Search Results

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

Page 96/300 | < Previous Page | 92 93 94 95 96 97 98 99 100 101 102 103  | Next Page >

  • Express any number as the sum of 4 prime numbers [Doubts]

    - by WarDoGG
    I was give a problem to express any number as sum of 4 prime numbers. Conditions: Not allowed to use any kind of database. Maximum execution time : 3 seconds Numbers only till 100,000 If the splitting is NOT possible, then return -1 What i did : using the sieve of eratosthenes, i calculated all prime numbers till the specified number. looked up a concept called goldbach conjecture which expresses an even number as the summation of 2 primes. However, i am stuck beyond that. Can anyone help me on this one as to what approach u might take ? The sieve of eratosthenes is taking 2 seconds to count primes till 100,000 :(((

    Read the article

  • Algorithm - combine multiple lists, resulting in unique list and retaining order

    - by hitch
    I want to combine multiple lists of items into a single list, retaining the overall order requirements. i.e.: 1: A C E 2: D E 3: B A D result: B A C D E above, starting with list 1, we have ACE, we then know that D must come before E, and from list 3, we know that B must come before A, and D must come after B and A. If there are conflicting orderings, the first ordering should be used. i.e. 1: A C E 2: B D E 3: F D result: A C B D E F 3 conflicts with 2, therefore requirements for 2 will be used. If ordering requirements mean an item must come before or after another, it doesn't matter if it comes immediately before or after, or at the start or end of the list, as long as overall ordering is maintained. This is being developed using VB.Net, so a LINQy solution (or any .Net solution) would be nice - otherwise pointers for an approach would be good.

    Read the article

  • Take the intersection of an arbitrary number of lists in python

    - by thepandaatemyface
    Suppose I have a list of lists of elements which are all the same (i'll use ints in this example) [range(100)[::4], range(100)[::3], range(100)[::2], range(100)[::1]] What would be a nice and/or efficient way to take the intersection of these lists (so you would get every element that is in each of the lists)? For the example that would be: [0, 12, 24, 36, 48, 60, 72, 84, 96]

    Read the article

  • Storing a bucket of numbers in an efficient data structure

    - by BlitzKrieg
    I have a buckets of numbers e.g. - 1 to 4, 5 to 15, 16 to 21, 22 to 34,.... I have roughly 600,000 such buckets. The range of numbers that fall in each of the bucket varies. I need to store these buckets in a suitable data structure so that the lookups for a number is as fast as possible. So my question is what is the suitable data structure and a sorting mechanism for this type of problem. Thanks in advance

    Read the article

  • Encoding arbitrary data into numbers?

    - by Pekka
    Is there a common method to encode and decode arbitrary data so the encoded end result consists of numbers only - like base64_encode but without the letters? Fictitious example: $encoded = numbers_encode("Mary had a little lamb"); echo $encoded; // outputs e.g. 122384337422394237423 (fictitious result) $decoded = numbers_decode("122384337422394237423"); echo $decoded; // outputs "Mary had a little lamb"

    Read the article

  • Find the element with the most "neighbors" in a sequence

    - by Bao An
    Assume that we have a sequence S with n elements <x1,x2,...,xn>. A pair of elements xi,xj are considered neighbors if |xi-xj|<d, with d a given distance between two neighbors. So how can find out the element that has most neighbors in the sequence? (A simply way is sorting the sequence and then calculating number of each element but it's time complexity is quite large): O(nlogn) May you please help me find a better way to reduce time complexity?

    Read the article

  • Why is my producer-consumer blocking?

    - by User007
    My code is here: http://pastebin.com/Fi3h0E0P Here is the output 0 Should we take order today (y or n): y Enter order number: 100 More customers (y or n): n Stop serving customers right now. Passing orders to cooker: There are total of 1 order(s) 1 Roger, waiter. I am processing order #100 The goal is waiter must take orders and then give them to the cook. The waiter has to wait cook finishes all pizza, deliver the pizza, and then take new orders. I asked how P-V work in my previous post here. I don't think it has anything to do with \n consuming? I tried all kinds of combination of wait(), but none work. Where did I make a mistake? The main part is here: //Producer process if(pid > 0) { while(1) { printf("0"); P(emptyShelf); // waiter as P finds no items on shelf; P(mutex); // has permission to use the shelf waiter_as_producer(); V(mutex); // cooker now can use the shelf V(orderOnShelf); // cooker now can pickup orders wait(); printf("2"); P(pizzaOnShelf); P(mutex); waiter_as_consumer(); V(mutex); V(emptyShelf); printf("3 "); } } if(pid == 0) { while(1) { printf("1"); P(orderOnShelf); // make sure there is an order on shelf P(mutex); //permission to work cooker_as_consumer(); // take order and put pizza on shelf printf("return from cooker"); V(mutex); //release permission printf("just released perm"); V(pizzaOnShelf); // pizza is now on shelf printf("after"); wait(); printf("4"); } } So I imagine this is the execution path: enter waiter_as_producer, then go to child process (cooker), then transfer the control back to parent, finish waiter_as_consumer, switch back to child. The two waits switch back to parent (like I said I tried all possible wait() combination...).

    Read the article

  • Reverse a singly linked list

    - by Madhan
    I would be wondered if there exists some logic to reverse the linked list using only two pointers. The following is used to reverse the single linked list using three pointers namely p, q, r: struct node { int data; struct node *link; }; void reverse() { struct node *p = first, *q = NULL, *r; while (p != NULL) { r = q; q = p; p = p->link; q->link = r; } q = first; } Is there any other alternate to reverse the linked list? what would be the best logic to reverse a singly linked list, in terms of time complexity?

    Read the article

  • Detecting one point's location compared to two other points.

    - by WizardOfOdds
    Hi all, you can check my profile, this is not homework. I've got an interesting little problem to solve in a very real software and I'm looking for an easy way to solve it. I've got two fixed points on screen (they're fixed, but I don't know beforehand their position) that are not at the same location. These two fixed points form an imaginary line. Now I've got a third point that is "on one side" of that line (it cannot be on the line). The user can grab the point (the user actually grab an object, whose I track by its center, which is the point I'm interested in) and drag it. But it cannot "cross" the imaginary line. What is the easiest way to detect if the user is crossing the imaginary line? Example: a c / / (c cannot be dragged here) / b Or: c b -------------- a (c cannot be dragged here) So what is an easy to detect if c is staying on the correct "side" of the line (I draw segments here, but it really can be thought of as a line). One way to detect this is to take the destination point d and see if segment (c,d) intersects with line (a,b), but isn't there an easier way? Can't I just do some 2D dot-product magic here and have basically a one or two liner solving my issue?

    Read the article

  • Android Compass orientation on unreliable (Low pass filter)

    - by madsleejensen
    Hi all Im creating an application where i need to position a ImageView depending on the Orientation of the device. I use the values from a MagneticField and Accelerometer Sensors to calculate the device orientation with SensorManager.getRotationMatrix(rotationMatrix, null, accelerometerValues, magneticFieldValues) SensorManager.getOrientation(rotationMatrix, values); double degrees = Math.toDegrees(values[0]); My problem is that the positioning of the ImageView is very sensitive to changes in the orientation. Making the imageview constantly jumping around the screen. (because the degrees change) I read that this can be because my device is close to things that can affect the magneticfield readings. But this is not the only reason it seems. I tried downloading some applications and found that the "3D compass" application remains extremely steady in its readings, i would like the same behavior in my application. I read that i can tweak the "noise" of my readings by adding a "Low pass filter", but i have no idea how to implement this (because of my lack of Math). Im hoping someone can help me creating a more steady reading on my device, Where a little movement to the device wont affect the current orientation. Right now i do a small if (Math.abs(lastReadingDegrees - newReadingDegrees) > 1) { updatePosition() } To filter abit of the noise. But its not working very well :)

    Read the article

  • question about polynomial multiplication

    - by davit-datuashvili
    i know that horners method for polynomial pultiplication is faster but here i dont know what is happening here is code public class horner{ public static final int n=10; public static final int x=7; public static void main(String[] args){ //non fast version int a[]=new int[]{1,2,3,4,5,6,7,8,9,10}; int xi=1; int y=a[0]; for (int i=1;i<n;i++){ xi=x*xi; y=y+a[i]*xi; } System.out.println(y); //fast method int y1=a[n-1]; for (int i=n-2;i>=0;i--){ y1=x*y+a[i]; } System.out.println(y1); } } result of this two methods are not same result of first method is 462945547 and result of second method is -1054348465 please help

    Read the article

  • Orbital equations, and power required to run them

    - by Adam Davis
    Due to a discussion on the SO IRC today, I'm curious about orbital mechanics, and The equations needed to solve orbital problems The computing power required to solve complex problems The question in particular is calculating when the Earth will plow into the Sun (or vice versa, depending on the frame of reference). I suspect that all the gravitational pulls within our solar system may need to be calculated, which makes me wonder what type of computer cluster is required, or can this be done on a single box? I don't have the experience to do a back of the napkin test here, but perhaps you do? Also, much thx to Gortok for the original inspiration (see comments).

    Read the article

  • flood fill algorithm

    - by user335593
    i want to implement the flood fill algorthm...so that when i get the x and y co-od of a point...it should start flooding from that point and fill till it finds a boundary but it is not filling the entire region...say a pentagon this is the code i am using void setpixel(struct fill fillcolor,int x,int y) { glColor3f(fillcolor.r,fillcolor.g,fillcolor.b); glBegin(GL_POINTS); glVertex2i(x,y); glEnd(); glFlush(); } struct fill getpixcol(int x,int y) { struct fill gotpixel; glReadPixels(x,y,1,1,GL_RGB,GL_UNSIGNED_BYTE,pick_col); gotpixel.r =(float) pick_col[0]/255.0; gotpixel.g =(float) pick_col[1]/255.0; gotpixel.b =(float) pick_col[2]/255.0; return(gotpixel); } void floodFill(int x, int y,struct fill fillcolor,struct fill boundarycolor) { struct fill tmp; // if ((x < 0) || (x >= 500)) return; // if ((y < 0) || (y >= 500)) return; tmp=getpixcol(x,y); while (tmp.r!=boundarycolor.r && tmp.g!=boundarycolor.g && tmp.b!=boundarycolor.b) { setpixel(fillcolor,x,y); setpixel(fillcolor,x+1,y); setpixel(fillcolor,x,y+1); setpixel(fillcolor,x,y-1); setpixel(fillcolor,x-1,y); floodFill(x-1,y+1,fillcolor,boundarycolor); floodFill(x-1,y,fillcolor,boundarycolor); floodFill(x-1,y-1,fillcolor,boundarycolor); floodFill(x,y+1,fillcolor,boundarycolor); floodFill(x,y-1,fillcolor,boundarycolor); floodFill(x+1,y+1,fillcolor,boundarycolor); floodFill(x+1,y,fillcolor,boundarycolor); floodFill(x+1,y-1,fillcolor,boundarycolor); } }

    Read the article

  • Point data structure for a sketching application

    - by bebraw
    I am currently developing a little sketching application based on HTML5 Canvas element. There is one particular problem I haven't yet managed to find a proper solution for. The idea is that the user will be able to manipulate existing stroke data (points) quite freely. This includes pushing point data around (ie. magnet tool) and manipulating it at whim otherwise (ie. altering color). Note that the current brush engine is able to shade by taking existing stroke data in count. It's a quick and dirty solution as it just iterates the points in the current stroke and checks them against a distance rule. Now the problem is how to do this in a nice manner. It is extremely important to be able to perform efficient queries that return all points within given canvas coordinate and radius. Other features, such as space usage, should be secondary to this. I don't mind doing some extra processing between strokes while the user is not painting. Any pointers are welcome. :)

    Read the article

  • How to scale rotated objects properly in Actionscript 3?

    - by Tom
    This is unfortunately a quite complex issue to explain, so please don't get discouraged by the wall of text - it's there for a reason. ;) I'm working on a transformation manager for flash, written with Actionscript 3. Users can place objects on the screen, for example a rectangle. This rectangle can then be selected and transformed: move, scale or rotate. Because flash by default rotates around the top left point of the object, and I want it to rotate around the center, I created a wrapper setup for each display object (eg. a rectangle). This is how the wrappers are setup: //the position wrapper makes sure that we do get the top left position when we access x and y var positionWrapper:Sprite = new Sprite(); positionWrapper.x = renderObject.x; positionWrapper.y = renderObject.y; //set the render objects location to center at the rotation wrappers top left renderObject.x = 0 - renderObject.width / 2; renderObject.y = 0 - renderObject.height / 2; //now create a rotation wrapper, at the center of the display object var rotationWrapper:Sprite = new Sprite(); rotationWrapper.x = renderObject.width / 2; rotationWrapper.y = renderObject.height / 2; //put the rotation wrapper inside the position wrapper and the render object inside the rotation wrapper positionWrapper.addChild(rotationWrapper); rotationWrapper.addChild(renderObject); Now, the x and y of the object can be accessed and set directly: mainWrapper.x or mainWrapper.y. The rotation can be set and accessed from the child of this main wrapper: mainWrapper.getChildAt(0).rotation. Finally, the width and height of the display object can be retreived and set by getting the child of the rotation wrapper and accessing the display object directly. An example on how I access them: //get wrappers and render object var positionWrapper:Sprite = currentSelection["render"]; var rotationWrapper:Sprite = positionWrapper.getChildAt(0) as Sprite; var renderObject:DisplayObject = rotationWrapper.getChildAt(0); This works perfectly for all initial transformations: moving, scaling and rotating. However, the problem arises when you first rotate an object (eg. 45 degrees) and then scale it. The scaled object is getting out of shape and doesn't scale as it should. This for example happens when you scale to the left. Scaling left is basically adding n width to the object and then reduce the x coord of the position wrapper by n too: renderObject.width -= diffX; positionWrapper.x += diffX; This works when the object is not rotated. However, when it is, the position wrapper won't be rotated as it is a parent of the rotation wrapper. This will make the position wrapper move left horizontally while the width of the object is increased diagonally. I hope this makes any sense, if not, please tell me and I'll try to elaborate more. Now, to the question: should I use a different kind of setup, system or structure? Should I maybe use matrixes, if so, how would you keep a static width/height after rotation? Or how do I fix my current wrapper system for scaling after rotation? Any help is appreciated.

    Read the article

  • Graph diffing and versioning tool

    - by hashable
    I am working with a team that edits large DAGs represented as single files. Currently we are unable to work with multiple users concurrently modifying the DAG. Is there a tool (somewhat like the Eclipse SVN plugin) that can do do revision control on the file (manage timestamps/revision stamps) to identify incoming/outgoing/conflicting changes (Node/Link insertion/deletion/modification) and merge changes just like programmers do with source code files? The system should be able to do dependency management also. E.g. an incoming Link must not be accepted when one of the two Nodes is absent. That is, it should not "break" the existing DAG by allowing partial updates. If there is a framework to do this using generic "Node" and "Link" interfaces? Note: I am aware of Protege and its plugins. They currently do not satisfy my requirements.

    Read the article

  • Single dimension peak fitting

    - by bufferz
    I have a single dimensional array of floating point values (c# doubles FYI) and I need to find the "peak" of the values ... as if graphed. I can't just take the highest value, as the peak is actually a plateau that has small fluctuations. This plateau is in the middle of a bunch of noise. I'm looking find a solution that would give me the center of this plateau. An example array might look like this: 1,2,1,1,2,1,3,2,4,4,4,5,6,8,8,8,8,7,8,7,9,7,5,4,4,3,3,2,2,1,1,1,1,1,2,1,1,1,1 where the peak is somewhere in the bolded section. Any ideas?

    Read the article

  • Better way to summarize data about stop times?

    - by Vimvq1987
    This question is close to this: http://stackoverflow.com/questions/2947963/find-the-period-of-over-speed Here's my table: Longtitude Latitude Velocity Time 102 401 40 2010-06-01 10:22:34.000 103 403 50 2010-06-01 10:40:00.000 104 405 0 2010-06-01 11:00:03.000 104 405 0 2010-06-01 11:10:05.000 105 406 35 2010-06-01 11:15:30.000 106 403 60 2010-06-01 11:20:00.000 108 404 70 2010-06-01 11:30:05.000 109 405 0 2010-06-01 11:35:00.000 109 405 0 2010-06-01 11:40:00.000 105 407 40 2010-06-01 11:50:00.000 104 406 30 2010-06-01 12:00:00.000 101 409 50 2010-06-01 12:05:30.000 104 405 0 2010-06-01 11:05:30.000 I want to summarize times when vehicle had stopped (velocity = 0), include: it had stopped since "when" to "when" in how much minutes, how many times it stopped and how much time it stopped. I wrote this query to do it: select longtitude, latitude, MIN(time), MAX(time), DATEDIFF(minute, MIN(Time), MAX(time)) as Timespan from table_1 where velocity = 0 group by longtitude,latitude select DATEDIFF(minute, MIN(Time), MAX(time)) as minute into #temp3 from table_1 where velocity = 0 group by longtitude,latitude select COUNT(*) as [number]from #temp select SUM(minute) as [totaltime] from #temp3 drop table #temp This query return: longtitude latitude (No column name) (No column name) Timespan 104 405 2010-06-01 11:00:03.000 2010-06-01 11:10:05.000 10 109 405 2010-06-01 11:35:00.000 2010-06-01 11:40:00.000 5 number 2 totaltime 15 You can see, it works fine, but I really don't like the #temp table. Is there anyway to query this without use a temp table? Thank you.

    Read the article

  • Byte-Pairing for data compression

    - by user1669533
    Question about Byte-Pairing for data compression. If byte pairing converts two byte values to a single byte value, splitting the file in half, then taking a gig file and recusing it 16 times shrinks it to 62,500,000. My question is, is byte-pairing really efficient? Is the creation of a 5,000,000 iteration loop, to be conservative, efficient? I would like some feed back on and some incisive opinions please. Dave, what I read was: "The US patent office no longer grants patents on perpetual motion machines, but has recently granted at least two patents on a mathematically impossible process: compression of truly random data." I was not inferring the Patent Office was actually considering what I am inquiring about. I was merely commenting on the notion of a "mathematically impossible process." If someone has, in some way created a method of having a "single" data byte as a placeholder of 8 individual bytes of data, that would be a consideration for a patent. Now, about the mathematically impossibility of an 8 to 1 compression method, it is not so much a mathematically impossibility, but a series of rules and conditions that can be created. As long as there is the rule of 8 or 16 bit representation of storing data on a medium, there are ways to manipulate data that mirrors current methods, or creation by a new way of thinking.

    Read the article

  • sorting a doubly linked list with merge sort.

    - by user329820
    Hi I have found this code in the internet and it was for arrays ,I want to change it for doubly linked list(instead of index we should use pointer) would you please help me that how can i change merge method(I have changed sort method by myself) also this is not my home work ,I love working with linked list!! public class MergeSort { private DoublyLinkedList LocalDoublyLinkedList; public MergeSort(DoublyLinkedList list) { LocalDoublyLinkedList = list; } public void sort() { if (LocalDoublyLinkedList.size() <= 1) { return; } DoublyLinkedList listOne = new DoublyLinkedList(); DoublyLinkedList listTwo = new DoublyLinkedList(); for (int x = 0; x < (LocalDoublyLinkedList.size() / 2); x++) { listOne.add(x, LocalDoublyLinkedList.getValue(x)); } for (int x = (LocalDoublyLinkedList.size() / 2) + 1; x < LocalDoublyLinkedList.size`(); x++) {` listTwo.add(x, LocalDoublyLinkedList.getValue(x)); } //Split the DoublyLinkedList again MergeSort sort1 = new MergeSort(listOne); MergeSort sort2 = new MergeSort(listTwo); sort1.sort(); sort2.sort(); merge(listOne, listTwo); } private void merge(DoublyLinkedList a, DoublyLinkedList b) { int x = 0; int y = 0; int z = 0; while (x < first.length && y < second.length) { if (first[x] < second[y]) { a[z] = first[x]; x++; } else { a[z] = second[y]; y++; } z++; } //copy remaining elements to the tail of a[]; for (int i = x; i < first.length; i++) { a[z] = first[i]; z++; } for (int i = y; i < second.length; i++) { a[z] = second[i]; z++; } } }

    Read the article

  • Lexographical sorting problem

    - by Shawn Mclean
    I'm doing a problem that says concatenate the words to generate the lexicographically lowest possible string. from a competition. Take for example this string: jibw ji jp bw jibw The actual output turns out to be: bw jibw jibw ji jp When I do sorting on this, I get: bw ji jibw jibw jp. Does this mean that this is not sorting? If it is sorting, does lexicographic sorting take into consideration pushing the shorter strings to the back or something? I've been doing some reading on lexigographical order and I dont see any point or scenarios on which this is used, do you have any?

    Read the article

< Previous Page | 92 93 94 95 96 97 98 99 100 101 102 103  | Next Page >