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  • Counting bits set in a .Net BitArray Class

    - by Sam
    I am implementing a library where I am extensively using the .Net BitArray class and need an equivalent to the Java BitSet.Cardinality() method, i.e. a method which returns the number of bits set. I was thinking of implementing it as an extension method for the BitArray class. The trivial implementation is to iterate and count the bits set (like below), but I wanted a faster implementation as I would be performing thousands of set operations and counting the answer. Is there a faster way than the example below? count = 0; for (int i = 0; i < mybitarray.Length; i++) { if (mybitarray [i]) count++; }

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  • Credit Card checksums and validations that do not require connection to the financial institution

    - by cjavapro
    The validations I know of are: Checksum the whole card number should add up to zero. (range is 0-9) Check the first digit(s) against the card type Check the length against the card type Check the CCV length against the card type (I think all the major types are 3 anyway) Of course make sure it is accepted card type as well as non expired. Are there any other validations :) (I expect many folks did not know about all of these) The reason I ask is because I overheard there was one to checksum number against expiration or CCV.. I just wanted to check.

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  • How to find kth minimal element in the union of two sorted arrays?

    - by Michael
    This is a homework question. They say it takes O(logN + logM) where N and M are the arrays lengths. Let's name the arrays a and b. Obviously we can ignore all a[i] and b[i] where i k. First let's compare a[k/2] and b[k/2]. Let b[k/2] a[k/2]. Therefore we can discard also all b[i], where i k/2. Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer. What is the next step?

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  • Creating objects makes the VM faster?

    - by Sudhir Jonathan
    Look at this piece of code: MessageParser parser = new MessageParser(); for (int i = 0; i < 10000; i++) { parser.parse(plainMessage, user); } For some reason, it runs SLOWER (by about 100ms) than for (int i = 0; i < 10000; i++) { MessageParser parser = new MessageParser(); parser.parse(plainMessage, user); } Any ideas why? The tests were repeated a lot of times, so it wasn't just random. How could creating an object 10000 times be faster than creating it once?

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  • Formula for popularity? (based on "like it", "comments", "views")

    - by paullb
    I have some pages on a website and I have to create an ordering based on "popularity"/"activity" The parameters that I have to use are: views to the page comments made on the page (there is a form at the bottom where uses can make comments) clicks made to the "like it" icon Are there any standards for what a formula for popularity would be? (if not opinions are good too) (initially I thought of views + 10*comments + 10*likeit)

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  • Recursion - Ship Battle

    - by rgorrosini
    I'm trying to write a little ship battle game in java. It is 100% academic, I made it to practice recursion, so... I want to use it instead of iteration, even if it's simpler and more efficient in most some cases. Let's get down to business. These are the rules: Ships are 1, 2 or 3 cells wide and are placed horizontally only. Water is represented with 0, non-hit ship cells are 1, hit ship cells are 2 and sunken ships have all it's cells in 3. With those rules set, I'm using the following array for testing: int[][] board = new int[][] { {0, 1, 2, 0, 1, 0}, {0, 0, 1, 1, 1, 0}, {0, 3, 0, 0, 0, 0}, {0, 0, 2, 1, 2, 0}, {0, 0, 0, 1, 1, 1}, }; It works pretty good so far, and to make it more user-friendly I would like to add a couple of reports. these are the methods I need for them: Given the matrix, return the amount of ships in it. Same as a), but separating them by state (amount of non-hit ships, hit and sunken ones). I will need a hand with those reports, and I would like to get some ideas. Remember it must be done using recursion, I want to understand this, and the only way to go is practice! Thanks a lot for your time and patience :).

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  • Fast inter-process (inter-threaded) communications IPC on large multi-cpu system.

    - by IPC
    What would be the fastest portable bi-directional communication mechanism for inter-process communication where threads from one application need to communicate to multiple threads in another application on the same computer, and the communicating threads can be on different physical CPUs). I assume that it would involve a shared memory and a circular buffer and shared synchronization mechanisms. But shared mutexes are very expensive (and there are limited number of them too) to synchronize when threads are running on different physical CPUs.

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  • What are some practical uses of generating all permutations of a list, such as ['a', 'b', 'c'] ?

    - by Jian Lin
    I was asked by somebody in an interview for web front end job, to write a function that generates all permutation of a string, such as "abc" (or consider it ['a', 'b', 'c']). so the expected result from the function, when given ['a', 'b', 'c'], is abc acb bac bca cab cba Actually in my past 20 years of career, I have never needed to do something like that, especially when doing front end work for web programming. What are some practical use of this problem nowadays, in web programming, front end or back end, I wonder? As a side note, I kind of feel that expecting a result in 3 minutes might be "either he gets it or he doesn't", especially I was thinking of doing it by a procedural, non-recursive way at first. After the interview, I spent another 10 minutes and thought of how to do it using recursion, but expecting it to be solved within 3 minutes... may not be a good test of how qualified he is, especially for front end work.

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  • A data structure based on the R-Tree: creating new child nodes when a node is full, but what if I ha

    - by Tom
    I realize my title is not very clear, but I am having trouble thinking of a better one. If anyone wants to correct it, please do. I'm developing a data structure for my 2 dimensional game with an infinite universe. The data structure is based on a simple (!) node/leaf system, like the R-Tree. This is the basic concept: you set howmany childs you want a node (a container) to have maximum. If you want to add a leaf, but the node the leaf should be in is full, then it will create a new set of nodes within this node and move all current leafs to their new (more exact) node. This way, very populated areas will have a lot more subdivisions than a very big but rarely visited area. This works for normal objects. The only problem arises when I have more than maxChildsPerNode objects with the exact same X,Y location: because the node is full, it will create more exact subnodes, but the old leafs will all be put in the exact same node again because they have the exact same position -- resulting in an infinite loop of creating more nodes and more nodes. So, what should I do when I want to add more leafs than maxChildsPerNode with the exact same position to my tree? PS. if I failed to explain my problem, please tell me, so I can try to improve the explanation.

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  • Are fragments of hashes collision-resistent?

    - by Mark
    Let me see if someone would mind clearing up this elementary point about md5 and hashing. If you only use the first 4 bytes of an md5 hash, would that mean theoretically only 1 in 255^4 chance of collision. iow is that the intention with it (and other hash algorithms) - that you only have to use a small portion of the returned hash (say the hash is of a file of some size).

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  • sloving Algorithm notation

    - by neednewname
    Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

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  • Logic to mirror byte value around 128

    - by Kazar
    Hey, I have a need to mirror a byte's value around the centre of 128. So, example outputs of this function include: In 0 Out 255 In 255 Out 0 In 128 Out 128 In 127 Out 1 In 30 Out 225 In 225 Out 30 I'm driving myself nuts with this, I'm sure I'll kick myself when I read the answers. Cheers

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  • Porting Python algorithm to C++ - different solution

    - by cb0
    Hello, I have written a little brute string generation script in python to generate all possible combinations of an alphabet within a given length. It works quite nice, but for the reason I wan't it to be faster I try to port it to C++. The problem is that my C++ Code is creating far too much combination for one word. Heres my example in python: ./test.py gives me aaa aab aac aad aa aba .... while ./test (the c++ programm gives me) aaa aaa aaa aaa aa Here I also get all possible combinations, but I get them twice ore more often. Here is the Code for both programms: #!/usr/bin/env python import sys #Brute String Generator #Start it with ./brutestringer.py 4 6 "abcdefghijklmnopqrstuvwxyz1234567890" "" #will produce all strings with length 4 to 6 and chars from a to z and numbers 0 to 9 def rec(w, p, baseString): for c in "abcd": if (p<w - 1): rec(w, p + 1, baseString + "%c" % c) print baseString for b in range(3,4): rec(b, 0, "") And here the C++ Code #include <iostream> using namespace std; string chars="abcd"; void rec(int w,int b,string p){ unsigned int i; for(i=0;i<chars.size();i++){ if(b < (w-1)){ rec(w, (b+1), p+chars[i]); } cout << p << "\n"; } } int main () { int a=3, b=0; rec (a+1,b, ""); return 0; } Does anybody see my fault ? I don't have much experience with C++. Thanks indeed

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  • Constructing a tree using Python

    - by stealthspy
    I am trying to implement a unranked boolean retrieval. For this, I need to construct a tree and perform a DFS to retrieve documents. I have the leaf nodes but I am having difficulty to construct the tree. Eg: query = OR ( AND (maria sharapova) tennis) Result: OR | | AND tennis | | maria sharapova I traverse the tree using DFS and calculate the boolean equivalent of certain document ids to identify the required document from the corpus. Can someone help me with the design of this using python? I have parsed the query and retrieved the leaf nodes for now.

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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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  • 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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  • Did I implement this correctly?

    - by user146780
    I'm trying to implement line thickness as denoted here: start = line start = vector(x1, y1) end = line end = vector(x2, y2) dir = line direction = end - start = vector(x2-x1, y2-y1) ndir = normalized direction = dir*1.0/length(dir) perp = perpendicular to direction = vector(dir.x, -dir.y) nperp = normalized perpendicular = perp*1.0/length(perp) perpoffset = nperp*w*0.5 diroffset = ndir*w*0.5 p0, p1, p2, p3 = polygon points: p0 = start + perpoffset - diroffset p1 = start - perpoffset - diroffset p2 = end + perpoffset + diroffset p3 = end - perpoffset + diroffset I'v implemented this like so: void OGLENGINEFUNCTIONS::GenerateLinePoly(const std::vector<std::vector<GLdouble>> &input, std::vector<GLfloat> &output, int width) { output.clear(); float temp; float dirlen; float perplen; POINTFLOAT start; POINTFLOAT end; POINTFLOAT dir; POINTFLOAT ndir; POINTFLOAT perp; POINTFLOAT nperp; POINTFLOAT perpoffset; POINTFLOAT diroffset; POINTFLOAT p0, p1, p2, p3; for(int i = 0; i < input.size() - 1; ++i) { start.x = input[i][0]; start.y = input[i][1]; end.x = input[i + 1][0]; end.y = input[i + 1][1]; dir.x = end.x - start.x; dir.y = end.y - start.y; dirlen = sqrt((dir.x * dir.x) + (dir.y * dir.y)); ndir.x = dir.x * (1.0 / dirlen); ndir.y = dir.y * (1.0 / dirlen); perp.x = dir.x; perp.y = -dir.y; perplen = sqrt((perp.x * perp.x) + (perp.y * perp.y)); nperp.x = perp.x * (1.0 / perplen); nperp.y = perp.y * (1.0 / perplen); perpoffset.x = nperp.x * width * 0.5; perpoffset.y = nperp.y * width * 0.5; diroffset.x = ndir.x * width * 0.5; diroffset.y = ndir.x * width * 0.5; // p0 = start + perpoffset - diroffset //p1 = start - perpoffset - diroffset //p2 = end + perpoffset + diroffset // p3 = end - perpoffset + diroffset p0.x = start.x + perpoffset.x - diroffset.x; p0.y = start.y + perpoffset.y - diroffset.y; p1.x = start.x - perpoffset.x - diroffset.x; p1.y = start.y - perpoffset.y - diroffset.y; p2.x = end.x + perpoffset.x + diroffset.x; p2.y = end.y + perpoffset.y + diroffset.y; p3.x = end.x - perpoffset.x + diroffset.x; p3.y = end.y - perpoffset.y + diroffset.y; output.push_back(p0.x); output.push_back(p0.y); output.push_back(p1.x); output.push_back(p1.y); output.push_back(p2.x); output.push_back(p2.y); output.push_back(p3.x); output.push_back(p3.y); } } But right now the lines look perpendicular and wrong, it should be giving me quads to render which is what i'm rendering, but the points it is outputing are strange. Have I done this wrong? Thanks

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