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  • Merging k sorted linked lists - analysis

    - by Kotti
    Hi! I am thinking about different solutions for one problem. Assume we have K sorted linked lists and we are merging them into one. All these lists together have N elements. The well known solution is to use priority queue and pop / push first elements from every lists and I can understand why it takes O(N log K) time. But let's take a look at another approach. Suppose we have some MERGE_LISTS(LIST1, LIST2) procedure, that merges two sorted lists and it would take O(T1 + T2) time, where T1 and T2 stand for LIST1 and LIST2 sizes. What we do now generally means pairing these lists and merging them pair-by-pair (if the number is odd, last list, for example, could be ignored at first steps). This generally means we have to make the following "tree" of merge operations: N1, N2, N3... stand for LIST1, LIST2, LIST3 sizes O(N1 + N2) + O(N3 + N4) + O(N5 + N6) + ... O(N1 + N2 + N3 + N4) + O(N5 + N6 + N7 + N8) + ... O(N1 + N2 + N3 + N4 + .... + NK) It looks obvious that there will be log(K) of these rows, each of them implementing O(N) operations, so time for MERGE(LIST1, LIST2, ... , LISTK) operation would actually equal O(N log K). My friend told me (two days ago) it would take O(K N) time. So, the question is - did I f%ck up somewhere or is he actually wrong about this? And if I am right, why doesn't this 'divide&conquer' approach can't be used instead of priority queue approach?

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  • submatrix from a matrix

    - by Grv
    A matrix is of size n*n and it consists only 0 and 1 find the largest submatrix that consists of 1's only eg 10010 11100 11001 11110 largest sub matrix will be of 3*2 from row 2 to row 4 please answer with best space and time complexity

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  • Permutations of Varying Size

    - by waiwai933
    I'm trying to write a function in PHP that gets all permutations of all possible sizes. I think an example would be the best way to start off: $my_array = array(1,1,2,3); Possible permutations of varying size: 1 1 // * See Note 2 3 1,1 1,2 1,3 // And so forth, for all the sets of size 2 1,1,2 1,1,3 1,2,1 // And so forth, for all the sets of size 3 1,1,2,3 1,1,3,2 // And so forth, for all the sets of size 4 Note: I don't care if there's a duplicate or not. For the purposes of this example, all future duplicates have been omitted. What I have so far in PHP: function getPermutations($my_array){ $permutation_length = 1; $keep_going = true; while($keep_going){ while($there_are_still_permutations_with_this_length){ // Generate the next permutation and return it into an array // Of course, the actual important part of the code is what I'm having trouble with. } $permutation_length++; if($permutation_length>count($my_array)){ $keep_going = false; } else{ $keep_going = true; } } return $return_array; } The closest thing I can think of is shuffling the array, picking the first n elements, seeing if it's already in the results array, and if it's not, add it in, and then stop when there are mathematically no more possible permutations for that length. But it's ugly and resource-inefficient. Any pseudocode algorithms would be greatly appreciated. Also, for super-duper (worthless) bonus points, is there a way to get just 1 permutation with the function but make it so that it doesn't have to recalculate all previous permutations to get the next? For example, I pass it a parameter 3, which means it's already done 3 permutations, and it just generates number 4 without redoing the previous 3? (Passing it the parameter is not necessary, it could keep track in a global or static). The reason I ask this is because as the array grows, so does the number of possible combinations. Suffice it to say that one small data set with only a dozen elements grows quickly into the trillions of possible combinations and I don't want to task PHP with holding trillions of permutations in its memory at once.

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  • Is there any simple way to test two PNGs for equality?

    - by Mason Wheeler
    I've got a bunch of PNG images, and I'm looking for a way to identify duplicates. By duplicates I mean, specifically, two PNG files whose uncompressed image data are identical, not necessarily whose files are identical. This means I can't do something simple like compare CRC hash values. I figure this can actually be done reliably since PNGs use lossless compression, but I'm worried about speed. I know I can winnow things down a little by testing for equal dimensions first, but when it comes time to actually compare the images against each other, is there any way to do it reasonably efficiently? (ie. faster than the "double-for-loop checking pixel values against each other" brute-force method?)

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  • what is order notation f(n)=O(g(n))?

    - by Lopa
    2 questions: question 1: under what circumstances would this[O(f(n))=O(k.f(n))] be the most appropriate form of time-complexity analysis? question 2: working from mathematical definition of O notation, show that O(f(n))=O(k.f(n)), for positive constant k.? My view: For the first one I think it is average case and worst case form of time-complexity. am i right? and what else do i write in that? for the second one I think we need to define the function mathematically, so is the answer something like because the multiplication by a constant just corresponds to a readjustment of value of the arbitrary constant 'k' in definition of O.

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  • Time complexity of a certain program

    - by HokageSama
    In a discussion with my friend i am not able to predict correct and tight time complexity of a program. Program is as below. /* This Function reads input array "input" and update array "output" in such a way that B[i] = index value of nearest greater value from A[i], A[i+1] ... A[n], for all i belongs to [1, n] Time Complexity: ?? **/ void createNearestRightSidedLargestArr(int* input, int size, int* output){ if(!input || size < 1) return; //last element of output will always be zero, since no element is present on its right. output[size-1] = -1; int curr = size - 2; int trav; while(curr >= 0){ if(input[curr] < input[curr + 1]){ output[curr] = curr + 1; curr--; continue; } trav = curr + 1; while( input[ output [trav] ] < input[curr] && output [trav] != -1) trav = output[trav]; output[curr--] = output[trav]; } } I said time complexity is O(n^2) but my friend insists that this is not correct. What is the actual time complexity?

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  • big O notation algorithm

    - by niggersak
    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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  • recursively implementing 'minimum number of coins' in python

    - by user5198
    This problem is same as asked in here. Given a list of coins, their values (c1, c2, c3, ... cj, ...), and the total sum i. Find the minimum number of coins the sum of which is i (we can use as many coins of one type as we want), or report that it's not possible to select coins in such a way that they sum up to S. I"m just introduced to dynamic programming yesterday and I tried to make a code for it. # Optimal substructure: C[i] = 1 + min_j(C[i-cj]) cdict = {} def C(i, coins): if i <= 0: return 0 if i in cdict: return cdict[i] else: answer = 1 + min([C(i - cj, coins) for cj in coins]) cdict[i] = answer return answer Here, C[i] is the optimal solution for amount of money 'i'. And available coins are {c1, c2, ... , cj, ...} for the program, I've increased the recursion limit to avoid maximum recursion depth exceeded error. But, this program gives the right answer only someones and when a solution is not possible, it doesn't indicate that. What is wrong with my code and how to correct it?

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  • Mathematical attack on the Digital Signature Algorithm

    - by drelihan
    Does anybody know the mathematics behind an attack on DSA where modulus p has p-1 made up of only small factors. In reality, this would not happen as the key generator would guarantee that this is not so. There is much information on the web on generating good input paramters for DSA so that it is hard to crack but no information on how you find X if modulus p has p-1 made up of only small factors.

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  • Suggestions for duplicate file finder algorithm (using C)

    - by Andrei Ciobanu
    Hello, I wanted to write a program that test if two files are duplicates (have exactly the same content). First I test if the files have the same sizes, and if they have i start to compare their contents. My first idea, was to "split" the files into fixed size blocks, then start a thread for every block, fseek to startup character of every block and continue the comparisons in parallel. When a comparison from a thread fails, the other working threads are canceled, and the program exits out of the thread spawning loop. The code looks like this: dupf.h #ifndef __NM__DUPF__H__ #define __NM__DUPF__H__ #define NUM_THREADS 15 #define BLOCK_SIZE 8192 /* Thread argument structure */ struct thread_arg_s { const char *name_f1; /* First file name */ const char *name_f2; /* Second file name */ int cursor; /* Where to seek in the file */ }; typedef struct thread_arg_s thread_arg; /** * 'arg' is of type thread_arg. * Checks if the specified file blocks are * duplicates. */ void *check_block_dup(void *arg); /** * Checks if two files are duplicates */ int check_dup(const char *name_f1, const char *name_f2); /** * Returns a valid pointer to a file. * If the file (given by the path/name 'fname') cannot be opened * in 'mode', the program is interrupted an error message is shown. **/ FILE *safe_fopen(const char *name, const char *mode); #endif dupf.c #include <errno.h> #include <pthread.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <sys/types.h> #include <sys/stat.h> #include <unistd.h> #include "dupf.h" FILE *safe_fopen(const char *fname, const char *mode) { FILE *f = NULL; f = fopen(fname, mode); if (f == NULL) { char emsg[255]; sprintf(emsg, "FOPEN() %s\t", fname); perror(emsg); exit(-1); } return (f); } void *check_block_dup(void *arg) { const char *name_f1 = NULL, *name_f2 = NULL; /* File names */ FILE *f1 = NULL, *f2 = NULL; /* Streams */ int cursor = 0; /* Reading cursor */ char buff_f1[BLOCK_SIZE], buff_f2[BLOCK_SIZE]; /* Character buffers */ int rchars_1, rchars_2; /* Readed characters */ /* Initializing variables from 'arg' */ name_f1 = ((thread_arg*)arg)->name_f1; name_f2 = ((thread_arg*)arg)->name_f2; cursor = ((thread_arg*)arg)->cursor; /* Opening files */ f1 = safe_fopen(name_f1, "r"); f2 = safe_fopen(name_f2, "r"); /* Setup cursor in files */ fseek(f1, cursor, SEEK_SET); fseek(f2, cursor, SEEK_SET); /* Initialize buffers */ rchars_1 = fread(buff_f1, 1, BLOCK_SIZE, f1); rchars_2 = fread(buff_f2, 1, BLOCK_SIZE, f2); if (rchars_1 != rchars_2) { /* fread failed to read the same portion. * program cannot continue */ perror("ERROR WHEN READING BLOCK"); exit(-1); } while (rchars_1-->0) { if (buff_f1[rchars_1] != buff_f2[rchars_1]) { /* Different characters */ fclose(f1); fclose(f2); pthread_exit("notdup"); } } /* Close streams */ fclose(f1); fclose(f2); pthread_exit("dup"); } int check_dup(const char *name_f1, const char *name_f2) { int num_blocks = 0; /* Number of 'blocks' to check */ int num_tsp = 0; /* Number of threads spawns */ int tsp_iter = 0; /* Iterator for threads spawns */ pthread_t *tsp_threads = NULL; thread_arg *tsp_threads_args = NULL; int tsp_threads_iter = 0; int thread_c_res = 0; /* Thread creation result */ int thread_j_res = 0; /* Thread join res */ int loop_res = 0; /* Function result */ int cursor; struct stat buf_f1; struct stat buf_f2; if (name_f1 == NULL || name_f2 == NULL) { /* Invalid input parameters */ perror("INVALID FNAMES\t"); return (-1); } if (stat(name_f1, &buf_f1) != 0 || stat(name_f2, &buf_f2) != 0) { /* Stat fails */ char emsg[255]; sprintf(emsg, "STAT() ERROR: %s %s\t", name_f1, name_f2); perror(emsg); return (-1); } if (buf_f1.st_size != buf_f2.st_size) { /* File have different sizes */ return (1); } /* Files have the same size, function exec. is continued */ num_blocks = (buf_f1.st_size / BLOCK_SIZE) + 1; num_tsp = (num_blocks / NUM_THREADS) + 1; cursor = 0; for (tsp_iter = 0; tsp_iter < num_tsp; tsp_iter++) { loop_res = 0; /* Create threads array for this spawn */ tsp_threads = malloc(NUM_THREADS * sizeof(*tsp_threads)); if (tsp_threads == NULL) { perror("TSP_THREADS ALLOC FAILURE\t"); return (-1); } /* Create arguments for every thread in the current spawn */ tsp_threads_args = malloc(NUM_THREADS * sizeof(*tsp_threads_args)); if (tsp_threads_args == NULL) { perror("TSP THREADS ARGS ALLOCA FAILURE\t"); return (-1); } /* Initialize arguments and create threads */ for (tsp_threads_iter = 0; tsp_threads_iter < NUM_THREADS; tsp_threads_iter++) { if (cursor >= buf_f1.st_size) { break; } tsp_threads_args[tsp_threads_iter].name_f1 = name_f1; tsp_threads_args[tsp_threads_iter].name_f2 = name_f2; tsp_threads_args[tsp_threads_iter].cursor = cursor; thread_c_res = pthread_create( &tsp_threads[tsp_threads_iter], NULL, check_block_dup, (void*)&tsp_threads_args[tsp_threads_iter]); if (thread_c_res != 0) { perror("THREAD CREATION FAILURE"); return (-1); } cursor+=BLOCK_SIZE; } /* Join last threads and get their status */ while (tsp_threads_iter-->0) { void *thread_res = NULL; thread_j_res = pthread_join(tsp_threads[tsp_threads_iter], &thread_res); if (thread_j_res != 0) { perror("THREAD JOIN FAILURE"); return (-1); } if (strcmp((char*)thread_res, "notdup")==0) { loop_res++; /* Closing other threads and exiting by condition * from loop. */ while (tsp_threads_iter-->0) { pthread_cancel(tsp_threads[tsp_threads_iter]); } } } free(tsp_threads); free(tsp_threads_args); if (loop_res > 0) { break; } } return (loop_res > 0) ? 1 : 0; } The function works fine (at least for what I've tested). Still, some guys from #C (freenode) suggested that the solution is overly complicated, and it may perform poorly because of parallel reading on hddisk. What I want to know: Is the threaded approach flawed by default ? Is fseek() so slow ? Is there a way to somehow map the files to memory and then compare them ?

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  • Algorithm for Negating Sentences

    - by Kevin Dolan
    I was wondering if anyone was familiar with any attempts at algorithmic sentence negation. For example, given a sentence like "This book is good" provide any number of alternative sentences meaning the opposite like "This book is not good" or even "This book is bad". Obviously, accomplishing this with a high degree of accuracy would probably be beyond the scope of current NLP, but I'm sure there has been some work on the subject. If anybody knows of any work, care to point me to some papers?

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  • about Master theorem

    - by matin1234
    Hi this is the link http://www.cs.mcgill.ca/~cs251/OldCourses/1997/topic5/ is written that for T(n)<=2n+T(n/3)+T(n/3) the T(n) is not O(n) but with master theorem we can use case 3 and we can say that its T(n) is theta(n) please help me! thanks how can we prove that T(n) is not O(n)

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  • How to use Haar wavelet to detect LINES on an image?

    - by Ole Jak
    So I have Image like this I want to get something like this (I hevent drawn all lines I want but I hope you can get my idea) I want to use SURF ( (Speeded Up Robust Features) is a robust image descriptor, first presented by Herbert Bay et al. in 2006 ) or something that is based on sums of 2D Haar wavelet responses and makes an efficient use of integral images for finding all straight lines on image. I want to get relative to picture pixel coords start and end points of lines. So on this picture to find all lines between tiles and thouse 2 black lines on top. Is there any such Code Example (with lines search capability) to start from? I love C and C++ but any other readable code will probably work for me=)

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  • Stack and queue operations on the same array.

    - by Passonate Learner
    Hi. I've been thinking about a program logic, but I cannot draw a conclusion to my problem. Here, I've implemented stack and queue operations to a fixed array. int A[1000]; int size=1000; int top; int front; int rear; bool StackIsEmpty() { return (top==0); } bool StackPush( int x ) { if ( top >= size ) return false; A[top++] = x; return true; } int StackTop( ) { return A[top-1]; } bool StackPop() { if ( top <= 0 ) return false; A[--top] = 0; return true; } bool QueueIsEmpty() { return (front==rear); } bool QueuePush( int x ) { if ( rear >= size ) return false; A[rear++] = x; return true; } int QueueFront( ) { return A[front]; } bool QueuePop() { if ( front >= rear ) return false; A[front++] = 0; return true; } It is presumed(or obvious) that the bottom of the stack and the front of the queue is pointing at the same location, and vice versa(top of the stack points the same location as rear of the queue). For example, integer 1 and 2 is inside an array in order of writing. And if I call StackPop(), the integer 2 will be popped out, and if I call QueuePop(), the integer 1 will be popped out. My problem is that I don't know what happens if I do both stack and queue operations on the same array. The example above is easy to work out, because there are only two values involved. But what if there are more than 2 values involved? For example, if I call StackPush(1); QueuePush(2); QueuePush(4); StackPop(); StackPush(5); QueuePop(); what values will be returned in the order of bottom(front) from the final array? I know that if I code a program, I would receive a quick answer. But the reason I'm asking this is because I want to hear a logical explanations from a human being, not a computer.

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  • B-Tree Revision

    - by stan
    Hi, If we are looking for line intersections (horizontal and vertical lines only) and we have n lines with half of them vertical and no intersections then Sorting the list of line end points on y value will take N log N using mergesort Each insert delete and search of our data structue (assuming its a b-tree) will be < log n so the total search time will be N log N What am i missing here, if the time to sort using mergesort takes a time of N log N and insert and delete takes a time of < log n are we dropping the constant factor to give an overal time of N log N. If not then how comes < log n goes missing in total ONotation run time? Thanks

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  • How to detect a Triangle gesture with kinect?

    - by Akhilesh Mishra
    I am trying to implement a gesture recognition system which interprets the geometric gestures user makes and draws it on screen, I have some idea of how circle can be recognized, however i have no clue how to get started with triangle recognition. the data I have is X and Y coordinates of all points the gesture passed through. I get this data by tracking right hand. I found something online called Hough Transform , which is used for detecting lines but i am not sure whether it will work for discrete collection of points, Any ideas folks?

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  • Pong: How does the paddle know where the ball will hit?

    - by Roflcoptr
    After implementing Pacman and Snake I'm implementing the next very very classic game: Pong. The implementation is really simple, but I just have one little problem remaining. When one of the paddle (I'm not sure if it is called paddle) is controlled by the computer, I have trouble to position it at the correct position. The ball has a current position, a speed (which for now is constant) and a direction angle. So I could calculate the position where it will hit the side of the computer controlled paddle. And so Icould position the paddle right there. But however in the real game, there is a probability that the computer's paddle will miss the ball. How can I implement this probability? If I only use a probability of lets say 0.5 that the computer's paddle will hit the ball, the problem is solved, but I think it isn't that simple. From the original game I think the probability depends on the distance between the current paddle position and the position the ball will hit the border. Does anybody have any hints how exactly this is calculated?

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  • Efficiently storing a list of prime numbers

    - by eSKay
    This article says: Every prime number can be expressed as 30k±1, 30k±7, 30k±11, or 30k±13 for some k. That means we can use eight bits per thirty numbers to store all the primes; a million primes can be compressed to 33,334 bytes "That means we can use eight bits per thirty numbers to store all the primes" This "eight bits per thirty numbers" would be for k, correct? But each k value will not necessarily take-up just one bit. Shouldn't it be eight k values instead? "a million primes can be compressed to 33,334 bytes" I am not sure how this is true. We need to indicate two things: VALUE of k (can be arbitrarily large) STATE from one of the eight states (-13,-11,-7,-1,1,7,11,13) I am not following how 33,334 bytes was arrived at, but I can say one thing: as the prime numbers become larger and larger in value, we will need more space to store the value of k. How, then can we fix it at 33,334 bytes?

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  • Return a number between 0 and 4

    - by munchine
    How do I return a number between 0 and 4, depending the input number? For example if I pass it number 23 it will return 3. The number set should look like 0 5 10 15 20 .. 1 6 11 16 21 .. 2 7 12 17 22 .. 3 8 13 18 23 .. 4 9 14 19 24 What's the math for this?

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  • sorting using recursion

    - by user310587
    I have the following function to sort an array with even numbers in the front and odd numbers in the back. Is there a way to get it done without using any loops? //front is 0, back =array.length-1; arrangeArray (front, back); public static void arrangeArray (int front, int back) { if (front != back || front<back) { while (numbers [front]%2 == 0) front++; while (numbers[back]%2!=0) back--; if (front < back) { int oddnum = numbers [front]; numbers[front]= numbers[back]; numbers[back]=oddnum; arrangeArray (front+1, back-1); } } }

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