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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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• 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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• 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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• 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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• graph and all pairs shortest path in java

  - by Sandra

  I am writing a java program using Flyod-Warshall algorithm “All pairs shortest path”. I have written the following : a0 is the adjacency matrix of my graph, but has infinity instead of 0. vList is the list of vertexes and the cost for each edge is 1. Path[i][j] = k+1 means for going from I to j you first go to k then j int[][] path = new int[size][size]; for(int i = 0; i<path.length;i++) { for(int j = 0; j<path.length; j++) { if(adjM[i][j]==1) path[i][j]=j+1; } } //*************** for (int k = 0; k < vList.size(); k++) for (int i = 0; i < vList.size(); i++) for (int j = 0; j < vList.size(); j++) { if (a0[i][j]>a0[i][k]+ a0[k][j]) path[i][j] = k + 1; a0[i][j] = Math.min(a0[i][j], a0[i][k] + a0[k][j]); } After running this code, in the result a0 is correct, but path is not correct and I don’t know why!. Would you please help me?

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• Generate unique ID from multiple values with fault tolerance

  - by ojreadmore

  Given some values, I'd like to make a (pretty darn) unique result. $unique1 = generate(array('ab034', '981kja7261', '381jkfa0', 'vzcvqdx2993883i3ifja8', '0plnmjfys')); //now $unique1 == "sqef3452y"; I also need something that's pretty close to return the same result. In this case, 20% of the values is missing. $unique2 = generate(array('ab034', '981kja7261', '381jkfa0', 'vzcvqdx2993883i3ifja8')); //also $unique2 == "sqef3452y"; I'm not sure where to begin with such an algorithm but I have some assumptions. I assume that the more values given, the more accurate the resulting ID – in other words, using 20 values is better than 5. I also assume that a confidence factor can be calculated and adjusted. What would be nice to have is a weight factor where one can say 'value 1 is more important than value 3'. This would require a multidimensional array for input instead of one dimension. I just mashed on the keyboard for these values, but in practice they may be short or long alpha numeric values.

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• Reversing strings in a vector using for_each and bind

  - by fmuecke

  Hi! I was wandering how it's possible to reverese strings that are contained in a vector using a single for_each command just in one "simple" line. Yea, I know it is easy with a custom functor, but I can't accept, that it can't be done using bind (at least I couldn't do it). #include <vector> #include <string> #include <algorithm> std::vector<std::string> v; v.push_back("abc"); v.push_back("12345"); std::for_each(v.begin(), v.end(), /*call std::reverse for each element*/); Edit: Thanks a lot for those funtastic solutions. However, the solution for me was not to use the tr1::bind that comes with the Visual Studio 2008 feature pack/SP1. I don't know why it does not work like expected but that's the way it is (even MS admits that it's buggy). Maybe some hotfixes will help. With boost::bind everything works like desired and is so easy (but sometimes relly messy:)). I really should have tried boost::bind in the first place...

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• binary quicksort

  - by davit-datuashvili

  hi i want implement Binary quicksort algorithm from robert sedgewick book it looks like this public class quickb{ public static final int bitsword=32; public static void quicksortB(int a[],int l,int r,int d){ int i=l; int j=r-1; if (r<=l || d>bitsword) return ; while (j!=i) { while (digit(a[i],d)==0 && (i<j)) i++; while (digit(a[j],d)==1 && (j>i)) j++; int t=a[i]; a[i]=a[j]; a[j]=t; } if (digit(a[r-1],d)== 0) j++; quicksortB(a,l,j-1,d+1); quicksortB(a,j,r,d+1); } public static void main(String[]args){ int a[]=new int[]{4,7,3,9,8,2}; quicksortB(a,0,a.length-1,0); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } } public static int digit(int m,int d){ return (m>>d)&1; } } but it show me error: java.lang.ArrayIndexOutOfBoundsException: 6 at quickb.quicksortB(quickb.java:13) at quickb.main(quickb.java:32) what is wrong?

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• Iterative Reduction to Null Matrix

  - by user1459032

  Here's the problem: I'm given a matrix like Input: 1 1 1 1 1 1 1 1 1 At each step, I need to find a "second" matrix of 1's and 0's with no two 1's on the same row or column. Then, I'll subtract the second matrix from the original matrix. I will repeat the process until I get a matrix with all 0's. Furthermore, I need to take the least possible number of steps. I need to print all the "second" matrices in O(n) time. In the above example I can get to the null matrix in 3 steps by subtracting these three matrices in order: Expected output: 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1 0 0 I have coded an attempt, in which I am finding the first maximum value and creating the second matrices based on the index of that value. But for the above input I am getting 4 output matrices, which is wrong: My output: 1 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 My solution works for most of the test cases but fails for the one given above. Can someone give me some pointers on how to proceed, or find an algorithm that guarantees optimality? Test case that works: Input: 0 2 1 0 0 0 3 0 0 Output 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0

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• Generate regular expression to match strings from the list A, but not from list B

  - by Vlad

  I have two lists of strings ListA and ListB. I need to generate a regular expression that will match all strings in ListA and will not match any string in ListB. The strings could contain any combination of characters, numbers and punctuation. If a string appears on ListA it is guaranteed that it will not be in the ListB. If a string is not in either of these two lists I don't care what the result of the matching should be. The lists typically contain thousands of strings, and strings are fairly similar to each other. I know the trivial answer to this question, which is just generate a regular expression of the form (Str1)|(Str2)|(Str3) where StrN is the string from ListA. But I am looking for a more efficient way to do this. Ideal solution would be some sort of tool that will take two lists and generate a Java regular expression for this. Update 1: By "efficient", I mean to generate expression that is shorter than trivial solution. The ideal algorithm would generate the shorted possible expression. Here are some examples. ListA = { C10 , C15, C195 } ListB = { Bob, Billy } The ideal expression would be /^C1.+$/ Another example, note the third element of ListB ListA = { C10 , C15, C195 } ListB = { Bob, Billy, C25 } The ideal expression is /^C[^2]{1}.+$/ The last example ListA = { A , D ,E , F , H } ListB = { B , C , G , I } The ideal expression is the same as trivial solution which is /^(A|D|E|F|H)$/ Also, I am not looking for the ideal solution, anything better than trivial would help. I was thinking along the lines of generating the list of trivial solutions, and then try to merge the common substrings while watching that we don't wander into ListB territory. *Update 2: I am not particularly worried about the time it takes to generate the RegEx, anything under 10 minutes on the modern machine is acceptable

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• Fastest container or algorithm for unique reusable ids in C++

  - by gman

  I have a need for unique reusable ids. The user can choose his own ids or he can ask for a free one. The API is basically class IdManager { public: int AllocateId(); // Allocates an id void FreeId(int id); // Frees an id so it can be used again bool MarkAsUsed(int id); // Let's the user register an id. // returns false if the id was already used. }; Assume ids happen to start at 1 and progress, 2, 3, etc. This is not a requirement, just to help illustrate. IdManager mgr; mgr.MarkAsUsed(3); printf ("%d\n", mgr.AllocateId()); printf ("%d\n", mgr.AllocateId()); printf ("%d\n", mgr.AllocateId()); Would print 1 2 4 Because id 3 has already been declared used. What's the best container / algorithm to both remember which ids are used AND find a free id? If you want to know the a specific use case, OpenGL's glGenTextures, glBindTexture and glDeleteTextures are equivalent to AllocateId, MarkAsUsed and FreeId

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• Learning Java and logic using debugger. Did I cheat?

  - by centr0

  After a break from coding in general, my way of thinking logically faded (as if it was there to begin with...). I'm no master programmer. Intermediate at best. I decided to see if i can write an algorithm to print out the fibonacci sequence in Java. I got really frustrated because it was something so simple, and used the debugger to see what was going on with my variables. solved it in less than a minute with the help of the debugger. Is this cheating? When I read code either from a book or someone else's, I now find that it takes me a little more time to understand. If the alghorithm is complex (to me) i end up writing notes as to whats going on in the loop. A primitive debugger if you will. When you other programmers read code, do you also need to write things down as to whats the code doing? Or are you a genius and and just retain it?

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• Finding the closest match

  - by doublescorpio

  I Have an object with a set of parameters like: var obj = new {Param1 = 100; Param2 = 212; Param3 = 311; param4 = 11; Param5 = 290;} On the other side i have a list of object: var obj1 = new {Param1 = 1221; Param2 = 212; Param3 = 311; param4 = 11; Param5 = 290;} var obj3 = new {Param1 = 35; Param2 = 11; Param3 = 319; param4 = 211; Param5 = 790;} var obj4 = new {Param1 = 126; Param2 = 218; Param3 = 2; param4 = 6; Param5 = 190;} var obj5 = new {Param1 = 213; Param2 = 121; Param3 = 61; param4 = 11; Param5 = 29;} var obj7 = new {Param1 = 161; Param2 = 21; Param3 = 71; param4 = 51; Param5 = 232;} var obj9 = new {Param1 = 891; Param2 = 58; Param3 = 311; param4 = 21; Param5 = 590;} var obj11 = new {Param1 = 61; Param2 = 212; Param3 = 843; param4 = 89; Param5 = 210;} What is the best (easiest) algorithm to find the closest match for the first obj in the listed objects?

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• Given a vector of maximum 10 000 natural and distinct numbers, find 4 numbers(a, b, c, d) such that

  - by king_kong

  Hi, I solved this problem by following a straightforward but not optimal algorithm. I sorted the vector in descending order and after that substracted numbers from max to min to see if I get a + b + c = d. Notice that I haven't used anywhere the fact that elements are natural, distinct and 10 000 at most. I suppose these details are the key. Does anyone here have a hint over an optimal way of solving this? Thank you in advance! Later Edit: My idea goes like this: '<<quicksort in descending order>>' for i:=0 to count { // after sorting, loop through the array int d := v[i]; for j:=i+1 to count { int dif1 := d - v[j]; int a := v[j]; for k:=j+1 to count { if (v[k] > dif1) continue; int dif2 := dif1 - v[k]; b := v[k]; for l:=k+1 to count { if (dif2 = v[l]) { c := dif2; return {a, b, c, d} } } } } } What do you think?(sorry for the bad indentation)

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• Android - How to approach fall detection algorithm

  - by bobby123

  I want to be able to feature a fairly simple fall detection algorithm in my application. At the moment in onSensorChanged(), I am getting the absolute value of the current x,x,z values and subtracting SensorManager.GRAVITY_EARTH (9.8 m/s) from this. The resulting value has to be bigger than a threshold value 10 times in a row to set a flag saying a fall has been detected by the accelerometer, the threshold value is about 8m/s. Also I'm comparing the orientation of the phone as soon as the threshold has been passed and the orienation of it when the threshold is no longer being passed, this sets another flag saying the orientation sensor has detected a fall. When both flags are set, an event occurs to check is user ok, etc etc. My problem is with the threshold, when the phone is held straight up the absolute value of accelerometer is about 9.8 m/s, but when i hold it still at an angle it can be over 15m/s. This is causing other events to trigger the fall detection, and if i increase the threshold to avoid that, it won't detect falls. Can anyone give me some advice here with what possible values i should use or how to even improve my method? Many thanks.

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• The perverse hangman problem

  - by Shalmanese

  Perverse Hangman is a game played much like regular Hangman with one important difference: The winning word is determined dynamically by the house depending on what letters have been guessed. For example, say you have the board _ A I L and 12 remaining guesses. Because there are 13 different words ending in AIL (bail, fail, hail, jail, kail, mail, nail, pail, rail, sail, tail, vail, wail) the house is guaranteed to win because no matter what 12 letters you guess, the house will claim the chosen word was the one you didn't guess. However, if the board was _ I L M, you have cornered the house as FILM is the only word that ends in ILM. The challenge is: Given a dictionary, a word length & the number of allowed guesses, come up with an algorithm that either: a) proves that the player always wins by outputting a decision tree for the player that corners the house no matter what b) proves the house always wins by outputting a decision tree for the house that allows the house to escape no matter what. As a toy example, consider the dictionary: bat bar car If you are allowed 3 wrong guesses, the player wins with the following tree: Guess B NO -> Guess C, Guess A, Guess R, WIN YES-> Guess T NO -> Guess A, Guess R, WIN YES-> Guess A, WIN

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• Writing a recursive sorting algorithm of an array of integers

  - by 12345

  I am trying to write a recursive sorting algorithm for an array of integers. The following codes prints to the console: 3, 5, 2, 1, 1, 2, 6, 7, 8, 10, 20 The output should be sorted but somehow "it doesn't work". public static void main(String[] args) { int[] unsortedList = {20, 3, 1, 2, 1, 2, 6, 8, 10, 5, 7}; duplexSelectionSort(unsortedList, 0, unsortedList.length-1); for (int i = 0; i < unsortedList.length; i++) { System.out.println(unsortedList[i]); } } public static void duplexSelectionSort( int[] unsortedNumbers, int startIndex, int stopIndex) { int minimumIndex = 0; int maximumIndex = 0; if (startIndex < stopIndex) { int index = 0; while (index <= stopIndex) { if (unsortedNumbers[index] < unsortedNumbers[minimumIndex]) { minimumIndex = index; } if (unsortedNumbers[index] > unsortedNumbers[maximumIndex]) { maximumIndex = index; } index++; } swapEdges(unsortedNumbers, startIndex, stopIndex, minimumIndex, maximumIndex); duplexSelectionSort(unsortedNumbers, startIndex + 1, stopIndex - 1); } } public static void swapEdges( int[] listOfIntegers, int startIndex, int stopIndex, int minimumIndex, int maximumIndex) { if ((minimumIndex == stopIndex) && (maximumIndex == startIndex)) { swap(listOfIntegers, startIndex, stopIndex); } else { if (maximumIndex == startIndex) { swap(listOfIntegers, maximumIndex, stopIndex); swap(listOfIntegers, minimumIndex, startIndex); } else { swap(listOfIntegers, minimumIndex, startIndex); swap(listOfIntegers, maximumIndex, stopIndex); } } } public static void swap(int[] listOfIntegers, int index1, int index2) { int savedElementAtIndex1 = listOfIntegers[index1]; listOfIntegers[index1] = listOfIntegers[index2]; listOfIntegers[index2] = savedElementAtIndex1; }

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