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• Ideas Related to Subset Sum with 2,3 and more integers

  - by rolandbishop

  I've been struggling with this problem just like everyone else and I'm quite sure there has been more than enough posts to explain this problem. However in terms of understanding it fully, I wanted to share my thoughts and get more efficient solutions from all the great people in here related to Subset Sum problem. I've searched it over the Internet and there is actually a lot sources but I'm really willing to re-implement an algorithm or finding my own in order to understand fully. The key thing I'm struggling with is the efficiency considering the set size will be large. (I do not have a limit, just conceptually large). The two phases I'm trying to implement ideas on is finding two numbers that are equal to given integer T, finding three numbers and eventually K numbers. Some ideas I've though; For the two integer part I'm thing basically sorting the array O(nlogn) and for each element in the array searching for its negative value. (i.e if the array element is 3 searching for -3). Maybe a hash table inclusion could be better, providing a O(1) indexing the element? For the three or more integers I've found an amazing blog post;http://www.skorks.com/2011/02/algorithms-a-dropbox-challenge-and-dynamic-programming/. However even the author itself states that it is not applicable for large numbers. So I was for 2 and 3 and more integers what ideas could be applied for the subset problem. I'm struggling with setting up a dynamic programming method that will be efficient for the large inputs as well.

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• How do I create a graph from this datastructure?

  - by Shawn Mclean

  I took this data structure from this A* tutorial: public interface IHasNeighbours<N> { IEnumerable<N> Neighbours { get; } } public class Path<TNode> : IEnumerable<TNode> { public TNode LastStep { get; private set; } public Path<TNode> PreviousSteps { get; private set; } public double TotalCost { get; private set; } private Path(TNode lastStep, Path<TNode> previousSteps, double totalCost) { LastStep = lastStep; PreviousSteps = previousSteps; TotalCost = totalCost; } public Path(TNode start) : this(start, null, 0) { } public Path<TNode> AddStep(TNode step, double stepCost) { return new Path<TNode>(step, this, TotalCost + stepCost); } public IEnumerator<TNode> GetEnumerator() { for (Path<TNode> p = this; p != null; p = p.PreviousSteps) yield return p.LastStep; } IEnumerator IEnumerable.GetEnumerator() { return this.GetEnumerator(); } } I have no idea how to create a simple graph with. How do I add something like the following undirected graph using C#: Basically I'd like to know how to connect nodes. I have my own datastructures that I can already determine the neighbors and the distance. I'd now like to convert that into this posted datastructure so I can run it through the AStar algorithm. I was seeking something more like: Path<EdgeNode> startGraphNode = new Path<EdgeNode>(tempStartNode); startGraphNode.AddNeighbor(someOtherNode, distance);

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• Find all ways to insert zeroes into a bit pattern

  - by James

  I've been struggling to wrap my head around this for some reason. I have 15 bits that represent a number. The bits must match a pattern. The pattern is defined in the way the bits start out: they are in the most flush-right representation of that pattern. So say the pattern is 1 4 1. The bits will be: 000000010111101 So the general rule is, take each number in the pattern, create that many bits (1, 4 or 1 in this case) and then have at least one space separating them. So if it's 1 2 6 1 (it will be random): 001011011111101 Starting with the flush-right version, I want to generate every single possible number that meets that pattern. The # of bits will be stored in a variable. So for a simple case, assume it's 5 bits and the initial bit pattern is: 00101. I want to generate: 00101 01001 01010 10001 10010 10100 I'm trying to do this in Objective-C, but anything resembling C would be fine. I just can't seem to come up with a good recursive algorithm for this. It makes sense in the above example, but when I start getting into 12431 and having to keep track of everything it breaks down.

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• Ad distribution problem: an optimal solution?

  - by Mokuchan

  I'm asked to find a 2 approximate solution to this problem: You’re consulting for an e-commerce site that receives a large number of visitors each day. For each visitor i, where i € {1, 2 ..... n}, the site has assigned a value v[i], representing the expected revenue that can be obtained from this customer. Each visitor i is shown one of m possible ads A1, A2 ..... An as they enter the site. The site wants a selection of one ad for each customer so that each ad is seen, overall, by a set of customers of reasonably large total weight. Thus, given a selection of one ad for each customer, we will define the spread of this selection to be the minimum, over j = 1, 2 ..... m, of the total weight of all customers who were shown ad Aj. Example Suppose there are six customers with values 3, 4, 12, 2, 4, 6, and there are m = 3 ads. Then, in this instance, one could achieve a spread of 9 by showing ad A1 to customers 1, 2, 4, ad A2 to customer 3, and ad A3 to customers 5 and 6. The ultimate goal is to find a selection of an ad for each customer that maximizes the spread. Unfortunately, this optimization problem is NP-hard (you don’t have to prove this). So instead give a polynomial-time algorithm that approximates the maximum spread within a factor of 2. The solution I found is the following: Order visitors values in descending order Add the next visitor value (i.e. assign the visitor) to the Ad with the current lowest total value Repeat This solution actually seems to always find the optimal solution, or I simply can't find a counterexample. Can you find it? Is this a non-polinomial solution and I just can't see it?

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• minimum L sum in a mxn matrix - 2

  - by hilal

  Here is my first question about maximum L sum and here is different and hard version of it. Problem : Given a mxn *positive* integer matrix find the minimum L sum from 0th row to the m'th row . L(4 item) likes chess horse move Example : M = 3x3 0 1 2 1 3 2 4 2 1 Possible L moves are : (0 1 2 2), (0 1 3 2) (0 1 4 2) We should go from 0th row to the 3th row with minimum sum I solved this with dynamic-programming and here is my algorithm : 1. Take a mxn another Minimum L Moves Sum array and copy the first row of main matrix. I call it (MLMS) 2. start from first cell and look the up L moves and calculate it 3. insert it in MLMS if it is less than exists value 4. Do step 2. until m'th row 5. Choose the minimum sum in the m'th row Let me explain on my example step by step: M[ 0 ][ 0 ] sum(L1 = (0, 1, 2, 2)) = 5 ; sum(L2 = (0,1,3,2)) = 6; so MLMS[ 0 ][ 1 ] = 6 sum(L3 = (0, 1, 3, 2)) = 6 ; sum(L4 = (0,1,4,2)) = 7; so MLMS[ 2 ][ 1 ] = 6 M[ 0 ][ 1 ] sum(L5 = (1, 0, 1, 4)) = 6; sum(L6 = (1,3,2,4)) = 10; so MLMS[ 2 ][ 2 ] = 6 ... the last MSLS is : 0 1 2 4 3 6 6 6 6 Which means 6 is the minimum L sum that can be reach from 0 to the m. I think it is O(8*(m-1)*n) = O(m*n). Is there any optimal solution or dynamic-programming algorithms fit this problem? Thanks, sorry for long question

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• What is the complexity of this specialized sort

  - by ADB

  I would like to know the complexity (as in O(...) ) of the following sorting algorithm: there are B barrels that contains a total of N elements, spread unevenly across the barrels. the elements in each barrel are already sorted The sort takes combines all the elements from each barrel in a single sorted list: using an array of size B to store the last sorted element of each barrel (starting at 0) check each barrel (at the last stored index) and find the smallest element copy the element in the final sorted array, increment array counter increment the last sorted element for the barrel we picked from perform those steps N times or in pseudo: for i from 0 to N smallest = MAX_ELEMENT foreach b in B if bIndex[b] < b.length && b[bIndex[b]] < smallest smallest_barrel = b smallest = b[bIndex[b]] result[i] = smallest bIndex[smallest_barrel] += 1 I thought that the complexity would be O(n), but the problem I have with finding the complexity is that if B grows, it has a larger impact than if N grows because it adds another round in the B loop. But maybe that has no effect on the complexity?

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• How to convert between different currencies?

  - by sil3nt

  Hey there, this is part of a question i got in class, im at the final stretch but this has become a major problem. In it im given a certain value which is called the "gold value" and it is 40.5, this value changes in input. and i have these constants const int RUBIES_PER_DIAMOND = 5; // relative values. * const int EMERALDS_PER_RUBY = 2; const int GOLDS_PER_EMERALDS = 5; const int SILVERS_PER_GOLD = 4; const int COPPERS_PER_SILVER = 5; const int DIAMOND_VALUE = 50; // gold values. * const int RUBY_VALUE = 10; const int EMERALD_VALUE = 5; const float SILVER_VALUE = 0.25; const float COPPER_VALUE = 0.05; which means that basically for every diamond there are 5 rubies, and for every ruby there are 2 emeralds. So on and so forth. and the "gold value" for every diamond for example is 50 (diamond value = 50) this is how much one diamond is worth in golds. my problem is converting 40.5 into these diamonds and ruby values. I know the answer is 4rubies and 2silvers but how do i write the algorithm for this so that it gives the best estimate for every goldvalue that comes along?? please help!, im at my wits end

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• question about counting sort

  - by davit-datuashvili

  hi i have write following code which prints elements in sorted order only one big problem is that it use two additional array here is my code public class occurance{ public static final int n=5; public static void main(String[]args){ // n is maximum possible value what it should be in array suppose n=5 then array may be int a[]=new int[]{3,4,4,2,1,3,5};// as u see all elements are less or equal to n //create array a.length*n int b[]=new int[a.length*n]; int c[]=new int[b.length]; for (int i=0;i<b.length;i++){ b[i]=0; c[i]=0; } for (int i=0;i<a.length;i++){ if (b[a[i]]==1){ c[a[i]]=1; } else{ b[a[i]]=1; } } for (int i=0;i<b.length;i++){ if (b[i]==1) { System.out.println(i); } if (c[i]==1){ System.out.println(i); } } } } // 1 2 3 3 4 4 5 1.i have two question what is complexity of this algorithm?i mean running time 2. how put this elements into other array with sorted order? thanks

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• How can I evaluate the connectedness of my nodes?

  - by Travis Leleu

  I've got a space that has nodes that are all interconnected, based on a "similarity score". I would like to determine how "connected" a node is with the others. My purpose is to find nodes that are poorly connected to make sure that the backlink from the other node is prioritized. Perhaps an example would help. I've got a web page that links to my other pages based on a similarity score. Suppose I have the pages: A, B, C, ... A has a backlink from every other page, so it's very well connected. It also has links to all my other pages (each line in the graph is essentially bidirectional). B only has 1 backlink, from A. C has a link from A and D. I would like to make sure that the A-B link is prioritized over the A-C link (even if the similarity score between C and A is higher than B and A). In short, I would like to evaluate which nodes are least and best connected, so that I can mangle the results to my means. I believe this is Graph Connectedness, but I'm at a loss to develop a (simple) algorithm that will help me here. Simply counting the backlinks to a node may be a starting point -- but then how do I take the next step, which is to properly weight the links on the original node (A, in the example above)?

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• finding common prefix of array of strings

  - by bumperbox

  I have an array like this $sports = array( 'Softball - Counties', 'Softball - Eastern', 'Softball - North Harbour', 'Softball - South', 'Softball - Western' ); and i would like to find the longest common part of the string so in this instance, it would be 'Softball - '; I am thinking that I would follow the this process $i = 1; // loop to the length of the first string while ($i < strlen($sports[0]) { // grab the left most part up to i in length $match = substr($sports[0], 0, $i); // loop through all the values in array, and compare if they match foreach ($sports as $sport) { if ($match != substr($sport, 0, $i) { // didn't match, return the part that did match return substr($sport, 0, $i-1); } } // foreach // increase string length $i++; } // while // if you got to here, then all of them must be identical Questions is there a built in function or much simpler way of doing this ? for my 5 line array that is probably fine, but if i were to do several thousand line arrays, there would be a lot of overhead, so i would have to be move calculated with my starting values of $i, eg $i = halfway of string, if it fails, then $i/2 until it works, then increment $i by 1 until we succeed. so that we are doing the least number of comparisons to get a result If there a formula/algorithm out already out there for this kind of problem ? thanks alex

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• Need Help finding an appropriate task assignment algorithm for a college project involving coordinat

  - by Trif Mircea

  I am a long time lurker here and have found over time many answers regarding jquery and web development topics so I decided to ask a question of my own. This time I have to create a c++ project for college which should help manage the workflow of a company providing all kinds of services through in the field teams. The ideas I have so far are: client-server application; the server is a dispatcher where all the orders from clients get and the clients are mobile devices (PDAs) each team in the field having one a order from a client is a task. Each task is made up of a series of subtasks. You have a database with estimations on how long a task should take to complete you also know what tasks or subtasks each team on the field can perform based on what kind of specialists made up the team (not going to complicate the problem by adding needed materials, it is considered that if a member of a team can perform a subtask he has the stuff needed) Now knowing these factors, what would a good task assignment algorithm be? The criteria is: how many tasks can a team do, how many tasks they have in the queue, it could also be location, how far away are they from the place but I don't think I can implement that.. It needs to be efficient and also to adapt quickly is the human dispatcher manually assigns a task. Any help or leads would be really appreciated. Also I'm not 100% sure in the idea so if you have another way you would go about creating such an application please share, even if it just a quick outline. I have to write a theoretical part too so even if the ideas are far more complex that what i outlined that would be ok ; I'd write those and implement what I can. Edit: C++ is the only language I know unfortunately.

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• What algorithms compute directions from point A to point B on a map?

  - by A. Rex

  How do map providers (such as Google or Yahoo! Maps) suggest directions? I mean, they probably have real-world data in some form, certainly including distances but also perhaps things like driving speeds, presence of sidewalks, train schedules, etc. But suppose the data were in a simpler format, say a very large directed graph with edge weights reflecting distances. I want to be able to quickly compute directions from one arbitrary point to another. Sometimes these points will be close together (within one city) while sometimes they will be far apart (cross-country). Graph algorithms like Dijkstra's algorithm will not work because the graph is enormous. Luckily, heuristic algorithms like A* will probably work. However, our data is very structured, and perhaps some kind of tiered approach might work? (For example, store precomputed directions between certain "key" points far apart, as well as some local directions. Then directions for two far-away points will involve local directions to a key points, global directions to another key point, and then local directions again.) What algorithms are actually used in practice? PS. This question was motivated by finding quirks in online mapping directions. Contrary to the triangle inequality, sometimes Google Maps thinks that X-Z takes longer and is farther than using an intermediate point as in X-Y-Z. But maybe their walking directions optimize for another parameter, too? PPS. Here's another violation of the triangle inequality that suggests (to me) that they use some kind of tiered approach: X-Z versus X-Y-Z. The former seems to use prominent Boulevard de Sebastopol even though it's slightly out of the way. (Edit: this example doesn't work anymore, but did at the time of the original post. The one above still works as of early November 2009.)

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• Optimally place a pie slice in a rectangle.

  - by Lisa

  Given a rectangle (w, h) and a pie slice with start angle and end angle, how can I place the slice optimally in the rectangle so that it fills the room best (from an optical point of view, not mathematically speaking)? I'm currently placing the pie slice's center in the center of the rectangle and use the half of the smaller of both rectangle sides as the radius. This leaves plenty of room for certain configurations. Examples to make clear what I'm after, based on the precondition that the slice is drawn like a unit circle: A start angle of 0 and an end angle of PI would lead to a filled lower half of the rectangle and an empty upper half. A good solution here would be to move the center up by 1/4*h. A start angle of 0 and an end angle of PI/2 would lead to a filled bottom right quarter of the rectangle. A good solution here would be to move the center point to the top left of the rectangle and to set the radius to the smaller of both rectangle sides. This is fairly easy for the cases I've sketched but it becomes complicated when the start and end angles are arbitrary. I am searching for an algorithm which determines center of the slice and radius in a way that fills the rectangle best. Pseudo code would be great since I'm not a big mathematician.

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• Recursive code Sorting in VB

  - by Peter

  Ages old question: You have 2 hypothetical eggs, and a 100 story building to drop them from. The goal is to have the least number of guaranteed drops that will ensure you can find what floor the eggs break from the fall. You can only break 2 eggs. Using a 14 drop minimum method, I need help writing code that will allow me to calculate the following: Start with first drop attempt on 14th floor. If egg breaks then drop floors 1-13 to find the floor that causes break. ElseIf egg does not break then move up 13 floors to floor number 27 and drop again. If egg breaks then drop floors 15-26 starting on 15 working up to find the floor egg breaks on. ElseIf egg does not break then move up 12 floors to floor number 39 and drop again. etc. etc. The way this increases is as follows 14+13+12+11+10+9+8+7+6+5+4+3+2+1 So always adding to the previous value, by one less. I have never written a sorting algorithm before, and was curious how I might go about setting this up in a much more efficient way than a mile long of if then statements. My original idea was to store values for the floors in an array, and pull from that, using the index to move up or down and subtract or add to the variables. The most elegant solution would be a recursive function that handled this for any selected floor, 1-100, and ran the math, with an output that shows how many drops were needed in order to find that floor. Maximum is always 14, but some can be done in less.

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• How to store visited states in iterative deepening / depth limited search?

  - by colinfang

  Update: Search for the first solution. for a normal Depth First Search it is simple, just use a hashset bool DFS (currentState) = { if (myHashSet.Contains(currentState)) { return; } else { myHashSet.Add(currentState); } if (IsSolution(currentState) return true; else { for (var nextState in GetNextStates(currentState)) if (DFS(nextState)) return true; } return false; } However, when it becomes depth limited, i cannot simply do this bool DFS (currentState, maxDepth) = { if (maxDepth = 0) return false; if (myHashSet.Contains(currentState)) { return; } else { myHashSet.Add(currentState); } if (IsSolution(currentState) return true; else { for (var nextState in GetNextStates(currentState)) if (DFS(nextState, maxDepth - 1)) return true; } return false; } Because then it is not going to do a complete search (in a sense of always be able to find a solution if there is any) before maxdepth How should I fix it? Would it add more space complexity to the algorithm? Or it just doesn't require to memoize the state at all. Update: for example, a decision tree is the following: A - B - C - D - E - A | F - G (Goal) Starting from state A. and G is a goal state. Clearly there is a solution under depth 3. However, using my implementation under depth 4, if the direction of search happens to be A(0) -> B(1) -> C(2) -> D(3) -> E(4) -> F(5) exceeds depth, then it would do back track to A, however E is visited, it would ignore the check direction A - E - F - G

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• sum of square of each elements in the vector using for_each

  - by pierr

  Hi, As the function accepted by for_each take only one parameter (the element of the vector), I have to define a static int sum = 0 somewhere so that It can be accessed after calling the for_each . I think this is awkward. Any better way to do this (still use for_each) ? #include <algorithm> #include <vector> #include <iostream> using namespace std; static int sum = 0; void add_f(int i ) { sum += i * i; } void test_using_for_each() { int arr[] = {1,2,3,4}; vector<int> a (arr ,arr + sizeof(arr)/sizeof(arr[0])); for_each( a.begin(),a.end(), add_f); cout << "sum of the square of the element is " << sum << endl; } In Ruby, We can do it this way: sum = 0 [1,2,3,4].each { |i| sum += i*i} #local variable can be used in the callback function puts sum #=> 30 Would you please show more examples how for_each is typically used in practical programming (not just print out each element)? Is it possible use for_each simulate 'programming pattern' like map and inject in Ruby (or map /fold in Haskell). #map in ruby >> [1,2,3,4].map {|i| i*i} => [1, 4, 9, 16] #inject in ruby [1, 4, 9, 16].inject(0) {|aac ,i| aac +=i} #=> 30 EDIT: Thank you all. I have learned so much from your replies. We have so many ways to do the same single thing in C++ , which makes it a little bit difficult to learn. But it's interesting :)

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• Determining if an unordered vector<T> has all unique elements

  - by Hooked

  Profiling my cpu-bound code has suggested I that spend a long time checking to see if a container contains completely unique elements. Assuming that I have some large container of unsorted elements (with < and = defined), I have two ideas on how this might be done: The first using a set: template <class T> bool is_unique(vector<T> X) { set<T> Y(X.begin(), X.end()); return X.size() == Y.size(); } The second looping over the elements: template <class T> bool is_unique2(vector<T> X) { typename vector<T>::iterator i,j; for(i=X.begin();i!=X.end();++i) { for(j=i+1;j!=X.end();++j) { if(*i == *j) return 0; } } return 1; } I've tested them the best I can, and from what I can gather from reading the documentation about STL, the answer is (as usual), it depends. I think that in the first case, if all the elements are unique it is very quick, but if there is a large degeneracy the operation seems to take O(N^2) time. For the nested iterator approach the opposite seems to be true, it is lighting fast if X[0]==X[1] but takes (understandably) O(N^2) time if all the elements are unique. Is there a better way to do this, perhaps a STL algorithm built for this very purpose? If not, are there any suggestions eek out a bit more efficiency?

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• How to optimize shopping carts for minimal prices?

  - by tangens

  I have a list of items I want to buy. The items are offered by different shops and different prices. The shops have individual delivery costs. I'm looking for an optimal shopping strategy (and a java library supporting it) to purchase all of the items with a minimal total price. Example: Item1 is offered at Shop1 for $100, at Shop2 for $111. Item2 is offered at Shop1 for $90, at Shop2 for $85. Delivery cost of Shop1: $10 if total order < $150; $0 otherwise Delivery cost of Shop2: $5 if total order < $50; $0 otherwise If I buy Item1 and Item2 at Shop1 the total cost is $100 + $90 +$0 = $190. If I buy Item1 and Item2 at Shop2 the total cost is $111 + $85 +$0 = $196. If I buy Item1 at Shop1 and Item2 at Shop2 the total cost is $100 + $10 + $85 + $0 = 195. I get the minimal price if I order Item1 at Shop1 and Item2 at Shop2: $195 Question I need some hints which algorithms may help me to solve optimization problems of this kind for number of items about 100 and number of shops about 20. I already looked at apache-math and its optimization package, but I have no idea what algorithm to look for.

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• How can I find the common ancestor of two nodes in a binary tree?

  - by Siddhant

  The Binary Tree here is not a Binary Search Tree. Its just a Binary Tree. The structure could be taken as - struct node { int data; struct node *left; struct node *right; }; The maximum solution I could work out with a friend was something of this sort - Consider this binary tree (from http://lcm.csa.iisc.ernet.in/dsa/node87.html) : The inorder traversal yields - 8, 4, 9, 2, 5, 1, 6, 3, 7 And the postorder traversal yields - 8, 9, 4, 5, 2, 6, 7, 3, 1 So for instance, if we want to find the common ancestor of nodes 8 and 5, then we make a list of all the nodes which are between 8 and 5 in the inorder tree traversal, which in this case happens to be [4, 9, 2]. Then we check which node in this list appears last in the postorder traversal, which is 2. Hence the common ancestor for 8 and 5 is 2. The complexity for this algorithm, I believe is O(n) (O(n) for inorder/postorder traversals, the rest of the steps again being O(n) since they are nothing more than simple iterations in arrays). But there is a strong chance that this is wrong. :-) But this is a very crude approach, and I'm not sure if it breaks down for some case. Is there any other (possibly more optimal) solution to this problem?

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• Image Gurus: Optimize my Python PNG transparency function

  - by ozone

  I need to replace all the white(ish) pixels in a PNG image with alpha transparency. I'm using Python in AppEngine and so do not have access to libraries like PIL, imagemagick etc. AppEngine does have an image library, but is pitched mainly at image resizing. I found the excellent little pyPNG module and managed to knock up a little function that does what I need: make_transparent.py pseudo-code for the main loop would be something like: for each pixel: if pixel looks "quite white": set pixel values to transparent otherwise: keep existing pixel values and (assuming 8bit values) "quite white" would be: where each r,g,b value is greater than "240" AND each r,g,b value is within "20" of each other This is the first time I've worked with raw pixel data in this way, and although works, it also performs extremely poorly. It seems like there must be a more efficient way of processing the data without iterating over each pixel in this manner? (Matrices?) I was hoping someone with more experience in dealing with these things might be able to point out some of my more obvious mistakes/improvements in my algorithm. Thanks!

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• Determing if an unordered vector<T> has all unique elements

  - by Hooked

  Profiling my cpu-bound code has suggested I that spend a long time checking to see if a container contains completely unique elements. Assuming that I have some large container of unsorted elements (with < and = defined), I have two ideas on how this might be done: The first using a set: template <class T> bool is_unique(vector<T> X) { set<T> Y(X.begin(), X.end()); return X.size() == Y.size(); } The second looping over the elements: template <class T> bool is_unique2(vector<T> X) { typename vector<T>::iterator i,j; for(i=X.begin();i!=X.end();++i) { for(j=i+1;j!=X.end();++j) { if(*i == *j) return 0; } } return 1; } I've tested them the best I can, and from what I can gather from reading the documentation about STL, the answer is (as usual), it depends. I think that in the first case, if all the elements are unique it is very quick, but if there is a large degeneracy the operation seems to take O(N^2) time. For the nested iterator approach the opposite seems to be true, it is lighting fast if X[0]==X[1] but takes (understandably) O(N^2) time if all the elements are unique. Is there a better way to do this, perhaps a STL algorithm built for this very purpose? If not, are there any suggestions eek out a bit more efficiency?

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• Urgent help! how do i convert this?..

  - by sil3nt

  Hey there, this is part of a question i got in class, im at the final stretch but this has become a major problem. In it im given a certain value which is called the "gold value" and it is 40.5, this value changes in input. and i have these constants const int RUBIES_PER_DIAMOND = 5; // relative values. * const int EMERALDS_PER_RUBY = 2; const int GOLDS_PER_EMERALDS = 5; const int SILVERS_PER_GOLD = 4; const int COPPERS_PER_SILVER = 5; const int DIAMOND_VALUE = 50; // gold values. * const int RUBY_VALUE = 10; const int EMERALD_VALUE = 5; const float SILVER_VALUE = 0.25; const float COPPER_VALUE = 0.05; which means that basically for every diamond there are 5 rubies, and for every ruby there are 2 emeralds. So on and so forth. and the "gold value" for every diamond for example is 50 (diamond value = 50) this is how much one diamond is worth in golds. my problem is converting 40.5 into these diamonds and ruby values. I know the answer is 4rubies and 2silvers but how do i write the algorithm for this so that it gives the best estimate for every goldvalue that comes along?? please help!, im at my wits end

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• Modular Inverse and BigInteger division

  - by dano82

  I've been working on the problem of calculating the modular inverse of an large integer i.e. a^-1 mod n. and have been using BigInteger's built in function modInverse to check my work. I've coded the algorithm as shown in The Handbook of Applied Cryptography by Menezes, et al. Unfortunately for me, I do not get the correct outcome for all integers. My thinking is that the line q = a.divide(b) is my problem as the divide function is not well documented (IMO)(my code suffers similarly). Does BigInteger.divide(val) round or truncate? My assumption is truncation since the docs say that it mimics int's behavior. Any other insights are appreciated. This is the code that I have been working with: private static BigInteger modInverse(BigInteger a, BigInteger b) throws ArithmeticException { //make sure a >= b if (a.compareTo(b) < 0) { BigInteger temp = a; a = b; b = temp; } //trivial case: b = 0 => a^-1 = 1 if (b.equals(BigInteger.ZERO)) { return BigInteger.ONE; } //all other cases BigInteger x2 = BigInteger.ONE; BigInteger x1 = BigInteger.ZERO; BigInteger y2 = BigInteger.ZERO; BigInteger y1 = BigInteger.ONE; BigInteger x, y, q, r; while (b.compareTo(BigInteger.ZERO) == 1) { q = a.divide(b); r = a.subtract(q.multiply(b)); x = x2.subtract(q.multiply(x1)); y = y2.subtract(q.multiply(y1)); a = b; b = r; x2 = x1; x1 = x; y2 = y1; y1 = y; } if (!a.equals(BigInteger.ONE)) throw new ArithmeticException("a and n are not coprime"); return x2; }

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• Interview Q: sorting an almost sorted array (elements misplaced by no more than k)

  - by polygenelubricants

  I was asked this interview question recently: You're given an array that is almost sorted, in that each of the N elements may be misplaced by no more than k positions from the correct sorted order. Find a space-and-time efficient algorithm to sort the array. I have an O(N log k) solution as follows. Let's denote arr[0..n) to mean the elements of the array from index 0 (inclusive) to N (exclusive). Sort arr[0..2k) Now we know that arr[0..k) are in their final sorted positions... ...but arr[k..2k) may still be misplaced by k! Sort arr[k..3k) Now we know that arr[k..2k) are in their final sorted positions... ...but arr[2k..3k) may still be misplaced by k Sort arr[2k..4k) .... Until you sort arr[ik..N), then you're done! This final step may be cheaper than the other steps when you have less than 2k elements left In each step, you sort at most 2k elements in O(k log k), putting at least k elements in their final sorted positions at the end of each step. There are O(N/k) steps, so the overall complexity is O(N log k). My questions are: Is O(N log k) optimal? Can this be improved upon? Can you do this without (partially) re-sorting the same elements?

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• How to calculate the current index?

  - by niko

  Hi, I have written an algorithm which iteratively solves the problem. The first iteration consists of 6 steps and all the following iterations consist of 5 steps (first step is skipped). What I want to calculate is the current (local) step in the iteration from current global step. For example if there are 41 steps in total which means there are 8 iterations: indices from 1 to 6 belong to 1st iteration indices from 7 to 11 belong to second iteration ... For calculating the current iteration I have written the following code: if(currentStep <= 6) iteration = 1; else iteration = floor((currentStep - 7)/5) + 2; end The problem remains in calculating local steps. in first iteration the performed steps are: 1, 2, 3, 4, 5, 6 in all the following iterations the performing steps are 2, 3, 4, 5, 6 So what has to be done is to transform the array of global steps [1 2 3 4 5 6 7 8 9 10 11 12 13 ... 41] into array of local steps [1 2 3 4 5 6 2 3 4 5 6 2 3 ... 6]. I would appreciate if anyone could help in finding the solution to a given problem. Thank you!

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