Search Results

Search found 7490 results on 300 pages for 'algorithm analysis'.

Page 113/300 | < Previous Page | 109 110 111 112 113 114 115 116 117 118 119 120  | Next Page >

  • 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

  • Are there any well-known algorithms or computer models that computer scientists use to predict FIFA

    - by Khnle
    Occasionally I read news articles that mention about some computer models that computer scientists use to predict winners of some sporting events or the odds for betting which I think there must be a mathematical model behind it. I never bothered to think twice even though I am a "pseudo computer scientist" myself. With the 2010 FIFA World Cup just underway, and since I am also a "pseudo football/soccer player" myself, I just started to wonder about these calculations algorithms. For example, I know one factor is determining the strength of opponents, so that a win against a strong opponent can count more than a win against a weak opponent. But it now kind of gets in a circular loop, or at least how does one determine the strength of a team in the first place, before that team can be considered strong or weak? If it's based on a historical data then there's no way that could be accurate, because those players of the past are no longer on the fields so their impact is none (except maybe if they become coaches like Maradona) Anyway, long question short, if you're happen to be working in this field or have some knowledge, please shed some lights.

    Read the article

  • Print number series in java

    - by user1898282
    I have to print the series shown below in java: ***1*** **2*2** *3*3*3* 4*4*4*4 My current implementation is: public static void printSeries(int number,int numberOfCharsinEachLine){ String s="*"; for(int i=1;i<=number;i++){ int countOfs=(numberOfCharsinEachLine-(i)-(i-1))/2; if(countOfs<0){ System.out.println("Can't be done"); break; } for(int j=0;j<countOfs;j++){ System.out.print(s); } System.out.print(i); for(int k=1;k<i;k++){ System.out.print(s); System.out.print(i); } for(int j=0;j<countOfs;j++){ System.out.print(s); } System.out.println(); } } But there are lot of for loops, so I'm wondering whether this can be done in a better way or not?

    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

  • Count double palindromes in given int sequence

    - by jakubmal
    For a given int sequence check number of double palindromes, where by double palindrome we mean sequence of two same palindromes without break between them. So for example: in 1 0 1 1 0 1 we have 1 0 1 as a palindrome which appears 2 times without a break, in 1 0 1 5 1 0 1 we have 1 0 1 but it's separated (apart from the other palindromes in these sequences) Problem example test data is: 3 12 0 1 1 0 0 1 1 0 0 1 1 0 12 1 0 1 0 1 0 1 0 1 0 1 0 6 3 3 3 3 3 3 with answers 8 0 9 Manacher is obvious for the begging, but I'm not sure what to do next. Any ideas appreciated. Complexity should be below n^2 I guess. EDIT: int is here treated as single element of alphabet

    Read the article

  • select i th smallest element from array

    - by davit-datuashvili
    i have divide and conqurer method to find i th smalles element from array here is code public class rand_select{ public static int Rand_partition( int a[],int p,int q,int i){ //smallest in a[p..q] if ( p==q) return a[p]; int r=partition (a,p,q); int k=r-p+1; if (i==k) return a[r]; if (i<k){ return Rand_partition(a,p,r-1,i); } return Rand_partition(a,r-1,q,i-k); } public static void main(String[]args){ int a[]=new int []{6,10,13,15,8,3,2,12}; System.out.println(Rand_partition(a,0,a.length-1,7)); } public static int partition(int a[],int p,int q){ int m=a[0]; while ( p<q){ while (p<q && a[p++] <m){ p++; } while (q>p && a[q--]>m){ q--; } int t=a[p]; a[p]=a[q]; a[q]=t; } int k=0; for (int i=0;i<a.length;i++){ if ( a[i]==m){ k=i; } } return k; } } but here is problem java.lang.ArrayIndexOutOfBoundsException please help me

    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

  • Find a common element within N arrays

    - by kunjaan
    If I have N arrays, what is the best(Time complexity. Space is not important) way to find the common elements. You could just find 1 element and stop. Edit: The elements are all Numbers. Edit: These are unsorted. Please do not sort and scan. This is not a homework problem. Somebody asked me this question a long time ago. He was using a hash to solve the problem. I was thinking if SO has solved similar problems.

    Read the article

  • Finding number of different paths

    - by peiska
    I have a game that one player X wants to pass a ball to player Y, but he can be playing with more than one player and the others players can pass the ball to Y. I want to know how many different paths can the ball take from X to Y? for example if he is playing with 3 players there are 5 different paths, 4 players 16 paths, if he is playing with 20 players there are 330665665962404000 paths, and 40 players 55447192200369381342665835466328897344361743780 that the ball can take. the number max. of players that he can play with is 500. I was thinking in using Catalan Numbers? do you think is a correct approach to solve this? Can you give me some tips.

    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

  • 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

  • Heap Algorithmic Issue

    - by OberynMarDELL
    I am having this algorithmic problem that I want to discuss about. Its not about find a solution but about optimization in terms of runtime. So here it is: Suppose we have a race court of Length L and a total of N cars that participate on the race. The race rules are simple. Once a car overtakes an other car the second car is eliminated from the race. The race ends when no more overtakes are possible to happen. The tricky part is that the k'th car has a starting point x[k] and a velocity v[k]. The points are given in an ascending order, but the velocities may differ. What I've done so far: Given that a car can get overtaken only by its previous, I calculated the time that it takes for each car to reach its next one t = (x[i] - x[i+1])/(v[i] - v[i+1]) and I insert these times onto a min heap in O(n log n). So in theory I have to pop the first element in O(logn), find its previous, pop it as well , update its time and insert it in the heap once more, much like a priority queue. My main problem is how I can access specific points of a heap in O(log n) or faster in order to keep the complexity in O(n log n) levels. This program should be written on Haskell so I would like to keep things simple as far as possible EDIT: I Forgot to write the actual point of the race. The goal is to find the order in which cars exit the game

    Read the article

  • Fast ceiling of an integer division in C / C++

    - by andand
    Given integer values x and y, C and C++ returns as the quotient q = x/y the floor of the floating point valued equivalent. I'm interestd in a method of returning the ceiling instead? For example, ceil(10/5) = 2 and ceil(11/5) = 3. The obvious approach involves something like: q = x / y; if (q * y < x) ++q; This requires an extra comparison and multiplication; and other methods I've seen (used in fact) involve casting as a float or double. Is there a more direct method that avoids the additional multiplication (or a second division) and branch, and that also avoids casting as a floating point number?

    Read the article

  • Routing algorithm

    - by isaac
    Hello, I'm giving a presentation about computer routing and I want to make a good analogy with a real-world situation. However, I could not find it. Do you have in mind any of the situations like the computer routing. If yes, could you please provide me with it

    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

  • question about siftdown operation on heap

    - by davit-datuashvili
    i have following pseudo code which execute siftdown operation on heap array suppose is x void siftdown(int n) pre heap(2,n) && n>=0 post heap(1,n) i=1; loop /*invariant heap(1,n) except perhaps between i and it's (0,1,or 2) children*/ c=2*i; if (c>n) break; // c is left child of i if (c+1)<=n /* c+1 is rigth child of i if (x[c+1]<x[c]) c++ /* c is lesser child of i if (x[i]<=x[c]) break; swap(c,i) i=c; i have wrote following code is it correct? public class siftdown{ public static void main(String[]args){ int c; int n=9; int a[]=new int[]{19,100,17,2,7,3,36,1,25}; int i=1; while (i<n){ c=2*i; if (c>n) break; //c is the left child of i if (c+1<=n) //c+1 ir rigth child of i if (a[c+1]<a[c]) c++; if (a[i]<=a[c]) break; int t=a[c]; a[c]=a[i]; a[i]=t; i=c; } for (int j=0;j<a.length;j++){ System.out.println(a[j]); } } } // result is 19 2 17 1 7 3 36 100 25

    Read the article

  • finding middle element of an array

    - by senthil
    Hi all, I came cross a question in my interview. Question: Array of integers will be given as the input and you should find out the middle element when sorted , but without sorting. For Example. Input: 1,3,5,4,2 Output: 3 When you sort the given input array, it will be 1,2,3,4,5 where middle element is 3. You should find this in one pass without sorting. Any solutions for this?

    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

  • bidirectional buble sort

    - by davit-datuashvili
    Here is the code I'm using for shacker sort (or bidirectional buble sort). Something is wrong; the error is java.lang.ArrayIndexOutOfBoundsException. Can anybody help me? public class bidirectional{ public static void main(String[]args){ int x[]=new int[]{12,9,4,99,120,1,3,10}; int j; int n=x.length; int st=-1; while (st<n){ st++; n--; for (j=st;j<n;j++) { if (x[j]>x[j+1]) { int t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } for (j=n;--j>=st;) { if (x[j]>x[j+1]) { int t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } } for (int k=0;k<x.length;k++) { System.out.println(x[k]); } } }

    Read the article

  • Separating text and graphics in an image

    - by avd
    I dont know whether should I post this question here or not? But if someone knows it, please answer? What are the algorithms for determining which region in an image is text and which one is graphic? Means how to separate such regions? (figure or diagram)

    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

  • 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

< Previous Page | 109 110 111 112 113 114 115 116 117 118 119 120  | Next Page >