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  • Reverse factorial

    - by dada
    Well, we all know that if N is given it's easy to calculate N!. But what about reversing? N! is given and you are about to find N - Is that possible ? I'm curious.

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  • randomized quicksort: probability of two elements comparison?

    - by bantu
    I am reading "Probability and Computing" by M.Mitzenmacher, E.Upfal and I have problems understanding how the probability of comparison of two elements is calculated. Input: the list (y1,y2,...,YN) of numbers. We are looking for pivot element. Question: what is probability that two elements yi and yj (ji) will be compared? Answer (from book): yi and yj will be compared if either yi or yj will be selected as pivot in first draw from sequence (yi,yi+1,...,yj-1,yj). So the probablity is: 2/(y-i+1). The problem for me is initial claim: for example, picking up yi in the first draw from the whole list will cause the comparison with yj (and vice-versa) and the probability is 2/n. So, rather the "reverse" claim is true -- none of the (yi+1,...,yj-1) elements can be selected beforeyi or yj, but the "pool" size is not fixed (in first draw it is n for sure, but on the second it is smaller). Could someone please explain this, how the authors come up with such simplified conclusion? Thank you in advance

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  • C# - split String into smaller Strings by length variable

    - by tyndall
    I'd like to break apart a String by a certain length variable. It needs to bounds check so as not explode when the last section of string is not as long as or longer than the length. Looking for the most succinct (yet understandable) version. Example: string x = "AAABBBCC"; string[] arr = x.SplitByLength(3); // arr[0] -> "AAA"; // arr[1] -> "BBB"; // arr[2] -> "CC"

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  • What's the best general programming book to review basic development concepts?

    - by Charles S.
    I'm looking for for a programming book that reviews basic concepts like implementing linked lists, stacks, queues, hash tables, tree traversals, search algorithms, etc. etc. Basically, I'm looking for a review of everything I learned in college but have forgotten. I prefer something written in the last few years that includes at least a decent amount of code in object-oriented languages. This is to study for job interview questions but I already have the "solving interview questions" books. I'm looking for something with a little more depth and explanation. Any good recommendations?

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  • C++ string array binary search

    - by Jose Vega
    string Haystack[] = { "Alabama", "Alaska", "American Samoa", "Arizona", "Arkansas", "California", "Colorado", "Connecticut", "Delaware", "District of Columbia", "Florida", "Georgia", "Guam", "Hawaii", "Idaho", "Illinois", "Indiana", "Iowa", "Kansas", "Kentucky", "Louisiana", "Maine", "Maryland", "Massachusetts", "Michigan", "Minnesota", "Mississippi", "Missouri", "Montana", "Nebraska", "Nevada", "New Hampshire", "New Jersey", "New Mexico", "New York", "North Carolina", "North Dakota", "Northern Mariana Islands", "Ohio", "Oklahoma", "Oregon", "Pennsylvania", "Puerto Rico", "Rhode Island", "South Carolina", "South Dakota", "Tennessee", "Texas", "US Virgin Islands", "Utah", "Vermont", "Virginia", "Washington", "West Virginia", "Wisconsin", "Wyoming"}; string Needle = "Virginia"; if(std::binary_search(Haystack, Haystack+56, Needle)) cout<<"Found"; If I also wanted to find the location of the needle in the string array, is there an "easy" way to find out?

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  • KD-Trees and missing values (vector comparison)

    - by labratmatt
    I have a system that stores vectors and allows a user to find the n most similar vectors to the user's query vector. That is, a user submits a vector (I call it a query vector) and my system spits out "here are the n most similar vectors." I generate the similar vectors using a KD-Tree and everything works well, but I want to do more. I want to present a list of the n most similar vectors even if the user doesn't submit a complete vector (a vector with missing values). That is, if a user submits a vector with three dimensions, I still want to find the n nearest vectors (stored vectors are of 11 dimensions) I have stored. I have a couple of obvious solutions, but I'm not sure either one seem very good: Create multiple KD-Trees each built using the most popular subset of dimensions a user will search for. That is, if a user submits a query vector of thee dimensions, x, y, z, I match that query to my already built KD-Tree which only contains vectors of three dimensions, x, y, z. Ignore KD-Trees when a user submits a query vector with missing values and compare the query vector to the vectors (stored in a table in a DB) one by one using something like a dot product. This has to be a common problem, any suggestions? Thanks for the help.

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  • How To Generate Parameter Set for the Diffie-Hellman Key Agreement Algorithm in Android

    - by sebby_zml
    Hello everyone, I am working on mobile/server security related project. I am now stuck in generating a Diffie-Hellman key agreement part. It works fine in server side program but it is not working in mobile side. Thus, I assume that it is not compactible with Android. I used the following class to get the parameters. It returns a comma-separated string of 3 values. The first number is the prime modulus P. The second number is the base generator G. The third number is bit size of the random exponent L. My question is is there anything wrong with the code or it is not compactible for android?What kind of changes should I do? Your suggestion and guidance would be very much help for me. Thanks a lot in advance. public static String genDhParams() { try { // Create the parameter generator for a 1024-bit DH key pair AlgorithmParameterGenerator paramGen = AlgorithmParameterGenerator.getInstance("DH"); paramGen.init(1024); // Generate the parameters AlgorithmParameters params = paramGen.generateParameters(); DHParameterSpec dhSpec = (DHParameterSpec)params.getParameterSpec(DHParameterSpec.class); // Return the three values in a string return ""+dhSpec.getP()+","+dhSpec.getG()+","+dhSpec.getL(); } catch (NoSuchAlgorithmException e) { } catch (InvalidParameterSpecException e) { } return null; } Regards, Sebby

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  • open source smooth particle hydrodynamics

    - by user325181
    Anyone know of any open source libraries for particle based large scale smooth particle hydrodynamics. I am looking for a easier way of simulating large scale planetary body impacts with rotation. I was also wondering if you had any ideas on how to visualize the output from said simulation. I have tried using IBM graphviz, but it is very difficult to work with. Any pointers would be appreciated. Thanks!

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  • curious ill conditioned numerical problem

    - by aaa
    hello. somebody today showed me this curious ill conditioned problem (apparently pretty famous), which looks relatively simple ƒ = (333.75 - a^2)b^6 + a^2 (11a^2 b^2 - 121b^4 - 2) + 5.5b^8 + a/(2^b) where a = 77617 and b = 33096 can you determine correct answer?

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  • Pruning: When to Stop?

    - by cam
    When does pruning stop being efficient in a depth-first search? I've been working on an efficient method to solve the N-Queens problem and I'm looking at pruning for the first time. I've implemented it for the first two rows, but when does it stop being efficient? How far should I prune to?

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  • Deletion procedure for a Binary Search Tree

    - by Metz
    Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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  • Url shortening algorithm

    - by Bozho
    Now, this is not strictly about URL shortening, but my purpose is such anyway, so let's view it like that. Of course the steps to URL shortening are: Take the full URL Generate a unique short string to be the key for the URL Store the URL and the key in a database (a key-value store would be a perfect match here) Now, about the 2nd point. Here's what I've come up with: ByteArrayOutputStream baos = new ByteArrayOutputStream(); DataOutputStream dos = new DataOutputStream(baos); UUID uuid = UUID.randomUUID(); dos.writeLong(uuid.getMostSignificantBits()); String encoded = new String(Base64.encodeBase64(baos.toByteArray()), "ISO-8859-1"); String shortUrl = StringUtils.left(6); // returns the leftmost 6 characters // check if exists in database, repeat until it does not I wonder if this is good enough. Is it?

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  • How to find the insertion point in an array using binary search?

    - by ????
    The basic idea of binary search in an array is simple, but it might return an "approximate" index if the search fails to find the exact item. (we might sometimes get back an index for which the value is larger or smaller than the searched value). For looking for the exact insertion point, it seems that after we got the approximate location, we might need to "scan" to left or right for the exact insertion location, so that, say, in Ruby, we can do arr.insert(exact_index, value) I have the following solution, but the handling for the part when begin_index >= end_index is a bit messy. I wonder if a more elegant solution can be used? (this solution doesn't care to scan for multiple matches if an exact match is found, so the index returned for an exact match may point to any index that correspond to the value... but I think if they are all integers, we can always search for a - 1 after we know an exact match is found, to find the left boundary, or search for a + 1 for the right boundary.) My solution: DEBUGGING = true def binary_search_helper(arr, a, begin_index, end_index) middle_index = (begin_index + end_index) / 2 puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " + "begin_index = #{begin_index}, end_index = #{end_index}, " + "middle_index = #{middle_index}" if DEBUGGING if arr[middle_index] == a return middle_index elsif begin_index >= end_index index = [begin_index, end_index].min return index if a < arr[index] && index >= 0 #careful because -1 means end of array index = [begin_index, end_index].max return index if a < arr[index] && index >= 0 return index + 1 elsif a > arr[middle_index] return binary_search_helper(arr, a, middle_index + 1, end_index) else return binary_search_helper(arr, a, begin_index, middle_index - 1) end end # for [1,3,5,7,9], searching for 6 will return index for 7 for insertion # if exact match is found, then return that index def binary_search(arr, a) puts "\nSearching for #{a} in #{arr}" if DEBUGGING return 0 if arr.empty? result = binary_search_helper(arr, a, 0, arr.length - 1) puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING return result end arr = [1,3,5,7,9] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [] b = 60 arr.insert(binary_search(arr, b), b) p arr and result: Searching for 6 in [1, 3, 5, 7, 9] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9] Searching for 6 in [1, 3, 5, 7, 9, 11] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9, 11] Searching for 60 in [1, 3, 5, 7, 9] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4 the result is 5, the index for value nil [1, 3, 5, 7, 9, 60] Searching for 60 in [1, 3, 5, 7, 9, 11] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5 the result is 6, the index for value nil [1, 3, 5, 7, 9, 11, 60] Searching for -60 in [1, 3, 5, 7, 9] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9] Searching for -60 in [1, 3, 5, 7, 9, 11] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9, 11] Searching for -60 in [1] a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 0, the index for value 1 [-60, 1] Searching for 60 in [1] a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 1, the index for value nil [1, 60] Searching for 60 in [] [60]

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  • Code bacteria: evolving mathematical behavior

    - by Stefano Borini
    It would not be my intention to put a link on my blog, but I don't have any other method to clarify what I really mean. The article is quite long, and it's in three parts (1,2,3), but if you are curious, it's worth the reading. A long time ago (5 years, at least) I programmed a python program which generated "mathematical bacteria". These bacteria are python objects with a simple opcode-based genetic code. You can feed them with a number and they return a number, according to the execution of their code. I generate their genetic codes at random, and apply an environmental selection to those objects producing a result similar to a predefined expected value. Then I let them duplicate, introduce mutations, and evolve them. The result is quite interesting, as their genetic code basically learns how to solve simple equations, even for values different for the training dataset. Now, this thing is just a toy. I had time to waste and I wanted to satisfy my curiosity. however, I assume that something, in terms of research, has been made... I am reinventing the wheel here, I hope. Are you aware of more serious attempts at creating in-silico bacteria like the one I programmed? Please note that this is not really "genetic algorithms". Genetic algorithms is when you use evolution/selection to improve a vector of parameters against a given scoring function. This is kind of different. I optimize the code, not the parameters, against a given scoring function.

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  • VBA compare and sort strings with quirky characters

    - by Smandoli
    I am comparing text values from two DAO recordsets in MS Access. I sort on the text field, then go through both recordsets comparing the values from each. The sets are substantially different and while they're mostly alpha-numeric, spaces and symbols like hyphens and periods are very common. My program depends on predictable sorting and fool-proof comparing. But unfortunately, the sort will rank two values differently than the comparison function. StrComp is the obvious first choice: varResult = StrComp(Val_1, Val_2) RFA-300 14.9044 RFA300 14-2044 But for the two pairs above, StrComp returns a different value than one would expect based on the sort. Including vbTextCompare or vbBinaryCompare affects StrComp's result, but not so as to solve the problem. Note the values must always be compared as strings. Of course I make sure that "14-2044" and "14.9044" aren't evaluated as -2030 and ~15. That's not the cause of my problem. I learned API-based functions are more reliable for quirky texts, so I tried these: varResult = CompareString(LOCALE_SYSTEM_DEFAULT, _ SORT_STRINGSORT, strVal_2, -1, strVal_1, -1) varResult = CompareString(LOCALE_SYSTEM_DEFAULT, _ NORM_IGNOREWIDTH, strVal_2, -1, strVal_1, -1) The first one returns the opposite of StrComp. The second one returns the same as StrComp. But neither yields a result that is consistent with the sort order. (NORM_IGNOREWIDTH is probably not relevant, but I needed a place-holder substitute and it looked as good as any.) UPDATE: This is a complete rewrite of the original post, deleting all the info about why I really need this -- just take my word for it and enjoy the brevity.

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  • Problem implementing sorting algorithm in C with an array of structs

    - by dilog
    Well here is my little problem, first my code: struct alumn { char name[100]; char lastname[100]; int par; int nota; }; typedef struct alumn alumn; int bubble(alumn **arr, int length) { int i,j; alumn *temp; for (i=0; i<=length-2; i++) { for (j=i+1; j<=length-1;j++) { if ((*arr)[i].nota > (*arr)[j].nota) { temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } } } int main(int argc, char **argv) { alumn *alumns; ... here goes some other code ... bubble(&alumns,totalAlumns); return 0; } My problem is that this algorith is not sorting anything. I'm having a hard time doing the swap, i tried everything but nothing works :( . Any help??? struct alumn { char name[100]; char lastname[100]; int par; int nota; }; typedef struct alumn alumn; int bubble(alumn **arr, int length) { int i,j; alumn *temp; for (i=0; i<=length-2; i++) { for (j=i+1; j<=length-1;j++) { if ((*arr)[i].nota > (*arr)[j].nota) { temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } } } int main(int argc, char **argv) { alumn *alumns; ... here goes some other code ... bubble(&alumns,totalAlumns); return 0; } My problem is that this algorith is not sorting anything. I'm having a hard time doing the swap, i tried everything but nothing works :( . Any help???

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  • Reduce number of points in line

    - by culebrón
    I'm searching for algorithms to reduce the LOD of polylines, lines (looped or not) of nodes. In simple words, I want to take hi-resolution coastline data and be able to reduce its LOD hundred- or thousandfold to render it in small-scale. I found polygon reduction algorithms (but they require triangles) and Laplacian smoothing, but that doesn't seem exactly what I need.

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  • Need Urgent Help! Find a repeated numbers out of 3 boxes

    - by james1
    Hi guys, I really need help with finding a repeated number out of 3 boxes. Let's say that i have 3 boxes, each box contain 10 piece of numbered paper (1 - 10) but there is a number the same in all 3 boxes eg: box1 has number 4 and box2 has number 4 and box3 also has number 4. How to find that repeated number in java with an efficient/fastest way possible? Thanks in advance!

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  • Clojure - tail recursive sieve of Eratosthenes

    - by Konrad Garus
    I have this implementation of the sieve of Eratosthenes in Clojure: (defn sieve [n] (loop [last-tried 2 sift (range 2 (inc n))] (if (or (nil? last-tried) (> last-tried n)) sift (let [filtered (filter #(or (= % last-tried) (< 0 (rem % last-tried))) sift)] (let [next-to-try (first (filter #(> % last-tried) filtered))] (recur next-to-try filtered)))))) For larger n (like 20000) it ends with stack overflow. Why doesn't tail call elimination work here? How to fix it?

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  • How to combine elements of a list

    - by Addie
    I'm working in c#. I have a sorted List of structures. The structure has a DateTime object which stores month and year and an integer which stores a value. The list is sorted by date. I need to traverse the list and combine it so that I only have one instance of the structure per date. For example: My initial list would look like this: { (Apr10, 3), (Apr10, 2), (Apr10, -3), (May10, 1), (May10, 1), (May10, -3), (Jun10, 3) } The resulting list should look like this: { (Apr10, 2), (May10, -1), (Jun10, 3) } I'm looking for a simple / efficient solution. The struct is: class CurrentTrade { public DateTime date; public int dwBuy; } The list is: private List<CurrentTrade> FillList

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  • Array sorting efficiency... Beginner need advice

    - by SoleSoft
    I'll start by saying I am very much a beginner programmer, this is essentially my first real project outside of using learning material. I've been making a 'Simon Says' style game (the game where you repeat the pattern generated by the computer) using C# and XNA, the actual game is complete and working fine but while creating it, I wanted to also create a 'top 10' scoreboard. The scoreboard would record player name, level (how many 'rounds' they've completed) and combo (how many buttons presses they got correct), the scoreboard would then be sorted by combo score. This led me to XML, the first time using it, and I eventually got to the point of having an XML file that recorded the top 10 scores. The XML file is managed within a scoreboard class, which is also responsible for adding new scores and sorting scores. Which gets me to the point... I'd like some feedback on the way I've gone about sorting the score list and how I could have done it better, I have no other way to gain feedback =(. I know .NET features Array.Sort() but I wasn't too sure of how to use it as it's not just a single array that needs to be sorted. When a new score needs to be entered into the scoreboard, the player name and level also have to be added. These are stored within an 'array of arrays' (10 = for 'top 10' scores) scoreboardComboData = new int[10]; // Combo scoreboardTextData = new string[2][]; scoreboardTextData[0] = new string[10]; // Name scoreboardTextData[1] = new string[10]; // Level as string The scoreboard class works as follows: - Checks to see if 'scoreboard.xml' exists, if not it creates it - Initialises above arrays and adds any player data from scoreboard.xml, from previous run - when AddScore(name, level, combo) is called the sort begins - Another method can also be called that populates the XML file with above array data The sort checks to see if the new score (combo) is less than or equal to any recorded scores within the scoreboardComboData array (if it's greater than a score, it moves onto the next element). If so, it moves all scores below the score it is less than or equal to down one element, essentially removing the last score and then places the new score within the element below the score it is less than or equal to. If the score is greater than all recorded scores, it moves all scores down one and inserts the new score within the first element. If it's the only score, it simply adds it to the first element. When a new score is added, the Name and Level data is also added to their relevant arrays, in the same way. What a tongue twister. Below is the AddScore method, I've added comments in the hope that it makes things clearer O_o. You can get the actual source file HERE. Below the method is an example of the quickest way to add a score to follow through with a debugger. public static void AddScore(string name, string level, int combo) { // If the scoreboard has not yet been filled, this adds another 'active' // array element each time a new score is added. The actual array size is // defined within PopulateScoreBoard() (set to 10 - for 'top 10' if (totalScores < scoreboardComboData.Length) totalScores++; // Does the scoreboard even need sorting? if (totalScores > 1) { for (int i = totalScores - 1; i > - 1; i--) { // Check to see if score (combo) is greater than score stored in // array if (combo > scoreboardComboData[i] && i != 0) { // If so continue to next element continue; } // Check to see if score (combo) is less or equal to element 'i' // score && that the element is not the last in the // array, if so the score does not need to be added to the scoreboard else if (combo <= scoreboardComboData[i] && i != scoreboardComboData.Length - 1) { // If the score is lower than element 'i' and greater than the last // element within the array, it needs to be added to the scoreboard. This is achieved // by moving each element under element 'i' down an element. The new score is then inserted // into the array under element 'i' for (int j = totalScores - 1; j > i; j--) { // Name and level data are moved down in their relevant arrays scoreboardTextData[0][j] = scoreboardTextData[0][j - 1]; scoreboardTextData[1][j] = scoreboardTextData[1][j - 1]; // Score (combo) data is moved down in relevant array scoreboardComboData[j] = scoreboardComboData[j - 1]; } // The new Name, level and score (combo) data is inserted into the relevant array under element 'i' scoreboardTextData[0][i + 1] = name; scoreboardTextData[1][i + 1] = level; scoreboardComboData[i + 1] = combo; break; } // If the method gets the this point, it means that the score is greater than all scores within // the array and therefore cannot be added in the above way. As it is not less than any score within // the array. else if (i == 0) { // All Names, levels and scores are moved down within their relevant arrays for (int j = totalScores - 1; j != 0; j--) { scoreboardTextData[0][j] = scoreboardTextData[0][j - 1]; scoreboardTextData[1][j] = scoreboardTextData[1][j - 1]; scoreboardComboData[j] = scoreboardComboData[j - 1]; } // The new number 1 top name, level and score, are added into the first element // within each of their relevant arrays. scoreboardTextData[0][0] = name; scoreboardTextData[1][0] = level; scoreboardComboData[0] = combo; break; } // If the methods get to this point, the combo score is not high enough // to be on the top10 score list and therefore needs to break break; } } // As totalScores < 1, the current score is the first to be added. Therefore no checks need to be made // and the Name, Level and combo data can be entered directly into the first element of their relevant // array. else { scoreboardTextData[0][0] = name; scoreboardTextData[1][0] = level; scoreboardComboData[0] = combo; } } } Example for adding score: private static void Initialize() { scoreboardDoc = new XmlDocument(); if (!File.Exists("Scoreboard.xml")) GenerateXML("Scoreboard.xml"); PopulateScoreBoard("Scoreboard.xml"); // ADD TEST SCORES HERE! AddScore("EXAMPLE", "10", 100); AddScore("EXAMPLE2", "24", 999); PopulateXML("Scoreboard.xml"); } In it's current state the source file is just used for testing, initialize is called within main and PopulateScoreBoard handles the majority of other initialising, so nothing else is needed, except to add a test score. I thank you for your time!

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