Search Results

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

Page 94/300 | < Previous Page | 90 91 92 93 94 95 96 97 98 99 100 101  | Next Page >

  • Texture coordintes for a polygon and a square texture

    - by user146780
    basically I have a texture. I also have a lets say octagon (or any polygon). I find that octagon's bounding box. Let's say my texture is the size of the octagon's bounding box. How could I figure out the texture coordinates so that the texture maps to it. To clarify, lets say you had a square of tin foil and cut the octagon out you'd be left with a tin foil textured polygon.I'm just not sure how to figure it out for an arbitrary polygon. Thanks

    Read the article

  • mod,prime -> inverse possible

    - by Piet
    Hi all. I was wondering if one can do the following: We have: X is a product of N-primes, thus I assume unique. C is a constant. We can assure that C is a number that is part of the N-primes or not. Whichever will work best. Thus: X mod C = Z We have Z and C and we know that X was a product of N-primes, where N is restricted lets say first 100 primes. Is there anyway we can get back X?

    Read the article

  • ListAdapters and WrapperListAdapter algorithm

    - by Matty F
    This logic is written in a function with signature private void showDialog(final AdapterView<? extends Adapter> parent, String title, String message, final Tag subject) Is there a better way of doing this? // refresh adapter SimpleCursorAdapter adapter; if (parent.getAdapter() instanceof WrapperListAdapter) { adapter = (SimpleCursorAdapter) ((WrapperListAdapter) parent.getAdapter()).getWrappedAdapter(); } else { adapter = (SimpleCursorAdapter) parent.getAdapter(); } adapter.getCursor().requery(); adapter.notifyDataSetChanged(); Also, is there any point in having AdapterView<? extends Adapter> in the signature and not just AdapterView<?>?

    Read the article

  • UVA Online Judge 3n+1 : Right answer is Wrong answer

    - by Samuraisoulification
    Ive been toying with this problem for more than a week now, I have optimized it a lot, I seem to be getting the right answer, since it's the same as when I compare it to other's answers that got accepted, but I keep getting wrong answer. Im not sure what's going on! Anyone have any advice? I think it's a problem with the input or the output, cause Im not exactly sure how this judge thing works. So if anyone could pinpoint the problem, and also give me any advice on my code, Id be very appreciative!!! #include <iostream> #include <cstdlib> #include <stdio.h> #include <vector> using namespace std; class Node{ // node for each number that has teh cycles and number private: int number; int cycles; bool cycleset; // so it knows whether to re-set the cycle public: Node(int num){ number = num; cycles = 0; cycleset = false; } int getnumber(){ return number; } int getcycles(){ return cycles; } void setnumber(int num){ number = num; } void setcycles(int num){ cycles = num; cycleset = true; } bool cycled(){ return cycleset; } }; class Cycler{ private: vector<Node> cycleArray; int biggest; int cycleReal(unsigned int number){ // actually cycles through the number int cycles = 1; if (number != 1) { if (number < 1000000) { // makes sure it's in vector bounds if (!cycleArray[number].cycled()) { // sees if it's been cycled if (number % 2 == 0) { cycles += this->cycleReal((number / 2)); } else { cycles += this->cycleReal((3 * number) + 1); } } else { // if cycled get the number of cycles and don't re-calculate, ends recursion cycles = cycleArray[number].getcycles(); } } else { // continues recursing if it's too big for the vector if (number % 2 == 0) { cycles += this->cycleReal((number / 2)); } else { cycles += this->cycleReal((3 * number) + 1); } } } if(number < 1000000){ // sets cycles table for the number in the vector if (!cycleArray[number].cycled()) { cycleArray[number].setcycles(cycles); } } return cycles; } public: Cycler(){ biggest = 0; for(int i = 0; i < 1000000; i++){ // initialize the vector, set the numbers Node temp(i); cycleArray.push_back(temp); } } int cycle(int start, int end){ // cycles thorugh the inputted numbers. int size = 0; for(int i = start; i < end ; i++){ size = this->cycleReal(i); if(size > biggest){ biggest = size; } } int temp = biggest; biggest = 0; return temp; } int getBiggest(){ return biggest; } }; int main() { Cycler testCycler; int i, j; while(cin>>i>>j){ //read in untill \n int biggest = 0; if(i > j){ biggest = testCycler.cycle(j, i); }else{ biggest = testCycler.cycle(i, j); } cout << i << " " << j << " " << biggest << endl; } return 0; }

    Read the article

  • 3-way quicksort, question

    - by peiska
    I am trying to understand the 3-way radix Quicksort, and i dont understand why the the CUTOFF variable there? and the insertion method? public class Quick3string { private static final int CUTOFF = 15; // cutoff to insertion sort // sort the array a[] of strings public static void sort(String[] a) { // StdRandom.shuffle(a); sort(a, 0, a.length-1, 0); assert isSorted(a); } // return the dth character of s, -1 if d = length of s private static int charAt(String s, int d) { assert d >= 0 && d <= s.length(); if (d == s.length()) return -1; return s.charAt(d); } // 3-way string quicksort a[lo..hi] starting at dth character private static void sort(String[] a, int lo, int hi, int d) { // cutoff to insertion sort for small subarrays if (hi <= lo + CUTOFF) { insertion(a, lo, hi, d); return; } int lt = lo, gt = hi; int v = charAt(a[lo], d); int i = lo + 1; while (i <= gt) { int t = charAt(a[i], d); if (t < v) exch(a, lt++, i++); else if (t > v) exch(a, i, gt--); else i++; } // a[lo..lt-1] < v = a[lt..gt] < a[gt+1..hi]. sort(a, lo, lt-1, d); if (v >= 0) sort(a, lt, gt, d+1); sort(a, gt+1, hi, d); } // sort from a[lo] to a[hi], starting at the dth character private static void insertion(String[] a, int lo, int hi, int d) { for (int i = lo; i <= hi; i++) for (int j = i; j > lo && less(a[j], a[j-1], d); j--) exch(a, j, j-1); } // exchange a[i] and a[j] private static void exch(String[] a, int i, int j) { String temp = a[i]; a[i] = a[j]; a[j] = temp; } // is v less than w, starting at character d private static boolean less(String v, String w, int d) { assert v.substring(0, d).equals(w.substring(0, d)); return v.substring(d).compareTo(w.substring(d)) < 0; } // is the array sorted private static boolean isSorted(String[] a) { for (int i = 1; i < a.length; i++) if (a[i].compareTo(a[i-1]) < 0) return false; return true; } public static void main(String[] args) { // read in the strings from standard input String[] a = StdIn.readAll().split("\\s+"); int N = a.length; // sort the strings sort(a); // print the results for (int i = 0; i < N; i++) StdOut.println(a[i]); } } from http://www.cs.princeton.edu/algs4/51radix/Quick3string.java.html

    Read the article

  • Learning Algorithms and Data Structures Fundamentals

    - by valya
    Can you recommend me a book or (better!) a site with many hard problems and exercises about data structures? I'm already answering project Euler questions, but these questions are about interesting, but uncommon algorithms. I hardly used even a simple tree. Maybe there is a site with exercises like: hey, you need to calculate this: ... . Do it using a tree. Now do it using a zipper. Upload your C (Haskell, Lisp, even Pascal or Fortress go) solution. Oh, your solution is so slow! Self-education is very hard then you trying to learn very common, fundamental things. How can I help myself with them without attending to courses or whatever?

    Read the article

  • Number of different elements in an array.

    - by AB
    Is it possible to compute the number of different elements in an array in linear time and constant space? Let us say it's an array of long integers, and you can not allocate an array of length sizeof(long). P.S. Not homework, just curious. I've got a book that sort of implies that it is possible.

    Read the article

  • problem with evolutionary algorithms degrading into simulated annealing: mutation too small?

    - by Schnalle
    i have a problem understanding evolutionary algorithms. i tried using this technique several times, but i always ran into the same problem: degeneration into simulated annealing. lets say my initial population, with fitness in brackets, is: A (7), B (9), C (14), D (19) after mating and mutation i have following children: AB (8.3), AC (12.2), AD (14.1), BC(11), BD (14.7), CD (17) after elimination of the weakest, we get A, AB, B, AC next turn, AB will mate again with a result around 8, pushing AC out. next turn, AB again, pushing B out (assuming mutation changes fitness mostly in the 1 range). now, after only a few turns the pool is populated with the originally fittest candidates (A, B) and mutations of those two (AB). this happens regardless of the size of the initial pool, it just takes a bit longer. say, with an initial population of 50 it takes 50 turns, then all others are eliminated, turning the whole setup in a more complicated simulated annealing. in the beginning i also mated canditates with themselves, worsening the problem. so, what do i miss? are my mutation rates simply too small and will it go away if i increase them? here's the project i'm using it for: http://stefan.schallerl.com/simuan-grid-grad/ yeah, the code is buggy and the interface sucks, but i'm too lazy to fix it right now - and be careful, it may lock up your browser. better use chrome, even thought firefox is not slower than chrome for once (probably the tracing for the image comparison pays off, yay!). if anyone is interested, the code can be found here. here i just dropped the ev-alg idea and went for simulated annealing. ps: i'm not even sure about simulated annealing - it is like evolutionary algorithms, just with a population size of one, right?

    Read the article

  • STL map--> sort by value?

    - by Charlie Epps
    Hi I wonder how can I implement the STL map sorting by value. For example, I have a map m map<int, int>; m[1] = 10; m[2] = 5; m[4] = 6; m[6] = 1; and then.. I'd like to sort that with the m's value. So, if I print the map, I'd like to get the result like m[6] = 1 m[2] = 5 m[4] = 6 m[1] = 10 this. How can I sort like this way? Is there any way that I can deal with the key and value with sorted values?

    Read the article

  • Why does C qicksort function implementation works much slower (tape comparations, tape swapping) than bobble sort function?

    - by Artur Mustafin
    I'm going to implement a toy tape "mainframe" for a students, showing the quickness of "quicksort" class functions (recursive or not, does not really matters, due to the slow hardware, and well known stack reversal techniques) comparatively to the "bubblesort" function class, so, while I'm clear about the hardware implementation ans controllers, i guessed that quicksort function is much faster that other ones in terms of sequence, order and comparation distance (it is much faster to rewind the tape from the middle than from the very end, because of different speed of rewind). Unfortunately, this is not the true, this simple "bubble" code shows great improvements comparatively to the "quicksort" functions in terms of comparison distances, direction and number of comparisons and writes. So I have 3 questions: Does I have mistaken in my implememtation of quicksort function? Does I have mistaken in my implememtation of bubblesoft function? If not, why the "bubblesort" function is works much faster in (comparison and write operations) than "quicksort" function? I already have a "quicksort" function: void quicksort(float *a, long l, long r, const compare_function& compare) { long i=l, j=r, temp, m=(l+r)/2; if (l == r) return; if (l == r-1) { if (compare(a, l, r)) { swap(a, l, r); } return; } if (l < r-1) { while (1) { i = l; j = r; while (i < m && !compare(a, i, m)) i++; while (m < j && !compare(a, m, j)) j--; if (i >= j) { break; } swap(a, i, j); } if (l < m) quicksort(a, l, m, compare); if (m < r) quicksort(a, m, r, compare); return; } } and the kind of my own implememtation of the "bubblesort" function: void bubblesort(float *a, long l, long r, const compare_function& compare) { long i, j, k; if (l == r) { return; } if (l == r-1) { if (compare(a, l, r)) { swap(a, l, r); } return; } if (l < r-1) { while(l < r) { i = l; j = l; while (i < r) { i++; if (!compare(a, j, i)) { continue; } j = i; } if (l < j) { swap(a, l, j); } l++; i = r; k = r; while(l < i) { i--; if (!compare(a, i, k)) { continue; } k = i; } if (k < r) { swap(a, k, r); } r--; } return; } } I have used this sort functions in a test sample code, like this: #include <stdio.h> #include <stdlib.h> #include <math.h> #include <conio.h> long swap_count; long compare_count; typedef long (*compare_function)(float *, long, long ); typedef void (*sort_function)(float *, long , long , const compare_function& ); void init(float *, long ); void print(float *, long ); void sort(float *, long, const sort_function& ); void swap(float *a, long l, long r); long less(float *a, long l, long r); long greater(float *a, long l, long r); void bubblesort(float *, long , long , const compare_function& ); void quicksort(float *, long , long , const compare_function& ); void main() { int n; printf("n="); scanf("%d",&n); printf("\r\n"); long i; float *a = (float *)malloc(n*n*sizeof(float)); sort(a, n, &bubblesort); print(a, n); sort(a, n, &quicksort); print(a, n); free(a); } long less(float *a, long l, long r) { compare_count++; return *(a+l) < *(a+r) ? 1 : 0; } long greater(float *a, long l, long r) { compare_count++; return *(a+l) > *(a+r) ? 1 : 0; } void swap(float *a, long l, long r) { swap_count++; float temp; temp = *(a+l); *(a+l) = *(a+r); *(a+r) = temp; } float tg(float x) { return tan(x); } float ctg(float x) { return 1.0/tan(x); } void init(float *m,long n) { long i,j; for (i = 0; i < n; i++) { for (j=0; j< n; j++) { m[i + j*n] = tg(0.2*(i+1)) + ctg(0.3*(j+1)); } } } void print(float *m, long n) { long i, j; for(i = 0; i < n; i++) { for(j = 0; j < n; j++) { printf(" %5.1f", m[i + j*n]); } printf("\r\n"); } printf("\r\n"); } void sort(float *a, long n, const sort_function& sort) { long i, sort_compare = 0, sort_swap = 0; init(a,n); for(i = 0; i < n*n; i+=n) { if (fmod (i / n, 2) == 0) { compare_count = 0; swap_count = 0; sort(a, i, i+n-1, &less); if (swap_count == 0) { compare_count = 0; sort(a, i, i+n-1, &greater); } sort_compare += compare_count; sort_swap += swap_count; } } printf("compare=%ld\r\n", sort_compare); printf("swap=%ld\r\n", sort_swap); printf("\r\n"); }

    Read the article

  • circuit/block-diagram drawing

    - by JCLL
    I'm looking for either algorithms or visualization tool for (nice) circuit/block-diagram drawing. I am also interested in a general formulation of the problem. By "circuit drawing", I mean the capability of exploring place & route for block-diagrams (rectangles) with I/O ports and their connections (wires). These block-diagrams can be hierarchical i.e some blocks may have some nested internal sub-structure etc. This topic is strongly related to classical graph-drawing, with the supplemental constraint of the need to take ports location into account, and possibly the shape of the blocks (rectangle of various sizes). Graphviz tools do not respond to the problem (at least my previous experiments have not been satisfactory). Force-directed algorithms retain my attention, but I have just found papers on classical directed graphs. Any hints ?

    Read the article

  • Generating a URL pattern when provided a set of 5 or so URLs

    - by ryan
    Provided with a set of URLs, I need to generate a pattern, For example: http://www.buy.com/prod/disney-s-star-struck/q/loc/109/213724402.html http://www.buy.com/prod/samsung-f2380-23-widescreen-1080p-lcd-monitor-150-000-1-dc-8ms-1920-x/q/loc/101/211249863.html http://www.buy.com/prod/panasonic-nnh765wf-microwave-oven-countertop-1-6-ft-1250w-panasonic/q/loc/66357/202045865.html http://www.buy.com/prod/escape-by-calvin-klein-for-women-3-4-oz-edp-spray/q/loc/66740/211210860.html http://www.buy.com/prod/v-touch-8gb-mp3-mp4-2-8-touch-screen-2mp-camera-expandable-minisd-w/q/loc/111/211402014.html Pattern is http://www.buy.com/prod/[^~]/q/loc/[^~].html

    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

  • Sparse (Pseudo) Infinite Grid Data Structure for Web Game

    - by Ming
    I'm considering trying to make a game that takes place on an essentially infinite grid. The grid is very sparse. Certain small regions of relatively high density. Relatively few isolated nonempty cells. The amount of the grid in use is too large to implement naively but probably smallish by "big data" standards (I'm not trying to map the Internet or anything like that) This needs to be easy to persist. Here are the operations I may want to perform (reasonably efficiently) on this grid: Ask for some small rectangular region of cells and all their contents (a player's current neighborhood) Set individual cells or blit small regions (the player is making a move) Ask for the rough shape or outline/silhouette of some larger rectangular regions (a world map or region preview) Find some regions with approximately a given density (player spawning location) Approximate shortest path through gaps of at most some small constant empty spaces per hop (it's OK to be a bad approximation often, but not OK to keep heading the wrong direction searching) Approximate convex hull for a region Here's the catch: I want to do this in a web app. That is, I would prefer to use existing data storage (perhaps in the form of a relational database) and relatively little external dependency (preferably avoiding the need for a persistent process). Guys, what advice can you give me on actually implementing this? How would you do this if the web-app restrictions weren't in place? How would you modify that if they were? Thanks a lot, everyone!

    Read the article

  • Best data-structure to use for two ended sorted list

    - by fmark
    I need a collection data-structure that can do the following: Be sorted Allow me to quickly pop values off the front and back of the list Remain sorted after I insert a new value Allow a user-specified comparison function, as I will be storing tuples and want to sort on a particular value Thread-safety is not required Optionally allow efficient haskey() lookups (I'm happy to maintain a separate hash-table for this though) My thoughts at this stage are that I need a priority queue and a hash table, although I don't know if I can quickly pop values off both ends of a priority queue. I'm interested in performance for a moderate number of items (I would estimate less than 200,000). Another possibility is simply maintaining an OrderedDictionary and doing an insertion sort it every-time I add more data to it. Furthermore, are there any particular implementations in Python. I would really like to avoid writing this code myself.

    Read the article

  • Maximum number of characters using keystrokes A, Ctrl+A, Ctrl+C and Ctrl+V

    - by munda
    This is an interview question from google. I am not able to solve it by myself. Can somebody shed some light? Write a program to print the sequence of keystrokes such that it generates the maximum number of character 'A's. You are allowed to use only 4 keys: A, Ctrl+A, Ctrl+C and Ctrl+V. Only N keystrokes are allowed. All Ctrl+ characters are considered as one keystroke, so Ctrl+A is one keystroke. For example, the sequence A, Ctrl+A, Ctrl+C, Ctrl+V generates two A's in 4 keystrokes. Ctrl+A is Select All Ctrl+C is Copy Ctrl+V is Paste I did some mathematics. For any N, using x numbers of A's , one Ctrl+A, one Ctrl+C and y Ctrl+V, we can generate max ((N-1)/2)2 number of A's. For some N M, it is better to use as many Ctrl+A's, Ctrl+C and Ctrl+V sequences as it doubles the number of A's. The sequence Ctrl+A, Ctrl+V, Ctrl+C will not overwrite the existing selection. It will append the copied selection to selected one.

    Read the article

  • F# - Facebook Hacker Cup - Double Squares

    - by Jacob
    I'm working on strengthening my F#-fu and decided to tackle the Facebook Hacker Cup Double Squares problem. I'm having some problems with the run-time and was wondering if anyone could help me figure out why it is so much slower than my C# equivalent. There's a good description from another post; Source: Facebook Hacker Cup Qualification Round 2011 A double-square number is an integer X which can be expressed as the sum of two perfect squares. For example, 10 is a double-square because 10 = 3^2 + 1^2. Given X, how can we determine the number of ways in which it can be written as the sum of two squares? For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2 + 3^2. You need to solve this problem for 0 = X = 2,147,483,647. Examples: 10 = 1 25 = 2 3 = 0 0 = 1 1 = 1 My basic strategy (which I'm open to critique on) is to; Create a dictionary (for memoize) of the input numbers initialzed to 0 Get the largest number (LN) and pass it to count/memo function Get the LN square root as int Calculate squares for all numbers 0 to LN and store in dict Sum squares for non repeat combinations of numbers from 0 to LN If sum is in memo dict, add 1 to memo Finally, output the counts of the original numbers. Here is the F# code (See code changes at bottom) I've written that I believe corresponds to this strategy (Runtime: ~8:10); open System open System.Collections.Generic open System.IO /// Get a sequence of values let rec range min max = seq { for num in [min .. max] do yield num } /// Get a sequence starting from 0 and going to max let rec zeroRange max = range 0 max /// Find the maximum number in a list with a starting accumulator (acc) let rec maxNum acc = function | [] -> acc | p::tail when p > acc -> maxNum p tail | p::tail -> maxNum acc tail /// A helper for finding max that sets the accumulator to 0 let rec findMax nums = maxNum 0 nums /// Build a collection of combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) let rec combos range = seq { let count = ref 0 for inner in range do for outer in Seq.skip !count range do yield (inner, outer) count := !count + 1 } let rec squares nums = let dict = new Dictionary<int, int>() for s in nums do dict.[s] <- (s * s) dict /// Counts the number of possible double squares for a given number and keeps track of other counts that are provided in the memo dict. let rec countDoubleSquares (num: int) (memo: Dictionary<int, int>) = // The highest relevent square is the square root because it squared plus 0 squared is the top most possibility let maxSquare = System.Math.Sqrt((float)num) // Our relevant squares are 0 to the highest possible square; note the cast to int which shouldn't hurt. let relSquares = range 0 ((int)maxSquare) // calculate the squares up front; let calcSquares = squares relSquares // Build up our square combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) for (sq1, sq2) in combos relSquares do let v = calcSquares.[sq1] + calcSquares.[sq2] // Memoize our relevant results if memo.ContainsKey(v) then memo.[v] <- memo.[v] + 1 // return our count for the num passed in memo.[num] // Read our numbers from file. //let lines = File.ReadAllLines("test2.txt") //let nums = [ for line in Seq.skip 1 lines -> Int32.Parse(line) ] // Optionally, read them from straight array let nums = [1740798996; 1257431873; 2147483643; 602519112; 858320077; 1048039120; 415485223; 874566596; 1022907856; 65; 421330820; 1041493518; 5; 1328649093; 1941554117; 4225; 2082925; 0; 1; 3] // Initialize our memoize dictionary let memo = new Dictionary<int, int>() for num in nums do memo.[num] <- 0 // Get the largest number in our set, all other numbers will be memoized along the way let maxN = findMax nums // Do the memoize let maxCount = countDoubleSquares maxN memo // Output our results. for num in nums do printfn "%i" memo.[num] // Have a little pause for when we debug let line = Console.Read() And here is my version in C# (Runtime: ~1:40: using System; using System.Collections.Generic; using System.Diagnostics; using System.IO; using System.Linq; using System.Text; namespace FBHack_DoubleSquares { public class TestInput { public int NumCases { get; set; } public List<int> Nums { get; set; } public TestInput() { Nums = new List<int>(); } public int MaxNum() { return Nums.Max(); } } class Program { static void Main(string[] args) { // Read input from file. //TestInput input = ReadTestInput("live.txt"); // As example, load straight. TestInput input = new TestInput { NumCases = 20, Nums = new List<int> { 1740798996, 1257431873, 2147483643, 602519112, 858320077, 1048039120, 415485223, 874566596, 1022907856, 65, 421330820, 1041493518, 5, 1328649093, 1941554117, 4225, 2082925, 0, 1, 3, } }; var maxNum = input.MaxNum(); Dictionary<int, int> memo = new Dictionary<int, int>(); foreach (var num in input.Nums) { if (!memo.ContainsKey(num)) memo.Add(num, 0); } DoMemoize(maxNum, memo); StringBuilder sb = new StringBuilder(); foreach (var num in input.Nums) { //Console.WriteLine(memo[num]); sb.AppendLine(memo[num].ToString()); } Console.Write(sb.ToString()); var blah = Console.Read(); //File.WriteAllText("out.txt", sb.ToString()); } private static int DoMemoize(int num, Dictionary<int, int> memo) { var highSquare = (int)Math.Floor(Math.Sqrt(num)); var squares = CreateSquareLookup(highSquare); var relSquares = squares.Keys.ToList(); Debug.WriteLine("Starting - " + num.ToString()); Debug.WriteLine("RelSquares.Count = {0}", relSquares.Count); int sum = 0; var index = 0; foreach (var square in relSquares) { foreach (var inner in relSquares.Skip(index)) { sum = squares[square] + squares[inner]; if (memo.ContainsKey(sum)) memo[sum]++; } index++; } if (memo.ContainsKey(num)) return memo[num]; return 0; } private static TestInput ReadTestInput(string fileName) { var lines = File.ReadAllLines(fileName); var input = new TestInput(); input.NumCases = int.Parse(lines[0]); foreach (var lin in lines.Skip(1)) { input.Nums.Add(int.Parse(lin)); } return input; } public static Dictionary<int, int> CreateSquareLookup(int maxNum) { var dict = new Dictionary<int, int>(); int square; foreach (var num in Enumerable.Range(0, maxNum)) { square = num * num; dict[num] = square; } return dict; } } } Thanks for taking a look. UPDATE Changing the combos function slightly will result in a pretty big performance boost (from 8 min to 3:45): /// Old and Busted... let rec combosOld range = seq { let rangeCache = Seq.cache range let count = ref 0 for inner in rangeCache do for outer in Seq.skip !count rangeCache do yield (inner, outer) count := !count + 1 } /// The New Hotness... let rec combos maxNum = seq { for i in 0..maxNum do for j in i..maxNum do yield i,j }

    Read the article

  • Is this a variation of the traveling salesman problem?

    - by Ville Koskinen
    I'm interested in a function of two word lists, which would return an order agnostic edit distance between them. That is, the arguments would be two lists of (let's say space delimited) words and return value would be the minimum sum of the edit (or Levenshtein) distances of the words in the lists. Distance between "cat rat bat" and "rat bat cat" would be 0. Distance between "cat rat bat" and "fat had bad" would be the same as distance between "rat bat cat" and "had fat bad", 4. In the case the number of words in the lists are not the same, the shorter list would be padded with 0-length words. My intuition (which hasn't been nurtured with computer science classes) does not find any other solution than to use brute force: |had|fat|bad| a solution ---+---+---+---+ +---+---+---+ cat| 2 | 1 | 2 | | | 1 | | ---+---+---+---+ +---+---+---+ rat| 2 | 1 | 2 | | 3 | | | ---+---+---+---+ +---+---+---+ bat| 2 | 1 | 1 | | | | 4 | ---+---+---+---+ +---+---+---+ Starting from the first row, pick a column and go to the next rows without ever revisiting a column you have already visited. Do this over and over again until you've tried all combinations. To me this sounds a bit like the traveling salesman problem. Is it, and how would you solve my particular problem?

    Read the article

  • Genetic/Evolutionary algorithms and local minima/maxima problem

    - by el.gringogrande
    I have run across several posts and articles that suggests using things like simulated annealing to avoid the local minima/maxima problem. I don't understand why this would be necessary if you started out with a sufficiently large random population. Is it just another check to insure that the initial population was, in fact, sufficiently large and random? Or are those techniques just an alternative to producing a "good" initial population?

    Read the article

< Previous Page | 90 91 92 93 94 95 96 97 98 99 100 101  | Next Page >