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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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  • 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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  • 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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  • 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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  • 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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  • Using MySQL as a job queue

    - by user237815
    I'd like to use MySQL as a job queue. Multiple machines will be producing and consuming jobs. Jobs need to be scheduled; some may run every hour, some every day, etc. It seems fairly straightforward: for each job, have a "nextFireTime" column, and have worker machines search for the job with the nextFireTime, change the status of the record to "inProcess", and then update the nextFireTime when the job ends. The problem comes in when a worker dies silently. It won't be able to update the nextFireTime or set the status back to "idle". Unfortunately, jobs can be long-running, so a reaper thread that looks for jobs that have been inProcess too long isn't an option. There's no timeout value that would work. Can anyone suggest a design pattern that would properly handle unreliable worker machines?

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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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  • Purpose of IF, ELSE, FOR macros ?

    - by psihodelia
    I have a source code of a library which has a lot of strange IF, ELSE, FOR, etc. macros for all common C-keywords instead of using just usual if,else,for,while keywords. These macros are defined like this: #define IF( a) if( increment_if(), a) where increment_if() function is defined so: static __inline void increment_if( void) { // If the "IF" operator comes just after an "ELSE", its counter // must not be incremented. ... //implementation } I don't really understand, what is the purpose of such macros? This library is for a real-time application and I suppose that using such macros must slow-down an application.

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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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  • 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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  • 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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  • Recursive breadth-first travel function in Java or C++?

    - by joejax
    Here is a java code for breadth-first travel: void breadthFirstNonRecursive(){ Queue<Node> queue = new java.util.LinkedList<Node>(); queue.offer(root); while(!queue.isEmpty()){ Node node = queue.poll(); visit(node); if (node.left != null) queue.offer(node.left); if (node.right != null) queue.offer(node.right); } } Is it possible to write a recursive function to do the same? At first, I thought this would be easy, so I came out with this: void breadthFirstRecursive(){ Queue<Node> q = new LinkedList<Node>(); breadthFirst(root, q); } void breadthFirst(Node node, Queue<Node> q){ if (node == null) return; q.offer(node); Node n = q.poll(); visit(n); if (n.left != null) breadthFirst(n.left, q); if (n.right != null) breadthFirst(n.right, q); } Then I found it doesn't work. It is actually does the same thing as this: void preOrder(Node node) { if (node == null) return; visit(node); preOrder(node.left); preOrder(node.right); } Has any one thought about this before?

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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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  • Calculate coordinates between startpoint and endpoint

    - by MikeB123
    Given.. 1 - The start-point GPS coordinates, 2 - The end-point GPS coordinates, 3 - The speed at which the object is travelling, 4 - Knowing that the trip trajectory will be a straight line... How can I calculate what my GPS coordinated will be in n minutes' time? That is to say, how can I calculate my position at a given time after the trip has started before the trip has ended?

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