Activation Permissions” click “Edit Default” select(or add then select) “Performance Monitor Users” group and allow Local and Remote Launch and Activation Permissions. ­Open Administrative Tools - Component Services: Under Console Root go to Component Services- Computers - My Computer - DCOM Config - highlight “Windows Management and Instrumentation” right click and select properties, Select the Security tab, Under “Launch and Activation Permissions” select Customize, then click edit, add the “Performance Users Group” and allow local and remote Remote Launch and Remote Activation privileges. I am able to connect remotely via WMI Explorer but when I perform this query: Select CommandLine, ProcessId FROM Win32_Process I get a valid result but every row has an empty CommandLine. If I add the user to the Administrators group and re-run the query, the CommandLine column contains the expected data. It seems there is a permission I am missing somewhere but I am not having much luck tracking it down. Many thanks in advance.

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