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  • Genetic Algorithms applied to Curve Fitting

    - by devoured elysium
    Let's imagine I have an unknown function that I want to approximate via Genetic Algorithms. For this case, I'll assume it is y = 2x. I'd have a DNA composed of 5 elements, one y for each x, from x = 0 to x = 4, in which, after a lot of trials and computation and I'd arrive near something of the form: best_adn = [ 0, 2, 4, 6, 8 ] Keep in mind I don't know beforehand if it is a linear function, a polynomial or something way more ugly, Also, my goal is not to infer from the best_adn what is the type of function, I just want those points, so I can use them later. This was just an example problem. In my case, instead of having only 5 points in the DNA, I have something like 50 or 100. What is the best approach with GA to find the best set of points? Generating a population of 100, discard the worse 20% Recombine the remaining 80%? How? Cutting them at a random point and then putting together the first part of ADN of the father with the second part of ADN of the mother? Mutation, how should I define in this kind of problem mutation? Is it worth using Elitism? Any other simple idea worth using around? Thanks

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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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  • Average function without overflow exception

    - by Ron Klein
    .NET Framework 3.5. I'm trying to calculate the average of some pretty large numbers. For instance: using System; using System.Linq; class Program { static void Main(string[] args) { var items = new long[] { long.MinValue + 100, long.MinValue + 200, long.MinValue + 300 }; try { var avg = items.Average(); Console.WriteLine(avg); } catch (OverflowException ex) { Console.WriteLine("can't calculate that!"); } Console.ReadLine(); } } Obviously, the mathematical result is 9223372036854775607 (long.MaxValue - 200), but I get an exception there. This is because the implementation (on my machine) to the Average extension method, as inspected by .NET Reflector is: public static double Average(this IEnumerable<long> source) { if (source == null) { throw Error.ArgumentNull("source"); } long num = 0L; long num2 = 0L; foreach (long num3 in source) { num += num3; num2 += 1L; } if (num2 <= 0L) { throw Error.NoElements(); } return (((double) num) / ((double) num2)); } I know I can use a BigInt library (yes, I know that it is included in .NET Framework 4.0, but I'm tied to 3.5). But I still wonder if there's a pretty straight forward implementation of calculating the average of integers without an external library. Do you happen to know about such implementation? Thanks!! UPDATE: The previous example, of three large integers, was just an example to illustrate the overflow issue. The question is about calculating an average of any set of numbers which might sum to a large number that exceeds the type's max value. Sorry about this confusion. I also changed the question's title to avoid additional confusion. Thanks all!!

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  • Using base64 encoding as a mechanism to detect changes

    - by Mikos
    Is it possible to detect changes in the base64 encoding of an object to detect the degree of changes in the object. Suppose I send a document attachment to several users and each makes changes to it and emails back to me, can I use the string distance between original base64 and the received base64s to detect which version has the most changes. Would that be a valid metric? If not, would there be any other metrics to quantify the deltas?

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  • fast similarity detection

    - by reinierpost
    I have a large collection of objects and I need to figure out the similarities between them. To be exact: given two objects I can compute their dissimilarity as a number, a metric - higher values mean less similarity and 0 means the objects have identical contents. The cost of computing this number is proportional to the size of the smaller object (each object has a given size). I need the ability to quickly find, given an object, the set of objects similar to it. To be exact: I need to produce a data structure that maps any object o to the set of objects no more dissimilar to o than d, for some dissimilarity value d, such that listing the objects in the set takes no more time than if they were in an array or linked list (and perhaps they actually are). Typically, the set will be very much smaller than the total number of objects, so it is really worthwhile to perform this computation. It's good enough if the data structure assumes a fixed d, but if it works for an arbitrary d, even better. Have you seen this problem before, or something similar to it? What is a good solution? To be exact: a straightforward solution involves computing the dissimilarities between all pairs of objects, but this is slow - O(n2) where n is the number of objects. Is there a general solution with lower complexity?

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  • 3 dimensional bin packing algorithms

    - by BuschnicK
    I'm faced with a 3 dimensional bin packing problem and am currently conducting some preliminary research as to which algorithms/heuristics are currently yielding the best results. Since the problem is NP hard I do not expect to find the optimal solution in every case, but I was wondering: 1) what are the best exact solvers? Branch and Bound? What problem instance sizes can I expect to solve with reasonable computing resources? 2) what are the best heuristic solvers? 3) What off-the-shelf solutions exist to conduct some experiments with?

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  • Visualizing volume of PCM samples

    - by genevincent
    I have several chunks of PCM audio (G.711) in my C++ application. I would like to visualize the different audio volume in each of these chunks. My first attempt was to calculate the average of the sample values for each chunk and use that as an a volume indicator, but this doesn't work well. I do get 0 for chunks with silence and differing values for chunks with audio, but the values only differ slighly and don't seem to resemble the actual volume. What would be a better algorithem calculate the volume ? I hear G.711 audio is logarithmic PCM. How should I take that into account ?

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  • Trend analysis using iterative value increments

    - by Dave Jarvis
    We have configured iReport to generate the following graph: The real data points are in blue, the trend line is green. The problems include: Too many data points for the trend line Trend line does not follow a Bezier curve (spline) The source of the problem is with the incrementer class. The incrementer is provided with the data points iteratively. There does not appear to be a way to get the set of data. The code that calculates the trend line looks as follows: import java.math.BigDecimal; import net.sf.jasperreports.engine.fill.*; /** * Used by an iReport variable to increment its average. */ public class MovingAverageIncrementer implements JRIncrementer { private BigDecimal average; private int incr = 0; /** * Instantiated by the MovingAverageIncrementerFactory class. */ public MovingAverageIncrementer() { } /** * Returns the newly incremented value, which is calculated by averaging * the previous value from the previous call to this method. * * @param jrFillVariable Unused. * @param object New data point to average. * @param abstractValueProvider Unused. * @return The newly incremented value. */ public Object increment( JRFillVariable jrFillVariable, Object object, AbstractValueProvider abstractValueProvider ) { BigDecimal value = new BigDecimal( ( ( Number )object ).doubleValue() ); // Average every 10 data points // if( incr % 10 == 0 ) { setAverage( ( value.add( getAverage() ).doubleValue() / 2.0 ) ); } incr++; return getAverage(); } /** * Changes the value that is the moving average. * @param average The new moving average value. */ private void setAverage( BigDecimal average ) { this.average = average; } /** * Returns the current moving average average. * @return Value used for plotting on a report. */ protected BigDecimal getAverage() { if( this.average == null ) { this.average = new BigDecimal( 0 ); } return this.average; } /** Helper method. */ private void setAverage( double d ) { setAverage( new BigDecimal( d ) ); } } How would you create a smoother and more accurate representation of the trend line?

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  • Invert a stack, without using extra data structures?

    - by vks
    How would you invert a stack, without using extra data structures, like a second, or temporary, stack. Thus no stack1-stack2 or stack-queue-stack implementation in the answer. You just have access to push/pop feature of a standard stack. I think there is way to do it by keeping a global counter and using pointer manipulation. If I solve it myself, I will post it. If someone else figures it out, please post your solution.

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  • Modeling distribution of performance measurements

    - by peterchen
    How would you mathematically model the distribution of repeated real life performance measurements - "Real life" meaning you are not just looping over the code in question, but it is just a short snippet within a large application running in a typical user scenario? My experience shows that you usually have a peak around the average execution time that can be modeled adequately with a Gaussian distribution. In addition, there's a "long tail" containing outliers - often with a multiple of the average time. (The behavior is understandable considering the factors contributing to first execution penalty). My goal is to model aggregate values that reasonably reflect this, and can be calculated from aggregate values (like for the Gaussian, calculate mu and sigma from N, sum of values and sum of squares). In other terms, number of repetitions is unlimited, but memory and calculation requirements should be minimized. A normal Gaussian distribution can't model the long tail appropriately and will have the average biased strongly even by a very small percentage of outliers. I am looking for ideas, especially if this has been attempted/analysed before. I've checked various distributions models, and I think I could work out something, but my statistics is rusty and I might end up with an overblown solution. Oh, a complete shrink-wrapped solution would be fine, too ;) Other aspects / ideas: Sometimes you get "two humps" distributions, which would be acceptable in my scenario with a single mu/sigma covering both, but ideally would be identified separately. Extrapolating this, another approach would be a "floating probability density calculation" that uses only a limited buffer and adjusts automatically to the range (due to the long tail, bins may not be spaced evenly) - haven't found anything, but with some assumptions about the distribution it should be possible in principle. Why (since it was asked) - For a complex process we need to make guarantees such as "only 0.1% of runs exceed a limit of 3 seconds, and the average processing time is 2.8 seconds". The performance of an isolated piece of code can be very different from a normal run-time environment involving varying levels of disk and network access, background services, scheduled events that occur within a day, etc. This can be solved trivially by accumulating all data. However, to accumulate this data in production, the data produced needs to be limited. For analysis of isolated pieces of code, a gaussian deviation plus first run penalty is ok. That doesn't work anymore for the distributions found above. [edit] I've already got very good answers (and finally - maybe - some time to work on this). I'm starting a bounty to look for more input / ideas.

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  • Project Euler Question 14 (Collatz Problem)

    - by paradox
    The following iterative sequence is defined for the set of positive integers: n -n/2 (n is even) n -3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 40 20 10 5 16 8 4 2 1 It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? NOTE: Once the chain starts the terms are allowed to go above one million. I tried coding a solution to this in C using the bruteforce method. However, it seems that my program stalls when trying to calculate 113383. Please advise :) #include <stdio.h> #define LIMIT 1000000 int iteration(int value) { if(value%2==0) return (value/2); else return (3*value+1); } int count_iterations(int value) { int count=1; //printf("%d\n", value); while(value!=1) { value=iteration(value); //printf("%d\n", value); count++; } return count; } int main() { int iteration_count=0, max=0; int i,count; for (i=1; i<LIMIT; i++) { printf("Current iteration : %d\n", i); iteration_count=count_iterations(i); if (iteration_count>max) { max=iteration_count; count=i; } } //iteration_count=count_iterations(113383); printf("Count = %d\ni = %d\n",max,count); }

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  • Implementing PageRank using MapReduce

    - by Nick D.
    Hello, I'm trying to get my head around an issue with the theory of implementing the PageRank with MapReduce. I have the following simple scenario with three nodes: A B C. The adjacency matrix is here: A { B, C } B { A } The PageRank for B for example is equal to: (1-d)/N + d ( PR(A) / C(A) ) N = number of incoming links to B PR(A) = PageRank of incoming link A C(A) = number of outgoing links from page A I am fine with all the schematics and how the mapper and reducer would work but I cannot get my head around how at the time of calculation by the reducer, C(A) would be known. How will the reducer, when calculating the PageRank of B by aggregating the incoming links to B will know the number of outgoing links from each page. Does this require a lookup in some external data source?

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  • Interview question: Develop an application that can display trail period expires after 30 days witho

    - by Algorist
    Hi, I saw this question in a forum about how an application can be developed that can keep track of the installation date and show trail period expired after 30 days of usage. The only constraint is not to use the external storage of any kind. Question: How to achieve this? Thanks Bala --Edit I think its easy to figure out the place to insert a question work. Anyway, I will write the question clearly. "external storage" means don't use any kind of storage like file, registry, network or anything. You only have your program.

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  • Removing the obstacle that yields the best path from a map after A* traversal

    - by David Titarenco
    I traverse a 16x16 maze using my own A* implementation. This is exactly what my program does: http://www.screenjelly.com/watch/fDQh98zMP0c?showTab=share All is well. However, after the traversal, I would like to find out what wall would give me the best alternative path. Apart from removing every block and re-running A* on the maze, what's a clever solution? I was thinking give every wall node (ignored by A*), a tentative F-value, and change the node structure to also have a n-sized list of node *tentative_parent where n is the number of walls in the maze. Could this be viable?

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  • PHP: Script for generating Crossword game?

    - by Prashant
    I need an script for generating crossword game. I have a list of 8 words for which I wnat to generate a crossword game, let's say for 15 column and 15 row. I am not getting the concept of this problem. How to generate this using PHP ?? Can anyone tell me how to do that ??

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  • shortest digest of a string

    - by meta
    [Description] Given a string of char type, find a shortest digest, which is defined as: a shortest sub-string which contains all the characters in the original string. [Example] A = "aaabedacd" B = "bedac" is the answer. [My solution] Define an integer table with 256 elements, which is used to record the occurring times for each kind of character in the current sub-string. Scan the whole string, statistic the total kinds of character in the given string by using the above table. Use two pointers, start, end, which are initially pointing to the start and (start + 1) of the given string. The current kinds of character is 1. Expand sub-string[start, end) at the end until it contains all kinds of character. Update the shortest digest if possible. Contract sub-string[start, end] at the start by one character each time, try to restore its digest property if necessary by step 4. The time cost is O(n), and the extra space cost is constant. Any better solution without extra space?

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  • Java code optimization on matrix windowing computes in more time

    - by rano
    I have a matrix which represents an image and I need to cycle over each pixel and for each one of those I have to compute the sum of all its neighbors, ie the pixels that belong to a window of radius rad centered on the pixel. I came up with three alternatives: The simplest way, the one that recomputes the window for each pixel The more optimized way that uses a queue to store the sums of the window columns and cycling through the columns of the matrix updates this queue by adding a new element and removing the oldes The even more optimized way that does not need to recompute the queue for each row but incrementally adjusts a previously saved one I implemented them in c++ using a queue for the second method and a combination of deques for the third (I need to iterate through their elements without destructing them) and scored their times to see if there was an actual improvement. it appears that the third method is indeed faster. Then I tried to port the code to Java (and I must admit that I'm not very comfortable with it). I used ArrayDeque for the second method and LinkedLists for the third resulting in the third being inefficient in time. Here is the simplest method in C++ (I'm not posting the java version since it is almost identical): void normalWindowing(int mat[][MAX], int cols, int rows, int rad){ int i, j; int h = 0; for (i = 0; i < rows; ++i) { for (j = 0; j < cols; j++) { h = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { for (int rx =- rad; rx <= rad; rx++) { int x = j + rx; if (x >= 0 && x < cols) { h += mat[y][x]; } } } } } } } Here is the second method (the one optimized through columns) in C++: void opt1Windowing(int mat[][MAX], int cols, int rows, int rad){ int i, j, h, y, col; queue<int>* q = NULL; for (i = 0; i < rows; ++i) { if (q != NULL) delete(q); q = new queue<int>(); h = 0; for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][rx]; } } q->push(mem); h += mem; } } for (j = 1; j < cols; j++) { col = j + rad; if (j - rad > 0) { h -= q->front(); q->pop(); } if (j + rad < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][col]; } } q->push(mem); h += mem; } } } } And here is the Java version: public static void opt1Windowing(int [][] mat, int rad){ int i, j = 0, h, y, col; int cols = mat[0].length; int rows = mat.length; ArrayDeque<Integer> q = null; for (i = 0; i < rows; ++i) { q = new ArrayDeque<Integer>(); h = 0; for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][rx]; } } q.addLast(mem); h += mem; } } j = 0; for (j = 1; j < cols; j++) { col = j + rad; if (j - rad > 0) { h -= q.peekFirst(); q.pop(); } if (j + rad < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][col]; } } q.addLast(mem); h += mem; } } } } I recognize this post will be a wall of text. Here is the third method in C++: void opt2Windowing(int mat[][MAX], int cols, int rows, int rad){ int i = 0; int j = 0; int h = 0; int hh = 0; deque< deque<int> *> * M = new deque< deque<int> *>(); for (int ry = 0; ry <= rad; ry++) { if (ry < rows) { deque<int> * q = new deque<int>(); M->push_back(q); for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int val = mat[ry][rx]; q->push_back(val); h += val; } } } } deque<int> * C = new deque<int>(M->front()->size()); deque<int> * Q = new deque<int>(M->front()->size()); deque<int> * R = new deque<int>(M->size()); deque< deque<int> *>::iterator mit; deque< deque<int> *>::iterator mstart = M->begin(); deque< deque<int> *>::iterator mend = M->end(); deque<int>::iterator rit; deque<int>::iterator rstart = R->begin(); deque<int>::iterator rend = R->end(); deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); for (mit = mstart, rit = rstart; mit != mend, rit != rend; ++mit, ++rit) { deque<int>::iterator pit; deque<int>::iterator pstart = (* mit)->begin(); deque<int>::iterator pend = (* mit)->end(); for(cit = cstart, pit = pstart; cit != cend && pit != pend; ++cit, ++pit) { (* cit) += (* pit); (* rit) += (* pit); } } for (i = 0; i < rows; ++i) { j = 0; if (i - rad > 0) { deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); deque<int>::iterator pit; deque<int>::iterator pstart = (M->front())->begin(); deque<int>::iterator pend = (M->front())->end(); for(cit = cstart, pit = pstart; cit != cend; ++cit, ++pit) { (* cit) -= (* pit); } deque<int> * k = M->front(); M->pop_front(); delete k; h -= R->front(); R->pop_front(); } int row = i + rad; if (row < rows && i > 0) { deque<int> * newQ = new deque<int>(); M->push_back(newQ); deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); int rx; int tot = 0; for (rx = 0, cit = cstart; rx <= rad; rx++, ++cit) { if (rx < cols) { int val = mat[row][rx]; newQ->push_back(val); (* cit) += val; tot += val; } } R->push_back(tot); h += tot; } hh = h; copy(C->begin(), C->end(), Q->begin()); for (j = 1; j < cols; j++) { int col = j + rad; if (j - rad > 0) { hh -= Q->front(); Q->pop_front(); } if (j + rad < cols) { int val = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { val += mat[y][col]; } } hh += val; Q->push_back(val); } } } } And finally its Java version: public static void opt2Windowing(int [][] mat, int rad){ int cols = mat[0].length; int rows = mat.length; int i = 0; int j = 0; int h = 0; int hh = 0; LinkedList<LinkedList<Integer>> M = new LinkedList<LinkedList<Integer>>(); for (int ry = 0; ry <= rad; ry++) { if (ry < rows) { LinkedList<Integer> q = new LinkedList<Integer>(); M.addLast(q); for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int val = mat[ry][rx]; q.addLast(val); h += val; } } } } int firstSize = M.getFirst().size(); int mSize = M.size(); LinkedList<Integer> C = new LinkedList<Integer>(); LinkedList<Integer> Q = null; LinkedList<Integer> R = new LinkedList<Integer>(); for (int k = 0; k < firstSize; k++) { C.add(0); } for (int k = 0; k < mSize; k++) { R.add(0); } ListIterator<LinkedList<Integer>> mit; ListIterator<Integer> rit; ListIterator<Integer> cit; ListIterator<Integer> pit; for (mit = M.listIterator(), rit = R.listIterator(); mit.hasNext();) { Integer r = rit.next(); int rsum = 0; for (cit = C.listIterator(), pit = (mit.next()).listIterator(); cit.hasNext();) { Integer c = cit.next(); Integer p = pit.next(); rsum += p; cit.set(c + p); } rit.set(r + rsum); } for (i = 0; i < rows; ++i) { j = 0; if (i - rad > 0) { for(cit = C.listIterator(), pit = M.getFirst().listIterator(); cit.hasNext();) { Integer c = cit.next(); Integer p = pit.next(); cit.set(c - p); } M.removeFirst(); h -= R.getFirst(); R.removeFirst(); } int row = i + rad; if (row < rows && i > 0) { LinkedList<Integer> newQ = new LinkedList<Integer>(); M.addLast(newQ); int rx; int tot = 0; for (rx = 0, cit = C.listIterator(); rx <= rad; rx++) { if (rx < cols) { Integer c = cit.next(); int val = mat[row][rx]; newQ.addLast(val); cit.set(c + val); tot += val; } } R.addLast(tot); h += tot; } hh = h; Q = new LinkedList<Integer>(); Q.addAll(C); for (j = 1; j < cols; j++) { int col = j + rad; if (j - rad > 0) { hh -= Q.getFirst(); Q.pop(); } if (j + rad < cols) { int val = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { val += mat[y][col]; } } hh += val; Q.addLast(val); } } } } I guess that most is due to the poor choice of the LinkedList in Java and to the lack of an efficient (not shallow) copy method between two LinkedList. How can I improve the third Java method? Am I doing some conceptual error? As always, any criticisms is welcome. UPDATE Even if it does not solve the issue, using ArrayLists, as being suggested, instead of LinkedList improves the third method. The second one performs still better (but when the number of rows and columns of the matrix is lower than 300 and the window radius is small the first unoptimized method is the fastest in Java)

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  • C# searching for new Tool for the tool box, how to template this code

    - by Nix
    All i have something i have been trying to do for a while and have yet to find a good strategy to do it, i am not sure C# can even support what i am trying to do. Example imagine a template like this, repeated in manager code overarching cocept function Returns a result consisting of a success flag and error list. public Result<Boolean> RemoveLocation(LocationKey key) { List<Error> errorList = new List<Error>(); Boolean result = null; try{ result = locationDAO.RemoveLocation(key); }catch(UpdateException ue){ //Error happened less pass this back to the user! errorList = ue.ErrorList; } return new Result<Boolean>(result, errorList); } Looking to turn it into a template like the below where Do Something is some call (preferably not static) that returns a Boolean. I know i could do this in a stack sense, but i am really looking for a way to do it via object reference. public Result<Boolean> RemoveLocation(LocationKey key) { var magic = locationDAO.RemoveLocation(key); return ProtectedDAOCall(magic); } public Result<Boolean> CreateLocation(LocationKey key) { var magic = locationDAO.CreateLocation(key); return ProtectedDAOCall(magic); } public Result<Boolean> ProtectedDAOCall(Func<..., bool> doSomething) { List<Error> errorList = new List<Error>(); Boolean result = null; try{ result = doSomething(); }catch(UpdateException ue){ //Error happened less pass this back to the user! errorList = ue.ErrorList; } return new Result<Boolean>(result, errorList); } If there is any more information you may need let me know. I am interested to see what someone else can come up with. Marc solution applied to the code above public Result<Boolean> CreateLocation(LocationKey key) { LocationDAO locationDAO = new LocationDAO(); return WrapMethod(() => locationDAO.CreateLocation(key)); } public Result<Boolean> RemoveLocation(LocationKey key) { LocationDAO locationDAO = new LocationDAO(); return WrapMethod(() => locationDAO.RemoveLocation(key)); } static Result<T> WrapMethod<T>(Func<Result<T>> func) { try { return func(); } catch (UpdateException ue) { return new Result<T>(default(T), ue.Errors); } }

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  • Time complexity for Search and Insert operation in sorted and unsorted arrays that includes duplicat

    - by iecut
    1-)For sorted array I have used Binary Search. We know that the worst case complexity for SEARCH operation in sorted array is O(lg N), if we use Binary Search, where N are the number of items in an array. What is the worst case complexity for the search operation in the array that includes duplicate values, using binary search?? Will it be the be the same O(lg N)?? Please correct me if I am wrong!! Also what is the worst case for INSERT operation in sorted array using binary search?? My guess is O(N).... is that right?? 2-) For unsorted array I have used Linear search. Now we have an unsorted array that also accepts duplicate element/values. What are the best worst case complexity for both SEARCH and INSERT operation. I think that we can use linear search that will give us O(N) worst case time for both search and delete operations. Can we do better than this for unsorted array and does the complexity changes if we accepts duplicates in the array.

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  • How to find the longest contiguous subsequence whose reverse is also a subsequence

    - by iecut
    Suppose I have a sequence x1,x2,x3.....xn, and I want to find the longest contiguous subsequence xi,xi+1,xi+2......xi+k, whose reverse is also a subsequence of the given sequence. And if there are multiple such subsequences, then I also have to find the first. ex:- consider the sequences: abcdefgedcg here i=3 and k=2 aabcdddd here i=5, k=3 I tried looking at the original longest common subsequence problem, but that is used to compare the two sequences to find the longest common subsequence.... but here is only one sequence from which we have to find the subsequences. Please let me know what is the best way to approach this problem, to find the optimal solution.

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  • Visualization of Nelder-Mead algorithm in gnuplot

    - by gorczas
    Hi, does anyone know how I can achieve drawing triangle on level sets of some 3d function (something like on this image in gnuplot? When I tried doing this after reading some tutorials: gnuplot> set border 15 front linetype -1 linewidth 1.000 gnuplot> set logscale z 10 gnuplot> set view map gnuplot> set isosamples 60, 60 gnuplot> unset surface gnuplot> set contour base gnuplot> unset clabel gnuplot> set style data lines gnuplot> set ticslevel 0 gnuplot> set noztics gnuplot> set title "Trwa symulacja" gnuplot> set xlabel "x" gnuplot> set xrange [ * : * ] noreverse nowriteback gnuplot> set ylabel "y" gnuplot> set zlabel "" gnuplot> set yrange [ * : * ] noreverse nowriteback gnuplot> set zrange [ * : * ] noreverse nowriteback gnuplot> splot [-10.5:10.5] [-10.5:10.5] x**2 +y**2 with lines lc rgb "#000000" notitle,\ >'-' with lines notitle input data ('e' ends) > 5.39703780733842 0.424994542694183 29.3086374551602 input data ('e' ends) > -4.80045950473308 -8.66307635892326 98.0933034571172 input data ('e' ends) > -3.56740563691939 3.31903046267993 23.7423461905216 input data ('e' ends) > 5.39703780733842 0.424994542694183 29.3086374551602 input data ('e' ends) > e But I'm still getting warning: "Cannot contour non grid data. Please use "set dgrid3d".".

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  • Need help with dynamic programming problem

    - by John Retallack
    I have the following problem : I am given a tree with N apples, for each apple I am given it's weight and height,I can pick apples up to a given height H,each time I pick an apple the height of every apple is increased with U(also given).I have to find out the maximum weight of apples I can pick. e.g: N=4 H=100 U=10 (height-eight) apple1: 91 10 apple2: 82 30 apple3: 93 5 apple4: 94 15 The answer is 45 : I first pick the apple with the weight of 15 then the one with the weight of 30. I would like to know if someone here could help me with giving me an hint on how I should approach this problem. Thank you.

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