Search Results

Search found 5070 results on 203 pages for 'algorithm'.

Page 95/203 | < Previous Page | 91 92 93 94 95 96 97 98 99 100 101 102  | Next Page >

  • question about quicksort

    - by davit-datuashvili
    i have write code of quicksort from programming pearls here is code public class Quick{ public static void quicksort(int x[], int l,int u) { if (l>=u) return ; int t=x[l]; int i=l; int j=u; do { i++; } while (i<=u && x[i]<t); do { j--; if (i>=j) break; } while ( x[j]>t); swap(x,i,j); swap(x, l,j); quicksort(x, l,j-1); quicksort(x, j+1,u); } public static void main(String[]args){ int x[]=new int[]{55,41,59,26,53,58,97,93}; quicksort(x,0,x.length-1); for (int i=0;i<x.length;i++){ System.out.println(x[i]); } } public static void swap(int x[], int i,int j){ int s=x[i]; x[i]=x[j]; x[j]=s; } } but it does not work here is output 59 41 55 26 53 97 58 93 any idea?

    Read the article

  • Simple Big O with lg(n) proof

    - by halohunter
    I'm attempting to guess and prove the Big O for: f(n) = n^3 - 7n^2 + nlg(n) + 10 I guess that big O is n^3 as it is the term with the largest order of growth However, I'm having trouble proving it. My unsuccesful attempt follows: f(n) <= cg(n) f(n) <= n^3 - 7n^2 + nlg(n) + 10 <= cn^3 f(n) <= n^3 + (n^3)*lg(n) + 10n^3 <= cn^3 f(n) <= N^3(11 + lg(n)) <= cn^3 so 11 + lg(n) = c But this can't be right because c must be constant. What am I doing wrong?

    Read the article

  • Length of Encrypted String

    - by Agnel Kurian
    I need to create a database column which will store a string encrypted using Triple DES. How do I determine the length of the encrypted string column? (Answers for algorithms other than Triple DES are also welcome.)

    Read the article

  • User submitted content filtering

    - by Jim
    Hey all, Does anyone have any ideas on what could be used as a way to filter untrustworthy user submitted content? Take Yelp for instance, they would need to prevent competitors writing business reviews on their competitors. They would need to prevent business owners favourably reviewing their own business, or forcing friends/family to do so. They would need to prevent poor quality reviews from affecting a businesses rating and so on. I can't think what they might use to do this: Prevent multiple users from the same IP reviewing certain things Prevent business owners reviewing their own business (maybe even other businesses in the same categories as their own?) Somehow determine what a review is about and what the actual intentions behind it are Other than the first and second points, I can't think of any clever/easy way to filter potentially harmful reviews from being made available, other than a human doing it. Obviously for a site the size of Yelp this wouldn't be feasible, so what parameters could they take into consideration? Even with human intervention, how would anyone know it was the owners best buddy writing a fake review without knowing the people? I'm using this as an example in a larger study on the subject of filtering user content automatically. Does anyone have any ideas how these systems may work and what they take into consideration? Thanks!

    Read the article

  • [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 :)

    Read the article

  • Solving a recurrence T(n) = 2T(n/2) + n^4

    - by user563454
    I am studying using the MIT Courseware and the CLRS book Introduction to Algorithms. Solving recurrence T(n) = 2T(n/2) + n4 (page 107) If I make a recurrence tree I get: level 0 n^4 level 1 2(n/2)^4 level 2 4(n/4)^4 level 3 8(n/8)^4 The tree has lg(n) levels. Therefore the recurrence is T(n) = Theta(lg(n)n^4)) But, If I use the Master method I get. Apply case 3: T(n) = Theta(n^4) If I apply the substitution method both seem to hold. Which one is ri?

    Read the article

  • Simple loop, which one I would get more performance and which one is recommended? defining a variable inside a loop or outside of it?

    - by Grego
    Variable outside of the loop int number = 0; for(int i = 0; i < 10000; i++){ number = 3 * i; printf("%d",number); } or Variable inside of the loop for(int i = 0; i < 10000; i++){ int number = 3 * i; printf("%d",number); } Which one is recommended and which one is better in performance? Edit: This is just an example to exhibit what I mean, All I wanna know is if defining a variable inside a loop and outside a loop means the same thing , or there's a difference.

    Read the article

  • 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?

    Read the article

  • Generate a valid array key from an URL string in PHP

    - by John Riche
    I have a PHP array with some predefined values: $aArray = array( 0 => 'value0', 1 => 'value1' ); I need to create a function where the string input will always return the same, valid, array key so that when I call: GiveMeAKey('http://www.google.com'); // May return 0 or 1 I receive always the same key (I don't care which one) from the array. Obvisously I can't store the relationship in a database and the string passed to the GiveMeAKey method can be any URL. I wonder if there is a way of doing that ?

    Read the article

  • How to calculate the state of a graph?

    - by zcb
    Given a graph G=(V,E), each node i is associated with 'Ci' number of objects. At each step, for every node i, the Ci objects will be taken away by the neighbors of i equally. After K steps, output the number of objects of the top five nodes which has the most objects. Some Constrains: |V|<10^5, |E|<2*10^5, K<10^7, Ci<1000 My current idea is: represent the transformation in each step with a matrix. This problem is converted to the calculation of the power of matrix. But this solution is much too slow considering |V| can be 10^5. Is there any faster way to do it?

    Read the article

  • How to optimize this simple function which translates input bits into words?

    - by psihodelia
    I have written a function which reads an input buffer of bytes and produces an output buffer of words where every word can be either 0x0081 for each ON bit of the input buffer or 0x007F for each OFF bit. The length of the input buffer is given. Both arrays have enough physical place. I also have about 2Kbyte free RAM which I can use for lookup tables or so. Now, I found that this function is my bottleneck in a real time application. It will be called very frequently. Can you please suggest a way how to optimize this function? I see one possibility could be to use only one buffer and do in-place substitution. void inline BitsToWords(int8 *pc_BufIn, int16 *pw_BufOut, int32 BufInLen) { int32 i,j,z=0; for(i=0; i<BufInLen; i++) { for(j=0; j<8; j++, z++) { pw_BufOut[z] = ( ((pc_BufIn[i] >> (7-j))&0x01) == 1? 0x0081: 0x007f ); } } } Please do not offer any compiler specific or CPU/Hardware specific optimization, because it is a multi-platform project.

    Read the article

  • 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'

    Read the article

  • How to read arbitrary number of values using std::copy?

    - by Miro Kropacek
    Hi, I'm trying to code opposite action to this: std::ostream outs; // properly initialized of course std::set<int> my_set; // ditto outs << my_set.size(); std::copy( my_set.begin(), my_set.end(), std::ostream_iterator<int>( outs ) ); it should be something like this: std::istream ins; std::set<int>::size_type size; ins >> size; std::copy( std::istream_iterator<int>( ins ), std::istream_iterator<int>( ins ) ???, std::inserter( my_set, my_set.end() ) ); But I'm stuck with the 'end' iterator -- input interators can't use std::advance and neither I can use two streams with the same source... Is there any elegant way how to solve this? Of course I can use for loop, but maybe there's something nicer :)

    Read the article

  • Find all numbers that appear in each of a set of lists

    - by Ankur
    I have several ArrayLists of Integer objects, stored in a HashMap. I want to get a list (ArrayList) of all the numbers (Integer objects) that appear in each list. My thinking so far is: Iterate through each ArrayList and put all the values into a HashSet This will give us a "listing" of all the values in the lists, but only once Iterate through the HashSet 2.1 With each iteration perform ArrayList.contains() 2.2 If none of the ArrayLists return false for the operation add the number to a "master list" which contains all the final values. If you can come up with something faster or more efficient, funny thing is as I wrote this I came up with a reasonably good solution. But I'll still post it just in case it is useful for someone else. But of course if you have a better way please do let me know.

    Read the article

  • How to convert from base-256 to base-N, where N is higher than 16?

    - by mark
    Dear ladies and sirs. I need to convert an array of bytes to another base, namely 85. In math terms the question is how to convert from base-256 to base-85 in the most efficient way? This question is inspired by my previous question - http://stackoverflow.com/questions/2827627/what-is-the-most-efficient-way-to-encode-an-arbitrary-guid-into-readable-ascii-3 Thanks.

    Read the article

  • 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.

    Read the article

  • Longest increasing subsequence

    - by davit-datuashvili
    I have written this code; is it correct? public static void main(String[] args){ int a[]=new int[]{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int a_b[]=new int[a.length]; a_b[0]=1; int int_max=0; int int_index=0; for (int i=0;i<a.length;i++){ for (int j=0;j<i;j++){ if (a[i]>a[j] && a_b[i]<(a_b[j]+1)){ a_b[i]=a_b[j]+1; } } if (a_b[i]>int_max){ int_max=a_b[i]; int_index=i; } } int k=int_max+1; int list[]=new int[k]; for (int i=int_index;i>0;i--){ if (a_b[i]==k-1){ list[k-1]=a[i]; k=a_b[i]; } } for (int g=0;g<list.length;g++){ System.out.println(list[g]); } }

    Read the article

  • 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?

    Read the article

  • Intersection() and Except() is too slow with large collections of custom objects

    - by Theo
    I am importing data from another database. My process is importing data from a remote DB into a List<DataModel> named remoteData and also importing data from the local DB into a List<DataModel> named localData. I am then using LINQ to create a list of records that are different so that I can update the local DB to match the data pulled from remote DB. Like this: var outdatedData = this.localData.Intersect(this.remoteData, new OutdatedDataComparer()).ToList(); I am then using LINQ to create a list of records that no longer exist in remoteData, but do exist in localData, so that I delete them from local database. Like this: var oldData = this.localData.Except(this.remoteData, new MatchingDataComparer()).ToList(); I am then using LINQ to do the opposite of the above to add the new data to the local database. Like this: var newData = this.remoteData.Except(this.localData, new MatchingDataComparer()).ToList(); Each collection imports about 70k records, and each of the 3 LINQ operation take between 5 - 10 minutes to complete. How can I make this faster? Here is the object the collections are using: internal class DataModel { public string Key1{ get; set; } public string Key2{ get; set; } public string Value1{ get; set; } public string Value2{ get; set; } public byte? Value3{ get; set; } } The comparer used to check for outdated records: class OutdatedDataComparer : IEqualityComparer<DataModel> { public bool Equals(DataModel x, DataModel y) { var e = string.Equals(x.Key1, y.Key1) && string.Equals(x.Key2, y.Key2) && ( !string.Equals(x.Value1, y.Value1) || !string.Equals(x.Value2, y.Value2) || x.Value3 != y.Value3 ); return e; } public int GetHashCode(DataModel obj) { return 0; } } The comparer used to find old and new records: internal class MatchingDataComparer : IEqualityComparer<DataModel> { public bool Equals(DataModel x, DataModel y) { return string.Equals(x.Key1, y.Key1) && string.Equals(x.Key2, y.Key2); } public int GetHashCode(DataModel obj) { return 0; } }

    Read the article

  • algorithms that destruct and copy_construct

    - by FredOverflow
    I am currently building my own toy vector for fun, and I was wondering if there is something like the following in the current or next standard or in Boost? template<class T> void destruct(T* begin, T* end) { while (begin != end) { begin -> ~T(); ++begin; } } template<class T> T* copy_construct(T* begin, T* end, T* dst) { while (begin != end) { new(dst) T(*begin); ++begin; ++dst; } return dst; }

    Read the article

  • Java array of arry [matrix] of an integer partition with fixed term

    - by user335209
    Hello, for my study purpose I need to build an array of array filled with the partitions of an integer with fixed term. That is given an integer, suppose 10 and given the fixed number of terms, suppose 5 I need to populate an array like this 10 0 0 0 0 9 0 0 0 1 8 0 0 0 2 7 0 0 0 3 ............ 9 0 0 1 0 8 0 0 1 1 ............. 7 0 1 1 0 6 0 1 1 1 ............ ........... 0 6 1 1 1 ............. 0 0 0 0 10 am pretty new to Java and am getting confused with all the for loops. Right now my code can do the partition of the integer but unfortunately it is not with fixed term public class Partition { private static int[] riga; private static void printPartition(int[] p, int n) { for (int i= 0; i < n; i++) System.out.print(p[i]+" "); System.out.println(); } private static void partition(int[] p, int n, int m, int i) { if (n == 0) printPartition(p, i); else for (int k= m; k > 0; k--) { p[i]= k; partition(p, n-k, n-k, i+1); } } public static void main(String[] args) { riga = new int[6]; for(int i = 0; i<riga.length; i++){ riga[i] = 0; } partition(riga, 6, 1, 0); } } the output I get it from is like this: 1 5 1 4 1 1 3 2 1 3 1 1 1 2 3 1 2 2 1 1 2 1 2 1 2 1 1 1 what i'm actually trying to understand how to proceed is to have it with a fixed terms which would be the columns of my array. So, am stuck with trying to get a way to make it less dynamic. Any help?

    Read the article

  • 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).

    Read the article

< Previous Page | 91 92 93 94 95 96 97 98 99 100 101 102  | Next Page >