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  • Avoiding dog-piling or thundering herd in a memcached expiration scenario

    - by Quintin Par
    I have the result of a query that is very expensive. It is the join of several tables and a map reduce job. This is cached in memcached for 15 minutes. Once the cache expires the queries are obviously run and the cache warmed again. But at the point of expiration the thundering herd problem issue can happen. One way to fix this problem, that I do right now is to run a scheduled task that kicks in the 14th minute. But somehow this looks very sub optimal to me. Another approach I like is nginx’s proxy_cache_use_stale updating; mechanism. The webserver/machine continues to deliver stale cache while a thread kicks in the moment expiration happens and updates the cache. Has someone applied this to memcached scenario though I understand this is a client side strategy? If it benefits, I use Django.

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  • A balanced binary search tree which is also a heap

    - by saeedn
    I'm looking for a data structure where each element in it has two keys. With one of them the structure is a BST and looking at the other one, data structure is a heap. With a little search, I found a structure called Treap. It uses the heap property with a random distribution on heap keys to make the BST balanced! What I want is a Balanced BST, which can be also a heap. The BST in Treap could be unbalanced if I insert elements with heap Key in the order of my choice. Is there such a data structure?

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  • Why can't RB-Tree be a list?

    - by Alex
    Hey everyone. I have a problem with the rb-trees. according to wikipedia, rb-tree needs to follow the following: A node is either red or black. The root is black. (This rule is used in some definitions and not others. Since the root can always be changed from red to black but not necessarily vice-versa this rule has little effect on analysis.) All leaves are black. Both children of every red node are black. Every simple path from a given node to any of its descendant leaves contains the same number of black nodes. As we know, an rb-tree needs to be balanced and has the height of O(log(n)). But, if we insert an increasing series of numbers (1,2,3,4,5...) and theoretically we will get a tree that will look like a list and will have the height of O(n) with all its nodes black, which doesn't contradict the rb-tree properties mentioned above. So, where am I wrong?? thanks.

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  • simplify expression k/m%n

    - by aaa
    hello. Simple question, is it possible to simplify (or replace division or modulo by less-expensive operation) (k/m)%n where variables are integers and operators are C style division and modulo operators. what about the case where m and n are constants (both or just one), not based 2? Thank you

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  • Simple/Basic steganography algorithms and methods

    - by tomp
    What are the basic and simpliest steganography algorithms and methods? I mean the steganography applied to images. How does simple program that hides data to images work? How does the program recognize the encrypted message in image without the source image? What are the main techniques used?

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  • How to deal with JSON output that might be an array, or might be a value

    - by Summer
    Hi - I'm getting JSON-encoded output from another organization's API. In many cases, the output can be either an array of objects (if there are many) or an object (if there's just one). Right now I'm writing tortured code like this: if ( is_array($json['candidateList']['candidate'][0]) ) { foreach ($json['candidateList']['candidate'] as $candidate) { // do something to each object } } else { // do something to the single object } How can I handle it so the "do something" part of my code only appears once and uses a standard syntax?

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  • longest common subsequence

    - by davit-datuashvili
    i have following code public class LCS1 { public static String lcs(String a,String b) { String x; String y; int alen=a.length(); int blen=b.length(); if (alen==0 || blen==0) { return ""; } else if (a.charAt(alen-1)==b.charAt(blen-1)) { return lcs(a.substring(0,alen-1),b.substring(0,blen-1)); } else { x=lcs(a,b.substring(0,blen-1)); y=lcs(a.substring(0,alen-1),b); } return (x.length()>y.length()) ? x : y; } public static void main(String[]args){ String a="computer"; String b="houseboat"; System.out.println(lcs(a,b)); } } it should return "out" but returns nothing what is problem?

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  • Reverse regular expressions to generate data

    - by Anton Gogolev
    In one of the StackOverflow Podcasts (the one where guys were discussing data generation for testing DBs -- either #11 or #12), Jeff mentioned something like "reverse regular expressions", which are used exactly for that purpose: given a regex, produce a string which will eventually match said regex. What is the correct term for this whole concept? Is this a well-known concept?

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  • K-Means Algorithm and java code

    - by Thandar
    Hi all, I need to calculate for grouping objects according to their size. I got k-means algorithms in java which calculate mostly for classifying according to their two or more features and the results are not satisfy for me.I only want to calculate for grouping objects based on one feature.Pseudocode or code would be helpful, too. Thanks u all for helping.

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  • Effecient data structure design

    - by Sway
    Hi there, I need to match a series of user inputed words against a large dictionary of words (to ensure the entered value exists). So if the user entered: "orange" it should match an entry "orange' in the dictionary. Now the catch is that the user can also enter a wildcard or series of wildcard characters like say "or__ge" which would also match "orange" The key requirements are: * this should be as fast as possible. * use the smallest amount of memory to achieve it. If the size of the word list was small I could use a string containing all the words and use regular expressions. however given that the word list could contain potentially hundreds of thousands of enteries I'm assuming this wouldn't work. So is some sort of 'tree' be the way to go for this...? Any thoughts or suggestions on this would be totally appreciated! Thanks in advance, Matt

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  • Implementing a hilbert map of the internet

    - by Martin
    In the XKCD comic 195 a design for a map of the internet address space is suggested using a hilbert curve so that items from a similar IPs will be clustered together. Given an IP address, how would I calculate the 2D coordinates (in the range zero to one) that this IP is located on such a map?

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  • Could I do this blind relative to absolute path conversion (for perforce depot paths) better?

    - by wonderfulthunk
    I need to "blindly" (i.e. without access to the filesystem, in this case the source control server) convert some relative paths to absolute paths. So I'm playing with dotdots and indices. For those that are curious I have a log file produced by someone else's tool that sometimes outputs relative paths, and for performance reasons I don't want to access the source control server where the paths are located to check if they're valid and more easily convert them to their absolute path equivalents. I've gone through a number of (probably foolish) iterations trying to get it to work - mostly a few variations of iterating over the array of folders and trying delete_at(index) and delete_at(index-1) but my index kept incrementing while I was deleting elements of the array out from under myself, which didn't work for cases with multiple dotdots. Any tips on improving it in general or specifically the lack of non-consecutive dotdot support would be welcome. Currently this is working with my limited examples, but I think it could be improved. It can't handle non-consecutive '..' directories, and I am probably doing a lot of wasteful (and error-prone) things that I probably don't need to do because I'm a bit of a hack. I've found a lot of examples of converting other types of relative paths using other languages, but none of them seemed to fit my situation. These are my example paths that I need to convert, from: //depot/foo/../bar/single.c //depot/foo/docs/../../other/double.c //depot/foo/usr/bin/../../../else/more/triple.c to: //depot/bar/single.c //depot/other/double.c //depot/else/more/triple.c And my script: begin paths = File.open(ARGV[0]).readlines puts(paths) new_paths = Array.new paths.each { |path| folders = path.split('/') if ( folders.include?('..') ) num_dotdots = 0 first_dotdot = folders.index('..') last_dotdot = folders.rindex('..') folders.each { |item| if ( item == '..' ) num_dotdots += 1 end } if ( first_dotdot and ( num_dotdots > 0 ) ) # this might be redundant? folders.slice!(first_dotdot - num_dotdots..last_dotdot) # dependent on consecutive dotdots only end end folders.map! { |elem| if ( elem !~ /\n/ ) elem = elem + '/' else elem = elem end } new_paths << folders.to_s } puts(new_paths) end

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  • True random number generator

    - by goldenmean
    Sorry for this not being a "real" question, but Sometime back i remember seeing a post here about randomizing a randomizer randomly to generate truly random numbers, not just pseudo random. I dont see it if i search for it. Does anybody know about that article?

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  • Is incrementing in a loop exponential time?

    - by user356106
    I've a simple but confusing doubt about whether the program below runs in exponential time. The question is : given a +ve integer as input, print it out. The catch is that you deliberately do this in a loop, like this: int input,output=0; cininput; while(input--) ++output; // Takes time proportional to the value of input cout<< output; I'm claiming that this problem runs in exponential time. Because, the moment you increase the # of bits in input by 1, the program takes double the amount of time to execute. Put another way, to print out log2(input) bits, it takes O(input) time. Is this reasoning right?

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  • Extracting a given number of the highest values in a List

    - by James P.
    I'm seeking to display a fixed number of items on a web page according to their respective weight (represented by an Integer). The List where these items are found can be of virtually any size. The first solution that comes to mind is to do a Collections.sort() and to get the items one by one by going through the List. Is there a more elegant solution though that could be used to prepare, say, the top eight items?

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  • Java - Removing duplicates in an ArrayList

    - by Will
    I'm working on a program that uses an ArrayList to store Strings. The program prompts the user with a menu and allows the user to choose an operation to perform. Such operations are adding Strings to the List, printing the entries etc. What I want to be able to do is create a method called removeDuplicates().This method will search the ArrayList and remove any duplicated values. I want to leave one instance of the duplicated value(s) within the list. I also want this method to return the total number of duplicates removed. I've been trying to use nested loops to accomplish this but I've been running into trouble because when entries get deleted, the indexing of the ArrayList gets altered and things don't work as they should. I know conceptually what I need to do but I'm having trouble implementing this idea in code. Here is some pseudo code: start with first entry; check each subsequent entry in the list and see if it matches the first entry; remove each subsequent entry in the list that matches the first entry; after all entries have been examined, move on to the second entry; check each entry in the list and see if it matches the second entry; remove each entry in the list that matches the second entry; repeat for entry in the list Here's the code I have so far: public int removeDuplicates() { int duplicates = 0; for ( int i = 0; i < strings.size(); i++ ) { for ( int j = 0; j < strings.size(); j++ ) { if ( i == j ) { // i & j refer to same entry so do nothing } else if ( strings.get( j ).equals( strings.get( i ) ) ) { strings.remove( j ); duplicates++; } } } return duplicates; }

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  • BFS algorithm problem

    - by Gorkamorka
    The problem is as follows: A wanderer begins on the grid coordinates (x,y) and wants to reach the coordinates (0,0). From every gridpoint, the wanderer can go 8 steps north OR 3 steps south OR 5 steps east OR 6 steps west (8N/3S/5E/6W). How can I find the shortest route from (X,Y) to (0,0) using breadth-first search? Clarifications: Unlimited grid Negative coordinates are allowed A queue (linked list or array) must be used No obstacles present

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  • C++ priority queue structure used ?

    - by John Retallack
    While searching for some functions in C++ STL documentation I read that push and pop for priority queues needs constant time. "Constant (in the priority_queue). Although notice that push_heap operates in logarithmic time." My question is what kind of data structure is used to mantain a priority queue with O(1) for push and pop ?

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  • Sort months ( with strings ) algorithm

    - by Oscar Reyes
    I have this months array: ["January", "March", "December" , "October" ] And I want to have it sorted like this: ["January", "March", "October", "December" ] I'm currently thinking in a "if/else" horrible cascade but I wonder if there is some other way to do this. The bad part is that I need to do this only with "string" ( that is, without using Date object or anything like that ) What would be a good approach?

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  • Priority queue with dynamic item priorities.

    - by sean
    I need to implement a priority queue where the priority of an item in the queue can change and the queue adjusts itself so that items are always removed in the correct order. I have some ideas of how I could implement this but I'm sure this is quite a common data structure so I'm hoping I can use an implementation by someone smarter than me as a base. Can anyone tell me the name of this type of priority queue so I know what to search for or, even better, point me to an implementation?

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