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  • in-place permutation of a array follows this rule

    - by Mgccl
    Suppose there is an array, we want to find everything in the odd index, and move it to the end. Everything in the even index move it to the beginning. The relative order of all odd index items and all even index items are preserved. Suppose the values of the array, a[i] = i, n is even. Then we have. 0,1,2,3,4,5,...,n-1 after the operation 0,2,4,6,...,n-2,1,3,5,7,...,n-1 Can this be done in-place and in O(n) time?

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  • How to select number of lines from large text files?

    - by MiNdFrEaK
    I was wondering how to select number of lines from a certain text file. As an example: I have a text file containing the following lines: branch 27 : rect id 23400 rect: -115.475609 -115.474907 31.393650 31.411301 branch 28 : rect id 23398 rect: -115.474907 -115.472282 31.411301 31.417351 branch 29 : rect id 23396 rect: -115.472282 -115.468033 31.417351 31.427151 branch 30 : rect id 23394 rect: -115.468033 -115.458733 31.427151 31.438181 Non-Leaf Node: level=1 count=31 address=53 branch 0 : rect id 42 rect: -115.768539 -106.251556 31.425039 31.717550 branch 1 : rect id 50 rect: -109.559479 -106.009361 31.296721 31.775299 branch 2 : rect id 51 rect: -110.937401 -106.226143 31.285870 31.771971 branch 3 : rect id 54 rect: -109.584412 -106.069092 31.285240 31.775230 branch 4 : rect id 56 rect: -109.570961 -106.000954 31.296721 31.780769 branch 5 : rect id 58 rect: -115.806213 -106.366188 31.400450 31.687519 branch 6 : rect id 59 rect: -113.173859 -106.244057 31.297440 31.627750 branch 7 : rect id 60 rect: -115.811478 -106.278252 31.400450 31.679470 branch 8 : rect id 61 rect: -109.953888 -106.020111 31.325319 31.775270 branch 9 : rect id 64 rect: -113.070969 -106.015968 31.331841 31.704750 branch 10 : rect id 68 rect: -113.065689 -107.034576 31.326300 31.770809 branch 11 : rect id 71 rect: -112.333344 -106.059860 31.284081 31.662920 branch 12 : rect id 73 rect: -115.071083 -106.309677 31.267879 31.466850 branch 13 : rect id 74 rect: -116.094414 -106.286308 31.236290 31.424770 branch 14 : rect id 75 rect: -115.423264 -106.286308 31.229691 31.415510 branch 15 : rect id 76 rect: -116.111656 -106.313110 31.259390 31.478300 branch 16 : rect id 77 rect: -116.247467 -106.309677 31.240231 31.451799 branch 17 : rect id 78 rect: -116.170792 -106.094543 31.156429 31.391781 branch 18 : rect id 79 rect: -116.225723 -106.292709 31.239960 31.442850 branch 19 : rect id 80 rect: -116.268013 -105.769913 31.157240 31.378111 branch 20 : rect id 82 rect: -116.215424 -105.827202 31.198441 31.383421 branch 21 : rect id 83 rect: -116.095734 -105.826439 31.197460 31.373819 branch 22 : rect id 84 rect: -115.423264 -105.815018 31.182640 31.368891 branch 23 : rect id 85 rect: -116.221527 -105.776512 31.160931 31.389830 branch 24 : rect id 86 rect: -116.203369 -106.473831 31.168350 31.367611 branch 25 : rect id 87 rect: -115.727631 -106.501587 31.189100 31.395941 branch 26 : rect id 88 rect: -116.237289 -105.790756 31.164780 31.358959 branch 27 : rect id 89 rect: -115.791344 -105.990044 31.072620 31.349529 branch 28 : rect id 90 rect: -115.736847 -106.495079 31.187969 31.376900 branch 29 : rect id 91 rect: -115.721710 -106.000130 31.160351 31.354601 branch 30 : rect id 92 rect: -115.792236 -106.000793 31.166620 31.378811 Leaf Node: level=0 count=21 address=42 branch 0 : rect id 18312 rect: -106.412270 -106.401367 31.704750 31.717550 branch 1 : rect id 18288 rect: -106.278252 -106.253387 31.520321 31.548361 I just want those lines which are in between Non-Leaf Node level=1 to Leaf Node Level=0 and also there are a lot of segments like this and I need them all.

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  • Given a few strings, how many strings can be lexicographically least by modifying the alphabet?

    - by Jackson W
    Number of strings can be huge as in 30000. Given N strings, output which ones can be lexicographically least after modifying the english alphabet. e.g. acdbe...... for example if the strings were: omm moo mom ommnom "mom" is already lexicographically least with the original english alphabet. we can make the word "omm" least by switching "m" and "o" in the alphabet ("abcdefghijklonmpqrstuvwxyz"). the other ones you cant make lexicographically last, no matter what you do. any fast way to do this? I have no ways to approach this except try every single possible alphabet

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  • Can my tortoise vs. hare race be improved?

    - by FredOverflow
    Here is my code for detecting cycles in a linked list: do { hare = hare.next(); if (hare == back) return; hare = hare.next(); if (hare == back) return; tortoise = tortoise.next(); } while (tortoise != hare); throw new AssertionError("cyclic linkage"); Is there a way to get rid of the code duplication inside the loop? Am I right in assuming that I don't need a check after making the tortoise take a step forward? As I see it, the tortoise can never reach the end of the list before the hare (contrary to the fable). Any other ways to simplify/beautify this code?

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  • Faster way to compare two sets of points in N-dimensional space?

    - by Amit
    List1 contains a high number (~7^10) of N-dimensional points (N <=10), List2 contains the same or fewer number of N-dimensional points (N <=10). My task is this: I want to check which point in List2 is closest (euclidean distance) to a point in List1 for every point in List1 and subsequently perform some operation on it. I have been doing it the simple- the nested loop way when I didn't have more than 50 points in List1, but with 7^10 points, this obviously takes up a lot of time. What is the fastest way to do this? Any concepts from Computational Geometry might help?

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  • Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

    - by hilal
    Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

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  • Find recipes that can be cooked from provided ingridients

    - by skaurus
    Sorry for bad English :( Suppose i can preliminary organize recipes and ingredients data in any way. How can i effectively conduct search of recipes by user-provided ingredients, preferably sorted by max match - so, first going recipes that use maximum of provided ingridients and do not contain any other ingrs, after them recipes that uses less of provided set and still not any other ingrs, after them recipes with minimum additional requirements and so on? All i can think about is represent recipe ingridients like bitmasks, and compare required bitmask with all recipes, but it is obviously a bad way to go. And related things like Levenstein distance i don't see how to use here. I believe it should be quite common task...

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  • Tracking/Counting Word Frequency

    - by Joel Martinez
    I'd like to get some community consensus on a good design to be able to store and query word frequency counts. I'm building an application in which I have to parse text inputs and store how many times a word has appeared (over time). So given the following inputs: "To Kill a Mocking Bird" "Mocking a piano player" Would store the following values: Word Count ------------- To 1 Kill 1 A 2 Mocking 2 Bird 1 Piano 1 Player 1 And later be able to quickly query for the count value of a given arbitrary word. My current plan is to simply store the words and counts in a database, and rely on caching word count values ... But I suspect that I won't get enough cache hits to make this a viable solution long term. Can anyone suggest algorithms, or data structures, or any other idea that might make this a well-performing solution?

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  • parsing of mathematical expressions

    - by gcc
    (in c90) (linux) input: sqrt(2 - sin(3*A/B)^2.5) + 0.5*(C*~(D) + 3.11 +B) a b /*there are values for a,b,c,d */ c d input: cos(2 - asin(3*A/B)^2.5) +cos(0.5*(C*~(D)) + 3.11 +B) a b /*there are values for a,b,c,d */ c d input: sqrt(2 - sin(3*A/B)^2.5)/(0.5*(C*~(D)) + sin(3.11) +ln(B)) /*max lenght of formula is 250 characters*/ a b /*there are values for a,b,c,d */ c /*each variable with set of floating numbers*/ d As you can see infix formula in the input depends on user. My program will take a formula and n-tuples value. Then it calculate the results for each value of a,b,c and d. If you wonder I am saying ;outcome of program is graph. /sometimes,I think i will take input and store in string. then another idea is arise " I should store formula in the struct" but i don't know how I can construct the code on the base of structure./ really, I don't know way how to store the formula in program code so that I can do my job. can you show me? /* a,b,c,d is letters cos,sin,sqrt,ln is function*/

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  • Compact data structure for storing a large set of integral values

    - by Odrade
    I'm working on an application that needs to pass around large sets of Int32 values. The sets are expected to contain ~1,000,000-50,000,000 items, where each item is a database key in the range 0-50,000,000. I expect distribution of ids in any given set to be effectively random over this range. The operations I need on the set are dirt simple: Add a new value Iterate over all of the values. There is a serious concern about the memory usage of these sets, so I'm looking for a data structure that can store the ids more efficiently than a simple List<int>or HashSet<int>. I've looked at BitArray, but that can be wasteful depending on how sparse the ids are. I've also considered a bitwise trie, but I'm unsure how to calculate the space efficiency of that solution for the expected data. A Bloom Filter would be great, if only I could tolerate the false negatives. I would appreciate any suggestions of data structures suitable for this purpose. I'm interested in both out-of-the-box and custom solutions. EDIT: To answer your questions: No, the items don't need to be sorted By "pass around" I mean both pass between methods and serialize and send over the wire. I clearly should have mentioned this. There could be a decent number of these sets in memory at once (~100).

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  • How to do "map chunks", like terraria or minecraft maps?

    - by O'poil
    Due to performance issues, I have to cut my maps into chunks. I manage the maps in this way: listMap[x][y] = new Tile (x,y); I tried in vain to cut this list for several "chunk" to avoid loading all the map because the fps are not very high with large map. And yet, when I update or Draw I do it with a little tile range. Here is how I proceed: foreach (List<Tile> list in listMap) { foreach (Tile leTile in list) { if ((leTile.Position.X < screenWidth + hero.Pos.X) && (leTile.Position.X > hero.Pos.X - tileSize) && (leTile.Position.Y < screenHeight + hero.Pos.Y) && (leTile.Position.Y > hero.Pos.Y - tileSize) ) { leTile.Draw(spriteBatch, gameTime); } } } (and the same thing, for the update method). So I try to learn with games like minecraft or terraria, and any two manages maps much larger than mine, without the slightest drop of fps. And apparently, they load "chunks". What I would like to understand is how to cut my list in Chunk, and how to display depending on the position of my character. I try many things without success. Thank you in advance for putting me on the right track! Ps : Again, sorry for my English :'( Pps : I'm not an experimented developer ;)

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  • deteminant of matrix

    - by davit-datuashvili
    suppose there is given two dimensional array int a[][]=new int[4][4]; i am trying to find determinant of matrices please help i know how find it mathematical but i am trying to find it in programaticaly

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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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  • 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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  • How does Photoshop (Or drawing programs) blit?

    - by user146780
    I'm getting ready to make a drawing application in Windows. I'm just wondering, do drawing programs have a memory bitmap which they lock, then set each pixel, then blit? I don't understand how Photoshop can move entire layers without lag or flicker without using hardware acceleration. Also in a program like Expression Design, I could have 200 shapes and move them around all at once with no lag. I'm really wondering how this can be done without GPU help. I don't think super efficient algorithms could justify that? Thanks

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  • What are the advantages of a rebase over a merge in git?

    - by eSKay
    In this article, the author explains rebasing with this diagram: Rebase: If you have not yet published your branch, or have clearly communicated that others should not base their work on it, you have an alternative. You can rebase your branch, where instead of merging, your commit is replaced by another commit with a different parent, and your branch is moved there. while a normal merge would have looked like this: So, if you rebase, you are just losing a history state (which would be garbage collected sometime in the future). So, why would someone want to do a rebase at all? What am I missing here?

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  • Function for creating color wheels

    - by lbrandy
    This is something I've pseudo-solved many times and never quite found a solution that's stuck with me. The problem is to come up with a way to generate N colors that are as distinguishable as possible where N is a parameter.

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  • Sort string based upon the count of characters Options

    - by prp
    Sample Data : input : "abcdacdc" Output : "cadb" here we have to sort strings in order of count of characters. If the count is same for characters. maintain the original order of the characters from input string. my approach: i have used array of 26 for maintaining occurrence of all characters and sort it then print it.But while doing so i am not able to maintain order in case if two characters have same count. please suggest any improvement or any other algo.

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  • What is the optimum way to select the most dissimilar individuals from a population?

    - by Aaron D
    I have tried to use k-means clustering to select the most diverse markers in my population, for example, if we want to select 100 lines I cluster the whole population to 100 clusters then select the closest marker to the centroid from each cluster. The problem with my solution is it takes too much time (probably my function needs optimization), especially when the number of markers exceeds 100000. So, I will appreciate it so much if anyone can show me a new way to select markers that maximize diversity in my population and/or help me optimize my function to make it work faster. Thank you # example: library(BLR) data(wheat) dim(X) mdf<-mostdiff(t(X), 100,1,nstart=1000) Here is the mostdiff function that i used: mostdiff <- function(markers, nClust, nMrkPerClust, nstart=1000) { transposedMarkers <- as.array(markers) mrkClust <- kmeans(transposedMarkers, nClust, nstart=nstart) save(mrkClust, file="markerCluster.Rdata") # within clusters, pick the markers that are closest to the cluster centroid # turn the vector of which markers belong to which clusters into a list nClust long # each element of the list is a vector of the markers in that cluster clustersToList <- function(nClust, clusters) { vecOfCluster <- function(whichClust, clusters) { return(which(whichClust == clusters)) } return(apply(as.array(1:nClust), 1, vecOfCluster, clusters)) } pickCloseToCenter <- function(vecOfCluster, whichClust, transposedMarkers, centers, pickHowMany) { clustSize <- length(vecOfCluster) # if there are fewer than three markers, the center is equally distant from all so don't bother if (clustSize < 3) return(vecOfCluster[1:min(pickHowMany, clustSize)]) # figure out the distance (squared) between each marker in the cluster and the cluster center distToCenter <- function(marker, center){ diff <- center - marker return(sum(diff*diff)) } dists <- apply(transposedMarkers[vecOfCluster,], 1, distToCenter, center=centers[whichClust,]) return(vecOfCluster[order(dists)[1:min(pickHowMany, clustSize)]]) } }

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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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  • Setting last N bits in an array

    - by Martin
    I'm sure this is fairly simple, however I have a major mental block on it, so I need a little help here! I have an array of 5 integers, the array is already filled with some data. I want to set the last N bits of the array to be random noise. [int][int][int][int][int] set last 40 bits [unchanged][unchanged][unchanged][24 bits of old data followed 8 bits of randomness][all random] This is largely language agnostic, but I'm working in C# so bonus points for answers in C#

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