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Search found 6716 results on 269 pages for 'distributed algorithm'.

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  • Get `n` random values between 2 numbers having average `x`

    - by Somnath Muluk
    I want to get n random numbers(e.g n=16)(whole numbers) between 1 to 5(including both) so that average is x. x can be any value between (1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5). I am using PHP. e.g. Suppose I have average x= 3. Then required 16 whole numbers between 1 to 5(including both). like (1,5,3,3,3,3,2,4,2,4,1,5,1,5,3,3) Update: if x=3.5 means average of 16 numbers should be between 3.5 to 4. and if x=4 means average of 16 numbers should be between 4 to 4.5 and if x=5 means all numbers are 5

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  • Need help with basic optimization problem

    - by ??iu
    I know little of optimization problems, so hopefully this will be didactic for me: rotors = [1, 2, 3, 4...] widgets = ['a', 'b', 'c', 'd' ...] assert len(rotors) == len(widgets) part_values = [ (1, 'a', 34), (1, 'b', 26), (1, 'c', 11), (1, 'd', 8), (2, 'a', 5), (2, 'b', 17), .... ] Given a fixed number of widgets and a fixed number of rotors, how can you get a series of widget-rotor pairs that maximizes the total value where each widget and rotor can only be used once?

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  • Determining the order of a list of numbers (possibly without sorting)

    - by Victor Liu
    I have an array of unique integers (e.g. val[i]), in arbitrary order, and I would like to populate another array (ord[i]) with the the sorted indexes of the integers. In other words, val[ord[i]] is in sorted order for increasing i. Right now, I just fill in ord with 0, ..., N, then sort it based on the value array, but I am wondering if we can be more efficient about it since ord is not populated to begin with. This is more of a question out of curiousity; I don't really care about the extra overhead from having to prepopulate a list and then sort it (it's small, I use insertion sort). This may be a silly question with an obvious answer, but I couldn't find anything online.

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  • Determine coordinates of rotated rectangle

    - by MathieuK
    I'm creating an utility application that should detect and report the coordinates of the corners of a transparent rectangle (alpha=0) within an image. So far, I've set up a system with Javascript + Canvas that displays the image and starts a floodfill-like operation when I click inside the transparent rectangle in the image. It correctly determines the bounding box of the floodfill operation and as such can provide me the correct coordinates. Here's my implementation so-far: http://www.scriptorama.nl/image/ (works in recent Firefox / Safari ). However, the bounding box approach breaks down then the transparent rectangle is rotated (CW or CCW) as the resulting bounding box no longer properly represents the proper width and height. I've tried to come up with a few alternatives to detect to corners, but have not been able to think up a proper solution. So, does anyone have any suggestions on how I might approach this so I can properly detect the coordinates of 4 corners of the rotated rectangle?

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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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  • Finding cities close to one another using longitude and latitude

    - by Jamie
    Each user in my db is associated to a city (with it's longitude and latitude) How would I go about finding out which cities are close to one another? i.e. in England, Cambridge is fairly close to London. So If I have a user who lives in Cambridge. Users close to them would be users living in close surrounding cities, such as London, Hertford etc. Any ideas how I could go about this? And also, how would I define what is close? i.e. in the UK close would be much closer than if it were in the US as the US is far more spread out. Ideas and suggestions. Also, do you know any services that provide this sort of functionality? Thanks

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  • generic binary Search in c#

    - by Pro_Zeck
    Below is my Generic Binary Search it works ok with the intgers type array it finds all the elements in it . But the Problem Arises when i use a string array to find any string data. It runs ok for the first index and last index elements but i cant find the middle elements. Stringarray = new string[] { "b", "a", "ab", "abc", "c" }; public static void BinarySearch<T>(T[] array, T searchFor, Comparer<T> comparer) { int high, low, mid; high = array.Length - 1; low = 0; if (array[0].Equals(searchFor)) Console.WriteLine("Value {0} Found At Index {1}",array[0],0); else if (array[high].Equals(searchFor)) Console.WriteLine("Value {0} Found At Index {1}", array[high], high); else { while (low <= high) { mid = (high + low) / 2; if (comparer.Compare(array[mid], searchFor) == 0) { Console.WriteLine("Value {0} Found At Index {1}", array[mid], mid); break; } else { if (comparer.Compare(searchFor, array[mid]) > 0) high = mid + 1; else low = mid + 1; } } if (low > high) { Console.WriteLine("Value Not Found In the Collection"); } } }

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  • Another Nim's Game Variant

    - by Terry Smith
    Given N binary sequence Example : given one sequence 101001 means player 0 can only choose a position with 0 element and cut the sequence from that position {1,101,1010} player 1 can only choose a position with 1 element ans cut the sequence from that position {null,10,101000} now player 0 and player 1 take turn cutting the sequence, on each turn they can cut any one non-null sequence, if a player k can't make a move because there's no more k element on any sequence, he lose. Assume both player play optimally, who will win ? I tried to solve this problem with grundy but i'm unable to reduce the sequence to a grundy number because it both player don't have the same option to move. Can anyone give me a hint to solve this problem ?

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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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  • An O(1) Sort ~~~

    - by FlySwat
    Before you stone me for being a heretic, There is a sort that proclaims to be O(1), called "Bead Sort" (http://en.wikipedia.org/wiki/Bead_sort) , however that is pure theory, when actually applied I found that it was actually O(N * M), which is pretty pathetic. That said, Lets list out some of the fastest sorts, and their best case and worst case speed in Big O notation. ~~ FlySwat ~~

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  • Generate encoding String according to creation order.

    - by Tony
    I need to generate encoding String for each item I inserted into the database. for example: x00001 for the first item x00002 for the sencond item x00003 for the third item The way I chose to do this is counting the rows. Before I insert the third item, I count against the database, I know there're already 2 rows, so the next encoding is ended with 3. But there is a problem. If I delete the second item, the forth item will not be the x00004,but x00003. I can add additional columns to table, to store the next encoding, I don't know if there's other better solutions ?

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  • What Software Engineering Areas should be stressed upon while Interviewing Candidate for Fulltime So

    - by Rachel
    Hi, This question is somewhat related to other posts which I found on Stackoverflow but not exactly and so am prompted to ask about it. I know we must ask for Data-Structures and Algorithms but what specific data-structures or Algorithms or other CS Concepts should be asked while interviewing Sr. Software Engineering Fulltime Position as compared with Software Engineering Position. Thanks.

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  • Generating different randoms valid for a day on different independent devices?

    - by Pentium10
    Let me describe the system. There are several mobile devices, each independent from each other, and they are generating content for the same record id. I want to avoid generating the same content for the same record on different devices, for this I though I would use a random and make it so too cluster the content pool based on these randoms. Suppose you have choices from 1 to 100. Day 1 Device#1 will choose for the record#33 between 1-10 Device#2 will choose for the record#33 between 40-50 Device#3 will choose for the record#33 between 50-60 Device#1 will choose for the record#55 between 40-50 Device#2 will choose for the record#55 between 1-10 Device#3 will choose for the record#55 between 10-20 Device#1 will choose for the record#11 between 1-10 Device#2 will choose for the record#22 between 1-10 Device#3 will choose for the record#99 between 1-10 Day 2 Device#1 will choose for the record#33 between 90-100 Device#2 will choose for the record#33 between 1-10 Device#3 will choose for the record#33 between 50-60 They don't have access to a central server. Data available for each of them: IMEI (unique per mobile) Date of today (same on all devices) Record id (same on all devices) What do you think, how is it possible? ps. tags can be edited

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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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  • 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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  • 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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  • Formula for popularity? (based on "like it", "comments", "views")

    - by paullb
    I have some pages on a website and I have to create an ordering based on "popularity"/"activity" The parameters that I have to use are: views to the page comments made on the page (there is a form at the bottom where uses can make comments) clicks made to the "like it" icon Are there any standards for what a formula for popularity would be? (if not opinions are good too) (initially I thought of views + 10*comments + 10*likeit)

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  • Is this linear search implementation actually useful?

    - by Helper Method
    In Matters Computational I found this interesting linear search implementation (it's actually my Java implementation ;-)): public static int linearSearch(int[] a, int key) { int high = a.length - 1; int tmp = a[high]; // put a sentinel at the end of the array a[high] = key; int i = 0; while (a[i] != key) { i++; } // restore original value a[high] = tmp; if (i == high && key != tmp) { return NOT_CONTAINED; } return i; } It basically uses a sentinel, which is the searched for value, so that you always find the value and don't have to check for array boundaries. The last element is stored in a temp variable, and then the sentinel is placed at the last position. When the value is found (remember, it is always found due to the sentinel), the original element is restored and the index is checked if it represents the last index and is unequal to the searched for value. If that's the case, -1 (NOT_CONTAINED) is returned, otherwise the index. While I found this implementation really clever, I wonder if it is actually useful. For small arrays, it seems to be always slower, and for large arrays it only seems to be faster when the value is not found. Any ideas?

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  • Picking apples off a tree

    - by John Retallack
    I have the following problem: I am given a tree with N apples, for each apple I am given it's weight and height. I can pick apples up to a given height H, each time I pick an apple the height of every apple is increased with U. I have to find out the maximum weight of apples I can pick. 1 = N = 100000 0 < {H, U, apples' weight and height, maximum weight} < 231 Example: N=4 H=100 U=10 height weight 82 30 91 10 93 5 94 15 The answer is 45: first pick the apple with the weight of 15 then the one with the weight of 30. Could someone help me approach this problem?

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  • How to detect if a certain range resides (partly) within an other range?

    - by Tom
    Lets say I've got two squares and I know their positions, a red and blue square: redTopX; redTopY; redBotX; redBotY; blueTopX; blueTopY; blueBotX; blueBotY; Now, I want to check if square blue resides (partly) within (or around) square red. This can happen in a lot of situations, as you can see in this image I created to illustrate my situation better: Note that there's always only one blue and one red square, I just added multiple so I didn't have to redraw 18 times. My original logic was simple, I'd check all corners of square blue and see if any of them are inside square red: if ( ((redTopX >= blueTopX) && (redTopY >= blueTopY) && (redTopX <= blueBotX) && (redTopY <= blueBotY)) || //top left ((redBotX >= blueTopX) && (redTopY >= blueTopY) && (redBotX <= blueBotX) && (redTopY <= blueBotY)) || //top right ((redTopX >= blueTopX) && (redBotY >= blueTopY) && (redTopX <= blueBotX) && (redBotY <= blueBotY)) || //bottom left ((redBotX >= blueTopX) && (redBotY >= blueTopY) && (redBotX <= blueBotX) && (redBotY <= blueBotY)) //bottom right ) { //blue resides in red } Unfortunately, there are a couple of flaws in this logic. For example, what if red surrounds blue (like in situation 1)? I thought this would be pretty easy but am having trouble coming up with a good way of covering all these situations.. can anyone help me out here? Regards, Tom

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