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  • Print number series in java

    - by user1898282
    I have to print the series shown below in java: ***1*** **2*2** *3*3*3* 4*4*4*4 My current implementation is: public static void printSeries(int number,int numberOfCharsinEachLine){ String s="*"; for(int i=1;i<=number;i++){ int countOfs=(numberOfCharsinEachLine-(i)-(i-1))/2; if(countOfs<0){ System.out.println("Can't be done"); break; } for(int j=0;j<countOfs;j++){ System.out.print(s); } System.out.print(i); for(int k=1;k<i;k++){ System.out.print(s); System.out.print(i); } for(int j=0;j<countOfs;j++){ System.out.print(s); } System.out.println(); } } But there are lot of for loops, so I'm wondering whether this can be done in a better way or not?

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  • question about permutation problem

    - by davit-datuashvili
    i have posted similar problem here http://stackoverflow.com/questions/2920315/permutation-of-array but i want following we know that with length n there is n! possible permutation from which one such that all element are in order they are in sorted variant so i want break permutation when array is in order and print result but something is wrong i think that problem is repeated of permutation here is my code import java.util.*; public class permut{ public static Random r=new Random(); public static void display(int a[],int n){ for (int i=0;i<n;i++){ System.out.println(a[i]); } } public static void Permut(int a[],int n){ int j=0; int k=0; while (j<fact(n)){ int s=r.nextInt(n); for (int i=0;i<n;i++){ k=a[i]; a[i]=a[s]; a[s]=k; } j++; if (sorted(a,n)) display(a,n); break; } } public static void main(String[]args){ int a[]=new int[]{3,4,1,2}; int n=a.length; Permut(a,n); } public static int fact(int n){ if (n==0 || (n==1) ) return 1; return n*fact(n-1); } public static boolean sorted(int a[],int n ){ boolean flag=false; for (int i=0;i<n-1;i++){ if (a[i]<a[i+1]){ flag=true; } else{ flag=false; } } return flag; } } can anybody help me? result is nothing

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  • Optimizing / simplifying a path

    - by user146780
    Say I have a path with 150 nodes / verticies. How could I simplify if so that for example a straight line with 3 verticies, would remove the middle one since it does nothing to add to the path. Also how could I avoid destroying sharp corners? And how could I remove tiny variations and have smooth curves remaining. Thanks

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  • Special simple random number generator

    - by psihodelia
    How to create a function, which on every call generates a random integer number? This number must be most random as possible (according to uniform distribution). It is only allowed to use one static variable and at most 3 elementary steps, where each step consists of only one basic arithmetic operation of arity 1 or 2. Example: int myrandom(void){ static int x; x = some_step1; x = some_step2; x = some_step3; return x; } Basic arithmetic operations are +,-,%,and, not, xor, or, left shift, right shift, multiplication and division. Of course, no rand(), random() or similar staff is allowed.

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  • Collision free hash function for a specific data structure

    - by Max
    Is it possible to create collision free hash function for a data structure with specific properties. The datastructure is int[][][] It contains no duplicates The range of integers that are contained in it is defined. Let's say it's 0..1000, the maximal integer is definitely not greater than 10000. Big problem is that this hash function should also be very fast. Is there a way to create such a hash function? Maybe at run time depending on the integer range?

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  • [C++] std::tring manipulation: whitespace, "newline escapes '\'" and comments #

    - by rubenvb
    Kind of looking for affirmation here. I have some hand-written code, which I'm not shy to say I'm proud of, which reads a file, removes leading whitespace, processes newline escapes '\' and removes comments starting with #. It also removes all empty lines (also whitespace-only ones). Any thoughts/recommendations? I could probably replace some std::cout's with std::runtime_errors... but that's not a priority here :) const int RecipeReader::readRecipe() { ifstream is_recipe(s_buffer.c_str()); if (!is_recipe) cout << "unable to open file" << endl; while (getline(is_recipe, s_buffer)) { // whitespace+comment removeLeadingWhitespace(s_buffer); processComment(s_buffer); // newline escapes + append all subsequent lines with '\' processNewlineEscapes(s_buffer, is_recipe); // store the real text line if (!s_buffer.empty()) v_s_recipe.push_back(s_buffer); s_buffer.clear(); } is_recipe.close(); return 0; } void RecipeReader::processNewlineEscapes(string &s_string, ifstream &is_stream) { string s_temp; size_t sz_index = s_string.find_first_of("\\"); while (sz_index <= s_string.length()) { if (getline(is_stream,s_temp)) { removeLeadingWhitespace(s_temp); processComment(s_temp); s_string = s_string.substr(0,sz_index-1) + " " + s_temp; } else cout << "Error: newline escape '\' found at EOF" << endl; sz_index = s_string.find_first_of("\\"); } } void RecipeReader::processComment(string &s_string) { size_t sz_index = s_string.find_first_of("#"); s_string = s_string.substr(0,sz_index); } void RecipeReader::removeLeadingWhitespace(string &s_string) { const size_t sz_length = s_string.size(); size_t sz_index = s_string.find_first_not_of(" \t"); if (sz_index <= sz_length) s_string = s_string.substr(sz_index); else if ((sz_index > sz_length) && (sz_length != 0)) // "empty" lines with only whitespace s_string.clear(); } Some extra info: std::string s_buffer is a class data member, so is std::vector v_s_recipe. Any comment is welcome :)

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  • F# insert/remove item from list

    - by Timothy
    How should I go about removing a given element from a list? As an example, say I have list ['A'; 'B'; 'C'; 'D'; 'E'] and want to remove the element at index 2 to produce the list ['A'; 'B'; 'D'; 'E']? I've already written the following code which accomplishes the task, but it seems rather inefficient to traverse the start of the list when I already know the index. let remove lst i = let rec remove lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ t else remove t (lst' @ [h]) remove lst [] let myList = ['A'; 'B'; 'C'; 'D'; 'E'] let newList = remove myList 2 Alternatively, how should I insert an element at a given position? My code is similar to the above approach and most likely inefficient as well. let insert lst i x = let rec insert lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ [x] @ lst else insert t (lst' @ [h]) insert lst [] let myList = ['A'; 'B'; 'D'; 'E'] let newList = insert myList 2 'C'

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  • Automatic images translation to 3d model

    - by farrakhov-bulat
    I'm quite interested in automatic images translation to 3d models. Not really for commercial product, but from the point of possible academic research and implementation. What I'd like to achieve is almost transparent for user process of transformation series of images (fewer is better) to 3d model which might be shown in flash/silverlight/javafx or similar. Consider online furniture store with 3d models of all items in stock. Kinda cool to have ability to see the product in 3d before purchasing it. I managed to find a few pieces of software, like insight3d, but it couldn't be used in my case I guess. So, are there any similar projects or tips for me? If it would require to write that piece of software - I'd really love to dig into research on this field.

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  • Find a common element within N arrays

    - by kunjaan
    If I have N arrays, what is the best(Time complexity. Space is not important) way to find the common elements. You could just find 1 element and stop. Edit: The elements are all Numbers. Edit: These are unsorted. Please do not sort and scan. This is not a homework problem. Somebody asked me this question a long time ago. He was using a hash to solve the problem. I was thinking if SO has solved similar problems.

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  • GWT Calendrical Calculations

    - by Kyle Hayes
    We have a GWT application that needs to display various holidays. Is there a library available to do these calendrical calculations? If not, we'll have to do our own that we can ingest a set of rules to. Cheers

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  • Why is Dictionary.First() so slow?

    - by Rotsor
    Not a real question because I already found out the answer, but still interesting thing. I always thought that hash table is the fastest associative container if you hash properly. However, the following code is terribly slow. It executes only about 1 million iterations and takes more than 2 minutes of time on a Core 2 CPU. The code does the following: it maintains the collection todo of items it needs to process. At each iteration it takes an item from this collection (doesn't matter which item), deletes it, processes it if it wasn't processed (possibly adding more items to process), and repeats this until there are no items to process. The culprit seems to be the Dictionary.Keys.First() operation. The question is why is it slow? Stopwatch watch = new Stopwatch(); watch.Start(); HashSet<int> processed = new HashSet<int>(); Dictionary<int, int> todo = new Dictionary<int, int>(); todo.Add(1, 1); int iterations = 0; int limit = 500000; while (todo.Count > 0) { iterations++; var key = todo.Keys.First(); var value = todo[key]; todo.Remove(key); if (!processed.Contains(key)) { processed.Add(key); // process item here if (key < limit) { todo[key + 13] = value + 1; todo[key + 7] = value + 1; } // doesn't matter much how } } Console.WriteLine("Iterations: {0}; Time: {1}.", iterations, watch.Elapsed); This results in: Iterations: 923007; Time: 00:02:09.8414388. Simply changing Dictionary to SortedDictionary yields: Iterations: 499976; Time: 00:00:00.4451514. 300 times faster while having only 2 times less iterations. The same happens in java. Used HashMap instead of Dictionary and keySet().iterator().next() instead of Keys.First().

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  • Structure of Astar (A*) graph search data in C#

    - by Shawn Mclean
    How do you structure you graphs/nodes in a graph search class? I'm basically creating a NavMesh and need to generate the nodes from 1 polygon to the other. The edge that joins both polygons will be the node. I'll then run A* on these Nodes to calculate the shortest path. I just need to know how to structure my classes and their properties? I know for sure I wont need to create a fully blown undirected graph with nodes and edges.

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  • Solving a recurrence T(n) = 2T(n/2) + n^4

    - by user563454
    I am studying using the MIT Courseware and the CLRS book Introduction to Algorithms. Solving recurrence T(n) = 2T(n/2) + n4 (page 107) If I make a recurrence tree I get: level 0 n^4 level 1 2(n/2)^4 level 2 4(n/4)^4 level 3 8(n/8)^4 The tree has lg(n) levels. Therefore the recurrence is T(n) = Theta(lg(n)n^4)) But, If I use the Master method I get. Apply case 3: T(n) = Theta(n^4) If I apply the substitution method both seem to hold. Which one is ri?

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  • Scaling vectors from a center point?

    - by user146780
    I'm trying to figure out if I have points that make for example a square: * * * * and let's say I know the center of this square. I want a formula that will make it for eample twice its size but from the center * * * * * * * * Therefore the new shape is twice as large and from the center of the polygon. It has to work for any shape not just squares. I'm looking more for the theory behind it more than the implementation. Thanks

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  • question about siftdown operation on heap

    - by davit-datuashvili
    i have following pseudo code which execute siftdown operation on heap array suppose is x void siftdown(int n) pre heap(2,n) && n>=0 post heap(1,n) i=1; loop /*invariant heap(1,n) except perhaps between i and it's (0,1,or 2) children*/ c=2*i; if (c>n) break; // c is left child of i if (c+1)<=n /* c+1 is rigth child of i if (x[c+1]<x[c]) c++ /* c is lesser child of i if (x[i]<=x[c]) break; swap(c,i) i=c; i have wrote following code is it correct? public class siftdown{ public static void main(String[]args){ int c; int n=9; int a[]=new int[]{19,100,17,2,7,3,36,1,25}; int i=1; while (i<n){ c=2*i; if (c>n) break; //c is the left child of i if (c+1<=n) //c+1 ir rigth child of i if (a[c+1]<a[c]) c++; if (a[i]<=a[c]) break; int t=a[c]; a[c]=a[i]; a[i]=t; i=c; } for (int j=0;j<a.length;j++){ System.out.println(a[j]); } } } // result is 19 2 17 1 7 3 36 100 25

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  • ruby convert hundredth seconds to timestamp optimization

    - by Nik
    Hey! I want to convert "123456" to "00:20:34.56" where the two digits to the right of the decimal point is in hundredth of a second. So 00:00:00.99 + 00:00:00.01 = 00:00:01.00 What I have: def to_hmsc(cent) h = (cent/360000).floor cent -= h*360000 m = (cent/6000).floor cent -= m*6000 s = (cent/100).floor cent -= s*100 "#{h}:#{m}:#{s}.#{s}" end does this: to_hmsc("123456") #= "0:20:34.56" Question 1: I mean,this is ruby, I find the part ' cent -=... ' rather clunky. Can you see any way to shorten the entire process? Question 2: This has been asked a million times before, but please share whatever you've got: what's the shortest way to add leading zero to the digits. so that to_hmsc("123456") #= "00:20:34.56" Thanks!

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  • bidirectional buble sort

    - by davit-datuashvili
    Here is the code I'm using for shacker sort (or bidirectional buble sort). Something is wrong; the error is java.lang.ArrayIndexOutOfBoundsException. Can anybody help me? public class bidirectional{ public static void main(String[]args){ int x[]=new int[]{12,9,4,99,120,1,3,10}; int j; int n=x.length; int st=-1; while (st<n){ st++; n--; for (j=st;j<n;j++) { if (x[j]>x[j+1]) { int t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } for (j=n;--j>=st;) { if (x[j]>x[j+1]) { int t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } } for (int k=0;k<x.length;k++) { System.out.println(x[k]); } } }

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  • select i th smallest element from array

    - by davit-datuashvili
    i have divide and conqurer method to find i th smalles element from array here is code public class rand_select{ public static int Rand_partition( int a[],int p,int q,int i){ //smallest in a[p..q] if ( p==q) return a[p]; int r=partition (a,p,q); int k=r-p+1; if (i==k) return a[r]; if (i<k){ return Rand_partition(a,p,r-1,i); } return Rand_partition(a,r-1,q,i-k); } public static void main(String[]args){ int a[]=new int []{6,10,13,15,8,3,2,12}; System.out.println(Rand_partition(a,0,a.length-1,7)); } public static int partition(int a[],int p,int q){ int m=a[0]; while ( p<q){ while (p<q && a[p++] <m){ p++; } while (q>p && a[q--]>m){ q--; } int t=a[p]; a[p]=a[q]; a[q]=t; } int k=0; for (int i=0;i<a.length;i++){ if ( a[i]==m){ k=i; } } return k; } } but here is problem java.lang.ArrayIndexOutOfBoundsException please help me

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  • List circular group membership from active directory

    - by KAPes
    We have 40K+ groups in our active directory and we are increasingly facing problem of circular nested groups which are creating problems for some applications. Does anyone know how to list down the full route through which a circular group membership exists ? e.g. G1 --> G2 --> G3 --> G4 --> G1 How do I list it down.

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  • Find all numbers that appear in each of a set of lists

    - by Ankur
    I have several ArrayLists of Integer objects, stored in a HashMap. I want to get a list (ArrayList) of all the numbers (Integer objects) that appear in each list. My thinking so far is: Iterate through each ArrayList and put all the values into a HashSet This will give us a "listing" of all the values in the lists, but only once Iterate through the HashSet 2.1 With each iteration perform ArrayList.contains() 2.2 If none of the ArrayLists return false for the operation add the number to a "master list" which contains all the final values. If you can come up with something faster or more efficient, funny thing is as I wrote this I came up with a reasonably good solution. But I'll still post it just in case it is useful for someone else. But of course if you have a better way please do let me know.

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  • Routing algorithm

    - by isaac
    Hello, I'm giving a presentation about computer routing and I want to make a good analogy with a real-world situation. However, I could not find it. Do you have in mind any of the situations like the computer routing. If yes, could you please provide me with it

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  • Reduce text length to fit cell width in a smart manner

    - by Andrei Ciobanu
    Hello, I am in project where we are building a simple web calendar using Java EE technologies. We define a table where every row is an employee, and every column represents an hour interval. The table width and column widths are adjustable. In every cell we have a text retrieved from a database, indicating what the employee is doing / should do in that time interval. The problem is that sometimes the text in cells is getting bigger than the actual cell. My task is to make the text more "readable" by reducing it's length in a "smart way" so that it can fit in the cell more "gracefully". For example if initially in a cell I have: "Writing documents", after the resize I should retrieve: "Wrtng. dcmnts" or "Writ. docum." so that the text can fit well. Is there a smart way to do it ? Or removing vocals / split the string in two is enough ?

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  • finding middle element of an array

    - by senthil
    Hi all, I came cross a question in my interview. Question: Array of integers will be given as the input and you should find out the middle element when sorted , but without sorting. For Example. Input: 1,3,5,4,2 Output: 3 When you sort the given input array, it will be 1,2,3,4,5 where middle element is 3. You should find this in one pass without sorting. Any solutions for this?

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  • Simple Big O with lg(n) proof

    - by halohunter
    I'm attempting to guess and prove the Big O for: f(n) = n^3 - 7n^2 + nlg(n) + 10 I guess that big O is n^3 as it is the term with the largest order of growth However, I'm having trouble proving it. My unsuccesful attempt follows: f(n) <= cg(n) f(n) <= n^3 - 7n^2 + nlg(n) + 10 <= cn^3 f(n) <= n^3 + (n^3)*lg(n) + 10n^3 <= cn^3 f(n) <= N^3(11 + lg(n)) <= cn^3 so 11 + lg(n) = c But this can't be right because c must be constant. What am I doing wrong?

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