Search Results

Search found 22520 results on 901 pages for 'great big al'.

Page 5/901 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Convert a raw string to an array of big-endian words with Ruby

    - by Zag zag..
    Hello, I would like to convert a raw string to an array of big-endian words. As example, here is a JavaScript function that do it well (by Paul Johnston): /* * Convert a raw string to an array of big-endian words * Characters >255 have their high-byte silently ignored. */ function rstr2binb(input) { var output = Array(input.length >> 2); for(var i = 0; i < output.length; i++) output[i] = 0; for(var i = 0; i < input.length * 8; i += 8) output[i>>5] |= (input.charCodeAt(i / 8) & 0xFF) << (24 - i % 32); return output; } I believe the Ruby equivalent can be String#unpack(format). However, I don't know what should be the correct format parameter. Thank you for any help. Regards

    Read the article

  • Big-Oh running time of code in Java (are my answers accurate

    - by Terry Frederick
    the Method hasTwoTrueValues returns true if at least two values in an array of booleans are true. Provide the Big-Oh running time for all three implementations proposed. // Version 1 public boolean has TwoTrueValues( boolean [ ] arr ) { int count = 0; for( int i = 0; i < arr. length; i++ ) if( arr[ i ] ) count++; return count >= 2; } // Version 2 public boolean hasTwoTrueValues( boolean [ ] arr ) { for( int i = 0; i < arr.length; i++ ) for( int j = i + 1; j < arr.length; j++ ) if( arr[ i ] && arr[ j ] ) return true; } // Version 3 public boolean hasTwoTrueValues( boolean [ ] arr ) { for( int i = 0; i < arr.length; i++ if( arr[ i ] ) for( int j = i + 1; j < arr.length; j++ ) if( arr[ j ] ) return true; return false; } For Version 1 I say the running time is O(n) Version 2 I say O(n^2) Version 3 I say O(n^2) I am really new to this Big Oh Notation so if my answers are incorrect could you please explain and help.

    Read the article

  • SYSLINUX 4.07 EDD 2013-07-25 Copyright (C) 1994-2013 H. Peter Anvin et al [duplicate]

    - by Aniel Arias
    This question already has an answer here: Not booting from USB or CD (SYSLINUX Message) 10 answers this what is happening, i downloaded (ubuntu-gnome-14.04.1-desktop) and (elementaryos-unstable-amd64.20140810) to try out in my laptop and i have use (unetbootin-windows-608) and (Universal-USB-Installer-1.9.5.5) but i get this message every time i try to boot from the usb (SYSLINUX 4.07 EDD 2013-07-25 Copyright (C) 1994-2013 H. Peter Anvin et al) however i tried in an old desktop that i have and it works although the installer gets stuck on most of the time at the part of reading partitions/hard drives so please i really need help with this. note: i did installed os x long time ago and i broke windows installation then fix it following some online tutorials just for FYI thanks please can somebody help to fix this problem, i have been looking on google but haven't found anything in concrete. please help

    Read the article

  • Error al instalar Visual Studio 2008 en Windows 7

    - by José Marcos García Espinosa
    Post/Install() failed in ISetupManager::InternalInstallManager() with HRESULT -2147023293. - Es el mensaje que aparece en la descripción del error o en los archivos de log. Algunos dicen que si buscas más a detalle encuentras el problema. Después de reintentar la instalación como 8 veces, desinstalando cada vez más cosas, te encuentras con que la solución es sencilla: ¡Desinstala Office! No sé por qué, pero creo que Microsoft asume que si eres un desarrollador, primero instalarás VS2008 y después Office; si lo haces al revés, se generan problemas de compatibilidad (ya que VS2008 instala herramientas de interoperabilidad/desarrollo/conexión con Office, Visual Studio Tools for Office) y la instalación no pasa a veces ni del .net Framework. A final de cuentas, creo, ésta es una medio sencilla.

    Read the article

  • Great Programmer Productivity - Accounting for 10,000 fold difference?

    - by TheImpact
    "A great lathe operator commands several times the wage of an average lathe operator, but a great writer of software code is worth 10,000 times the price of an average software writer." - Bill Gates Say there's a "great" software engineer and an "average" software engineer on the same team. How can you account for one engineer being 10,000 times more productive? I can't quite fathom this, given they're both taking on their share of features, bugs and investigations, and consistently deliver with quality. Would my description possibly justify them to be above "average"? "great"? In a corporation like Microsoft, what % of software engineers are "average"? What % "great"?

    Read the article

  • Suggested Web Application Framework and Database for Enterprise, “Big-Data” App?

    - by willOEM
    I have a web application that I have been developing for a small group within my company over the past few years, using Pipeline Pilot (plus jQuery and Python scripting) for web development and back-end computation, and Oracle 10g for my RDBMS. Users upload experimental genomic data, which is parsed into a database, and made available for querying, transformation, and reporting. Experimental data sets are large and have many layers of metadata. A given experimental data record might have a foreign key relationship with a table that describes this data point's assay. Assays can cover multiple genes, which can have multiple transcript, which can have multiple mutations, which can affect multiple signaling pathways, etc. Users need to approach this data from any point in those layers in the metadata. Since all data sets for a given data type can run over a billion rows, this results in some large, dynamic queries that are hard to predict. New data sets are added on a weekly basis (~1GB per set). Experimental data is never updated, but the associated metadata can be updated weekly for a few records and yearly for most others. For every data set insert the system sees, there will be between 10 and 100 selects run against it and associated data. It is okay for updates and inserts to run slow, so long as queries run quick and are as up-to-date as possible. The application continues to grow in size and scope and is already starting to run slower than I like. I am worried that we have about outgrown Pipeline Pilot, and perhaps Oracle (as the sole database). Would a NoSQL database or an OLAP system be appropriate here? What web application frameworks work well with systems like this? I'd like the solution to be something scalable, portable and supportable X-years down the road. Here is the current state of the application: Web Server/Data Processing: Pipeline Pilot on Windows Server + IIS Database: Oracle 10g, ~1TB of data, ~180 tables with several billion-plus row tables Network Storage: Isilon, ~50TB of low-priority raw data

    Read the article

  • Big numbers in C

    - by teehoo
    I need help working with very big numbers. According to Windows calc, the exponent 174^55 = 1.6990597648061509725749329578093e+123. How would I store this using C (c99 standard). int main(){ long long int x = 174^55; //result is 153 printf("%lld\n", x); } For those curious, it is for a school project where we are implementing the RSA cryptographic algorithm, which deals with exponentiating large numbers with large powers for encryption/decryption.

    Read the article

  • Big O, how do you calculate/approximate it?

    - by Sven
    Most people with a degree in CS will certainly know what Big O stands for. It helps us to measure how (in)efficient an algorithm really is and if you know in what category the problem you are trying to solve lays in you can figure out if it is still possible to squeeze out that little extra performance.* But I'm curious, how do you calculate or approximate the complexity of your algorithms? *: but as they say, don't overdo it, premature optimization is the root of all evil, and optimization without a justified cause should deserve that name as well.

    Read the article

  • How big is too big (for NTFS)

    - by BCS
    I have a program and as it's done now, it has a data directory with something like 10-30K files in it and it's starting to cause problems. Should I expect that to cause problems and my only solution to tweak my file structure or does that indicate other problems?

    Read the article

  • Big O and Little o

    - by hyperdude
    If algorithm A has complexity O(n) and algorithm B has complexity o(n^2), what, if anything, can we say about the relationship between A and B? Note: the complexity of A is expressed using big-Oh, and the complexity of B is expressed using little-Oh.

    Read the article

  • BIG DATA eBook - Now Available

    - by Javier Puerta
    The Big Data interactive e-book “Meeting the Challenge of Big Data: Part One” has just been released. It’s your “one-stop shop” for info about Big Data and the Oracle offering around it.The new e-book (available on your computer or iPad) is packed with multi-media resources to educate Oracle staff, customers, prospects and partners on the value of Big Data. It features videos, tutorials, podcasts, reports, white papers, datasheets, blogs, web links, a 3-D demo, and more. Go and get it here!

    Read the article

  • Big Data Sessions at Openworld 2012

    - by Jean-Pierre Dijcks
    If you are coming to San Francisco, and you are interested in all the aspects to big data, this Focus On Big Data is a must have document.  Some (other) highlights: A performance demo of a full rack Big Data Appliance in the engineered systems showcase A set of handson labs on how to go from a NoSQL DB to an effective analytics play on big data Much, much more See you all in a few weeks in SF!

    Read the article

  • Big Oh Notation - formal definition.

    - by aloh
    I'm reading a textbook right now for my Java III class. We're reading about Big-Oh and I'm a little confused by its formal definition. Formal Definition: "A function f(n) is of order at most g(n) - that is, f(n) = O(g(n)) - if a positive real number c and positive integer N exist such that f(n) <= c g(n) for all n = N. That is, c g(n) is an upper bound on f(n) when n is sufficiently large." Ok, that makes sense. But hold on, keep reading...the book gave me this example: "In segment 9.14, we said that an algorithm that uses 5n + 3 operations is O(n). We now can show that 5n + 3 = O(n) by using the formal definition of Big Oh. When n = 3, 5n + 3 <= 5n + n = 6n. Thus, if we let f(n) = 5n + 3, g(n) = n, c = 6, N = 3, we have shown that f(n) <= 6 g(n) for n = 3, or 5n + 3 = O(n). That is, if an algorithm requires time directly proportional to 5n + 3, it is O(n)." Ok, this kind of makes sense to me. They're saying that if n = 3 or greater, 5n + 3 takes less time than if n was less than 3 - thus 5n + n = 6n - right? Makes sense, since if n was 2, 5n + 3 = 13 while 6n = 12 but when n is 3 or greater 5n + 3 will always be less than or equal to 6n. Here's where I get confused. They give me another example: Example 2: "Let's show that 4n^2 + 50n - 10 = O(n^2). It is easy to see that: 4n^2 + 50n - 10 <= 4n^2 + 50n for any n. Since 50n <= 50n^2 for n = 50, 4n^2 + 50n - 10 <= 4n^2 + 50n^2 = 54n^2 for n = 50. Thus, with c = 54 and N = 50, we have shown that 4n^2 + 50n - 10 = O(n^2)." This statement doesn't make sense: 50n <= 50n^2 for n = 50. Isn't any n going to make the 50n less than 50n^2? Not just greater than or equal to 50? Why did they even mention that 50n <= 50n^2? What does that have to do with the problem? Also, 4n^2 + 50n - 10 <= 4n^2 + 50n^2 = 54n^2 for n = 50 is going to be true no matter what n is. And how in the world does picking numbers show that f(n) = O(g(n))? Please help me understand! :(

    Read the article

  • Big O Complexity of a method

    - by timeNomad
    I have this method: public static int what(String str, char start, char end) { int count=0; for(int i=0;i<str.length(); i++) { if(str.charAt(i) == start) { for(int j=i+1;j<str.length(); j++) { if(str.charAt(j) == end) count++; } } } return count; } What I need to find is: 1) What is it doing? Answer: counting the total number of end occurrences after EACH (or is it? Not specified in the assignment, point 3 depends on this) start. 2) What is its complexity? Answer: the first loops iterates over the string completely, so it's at least O(n), the second loop executes only if start char is found and even then partially (index at which start was found + 1). Although, big O is all about worst case no? So in the worst case, start is the 1st char & the inner iteration iterates over the string n-1 times, the -1 is a constant so it's n. But, the inner loop won't be executed every outer iteration pass, statistically, but since big O is about worst case, is it correct to say the complexity of it is O(n^2)? Ignoring any constants and the fact that in 99.99% of times the inner loop won't execute every outer loop pass. 3) Rewrite it so that complexity is lower. What I'm not sure of is whether start occurs at most once or more, if once at most, then method can be rewritten using one loop (having a flag indicating whether start has been encountered and from there on incrementing count at each end occurrence), yielding a complexity of O(n). In case though, that start can appear multiple times, which most likely it is, because assignment is of a Java course and I don't think they would make such ambiguity. Solving, in this case, is not possible using one loop... WAIT! Yes it is..! Just have a variable, say, inc to be incremented each time start is encountered & used to increment count each time end is encountered after the 1st start was found: inc = 0, count = 0 if (current char == start) inc++ if (inc > 0 && current char == end) count += inc This would also yield a complexity of O(n)? Because there is only 1 loop. Yes I realize I wrote a lot hehe, but what I also realized is that I understand a lot better by forming my thoughts into words...

    Read the article

  • big O notation algorithm

    - by niggersak
    Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required? . Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

    Read the article

  • Tricky Big-O complexity

    - by timeNomad
    public void foo (int n, int m) { int i = m; while (i > 100) i = i/3; for (int k=i ; k>=0; k--) { for (int j=1; j<n; j*=2) System.out.print(k + "\t" + j); System.out.println(); } } I figured the complexity would be O(logn). That is as a product of the inner loop, the outer loop -- will never be executed more than 100 times, so it can be omitted. What I'm not sure about is the while clause, should it be incorporated into the Big-O complexity? For very large i values it could make an impact, or arithmetic operations, doesn't matter on what scale, count as basic operations and can be omitted?

    Read the article

  • Database indexes and their Big-O notation

    - by miket2e
    I'm trying to understand the performance of database indexes in terms of Big-O notation. Without knowing much about it, I would guess that: Querying on a primary key or unique index will give you a O(1) lookup time. Querying on a non-unique index will also give a O(1) time, albeit maybe the '1' is slower than for the unique index (?) Querying on a column without an index will give a O(N) lookup time (full table scan). Is this generally correct ? Will querying on a primary key ever give worse performance than O(1) ? My specific concern is for SQLite, but I'd be interested in knowing to what extent this varies between different databases too.

    Read the article

  • Can someone help with big O notation?

    - by Dann
    void printScientificNotation(double value, int powerOfTen) { if (value >= 1.0 && value < 10.0) { System.out.println(value + " x 10^" + powerOfTen); } else if (value < 1.0) { printScientificNotation(value * 10, powerOfTen - 1); } else // value >= 10.0 { printScientificNotation(value / 10, powerOfTen + 1); } } I understand how the method goes but I cannot figure out a way to represent the method. For example, if value was 0.00000009 or 9e-8, the method will call on printScientificNotation(value * 10, powerOfTen - 1); eight times and System.out.println(value + " x 10^" + powerOfTen); once. So the it is called recursively by the exponent for e. But how do I represent this by big O notation? Thanks!

    Read the article

  • Big Data for Retail

    - by David Dorf
    Right up there with mobile, social, and cloud is the term "big data," which seems to be popping up lots in the press these days.  Companies like Google, Yahoo, and Facebook have popularized a new class of data technologies meant to solve the problem of processing large amounts of data quickly.  I first mentioned this in a posting back in March 2009.  Put simply, big data implies datasets so large they can't normally be processed using a standard transactional database.  The term "noSQL" is often used in this context as well. Actually, using parallel processing within the Oracle database combined with Exadata can achieve impressive results.  Look for more from Oracle at OpenWorld as hinted by Jean-Pierre Dijcks. McKinsey recently released a report on big data in which retail was specifically mentioned as an industry that can benefit from the new technologies.  I won't rehash that report because my friend Rama already did such a good job in his posting, Impact of "Big Data" on Retail. The presentation below does a pretty good job of framing the problem, although it doesn't really get into the available technologies (e.g. Exadata, Hadoop, Cassandra, etc.) and isn't retail specific. Determine the Right Analytic Database: A Survey of New Data Technologies So when a retailer asks me about big data, here's what I say:  Big data refers to a set of technologies for processing large volumes of structured and unstructured data.  Imagine collecting everything uttered by your customers on Facebook and Twitter and combining it with all the data you can find about the products you sell (e.g. reviews, images, demonstration videos), including competitive data.  Assuming you could process all that data, you could then personalize offers to specific customers based on their tastes, ensure prices are competitive, and implement better local assortments.  It's really not that far off.

    Read the article

  • How To Record Great Vocals at Home

    How to record great vocals You have a great voice; so that?s all you need to record great vocals isn?t it? If only it was as simple as that. Well, now it can be. Many things ranging from acoustic co... [Author: Fallon Morris - Computers and Internet - May 31, 2010]

    Read the article

  • Recurrence relation solution

    - by Travis
    I'm revising past midterms for a final exam this week and am trying to make sense of a solution my professor posted for one of past exams. (You can see the original pdf here, question #6). I'm given the original recurrence relation T(m)=3T(n/2) + n and am told T(1) = 1. I'm pretty sure the solution I've been given is wrong in a few places. The solution is as follows: Let n=2^m T(2^m) = 3T(2^(m-1)) + 2^m 3T(2^(m-1)) = 3^2*T(2^(m-2)) + 2^(m-1)*3 ... 3^(m-1)T(2) = T(1) + 2*3^(m-1) I'm pretty sure this last line is incorrect and they forgot to multiply T(1) by 3^m. He then (tries to) sum the expressions: T(2^m) = 1 + (2^m + 2^(m-1)*3 + ... + 2*3(m-1)) = 1 + 2^m(1 + (3/2)^1 + (3/2)^2 + ... + (3/2)^(m-1)) = 1 + 2^m((3/2)^m-1)*(1/2) = 1 + 3^m - 2^(m-1) = 1 + n^log 3 - n/2 Thus the algorithm is big Theta of (n^log 3). I'm pretty sure that he also got the summation wrong here. By my calculations this should be as follows: T(2^m) = 2^m + 3 * 2^(m-1) + 3^2 * 2^(m-2) + ... + 3^m (3^m because 3^m*T(1) = 3^m should be added, not 1) = 2^m * ((3/2)^1 + (3/2)^2 + ... + (3/2)^m) = 2^m * sum of (3/2)^i from i=0 to m = 2^m * ((3/2)^(m+1) - 1)/(3/2 - 1) = 2^m * ((3/2)^(m+1) - 1)/(1/2) = 2^(m+1) * 3^(m+1)/2^(m+1) - 2^(m+1) = 3^(m+1) - 2 * 2^m Replacing n = 2^m, and from that m = log n T(n) = 3*3^(log n) - 2*n n is O(3^log n), thus the runtime is big Theta of (3^log n) Does this seem right? Thanks for your help!

    Read the article

  • Fast Data - Big Data's achilles heel

    - by thegreeneman
    At OOW 2013 in Mark Hurd and Thomas Kurian's keynote, they discussed Oracle's Fast Data software solution stack and discussed a number of customers deploying Oracle's Big Data / Fast Data solutions and in particular Oracle's NoSQL Database.  Since that time, there have been a large number of request seeking clarification on how the Fast Data software stack works together to deliver on the promise of real-time Big Data solutions.   Fast Data is a software solution stack that deals with one aspect of Big Data, high velocity.   The software in the Fast Data solution stack involves 3 key pieces and their integration:  Oracle Event Processing, Oracle Coherence, Oracle NoSQL Database.   All three of these technologies address a high throughput, low latency data management requirement.   Oracle Event Processing enables continuous query to filter the Big Data fire hose, enable intelligent chained events to real-time service invocation and augments the data stream to provide Big Data enrichment. Extended SQL syntax allows the definition of sliding windows of time to allow SQL statements to look for triggers on events like breach of weighted moving average on a real-time data stream.    Oracle Coherence is a distributed, grid caching solution which is used to provide very low latency access to cached data when the data is too big to fit into a single process, so it is spread around in a grid architecture to provide memory latency speed access.  It also has some special capabilities to deploy remote behavioral execution for "near data" processing.   The Oracle NoSQL Database is designed to ingest simple key-value data at a controlled throughput rate while providing data redundancy in a cluster to facilitate highly concurrent low latency reads.  For example, when large sensor networks are generating data that need to be captured while analysts are simultaneously extracting the data using range based queries for upstream analytics.  Another example might be storing cookies from user web sessions for ultra low latency user profile management, also leveraging that data using holistic MapReduce operations with your Hadoop cluster to do segmented site analysis.  Understand how NoSQL plays a critical role in Big Data capture and enrichment while simultaneously providing a low latency and scalable data management infrastructure thru clustered, always on, parallel processing in a shared nothing architecture. Learn how easily a NoSQL cluster can be deployed to provide essential services in industry specific Fast Data solutions. See these technologies work together in a demonstration highlighting the salient features of these Fast Data enabling technologies in a location based personalization service. The question then becomes how do these things work together to deliver an end to end Fast Data solution.  The answer is that while different applications will exhibit unique requirements that may drive the need for one or the other of these technologies, often when it comes to Big Data you may need to use them together.   You may have the need for the memory latencies of the Coherence cache, but just have too much data to cache, so you use a combination of Coherence and Oracle NoSQL to handle extreme speed cache overflow and retrieval.   Here is a great reference to how these two technologies are integrated and work together.  Coherence & Oracle NoSQL Database.   On the stream processing side, it is similar as with the Coherence case.  As your sliding windows get larger, holding all the data in the stream can become difficult and out of band data may need to be offloaded into persistent storage.  OEP needs an extreme speed database like Oracle NoSQL Database to help it continue to perform for the real time loop while dealing with persistent spill in the data stream.  Here is a great resource to learn more about how OEP and Oracle NoSQL Database are integrated and work together.  OEP & Oracle NoSQL Database.

    Read the article

  • Reporting a WCF application's status to F5's Big IP products

    - by ng5000
    In a Windows Server 2003 environment with a self hosted .Net 3.5/WCF application, how can an application report its status to a BigIP Local Traffic Manager? Example: One of my services errors. My custom WCF application hosting software (written because Windows Server 2008 is not yet available and I'm using WCF TCP bindings) detects this and wants to report itself as down until it can recover the errant service. It needs to report itself as down to the BigIP LTM so that it is no longer sent client originated requests.

    Read the article

  • analysis Big Oh notation psuedocode

    - by tesshu
    I'm having trouble getting my head around algorithm analysis. I seem to be okay identifying linear or squared algorithms but am totally lost with nlogn or logn algorithms, these seem to stem mainly from while loops? Heres an example I was looking at: Algorithm Calculate(A,n) Input: Array A of size n t?0 for i?0 to n-1 do if A[i] is an odd number then Q.enqueue(A[i]) else while Q is not empty do t?t+Q.dequeue() while Q is not empty do t?t+Q.dequeue() return t My best guess is the for loop is executed n times, its nested while loop q times making NQ and the final while loop also Q times resulting in O(NQ +Q) which is linear? I am probably totally wrong. Any help would be much appreciated. thanks

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >