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  • 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).

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  • 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. :)

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  • Graph - strongly connected components

    - by HardCoder1986
    Hello! Is there any fast way to determine the size of the largest strongly connected component in a graph? I mean, like, the obvious approach would mean determining every SCC (could be done using two DFS calls, I suppose) and then looping through them and taking the maximum. I'm pretty sure there has to be some better approach if I only need to have the size of that component and only the largest one, but I can't think of a good solution. Any ideas? Thanks.

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  • return first non repeating character in a string

    - by Amm Sokun
    I had to solve this question in which given a string, i had to return the first non repeating character present in the string. I solved it using hashtable and wrote a method which takes a constant reference to the string and returns the first non repeating character. However, when there is no non repeating character present in the string, i return -1 and in the main program i check as follows char c = firstNonRepeating( word ); if (static_cast<int> (c) == -1) cout<<"no non repeating character present\n"; else cout<<c<<endl; is that the correct way to return -1 when required character is not present?

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  • Complex LINQ paging algorithm

    - by sharepointmonkey
    We have a list of projects that may or may not have a collection of subprojects. Our report needs to contain all the projects except those that are the parent project of a subproject. I need to page this into pages of, say, 25 rows. But if subprojects appear on that page then ALL the subprojects of that project must appear on the same page. So more than 25 items may appear if necessary. I've got as far as var pagedProjects = db.Projects.Where(x => !x.SubProjects.Any()).Skip( (pageNo -1) * pageSize).Take(pageSize); Obviously, this fails the second part of the requirements. As a further pain in the arse, I need to have a pager control on the report. So I'll need to be able to calculate the total number of pages. I could loop through the whole table of projects but the performance will suffer. Can anybody come up with a paged solution? EDIT - I should probably mention that SubProjects joins back onto Projects via a selfreferencing foreign key so the whole lot comes back as an IQueryable<Project>.

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  • Flag bit computation and detection

    - by Majid
    Hi all, In some code I'm working on I should take care of ten independent parameters which can take one of two values (0 or 1). This creates 2^10 distinct conditions. Some of the conditions never occur and can be left out, but those which do occur are still A LOT and making a switch to handle all cases is insane. I want to use 10 if statements instead of a huge switch. For this I know I should use flag bits, or rather flag bytes as the language is javascript and its easier to work with a 10 byte string with to represent a 10-bit binary. Now, my problem is, I don't know how to implement this. I have seen this used in APIs where multiple-selectable options are exposed with numbers 1, 2, 4, 8, ... , n^(n-1) which are decimal equivalents of 1, 10, 100, 1000, etc. in binary. So if we make call like bar = foo(7), bar will be an object with whatever options the three rightmost flags enable. I can convert the decimal number into binary and in each if statement check to see if the corresponding digit is set or not. But I wonder, is there a way to determine the n-th digit of a decimal number is zero or one in binary form, without actually doing the conversion?

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  • 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?

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  • Find subset with K elements that are closest to eachother

    - by Nima
    Given an array of integers size N, how can you efficiently find a subset of size K with elements that are closest to each other? Let the closeness for a subset (x1,x2,x3,..xk) be defined as: 2 <= N <= 10^5 2 <= K <= N constraints: Array may contain duplicates and is not guaranteed to be sorted. My brute force solution is very slow for large N, and it doesn't check if there's more than 1 solution: N = input() K = input() assert 2 <= N <= 10**5 assert 2 <= K <= N a = [] for i in xrange(0, N): a.append(input()) a.sort() minimum = sys.maxint startindex = 0 for i in xrange(0,N-K+1): last = i + K tmp = 0 for j in xrange(i, last): for l in xrange(j+1, last): tmp += abs(a[j]-a[l]) if(tmp > minimum): break if(tmp < minimum): minimum = tmp startindex = i #end index = startindex + K? Examples: N = 7 K = 3 array = [10,100,300,200,1000,20,30] result = [10,20,30] N = 10 K = 4 array = [1,2,3,4,10,20,30,40,100,200] result = [1,2,3,4]

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  • 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 :(((

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  • 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.

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  • 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++; } } }

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  • Re-adjusting a binary heap after removing the minimum element

    - by BeeBand
    After removing the minimum element in a binary heap, i.e. after removing the root, why is the last leaf then assigned to the root and sifted down? Why not take the lesser child of what used to be the root and just keep sifting up all the children? Isn't this the same amount of operations, so why is the "sift down" method preferred?

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  • 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?

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  • How to calculate this string-dissimilarity function efficiently?

    - by ybungalobill
    Hello, I was looking for a string metric that have the property that moving around large blocks in a string won't affect the distance so much. So "helloworld" is close to "worldhello". Obviously Levenshtein distance and Longest common subsequence don't fulfill this requirement. Using Jaccard distance on the set of n-grams gives good results but has other drawbacks (it's a pseudometric and higher n results in higher penalty for changing single character). [original research] As I thought about it, what I'm looking for is a function f(A,B) such that f(A,B)+1 equals the minimum number of blocks that one have to divide A into (A1 ... An), apply a permutation on the blocks and get B: f("hello", "hello") = 0 f("helloworld", "worldhello") = 1 // hello world -> world hello f("abba", "baba") = 2 // ab b a -> b ab a f("computer", "copmuter") = 3 // co m p uter -> co p m uter This can be extended for A and B that aren't necessarily permutations of each other: any additional character that can't be matched is considered as one additional block. f("computer", "combuter") = 3 // com uter -> com uter, unmatched: p and b. Observing that instead of counting blocks we can count the number of pairs of indices that are taken apart by a permutation, we can write f(A,B) formally as: f(A,B) = min { C(P) | P:|A|?|B|, P is bijective, ?i?dom(P) A[P(i)]=B[P(i)] } C(P) = |A| + |B| - |dom(P)| - |{ i | i,i+1?dom(P) and P(i)+1=P(i+1) }| - 1 The problem is... guess what... ... that I'm not able to calculate this in polynomial time. Can someone suggest a way to do this efficiently? Or perhaps point me to already known metric that exhibits similar properties?

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  • Trying to generate all sequences of specified numbers up to a max sum

    - by Stecy
    Hi, Given the following list of descending unique numbers (3,2,1) I wish to generate all the sequences consisting of those numbers up to a maximum sum. Let's say that the sum should be below 10. Then the sequences I'm after are: 3 3 3 3 3 2 1 3 3 2 3 3 1 1 1 3 3 1 1 3 3 1 3 3 3 2 2 2 3 2 2 1 1 3 2 2 1 3 2 2 3 2 1 1 1 1 3 2 1 1 1 3 2 1 1 3 2 1 3 2 3 1 1 1 1 1 1 3 1 1 1 1 1 3 1 1 1 1 3 1 1 1 3 1 1 3 1 3 2 2 2 2 1 2 2 2 2 2 2 2 1 1 1 2 2 2 1 1 2 2 2 1 2 2 2 2 2 1 1 1 1 1 2 2 1 1 1 1 2 2 1 1 1 2 2 1 1 2 2 1 2 2 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 2 1 1 1 2 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 I'm sure that there's a "standard" way to generate this. I thought of using linq but can't figure it out. Also, I am trying a stack based approach but I am still not successful. Any idea?

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  • Making a simple searchable directory of people and their skills in a day - Which technologies?

    - by gav
    Hi All, I am working with a small theatre company. Currently they have a list of people on paper with notes about their skills next to each one. I want to create a database / directory for them so that they can add, delete, update and search for people. It is a very simple and common scenario I know but the issue here is that I only have a day to build a working solution. The search has to be very simple At first I was thinking LAMP but I'd rather not have to create it all from scratch and host it myself. That lead me to Google Spreadsheet as a database, this has the advantage that they already use google docs for everything and if my front end goes tits up they can still get to the data. Presuming none of you can think of some existing software which does exactly what I want the next step is to make a front end for the database. You can create forms for Google Spreadsheets but they only let you add new entries, I can make a Google Gadget but that will only let me implement the search as the Google Visualisation API provides read only access. It's at this point I'm stuck, should I just create a Java Servlet front end for the Google Spreadsheet and use the Java API to add, search and update? I know this is a broad question but I'm just asking 'What would you do?' to implement this system with a day's development time? Gav

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  • 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); } }

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  • need suggestions about content filtering project

    - by serdar
    i'm thinking of designing and implementing a content filtering software as my graduation project. i want it to be a user contributed software. i mean, users can also add/categorize websites. it should be also a web project and extensions for browsers like chrome, firefox, ie.. my question is which programming language do you suggest for this project? i know that firefox extensions are javascript based maybe you can say use .net framework 3.5 because it's better in communication with extensions. sorry for my bad english.. btw any other suggessions about project will be good.. thx a lot.

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  • Python - How to find a correlation between two vectors ?

    - by psihodelia
    Given two vectors X and Y I have to find their correlation, i.e. their linear dependence/independence. Both vectors have equal dimension. A resulted answer should be a floating point number from [-1.0 .. 1.0]. Example: X=[-1, 2, 0] Y=[ 4, 2, -0.3] Find y=cor(X,Y) such that y belongs to [-1.0 .. 1.0]. It should be a simple construction involving a list-comprehension. No external library is allowed. UPDATE: ok, if dot product is enough, then here is my solution: nX = 1/(sum([x*x for x in X]) ** 0.5) nY = 1/(sum([y*y for y in Y]) ** 0.5) cor = sum([(x*nX)*(y*nY) for x,y in zip(X,Y) ]) right?

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  • How to scale JPEG image down so that text is clear as possible?

    - by Juha Syrjälä
    I have some JPEG images that I need scale down to about 80% of original size. Original image dimension are about 700px × 1000px. Images contain some computer generated text and possibly some graphics (similar to what you would find in corporate word documents). How to scale image so that the text is as legible as possible? Currently we are scaling the imaeg down using bicubic interpolation, but that makes the text blurry and foggy.

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  • 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]

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