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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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  • Echo mysql results in a loop?

    - by Roy D. Porter
    I am using turn.js to make a book. Every div within the 'deathnote' div becomes a new page. <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page </div> //ends book What I am doing is trying to get 3 'content' (content being a name and cause of death) divs onto 1 page, and then generate a new page. So here is what i want: <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"> //creates new page but leaves it open <div> CONTENT </div> <div> CONTENT </div> <div> CONTENT </div> </div> //ends the page </div> //ends book Seems simple enough, however the content is data from a MySQL DB, so i have to echo it in using PHP. Here is what i have so far <div id="deathnote"> <div style="background-image:url(images/coverpage.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <?php $pagecount = 0; $db = new mysqli('localhost', 'username', 'passw', 'DB'); if($db->connect_errno > 0){ die('Unable to connect to database [' . $db->connect_error . ']'); } $sql = <<<SQL SELECT * FROM `TABLE` SQL; if(!$result = $db->query($sql)){ die('There was an error running the query [' . $db->error . ']'); } //IGNORE ALL OF THE GARBAGE ABOVE. IT IS SIMPLE CONNECTING SCRIPT THAT I KNOW WORKS //THE METHOD I AM HAVING TROUBLE WITH IS BELOW $pagecount = 0; while($row = $result->fetch_assoc()){ //GETS THE VALUE (and makes sure it isn't nothing echo '<div style="background-image:url(images/paper.jpg);">'; //THIS OPENS A NEW PAGE while ($pagecount !== 3) { //KEEPS COUNT OF HOW MUCH CONTENT DIVS IS ON THE PAGE while($row = $result->fetch_assoc()){ //START A CONTENT DIV echo '<div class="content"><div class="name">' . $row['victim'] . '</div><div class="cod">' . $row['cod'] . '</div></div>'; //END A CONTENT DIV $pagecount++; //UP THE PAGE COUNT } } $pagecount=0; //PUT IT BACK TO 0 echo '</div>'; //END PAGE } $db->close(); ?> <div style="background-image:url(images/backpage.jpg);"></div> //BACK PAGE </div> At the moment i seem to be causing and infinite loop so the page won't load. The problem resides within the while loops. Any help is greatly appreciated. Thanks in advance guys. :)

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  • Way to store a large dictionary with low memory footprint + fast lookups (on Android)

    - by BobbyJim
    I'm developing an android word game app that needs a large (~250,000 word dictionary) available. I need: reasonably fast look ups e.g. constant time preferable, need to do maybe 200 lookups a second on occasion to solve a word puzzle and maybe 20 lookups within 0.2 second more often to check words the user just spelled. EDIT: Lookups are typically asking "Is in the dictionary?". I'd like to support up to two wildcards in the word as well, but this is easy enough by just generating all possible letters the wildcards could have been and checking the generated words (i.e. 26 * 26 lookups for a word with two wildcards). as it's a mobile app, using as little memory as possible and requiring only a small initial download for the dictionary data is top priority. My first naive attempts used Java's HashMap class, which caused an out of memory exception. I've looked into using the SQL lite databases available on android, but this seems like overkill. What's a good way to do what I need?

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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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  • Find common nodes from two linked lists using recursion

    - by Dan
    I have to write a method that returns a linked list with all the nodes that are common to two linked lists using recursion, without loops. For example, first list is 2 - 5 - 7 - 10 second list is 2 - 4 - 8 - 10 the list that would be returned is 2 - 10 I am getting nowhere with this.. What I have been think of was to check each value of the first list with each value of the second list recursively but the second list would then be cut by one node everytime and I cannot compare the next value in the first list with the the second list. I hope this makes sense... Can anyone help?

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  • arbitrary vire connection / search and replace

    - by fatai
    input :["vire_connection",[1, 2, [ 3, [ 4, "connect"]]], ["connect", [3 , 5] ] ] output:["vire_connection",[ 1, 2, [ 3, [ 4, [ 3, 5 ] ] ] ] ], [ [ 3 , 5] ] ] after connection ( simply copying [3,5] to other wanted position ) , remove connect word input :["vire_connection", [ [ [ ["connect", [ 3, 4 ] ] ] ] ], [ 2, "connect"]] output :["vire_connection",[[[[[3,4]]]]], [ 2, [ 3 , 4 ]]] after connection ( simply copying [3,4] to other wanted position ) , remove connect word how can I do ?

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  • O(log N) == O(1) - Why not?

    - by phoku
    Whenever I consider algorithms/data structures I tend to replace the log(N) parts by constants. Oh, I know log(N) diverges - but does it matter in real world applications? log(infinity) < 100 for all practical purposes. I am really curious for real world examples where this doesn't hold. To clarify: I understand O(f(N)) I am curious about real world examples where the asymptotic behaviour matters more than the constants of the actual performance. If log(N) can be replaced by a constant it still can be replaced by a constant in O( N log N). This question is for the sake of (a) entertainment and (b) to gather arguments to use if I run (again) into a controversy about the performance of a design.

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  • Javascript Number Random Scrambler

    - by stjowa
    Hi, I need a Javascript random number scrambler for my website. Seems simple, but I can not figure out how to do it. Can anyone help me out? I have the following array of numbers: 1 2 3 4 5 6 7 8 9 I would like to be able to have these numbers scrambled randomly. Like the following: 3 6 4 2 9 5 1 8 7 or 4 1 7 3 5 9 2 6 8 So, specifically, I would like a function that takes in an array of numbers (1 - n) and then returns that same array of numbers - scrambled randomly with different calls to the function. Maybe a noob function, but can't seem to figure it out. 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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  • 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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  • 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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  • Writing shorter code/algorithms, is more efficient (performance)?

    - by Carlos
    After coming across the code golf trivia around the site it is obvious people try to find ways to write code and algorithms as short as the possibly can in terms of characters, lines and total size, even if that means writing something like: n=input() while n>1:n=(n/2,n*3+1)[n%2];print n So as a beginner I start to wonder whether size actually matters :D. It is obviously a very subjective question highly dependent on the actual code being used, but what is the rule of thumb in the real world. In the case that size wont matter, how come then we don't focus more on performance rather than size?

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  • How do I remove the leaves of a binary tree?

    - by flopex
    I'm trying to remove all of the leaves. I know that leaves have no children, this is what I have so far. public void removeLeaves(BinaryTree n){ if (n.left == null && n.right == null){ n = null; } if (n.left != null) removeLeaves(n.left); if (n.right != null) removeLeaves(n.right); }

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  • Searching algorithmics: Parsing and processing a request

    - by James P.
    Say you were to create a search engine that can accept a query statement under the form of a String. The statement can be used to retrieve different types of objects with a given set of characteristics and possibly linked to other objects. In plain english or pseudo-code using an OOP approach, how would you go about parsing and processing statements as follows to get the series of desired objects ? get fruit with colour green get variety of apples, pears from Andy get strawberry with colour "deep red" and origin not Spain get total of sales of melons between 2010-10-10 and 2010-12-30 get last deliverydate of bananas from "Pete" and state not sold Hope the question is clear. If not I'll be more than happy to reformulate. P.S: This isn't homework ;)

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  • Output from OouraFFT correct sometimes but completely false other times. Why ?

    - by Yan
    Hi I am using Ooura FFT to compute the FFT of the accelerometer data in windows of 1024 samples. The code works fine, but then for some reason it produces very strange outputs, i.e. continuous spectrum with amplitudes of the order of 10^200. Here is the code: OouraFFT *myFFT=[[OouraFFT alloc] initForSignalsOfLength:1024 NumWindows:10]; // had to allocate it UIAcceleration *tempAccel = nil; double *input=(double *)malloc(1024 * sizeof(double)); double *frequency=(double *)malloc(1024*sizeof(double)); if (input) { //NSLog(@"%d",[array count]); for (int u=0; u<[array count]; u++) { tempAccel = (UIAcceleration *)[array objectAtIndex:u]; input[u]=tempAccel.z; //NSLog(@"%g",input[u]); } } myFFT.inputData=input; // specifies input data to myFFT [myFFT calculateWelchPeriodogramWithNewSignalSegment]; // calculates FFT for (int i=0;i<myFFT.dataLength;i++) // loop to copy output of myFFT, length of spectrumData is half of input data, so copy twice { if (i<myFFT.numFrequencies) { frequency[i]=myFFT.spectrumData[i]; // } else { frequency[i]=myFFT.spectrumData[myFFT.dataLength-i]; // copy twice } } for (int i=0;i<[array count];i++) { TransformedAcceleration *NewAcceleration=[[TransformedAcceleration alloc]init]; tempAccel=(UIAcceleration*)[array objectAtIndex:i]; NewAcceleration.timestamp=tempAccel.timestamp; NewAcceleration.x=tempAccel.x; NewAcceleration.y=tempAccel.z; NewAcceleration.z=frequency[i]; [newcurrentarray addObject:NewAcceleration]; // this does not work //[self replaceAcceleration:NewAcceleration]; //[NewAcceleration release]; [NewAcceleration release]; } TransformedAcceleration *a=nil;//[[TransformedAcceleration alloc]init]; // object containing fft of x,y,z accelerations for(int i=0; i<[newcurrentarray count]; i++) { a=(TransformedAcceleration *)[newcurrentarray objectAtIndex:i]; //NSLog(@"%d,%@",i,[a printAcceleration]); fprintf(fp,[[a printAcceleration] UTF8String]); //this is going wrong somewhow } fclose(fp); [array release]; [myFFT release]; //[array removeAllObjects]; [newcurrentarray release]; free(input); free(frequency);

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  • Count Occurence of Needle String in Haystack String, most optimally?

    - by Taranfx
    The Problem is simple Find "ABC" in "ABCDSGDABCSAGAABCCCCAAABAABC" Here is the solution I propose, I'm looking for any solutions that might be better than this one. public static void main(String[] args) { String haystack = "ABCDSGDABCSAGAABCCCCAAABAABC"; String needle = "ABC"; char [] needl = needle.toCharArray(); int needleLen = needle.length(); int found=0; char hay[] = haystack.toCharArray(); int index =0; int chMatched =0; for (int i=0; i<hay.length; i++){ if (index >= needleLen || chMatched==0) index=0; System.out.print("\nchar-->"+hay[i] + ", with->"+needl[index]); if(hay[i] == needl[index]){ chMatched++; System.out.println(", matched"); }else { chMatched=0; index=0; if(hay[i] == needl[index]){ chMatched++; System.out.print("\nchar->"+hay[i] + ", with->"+needl[index]); System.out.print(", matched"); }else continue; } if(chMatched == needleLen){ found++; System.out.println("found. Total ->"+found); } index++; } System.out.println("Result Found-->"+found); } It took me a while creating this one. Can someone suggest a better solution (if any) P.S. Drop the sysouts if they look messy to you.

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  • Garbage Collection in Java

    - by simion
    On the slides I am revising from it says the following: Live objects can be identified either by maintaining a count of the number of references to each object, or by tracing chains of references from the roots. Reference counting is expensive – it needs action every time a reference changes and it doesn’t spot cyclical structures, but it can reclaim space incrementally. Tracing involves identifying live objects only when you need to reclaim space – moving the cost from general access to the time at which the GC runs, typically only when you are out of memory. I understand the principles of why reference counting is expensive but do not understand what "doesn’t spot cyclical structures, but it can reclaim space incrementally." means. Could anyone help me out a little bit please? Thanks

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  • Make function non-recursive

    - by user69514
    I'm not sure how to make this function non-recursive. Any ideas?: void foo(int a, int b){ while( a < len && arr[a][b] != -1){ if(++a == len){ a = 0; b++; } } if( a == len){ size++; return; } if( a < (len-1)){ arr[a][b] = 1; arr[a][(b+1)] = 1; foo(a, b); arr[a][b] = -1; arr[a][(b+1)] = -1; } if( a < (len-1) && arr[(a+1)][b] == -1){ arr[a][b] = 0; arr[(a+1)][b] = 0; foo(a,b); arr[a][b] = -1; arr[(a+1)][b] = -1; } }

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