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  • How can I make a core-plot graph on the phone that doesn't auto-expand to fill the whole superview

    - by Robb
    I'm pretty sure I saw an example where the graph wasn't filling the whole iPhone screen, but I can't get that to happen in my app, nor in the Core-Plot Test app from Switch On The Code. I've added a subview to the original CPLayerHostingView in the sample, then changed the classes – original back to UIView, new subview to CPLayerHostingView, and I've reconnected the File's owner's view outlet to the new subview. When I create a graph with: graph = [[CPXYGraph alloc] initWithFrame: theSubviewOutlet.bounds]; … and step through the first stages of building up the layers the bounds are accurate (i.e. the same as in the .xib) however, when all the initialization is done, and the graph shows up, it fills the whole superview. Am I missing something obvious?

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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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  • 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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  • 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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  • 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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  • Find numbers that equals a sum in an array

    - by valli-R
    I want to find the first set of integers in an array X that the sum equals a given number N. For example: X = {5, 13, 24, 9, 3, 3} N = 28 Solution = {13, 9, 3, 3} Here what I have so far : WARNING, I know it uses global and it is bad,that's not the point of the question. <?php function s($index = 0, $total = 0, $solution = '') { global $numbers; global $sum; echo $index; if($total == 28) { echo '<br/>'.$solution.' = '.$sum.'<br/>'; } elseif($index < count($numbers) && $total != 28) { s($index + 1, $total, $solution); s($index + 1, $total + $numbers[$index], $solution.' '.$numbers[$index]); } } $numbers = array(5, 13, 24, 9, 3, 3); $sum = 28; s(); ?> I don't get how I can stop the process when it finds the solution.. I know I am not far from good solution.. Thanks in advance

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  • How to find longest common substring using trees?

    - by user384706
    The longest common substring problem according to wiki can be solved using a suffix tree. From wiki: The longest common substrings of a set of strings can be found by building a generalised suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it I don't get this. Example: if I have: ABCDE and XABCZ then the suffix tree is (some branches from XABCZ omitted due to space): The longest common substring is ABC but it is not I can not see how the description of wiki helps here. ABC is not the deepest internal nodes with leaf nodes. Any help to understand how this works?

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  • Given an even number of vertices, how to find an optimum set of pairs based on proximity?

    - by Alex Z
    The problem: We have a set of n vertices in 3D euclidean space, and there is an even number of these vertices. We want to pair them up based on their proximity. In other words, we'd like to be able to find a set of vertex pairs, where the vertices in each pair are as close as possible together. We want to minimise sacrificing the proximity between the vertices of any other pairs as much as possible in doing this. I am not looking for the most optimal solution (if it even strictly exists/can be done), just a reasonable one that can be computed relatively quickly. A relatively awful brute force approach involves choosing a vertex and looping through the rest to find its nearest neighbor and then repeating until there are none left. Of course as we near the end of the list the closest vertex could be very far away, but it is the only choice, therefore this can fail badly on the third point above.

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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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  • 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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  • Facebook Hacker Cup: Power Overwhelming

    - by marcog
    A lot of people at Facebook like to play Starcraft II™. Some of them have made a custom game using the Starcraft II™ map editor. In this game, you play as the noble Protoss defending your adopted homeworld of Shakuras from a massive Zerg army. You must do as much damage to the Zerg as possible before getting overwhelmed. You can only build two types of units, shield generators and warriors. Shield generators do no damage, but your army survives for one second per shield generator that you build. Warriors do one damage every second. Your army is instantly overrun after your shield generators expire. How many shield generators and how many warriors should you build to inflict the maximum amount of damage on the Zerg before your army is overrun? Because the Protoss value bravery, if there is more than one solution you should return the one that uses the most warriors. Constraints 1 = G (cost for one shield generator) = 100 1 = W (cost for one warrior) = 100 G + W = M (available funds) = 1000000000000 (1012)

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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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  • Are there Adaptive Replacement Cache patent-free alternatives?

    - by aleccolocco
    An open source high-performance project I'm working on needs to keep a cache of parsed/compiled files. A plain LRU or a plain LFU wouldn't fit. Plain LRU wouldn't work as there will be remote batch/spider processes hitting the service regularly. Plain LFU wouldn't work because content will age. ARC seems like the perfect solution but since IBM holds patents to it at least one open source project dropped it. Are there any (good enough) alternatives? EDIT: I'm not looking for exactly the same thing, just something that could handle those two situations. Perhaps some simple strategy with timestamps and sources. There have to be many programmers who faced this situation before. That's why the "good enough" bit.

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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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  • 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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  • 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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  • make tree in scheme

    - by ???
    (define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (define (mktree order items_list) (cond ((= (length items_list) 1) (make-tree (car items_list) '() '())) (else (insert2 order (car items_list) (mktree order (cdr items_list)))))) (define (insert2 order x t) (cond ((null? t) (make-tree x '() '())) ((order x (entry t)) (make-tree (entry t) (insert2 order x (left-branch t)) (right-branch t))) ((order (entry t) x ) (make-tree (entry t) (left-branch t) (insert2 order x (right-branch t)))) (else t))) The result is: (mktree (lambda (x y) (< x y)) (list 7 3 5 1 9 11)) (11 (9 (1 () (5 (3 () ()) (7 () ()))) ()) ()) But I'm trying to get: (7 (3 (1 () ()) (5 () ())) (9 () (11 () ()))) Where is the problem?

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  • Display relative time in hour, day, month and year

    - by JohnJohnGa
    I wrote a function toBeautyString(epoch) : String which given a epoch, return a string which will display the relative time from now in hour and minute For instance: // epoch: 1346140800 -> Tue, 28 Aug 2012 05:00:00 GMT // and now: 1346313600 -> Thu, 30 Aug 2012 08:00:00 GMT toBeautyString(1346140800) -> "2 days and 3 hours ago" I want now to extend this function to month and year, so it will be able to print: 2 years, 1 month, 3 days and 1 hour ago Only with epoch without any external libraries. The purpose of this function is to give to the user a better way to visualize the time in the past. I found this: Calculating relative time but the granularity is not enough.

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  • Is there "good" PRNG generating values without hidden state?

    - by actual
    I need some good pseudo random number generator that can be computed like a pure function from its previous output without any state hiding. Under "good" I mean: I must be able to parametrize generator in such way that running it for 2^n iterations with any parameters should cover all or almost all values between 0 and 2^n - 1, where n is the number of bits in output value. Combined generator output of n + p bits must cover all or almost all values between 0 and 2^(n + p) - 1 if I run it for 2^n iterations for every possible combination of its parameters, where p is the number of bits in parameters. For example, LCG can be computed like a pure function and it can meet first condition, but it can not meet second one. Say, we have 32-bit generator, m = 2^32 and it is constant, our p = 64 (two 32-bit parameters a and c), n + p = 96, so we must peek data by three ints from output to meet second condition. Unfortunately, condition can not be meet because of strictly alternating sequence of odd and even ints in output. To overcome this, hidden state must be introduced, but that makes function not pure and breaks first condition (period become much longer). Am I wanting too much?

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  • Explanation needed for sum of prime below n numbers

    - by Bala Krishnan
    Today I solved a problem given in Project Euler its problem no 10 and it took 7 hrs for my python program to show the result. But in that forum itself a person named lassevk posted solution for this and it took only 4 sec. And its not possible for me to post this question in that forum because its not discussion forum. So, think about this if you want to mark this question as non-constructive. marked = [0] * 2000000 value = 3 s = 2 while value < 2000000: if marked[value] == 0: s += value i = value while i < 2000000: marked[i] = 1 i += value value += 2 print s If any one understand this code please explain it simple as possible. Link to the Problem 10 question.

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