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  • SQL SERVER – Disk Space Monitoring – Detecting Low Disk Space on Server

    - by Pinal Dave
    A very common question I often receive is how to detect if the disk space is running low on SQL Server. There are two different ways to do the same. I personally prefer method 2 as that is very easy to use and I can use it creatively along with database name. Method 1: EXEC MASTER..xp_fixeddrives GO Above query will return us two columns, drive name and MB free. If we want to use this data in our query, we will have to create a temporary table and insert the data from this stored procedure into the temporary table and use it. Method 2: SELECT DISTINCT dovs.logical_volume_name AS LogicalName, dovs.volume_mount_point AS Drive, CONVERT(INT,dovs.available_bytes/1048576.0) AS FreeSpaceInMB FROM sys.master_files mf CROSS APPLY sys.dm_os_volume_stats(mf.database_id, mf.FILE_ID) dovs ORDER BY FreeSpaceInMB ASC GO The above query will give us three columns: drive logical name, drive letter and free space in MB. We can further modify above query to also include database name in the query as well. SELECT DISTINCT DB_NAME(dovs.database_id) DBName, dovs.logical_volume_name AS LogicalName, dovs.volume_mount_point AS Drive, CONVERT(INT,dovs.available_bytes/1048576.0) AS FreeSpaceInMB FROM sys.master_files mf CROSS APPLY sys.dm_os_volume_stats(mf.database_id, mf.FILE_ID) dovs ORDER BY FreeSpaceInMB ASC GO This will give us additional data about which database is placed on which drive. If you see a database name multiple times, it is because your database has multiple files and they are on different drives. You can modify above query one more time to even include the details of actual file location. SELECT DISTINCT DB_NAME(dovs.database_id) DBName, mf.physical_name PhysicalFileLocation, dovs.logical_volume_name AS LogicalName, dovs.volume_mount_point AS Drive, CONVERT(INT,dovs.available_bytes/1048576.0) AS FreeSpaceInMB FROM sys.master_files mf CROSS APPLY sys.dm_os_volume_stats(mf.database_id, mf.FILE_ID) dovs ORDER BY FreeSpaceInMB ASC GO The above query will now additionally include the physical file location as well. As I mentioned earlier, I prefer method 2 as I can creatively use it as per the business need. Let me know which method are you using in your production server. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL

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  • Something eating space on OS drive

    - by noquery
    I am facing low disk space issue from last few days. I checked with Restore,System Volume Information, $Recycled folders. But there is nothing which is occupying space. I had scanned my system for virus too. Total size of C: is 18 GB. But when I select all folders inside C: and query for used space, it shows 20+ gb space is used. I vacate space some how by deleting temp files, program's cache files, disk clean up etc up to (3 gb). And I ensured that no cache/temp files are recreated who can use the space again. Even after cleaning so much data, I am again facing low disk space issue. Something is eating disk space within 15-20 mins.

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  • Big-O complexity of c^n + n*(logn)^2 + (10*n)^c

    - by zebraman
    I need to derive the Big-O complexity of this expression: c^n + n*(log(n))^2 + (10*n)^c where c is a constant and n is a variable. I'm pretty sure I understand how to derive the Big-O complexity of each term individually, I just don't know how the Big-O complexity changes when the terms are combined like this. Ideas? Any help would be great, thanks.

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  • How Do Guns Work In Space? [Video]

    - by Jason Fitzpatrick
    Why don’t astronauts fall back to Earth? What happens if you shoot a gun in space? How big of a squirt gun would you need to put out the sun? Don’t end your day with these pressing questions unanswered. [via Boing Boing] HTG Explains: Why Do Hard Drives Show the Wrong Capacity in Windows? Java is Insecure and Awful, It’s Time to Disable It, and Here’s How What Are the Windows A: and B: Drives Used For?

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  • Freeing disk space on Ubuntu to use in Windows

    - by Alex
    I have 250Gb drive on a laptop, which has Windows 7 on a 122Gb ntfs partition (which has a "boot" flag on it) and Ubuntu 12.04.1 on a 110Gb extended partition, of which the root ext4 partition is 108Gb and the swap is 1.74Gb. You can see everything in the screenshot below. My question is: I want to diminish the size of the linux root partition and then use that space to increase the windows partition. How do I do that? Also, is it possible to increase the size of the swap partition and not do any damage? If so, how? I'm using GParted, and i'd say i'm pretty confident with it. Screenshot of my partitions

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  • Thunderbird 16.0.1 filling disk space

    - by Kris
    I'm on Ubuntu 12.04 with Thunderbird 16.0.1 and Kernel 3.6.0-030600rc4-generic. I used Thunderbird for quite a while and never had any problems with it. But now it seems to fill up my disk space very fast: watch -n 1 df -h . so Ubuntu started giving out warnings. First I removed some files but not much later it had filled up around 600 MB. It eats around 50 MB/min while I just download 10 emails or so via IMAP. This behaviour is new and seems to be some kind of bug. I don't want to delete my old mails, so what else could I do?

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  • Using Completed User Stories to Estimate Future User Stories

    - by David Kaczynski
    In Scrum/Agile, the complexity of a user story can be estimated in story points. After completing some user stories, a programmer or team of programmers can use those experiences to better estimate how much time it might take to complete a future user story. Is there a methodology for breaking down the complexity of user stories into quantifiable or quantifiable attributes? For example, User Story X requires a rich, new view in the GUI, but User Story X can perform most of its functionality using existing business logic on the server. On a scale of 1 to 10, User Story X has a complexity of 7 on the client and a complexity of 2 on the server. After User Story X is completed, someone asks how long would it take to complete User Story Y, which has a complexity of 3 on the client and 6 on the server. Looking at how long it took to complete User Story X, we can make an educated estimate on how long it might take to complete User Story Y. I can imagine some other details: The complexity of one attribute (such as complexity of client) could have sub-attributes, such as number of steps in a sequence, function points, etc. Several other attributes that could be considered as well, such as the programmer's familiarity with the system or the number of components/interfaces involved These attributes could be accumulated into some sort of user story checklist. To reiterate: is there an existing methodology for decomposing the complexity of a user story into complexity of attributes/sub-attributes, or is using completed user stories as indicators in estimating future user stories more of an informal process?

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  • Whats consuming HDD Space

    - by Umair Mustafa
    I have single partition of 92GB in which I installed Ubuntu 12.04. And for some Unknown reason a message pop ups saying that I only have 1GB of HDD space left. I ran command sudo du -hscx * on / and /home /home gave me this result 4.0K C:\nppdf32Log\debuglog.txt 0 convertedvideo.avi 176M Desktop 16K Documents 169M Downloads 4.0K examples.desktop 17M file.txt 4.0K Music 984K Pictures 4.0K Public 320K Red Hat 6.iso 2.5M syslog-ng_3.3.6.tar.gz 4.0K Templates 8.0K terminal.png 1.2M Thunderbird Attachments 698M ubuntu10.04LTS.iso 16K Ubuntu One 4.0K Untitled Folder 4.0K Videos 21G VirtualBox VMs 22G total And / gave me this result 81G home 0 initrd.img 0 initrd.img.old 833M lib 16K lost+found 68K media 4.0K mnt 260M opt du: cannot access `proc/8339/task/8339/fd/4': No such file or directory du: cannot access `proc/8339/task/8339/fdinfo/4': No such file or directory du: cannot access `proc/8339/fd/4': No such file or directory du: cannot access `proc/8339/fdinfo/4': No such file or directory 0 proc 640K root 908K run 8.6M sbin 4.0K selinux 4.0K srv 0 sys 148K tmp 3.3G usr 436M var 0 vmlinuz 0 vmlinuz.old 86G total If you look at the result returned by / it shows that /home is consuming 81GB but on the other hand /home returns only 22GB. I cant figure out whats consuming the HDD. I have not installed anything except Virtual Machines Perpetrator found using Disk Usage Analyzer

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  • Efficient solution for multiplayer space partioning?

    - by DevilWithin
    This question is a little tricky, but I will try to make it clear, Lets say I am building an online game, not in a mmo scale, but gladly supporting as many players as possible, in a authoritative server approach, and I want really big worlds with lots of AI simulated enemies. I am aware of a few strategies to save server's CPU by subdividing the space and not processing what doesn't need processing. I 've already split the world by regions, that will require loading times and small transitions, which i think is important to mantain the quality of gameplay when playing locally (alone or even with a couple of friends) because the players won't normally be in more than one or two regions. But even a region can become pretty big, and have a lot of NPC simulating at a time, how do I handle this without screwing the player's experience? Approaches like one server per region and alike are not in the table. I am mainly looking for data structures to hold hordes of enemies, and even peaceful NPC. To finalize the question, please note that vehicles exist, therefore its considerably fast to travel within a region, influencing the "when" to cull areas. Sorry for the confusing question, thanks

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  • Asymptotic complexity of a compiler

    - by Meinersbur
    What is the maximal acceptable asymptotic runtime of a general-purpose compiler? For clarification: The complexity of compilation process itself, not of the compiled program. Depending on the program size, for instance, the number of source code characters, statements, variables, procedures, basic blocks, intermediate language instructions, assembler instructions, or whatever. This is highly depending on your point of view, so this is a community wiki. See this from the view of someone who writes a compiler. Will the optimisation level -O4 ever be used for larger programs when one of its optimisations takes O(n^6)? Related questions: When is superoptimisation (exponential complexity or even incomputable) acceptable? What is acceptable for JITs? Does it have to be linear? What is the complexity of established compilers? GCC? VC? Intel? Java? C#? Turbo Pascal? LCC? LLVM? (Reference?) If you do not know what asymptotic complexity is: How long are you willing to wait until the compiler compiled your project? (scripting languages excluded)

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  • Time complexity of a certain program

    - by HokageSama
    In a discussion with my friend i am not able to predict correct and tight time complexity of a program. Program is as below. /* This Function reads input array "input" and update array "output" in such a way that B[i] = index value of nearest greater value from A[i], A[i+1] ... A[n], for all i belongs to [1, n] Time Complexity: ?? **/ void createNearestRightSidedLargestArr(int* input, int size, int* output){ if(!input || size < 1) return; //last element of output will always be zero, since no element is present on its right. output[size-1] = -1; int curr = size - 2; int trav; while(curr >= 0){ if(input[curr] < input[curr + 1]){ output[curr] = curr + 1; curr--; continue; } trav = curr + 1; while( input[ output [trav] ] < input[curr] && output [trav] != -1) trav = output[trav]; output[curr--] = output[trav]; } } I said time complexity is O(n^2) but my friend insists that this is not correct. What is the actual time complexity?

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  • Time complexity with bit cost

    - by Keyser
    I think I might have completely misunderstood bit cost analysis. I'm trying to wrap my head around the concept of studying an algorithm's time complexity with respect to bit cost (instead of unit cost) and it seems to be impossible to find anything on the subject. Is this considered to be so trivial that no one ever needs to have it explained to them? Well I do. (Also, there doesn't even seem to be anything on wikipedia which is very unusual). Here's what I have so far: The bit cost of multiplication and division of two numbers with n bits is O(n^2) (in general?) So, for example: int number = 2; for(int i = 0; i < n; i++ ){ number = i*i; } has a time complexity with respect to bit cost of O(n^3), because it does n multiplications (right?) But in a regular scenario we want the time complexity with respect to the input. So, how does that scenario work? The number of bits in i could be considered a constant. Which would make the time complexity the same as with unit cost except with a bigger constant (and both would be linear). Also, I'm guessing addition and subtraction can be done in constant time, O(1). Couldn't find any info on it but it seems reasonable since it's one assembler operation.

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  • Reducing Time Complexity in Java

    - by Koeneuze
    Right, this is from an older exam which i'm using to prepare my own exam in january. We are given the following method: public static void Oorspronkelijk() { String bs = "Dit is een boodschap aan de wereld"; int max = -1; char let = '*'; for (int i=0;i<bs.length();i++) { int tel = 1; for (int j=i+1;j<bs.length();j++) { if (bs.charAt(j) == bs.charAt(i)) tel++; } if (tel > max) { max = tel; let = bs.charAt(i); } } System.out.println(max + " keer " + let); } The questions are: what is the output? - Since the code is just an algorithm to determine the most occuring character, the output is "6 keer " (6 times space) What is the time complexity of this code? Fairly sure it's O(n²), unless someone thinks otherwise? Can you reduce the time complexity, and if so, how? Well, you can. I've received some help already and managed to get the following code: public static void Nieuw() { String bs = "Dit is een boodschap aan de wereld"; HashMap<Character, Integer> letters = new HashMap<Character, Integer>(); char max = bs.charAt(0); for (int i=0;i<bs.length();i++) { char let = bs.charAt(i); if(!letters.containsKey(let)) { letters.put(let,0); } int tel = letters.get(let)+1; letters.put(let,tel); if(letters.get(max)<tel) { max = let; } } System.out.println(letters.get(max) + " keer " + max); } However, I'm uncertain of the time complexity of this new code: Is it O(n) because you only use one for-loop, or does the fact we require the use of the HashMap's get methods make it O(n log n) ? And if someone knows an even better way of reducing the time complexity, please do tell! :)

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  • What is the complexity of this specialized sort

    - by ADB
    I would like to know the complexity (as in O(...) ) of the following sorting algorithm: there are B barrels that contains a total of N elements, spread unevenly across the barrels. the elements in each barrel are already sorted The sort takes combines all the elements from each barrel in a single sorted list: using an array of size B to store the last sorted element of each barrel (starting at 0) check each barrel (at the last stored index) and find the smallest element copy the element in the final sorted array, increment array counter increment the last sorted element for the barrel we picked from perform those steps N times or in pseudo: for i from 0 to N smallest = MAX_ELEMENT foreach b in B if bIndex[b] < b.length && b[bIndex[b]] < smallest smallest_barrel = b smallest = b[bIndex[b]] result[i] = smallest bIndex[smallest_barrel] += 1 I thought that the complexity would be O(n), but the problem I have with finding the complexity is that if B grows, it has a larger impact than if N grows because it adds another round in the B loop. But maybe that has no effect on the complexity?

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  • Time complexity of Sieve of Eratosthenes algorithm

    - by eSKay
    From Wikipedia: The complexity of the algorithm is O(n(logn)(loglogn)) bit operations. How do you arrive at that? That the complexity includes the loglogn term tells me that there is a sqrt(n) somewhere. Suppose I am running the sieve on the first 100 numbers (n = 100), assuming that marking the numbers as composite takes constant time (array implementation), the number of times we use mark_composite() would be something like n/2 + n/3 + n/5 + n/7 + ... + n/97 = O(n) And to find the next prime number (for example to jump to 7 after crossing out all the numbers that are multiples of 5), the number of operations would be O(n). So, the complexity would be O(n^2). Do you agree?

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  • Time complexity of a sorting algorithm

    - by Passonate Learner
    The two programs below get n integers from file and calculates the sum of ath to bth integers q(number of question) times. I think the upper program has worse time complexity than the lower, but I'm having problems calculating the time complexity of these two algorithms. [input sample] 5 3 5 4 3 2 1 2 3 3 4 2 4 [output sample] 7 5 9 Program 1: #include <stdio.h> FILE *in=fopen("input.txt","r"); FILE *out=fopen("output.txt","w"); int n,q,a,b,sum; int data[1000]; int main() int i,j; fscanf(in,"%d%d",&n,&q); for(i=1;i<=n;i++) fscanf(in,"%d",&data[i]); for i=0;i<q;i++) { fscanf(in,"%d%d",&a,&b); sum=0; for(j=a;j<=b;j++) sum+=data[j]; fprintf(out,"%d\n",sum); } return 0; } Program 2: #include <stdio.h> FILE *in=fopen("input.txt","r"); FILE *out=fopen("output.txt","w"); int n,q,a,b; int data[1000]; int sum[1000]; int main() { int i,j; fscanf(in,"%d%d",&n,&q); for(i=1;i<=n;i++) fscanf(in,"%d",&data[i]); for(i=1;i<=n;i++) sum[i]=sum[i-1]+data[i]; for(i=0;i<q;i++) { fscanf(in,"%d%d",&a,&b); fprintf(out,"%d\n",sum[b]-sum[a-1]); } return 0; } The programs below gets n integers from 1 to m and sorts them. Again, I cannot calculate the time complexity. [input sample] 5 5 2 1 3 4 5 [output sample] 1 2 3 4 5 Program: #include <stdio.h> FILE *in=fopen("input.txt","r") FILE *out=fopen("output.txt","w") int n,m; int data[1000]; int count[1000]; int main() { int i,j; fscanf(in,"%d%d",&n,&m); for(i=0;i<n;i++) { fscanf(in,"%d",&data[i]); count[data[i]]++ } for(i=1;i<=m;i++) { for(j=0;j<count[i];j++) fprintf(out,"%d ",i); } return 0; } It's ironic(or not) that I cannot calculate the time complexity of my own algorithms, but I have passions to learn, so please programming gurus, help me!

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  • Computational complexity of Fibonacci Sequence

    - by Juliet
    I understand Big-O notation, but I don't know how to calculate it for many functions. In particular, I've been trying to figure out the computational complexity of the naive version of the Fibonacci sequence: int Fib(int n) { if (n <= 1) return 1; else return Fib(n - 1) + Fib(n - 2); } What is the computational complexity of the Fibonnaci sequence and how is it calculated?

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  • The Space Invader – A Childhood to Adulthood Story [Comic]

    - by Asian Angel
    Did you ever wonder what life is like for the Invaders? Take a journey through time with this particular Invader as he grows from a child into adulthood and decides to join the war. Note: We have shown only the first panel here. You can view the entire comic story by visiting the link below. The Invader – EL Comics [via Neatorama] How to Get Pro Features in Windows Home Versions with Third Party Tools HTG Explains: Is ReadyBoost Worth Using? HTG Explains: What The Windows Event Viewer Is and How You Can Use It

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  • Pseudo-quicksort time complexity

    - by Ord
    I know that quicksort has O(n log n) average time complexity. A pseudo-quicksort (which is only a quicksort when you look at it from far enough away, with a suitably high level of abstraction) that is often used to demonstrate the conciseness of functional languages is as follows (given in Haskell): quicksort :: Ord a => [a] -> [a] quicksort [] = [] quicksort (p:xs) = quicksort [y | y<-xs, y<p] ++ [p] ++ quicksort [y | y<-xs, y>=p] Okay, so I know this thing has problems. The biggest problem with this is that it does not sort in place, which is normally a big advantage of quicksort. Even if that didn't matter, it would still take longer than a typical quicksort because it has to do two passes of the list when it partitions it, and it does costly append operations to splice it back together afterwards. Further, the choice of the first element as the pivot is not the best choice. But even considering all of that, isn't the average time complexity of this quicksort the same as the standard quicksort? Namely, O(n log n)? Because the appends and the partition still have linear time complexity, even if they are inefficient.

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  • Windows Web Server 2008 R2 Server Core local password complexity

    - by Dennis Allen
    How can I disable the local user account password complexity settings on Windows 2008 R2 "Server Core"? I am trying to migrate our windows 2003 web server to windows 2008 R2. I am trying to see if I can use the "Server Core" install, and it has been a very internet search intensive experience. What I can't find out how to do is to find out how to disable password complexity for local user accounts. While our user account generator currently creates nice strong passwords, there was a time when this was not the case and unfortunately forcing the users to change their password is not an option at this time. Any help greatly appreciated. Dennis

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  • Lost disk space in Windows 7, cannot find the missing

    - by Tsanders
    My hard drive is complaining it is low on disk space, but a strange thing seems to be happening: Explorer reports 10Gb of available space (on a 120 Gb hard disk), chkdsk in the command prompt does the same but if I use a disk space tool such as SpaceSniffer or WinDirStat, only 50Gb of data is found. My guess is that there somehow is a hold on a large block of disk space (but that's just a guess) because of a prior very large (40 Gb) download attempt that didn't complete. There isn't 40Gb of files on the drive (hidden or visible), yet Explorer insists that something is there. How can I claim back this hard disk drive (without formatting my hard disk)? SpaceMonger is providing a clue, reporting four unscannable folders which add up to 43Gb: C:\RRBackups C:\System Volume Information C:\Windows\Csc\v2.06 C:\Windows\System32\LogFiles\Wmi\RtBackup Does anybody know what these folders are for, and how I can claim back at least some space? Restore point claims about 4Gb, so that doesn't seem to be the main problem.

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  • No space left on device with encrypted disk that takes all space

    - by Yosef
    I use Ubuntu 11.04. There's no space left on device. I have encrypted the disk that takes up space (maybe it's good to disable it, but I don't know how). In shell, I get this message: No space left on device I run df -I: Filesystem Inodes IUsed IFree IUse% Mounted on /dev/sda1 3055616 602499 2453117 20% / none 210161 890 209271 1% /dev none 214789 8 214781 1% /dev/shm none 214789 53 214736 1% /var/run none 214789 3 214786 1% /var/lock /home/myuser/.Private 3055616 602499 2453117 20% /home/myuser df -I Edit: When I run only df: Filesystem 1K-blocks Used Available Use% Mounted on /dev/sda1 48060296 45618928 0 100% / none 1538340 684 1537656 1% /dev none 1547596 808 1546788 1% /dev/shm none 1547596 104 1547492 1% /var/run none 1547596 0 1547596 0% /var/lock /home/myuser/.Private 48060296 45618928 0 100% /home/myuser Edit: I thinking about few solution but I don't know which better and how exactly to do them: to enlarge partition size (I cant install gparted - no more disk space) remove encryption of partition - I really not need that

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  • Time complexity O() of isPalindrome()

    - by Aran
    I have this method, isPalindrome(), and I am trying to find the time complexity of it, and also rewrite the code more efficiently. boolean isPalindrome(String s) { boolean bP = true; for(int i=0; i<s.length(); i++) { if(s.charAt(i) != s.charAt(s.length()-i-1)) { bP = false; } } return bP; } Now I know this code checks the string's characters to see whether it is the same as the one before it and if it is then it doesn't change bP. And I think I know that the operations are s.length(), s.charAt(i) and s.charAt(s.length()-i-!)). Making the time-complexity O(N + 3), I think? This correct, if not what is it and how is that figured out. Also to make this more efficient, would it be good to store the character in temporary strings?

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