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  • A question about matrix manipulation

    - by appi
    Given a 1*N matrix or an array, how do I find the first 4 elements which have the same value and then store the index for those elements? PS: I'm just curious. What if we want to find the first 4 elements whose value differences are within a certain range, say below 2? For example, M=[10,15,14.5,9,15.1,8.5,15.5,9.5], the elements I'm looking for will be 15,14.5,15.1,15.5 and the indices will be 2,3,5,7.

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  • How to find same-value rectangular areas of a given size in a matrix most efficiently?

    - by neo
    My problem is very simple but I haven't found an efficient implementation yet. Suppose there is a matrix A like this: 0 0 0 0 0 0 0 4 4 2 2 2 0 0 4 4 2 2 2 0 0 0 0 2 2 2 1 1 0 0 0 0 0 1 1 Now I want to find all starting positions of rectangular areas in this matrix which have a given size. An area is a subset of A where all numbers are the same. Let's say width=2 and height=3. There are 3 areas which have this size: 2 2 2 2 0 0 2 2 2 2 0 0 2 2 2 2 0 0 The result of the function call would be a list of starting positions (x,y starting with 0) of those areas. List((2,1),(3,1),(5,0)) The following is my current implementation. "Areas" are called "surfaces" here. case class Dimension2D(width: Int, height: Int) case class Position2D(x: Int, y: Int) def findFlatSurfaces(matrix: Array[Array[Int]], surfaceSize: Dimension2D): List[Position2D] = { val matrixWidth = matrix.length val matrixHeight = matrix(0).length var resultPositions: List[Position2D] = Nil for (y <- 0 to matrixHeight - surfaceSize.height) { var x = 0 while (x <= matrixWidth - surfaceSize.width) { val topLeft = matrix(x)(y) val topRight = matrix(x + surfaceSize.width - 1)(y) val bottomLeft = matrix(x)(y + surfaceSize.height - 1) val bottomRight = matrix(x + surfaceSize.width - 1)(y + surfaceSize.height - 1) // investigate further if corners are equal if (topLeft == bottomLeft && topLeft == topRight && topLeft == bottomRight) { breakable { for (sx <- x until x + surfaceSize.width; sy <- y until y + surfaceSize.height) { if (matrix(sx)(sy) != topLeft) { x = if (x == sx) sx + 1 else sx break } } // found one! resultPositions ::= Position2D(x, y) x += 1 } } else if (topRight != bottomRight) { // can skip x a bit as there won't be a valid match in current row in this area x += surfaceSize.width } else { x += 1 } } } return resultPositions } I already tried to include some optimizations in it but I am sure that there are far better solutions. Is there a matlab function existing for it which I could port? I'm also wondering whether this problem has its own name as I didn't exactly know what to google for. Thanks for thinking about it! I'm excited to see your proposals or solutions :)

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  • Help to learn Image Search algorithm

    - by R Manu
    I am a beginner in image processing. I want to write an application in C++ or in C# for Searching an image in a list of images Searching for a particular feature (for e.g. face) in a list of images. Can anybody suggest where should I start from? What all should I learn before doing this? Where can I find the correct information regarding this?

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  • How to store and collect data for mining such information as most viewed for last 24 hours, last 7 d

    - by Kirzilla
    Hello, Let's imagine that we have high traffic project (a tube site) which should provide sorting using this options (NOT IN REAL TIME). Number of videos is about 200K and all information about videos is stored in MySQL. Number of daily video views is about 1.5KK. As instruments we have Hard Disk Drive (text files), MySQL, Redis. Views top viewed top viewed last 24 hours top viewed last 7 days top viewed last 30 days top rated last 365 days How should I store such information? The first idea is to log all visits to text files (single file per hour, for example visits_20080101_00.log). At the beginning of each hour calculate views per video for previous hour and insert this information into MySQL. Then recalculate totals (for last 24 hours) and update statistics in tables. At the beginning of every day we have to do the same but recalculate for last 7 days, last 30 days, last 365 days. This method seems to be very poor for me because we have to store information about last 365 days for each video to make correct calculations. Is there any other good methods? Probably, we have to choose another instruments for this? Thank you.

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  • [C++] std::tring manipulation: whitespace, "newline escapes '\'" and comments #

    - by rubenvb
    Kind of looking for affirmation here. I have some hand-written code, which I'm not shy to say I'm proud of, which reads a file, removes leading whitespace, processes newline escapes '\' and removes comments starting with #. It also removes all empty lines (also whitespace-only ones). Any thoughts/recommendations? I could probably replace some std::cout's with std::runtime_errors... but that's not a priority here :) const int RecipeReader::readRecipe() { ifstream is_recipe(s_buffer.c_str()); if (!is_recipe) cout << "unable to open file" << endl; while (getline(is_recipe, s_buffer)) { // whitespace+comment removeLeadingWhitespace(s_buffer); processComment(s_buffer); // newline escapes + append all subsequent lines with '\' processNewlineEscapes(s_buffer, is_recipe); // store the real text line if (!s_buffer.empty()) v_s_recipe.push_back(s_buffer); s_buffer.clear(); } is_recipe.close(); return 0; } void RecipeReader::processNewlineEscapes(string &s_string, ifstream &is_stream) { string s_temp; size_t sz_index = s_string.find_first_of("\\"); while (sz_index <= s_string.length()) { if (getline(is_stream,s_temp)) { removeLeadingWhitespace(s_temp); processComment(s_temp); s_string = s_string.substr(0,sz_index-1) + " " + s_temp; } else cout << "Error: newline escape '\' found at EOF" << endl; sz_index = s_string.find_first_of("\\"); } } void RecipeReader::processComment(string &s_string) { size_t sz_index = s_string.find_first_of("#"); s_string = s_string.substr(0,sz_index); } void RecipeReader::removeLeadingWhitespace(string &s_string) { const size_t sz_length = s_string.size(); size_t sz_index = s_string.find_first_not_of(" \t"); if (sz_index <= sz_length) s_string = s_string.substr(sz_index); else if ((sz_index > sz_length) && (sz_length != 0)) // "empty" lines with only whitespace s_string.clear(); } Some extra info: std::string s_buffer is a class data member, so is std::vector v_s_recipe. Any comment is welcome :)

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  • Distributing players to tables

    - by IVlad
    Consider N = 4k players, k tables and a number of clans such that each member can belong to one clan. A clan can contain at most k players. We want to organize 3 rounds of a game such that, for each table that seats exactly 4 players, no 2 players sitting there are part of the same clan, and, for the later rounds, no 2 players sitting there have sat at the same table before. All players play all rounds. How can we do this efficiently if N can be about ~80 large? I thought of this: for each table T: repeat until 4 players have been seated at T: pick a random player X that is not currently seated anywhere if X has not sat at the same table as anyone currently at T AND X is not from the same clan as anyone currently at T seat X at T break I am not sure if this will always finish or if it can get stuck even if there is a valid assignment. Even if this works, is there a better way to do it?

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  • question about in -place sort

    - by davit-datuashvili
    for example we have following array char data[]=new char[]{'A','S','O','R','T','I','N','G','E','X','A','M','P','L','E'}; and index array int a[]=new int[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}: void insitu(char data[],int a[],N){ for (int i=0;i<N;i++) { char v=data[i]; int j,int k; for (k=i;a[k]!=i;k=a[j];a[j]=j) { j=k;data[k]=data[a[k]; } data[k]=v; a[k]=k; } i have question what is initialize value of j? when i run this code it asks me to initialize j and what should do? }

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  • Find all A^x in a given range

    - by Austin Henley
    I need to find all monomials in the form AX that when evaluated falls within a range from m to n. It is safe to say that the base A is greater than 1, the power X is greater than 2, and only integers need to be used. For example, in the range 50 to 100, the solutions would be: 2^6 3^4 4^3 My first attempt to solve this was to brute force all combinations of A and X that make "sense." However this becomes too slow when used for very large numbers in a big range since these solutions are used in part of much more intensive processing. Here is the code: def monoSearch(min, max): base = 2 power = 3 while 1: while base**power < max: if base**power > min: print "Found " + repr(base) + "^" + repr(power) + " = " + repr(base**power) power = power + 1 base = base + 1 power = 3 if base**power > max: break I could remove one base**power by saving the value in a temporary variable but I don't think that would make a drastic effect. I also wondered if using logarithms would be better or if there was a closed form expression for this. I am open to any optimizations or alternatives to finding the solutions.

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  • question about Batcher odd-even sort

    - by davit-datuashvili
    hi i ave question about Batcher's odd-even sort i have following code public class Batcher{ public static void batchsort(int a[],int l,int r){ int n=r-l+1; for (int p=1;p<n;p+=p) for (int k=p;k>0;k/=2) for (int j=k%p;j+k<n;j+=(k+k)) for (int i=0;i<n-j-k;i++) if ((j+i)/(p+p)==(j+i+k)/(p+p)) exch(a,l+j+i,l+j+i+k); } public static void main(String[]args){ int a[]=new int[]{2,4,3,4,6,5,3}; batchsort(a,0,a.length-1); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } } public static void exch(int a[],int i,int j){ int t=a[i]; a[i]=a[j]; a[j]=t; } } //result is 3 3 4 4 5 2 6 what i missed ? hat is wrong?

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  • Finding whether a point lies inside a rectangle or not

    - by avd
    The rectangle can be oriented in any way...need not be axis aligned. Now I want to find whether a point lies inside the rectangle or not. One method I could think of was to rotate the rectangle and point coordinates to make the rectangle axis aligned and then by simply testing the coordinates of point whether they lies within that of rectangle's or not. The above method requires rotation and hence floating point operations. Is there any other efficient way to do this??

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  • please help me to implement such kind of sort

    - by davit-datuashvili
    hi i need to write sucj kind of sorting maybe it is similary to radix sort and also (this is not homework because i created it problem myself and please if anobody can help me) problem is like this suppose i have array int x[]=new int[]{4,5,3,2,1}; let write it into binary form 5 -0101 4- 0100 3-0011 2-0010 1-0001 i want to sort this elements by using bitwise operatos or check each bit and if less exchange it can anybody help me for example take 5 and 4 check first rightmost bit 0==0 so continue in the 1 index also 1==1 next the same 0=0 and last one 10 it means that first element is greater then second so exchange it please help me

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  • How can we find second maximum from array efficiently?

    - by Xinus
    Is it possible to find the second maximum number from an array of integers by traversing the array only once? As an example, I have a array of five integers from which I want to find second maximum number. Here is an attempt I gave in the interview: #define MIN -1 int main() { int max=MIN,second_max=MIN; int arr[6]={0,1,2,3,4,5}; for(int i=0;i<5;i++){ cout<<"::"<<arr[i]; } for(int i=0;i<5;i++){ if(arr[i]>max){ second_max=max; max=arr[i]; } } cout<<endl<<"Second Max:"<<second_max; int i; cin>>i; return 0; } The interviewer, however, came up with the test case int arr[6]={5,4,3,2,1,0};, which prevents it from going to the if condition the second time. I said to the interviewer that the only way would be to parse the array two times (two for loops). Does anybody have a better solution?

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  • F# insert/remove item from list

    - by Timothy
    How should I go about removing a given element from a list? As an example, say I have list ['A'; 'B'; 'C'; 'D'; 'E'] and want to remove the element at index 2 to produce the list ['A'; 'B'; 'D'; 'E']? I've already written the following code which accomplishes the task, but it seems rather inefficient to traverse the start of the list when I already know the index. let remove lst i = let rec remove lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ t else remove t (lst' @ [h]) remove lst [] let myList = ['A'; 'B'; 'C'; 'D'; 'E'] let newList = remove myList 2 Alternatively, how should I insert an element at a given position? My code is similar to the above approach and most likely inefficient as well. let insert lst i x = let rec insert lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ [x] @ lst else insert t (lst' @ [h]) insert lst [] let myList = ['A'; 'B'; 'D'; 'E'] let newList = insert myList 2 'C'

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  • What is the most efficient method to find x contiguous values of y in an array?

    - by Alec
    Running my app through callgrind revealed that this line dwarfed everything else by a factor of about 10,000. I'm probably going to redesign around it, but it got me wondering; Is there a better way to do it? Here's what I'm doing at the moment: int i = 1; while ( ( (*(buffer++) == 0xffffffff && ++i) || (i = 1) ) && i < desiredLength + 1 && buffer < bufferEnd ); It's looking for the offset of the first chunk of desiredLength 0xffffffff values in a 32 bit unsigned int array. It's significantly faster than any implementations I could come up with involving an inner loop. But it's still too damn slow.

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  • how to find out if a shape is passable

    - by jd
    I have a complex polygon (possibly concave) and a few of its edges marked as entry/exit points. there is a possibility that inside this polygon may lie one or more blockades of arbitrary shape. what approaches could I use to determine whether a path of certain width exists between a pair of entry/exit edges? having read through the question it looks like a homework type - it is not. I just wish to have a at least a few leads I could pursue, as this is new to me.

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  • Graph search problem with route restrictions

    - by Darcara
    I want to calculate the most profitable route and I think this is a type of traveling salesman problem. I have a set of nodes that I can visit and a function to calculate cost for traveling between nodes and points for reaching the nodes. The goal is to reach a fixed known score while minimizing the cost. This cost and rewards are not fixed and depend on the nodes visited before. The starting node is fixed. There are some restrictions on how nodes can be visited. Some simplified examples include: Node B can only be visited after A After node C has been visited, D or E can be visited. Visiting at least one is required, visiting both is permissible. Z can only be visited after at least 5 other nodes have been visited Once 50 nodes have been visited, the nodes A-M will no longer reward points Certain nodes can (and probably must) be visited multiple times Currently I can think of only two ways to solve this: a) Genetic Algorithms, with the fitness function calculating the cost/benefit of the generated route b) Dijkstra search through the graph, since the starting node is fixed, although the large number of nodes will probably make that not feasible memory wise. Are there any other ways to determine the best route through the graph? It doesn't need to be perfect, an approximated path is perfectly fine, as long as it's error acceptable. Would TSP-solvers be an option here?

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  • How would I use for_each to delete every value in an STL map?

    - by stusmith
    Suppose I have a STL map where the values are pointers, and I want to delete them all. How would I represent the following code, but making use of std::for_each? I'm happy for solutions to use Boost. for( stdext::hash_map<int, Foo *>::iterator ir = myMap.begin(); ir != myMap.end(); ++ir ) { delete ir->second; // delete all the (Foo *) values. } (I've found Boost's checked_delete, but I'm not sure how to apply that to the pair<int, Foo *> that the iterator represents). (Also, for the purposes of this question, ignore the fact that storing raw pointers that need deleting in an STL container isn't very sensible).

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  • Reduce text length to fit cell width in a smart manner

    - by Andrei Ciobanu
    Hello, I am in project where we are building a simple web calendar using Java EE technologies. We define a table where every row is an employee, and every column represents an hour interval. The table width and column widths are adjustable. In every cell we have a text retrieved from a database, indicating what the employee is doing / should do in that time interval. The problem is that sometimes the text in cells is getting bigger than the actual cell. My task is to make the text more "readable" by reducing it's length in a "smart way" so that it can fit in the cell more "gracefully". For example if initially in a cell I have: "Writing documents", after the resize I should retrieve: "Wrtng. dcmnts" or "Writ. docum." so that the text can fit well. Is there a smart way to do it ? Or removing vocals / split the string in two is enough ?

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  • Interview question: C program to sort a binary array in O(n)

    - by Zacky112
    I've comeup with the following program to do it, but it does not seem to work and goes into infinite loop. Its working is similar to quicksort. int main() { int arr[] = {1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1}; int N = 18; int *front, *last; front = arr; last = arr + N; while(front <= last) { while( (front < last) && (*front == 0) ) front++; while( (front < last) && (*last == 1) ) last--; if( front < last) { int temp = *front; *front = *last; *last = temp; front ++; last--; } } for(int i=0;i<N;i++) printf("%d ",arr[i]); return 0; }

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  • One Database Field to Hold Survey Answer

    - by yar
    Skipping the question of whether this is bad design (and the question of why I would want/need to do this), I'm just wondering if my 'math' is right... I have n things that need to be put in order (n is always less than 5): thing1 thing2 thing3 ... and I'd like to store these results in a single integer for thing. My initial thought is that (obviously the code will not look like this): thing = thing1 * 1 + thing2 * 2 + thing3 * 4 + thing5 * 8 will always give me a unique value, and that any value for thing will always translate back to its values for thing1...thingn. Is this correct?

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