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  • How can I do block-oriented disk I/O with Java? Or similar for a B+ tree

    - by Sanoj
    I would like to implement an B+ tree in Java and try to optimize it for disk based I/O. Is there an API for accessing individual disk blocks from Java? or is there an API that can do similar block-oriented access that fits my purpose? I would like to create something like Tokyo Cabinet in 100% Java. Is there anyone that knows what Java only databases like JavaDB is using in the back-end for this? I know that there are probably other languages than Java that can do this better, but I do this in a learning purpose only.

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  • HTML Markup in einem APEX Tree - ganz einfach per Plugin!

    - by carstenczarski
    Die APEX Tree Region kennt sicherlich jeder APEX-Entwickler. Und vielfach besteht der Bedarf, das Aussehen des APEX Tree mit Hilfe von HTML Markup zu beeinflussen. Leider ist es seit APEX 4.0 nicht mehr möglich, eigenes HTML-Markup in einen APEX-Tree aufzunehmen - aus Sicherheitsgründen (Schutz vor Cross-Site-Scripting) werden alle HTML Sonderzeichen maskiert. Wenn kein XSS-Risiko besteht (die vom Tree dargestellten Inhalte basieren nicht auf Benutzereingaben und werden komplett vom Entwickler bestimmt), kann dies mit wenigen Zeilen JavaScript und jQuery-Code erreicht werden. Damit es noch einfacher wird,  haben wir die Funktionalität für Sie in einem APEX-Plugin gekapselt. Und so funktioniert es: APEX Plugin "HTML Markup for APEX Tree Region" herunterladenhttp://apex-plugin.com/oracle-apex-plugins/dynamic-action-plugin/html-markup-for-apex-tree_174.html APEX Plugin in die Anwendung importieren APEX Tree Region erzeugen und eigene Ersetzungen für HTML-Sonderzeichen verwenden, also bspw."[" für "<", "]" für ">" und "§" für "&". Eine neue dynamische Aktion erzeugen, die beim Laden der Seite ausgeführt wird und mit Hilfe des Plugins die Ersetzungen im Tree durch die "richtigen" HTML-Sonderzeichen ersetzt. Fertig. Wie das Plugin wirkt, können Sie sich auf einer Demo-Seite ansehen.

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  • Tool to compute SHA256 Tree Hash

    - by Benjamin
    I've started using AWS Glacier, and noticed that it hashes the files using an algorithm called SHA-256 Tree Hash. To my surprise, this algorithm is different from SHA-256, so I can't use the tools I'm used to, to compare hashes and verify file integrity. Do you know a Windows tool, if possible integrated in the context menu, to compute the SHA-256 Tree Hash of a file? I'd also accept a Linux command-line tool, as a second choice :-)

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  • Binary Search Tree in C

    - by heapzero
    Hi, I'm a Python guy. Learning C language and I've been trying to implement Binary Search Tree in C. I wrote down the code, and I've been trying from few hours but, not able to get the output as expected. Please help! Please correct me. #include<stdlib.h> #include<stdio.h> typedef int ElementType; typedef struct TreeNode { ElementType element; struct TreeNode *left, *right; } TreeNode; TreeNode *createTree(){ //Create the root of tree TreeNode *tempNode; tempNode = malloc(sizeof(TreeNode)); tempNode->element = 0; tempNode->left = NULL; tempNode->right = NULL; return tempNode; } TreeNode *createNode(ElementType X){ //Create a new leaf node and return the pointer TreeNode *tempNode; tempNode = malloc(sizeof(TreeNode)); tempNode->element = X; tempNode->left = NULL; tempNode->right = NULL; return tempNode; } TreeNode *insertElement(TreeNode *node, ElementType X){ //insert element to Tree if(node==NULL){ return createNode(X); } else{ if(X < node->element){ node->left = insertElement(node->left, X); } else if(X > node->element){ node->right = insertElement(node->right, X); } else if(X == node->element){ printf("Oops! the element is already present in the tree."); } } } TreeNode *displayTree(TreeNode *node){ //display the full tree if(node==NULL){ return; } displayTree(node->left); printf("| %d ", node->element); displayTree(node->right); } main(){ //pointer to root of tree #2 TreeNode *TreePtr; TreeNode *TreeRoot; TreeNode *TreeChild; //Create the root of tree TreePtr = createTree(); TreeRoot = TreePtr; TreeRoot->element = 32; printf("%d\n",TreeRoot->element); insertElement(TreeRoot, 8); TreeChild = TreeRoot->left; printf("%d\n",TreeChild->element); insertElement(TreeRoot, 2); insertElement(TreeRoot, 7); insertElement(TreeRoot, 42); insertElement(TreeRoot, 28); insertElement(TreeRoot, 1); insertElement(TreeRoot, 4); insertElement(TreeRoot, 5); // the output is not as expected :( displayTree(TreeRoot); }

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  • Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

    - by Justin L.
    It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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  • How can I convert this PHP script to Ruby? (build tree from tabbed string)

    - by Jon Sunrays
    I found this script below online, and I'm wondering how I can do the same thing with a Ruby on Rails setup. So, first off, I ran this command: rails g model Node node_id:integer title:string Given this set up, how can I make a tree from a tabbed string like the following? <?php // Make sure to have "Academia" be root node with nodeID of 1 $data = " Social sciences Anthropology Biological anthropology Forensic anthropology Gene-culture coevolution Human behavioral ecology Human evolution Medical anthropology Paleoanthropology Population genetics Primatology Anthropological linguistics Synchronic linguistics (or Descriptive linguistics) Diachronic linguistics (or Historical linguistics) Ethnolinguistics Sociolinguistics Cultural anthropology Anthropology of religion Economic anthropology Ethnography Ethnohistory Ethnology Ethnomusicology Folklore Mythology Political anthropology Psychological anthropology Archaeology ...(goes on for a long time) "; //echo "Checkpoint 2\n"; $lines = preg_split("/\n/", $data); $parentids = array(0 => null); $db = new PDO("host", 'username', 'pass'); $sql = 'INSERT INTO `TreeNode` SET ParentID = ?, Title = ?'; $stmt = $db->prepare($sql); foreach ($lines as $line) { if (!preg_match('/^([\s]*)(.*)$/', $line, $m)) { continue; } $spaces = strlen($m[1]); //$level = intval($spaces / 4); //assumes four spaces per indent $level = strlen($m[1]); // if data is tab indented $title = $m[2]; $parentid = ($level > 0 ? $parentids[$level - 1] : 1); //All "roots" are children of "Academia" which has an ID of "1"; $rv = $stmt->execute(array($parentid, $title)); $parentids[$level] = $db->lastInsertId(); echo "inserted $parentid - " . $parentid . " title: " . $title . "\n"; } ?>

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  • lists searches in SYB or uniplate haskell

    - by Chris
    I have been using uniplate and SYB and I am trying to transform a list For instance type Tree = [DataA] data DataA = DataA1 [DataB] | DataA2 String | DataA3 String [DataA] deriving Show data DataB = DataB1 [DataA] | DataB2 String | DataB3 String [DataB] deriving Show For instance, I would like to traverse my tree and append a value to all [DataB] So my first thought was to do this: changeDataB:: Tree -> Tree changeDataB = everywhere(mkT changeDataB') chanegDataB'::[DataB] -> [DataB] changeDataB' <add changes here> or if I was using uniplate changeDataB:: Tree -> Tree changeDataB = transformBi changeDataB' chanegDataB'::[DataB] -> [DataB] changeDataB' <add changes here> The problem is that I only want to search on the full list. Doing either of these searches will cause a search on the full list and all of the sub-lists (including the empty list) The other problem is that a value in [DataB] may generate a [DataB], so I don't know if this is the same kind of solution as not searching chars in a string. I could pattern match on DataA1 and DataB3, but in my real application there are a bunch of [DataB]. Pattern matching on the parents would be extensive. The other thought that I had was to create a data DataBs = [DataB] and use that to transform on. That seems kind of lame, there must be a better solution.

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  • How do I force all Tree itemrenderers to refresh?

    - by Richard Haven
    I have item renderers in an mx.controls.Tree that I need to refresh on demand. I have code in the updateDisplayList that fires for only some of the visible nodes no matter what I do. I've tried triggering a change that they should all be listening for; I have tried clearing and resetting the dataProvider and the itemRenderer properties. private function forceCategoryTreeRefresh(event : Event = null) : void { trace("forceCategoryTreeRefresh"); var prevDataProvider : Object = CategoryTree.dataProvider; CategoryTree.dataProvider = null; CategoryTree.validateNow(); CategoryTree.dataProvider = prevDataProvider; var prevItemRenderer : IFactory = CategoryTree.itemRenderer; CategoryTree.itemRenderer = null; CategoryTree.itemRenderer = prevItemRenderer as IFactory; _categoriesChangeDispatcher.dispatchEvent(new Event(Event.CHANGE)); } The nodes refresh properly when I scroll them into view (e.g. the .data gets set), but I cannot force the ones that already exist to refresh or reset themselves. Any ideas?

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  • Architecture for Social Graph data that has a Time Frame Associated?

    - by Jay Stevens
    I am adding some "social" type features to an existing application. There are a limited # of node & edge types. Overall the data itself is relatively small (50,000 - 70,000 for each type of node) there will be a number of edges (relationships) between them (almost all directional). This, I know, is relatively easy to represent with an SDF store (such as BrightstarDB) or something like Microsoft's Trinity (or really many of the noSQL options). The thing that, I think, makes this a unique use case is that each relationship will have a timeframe associated with it (start and end dates). Right now, I'm thinking of just storing this in a relational structure and dealing with the headaches of "traversing the graph", but I'm looking for suggestions on a better approach (both in terms of data structure and server): Column ================ From_Node_ID Relationship To_Node_ID StartDate EndDate Any suggestions or thoughts are welcomed.

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  • Improving the running time of Breadth First Search and Adjacency List creation

    - by user45957
    We are given an array of integers where all elements are between 0-9. have to start from the 1st position and reach end in minimum no of moves such that we can from an index i move 1 position back and forward i.e i-1 and i+1 and jump to any index having the same value as index i. Time Limit : 1 second Max input size : 100000 I have tried to solve this problem use a single source shortest path approach using Breadth First Search and though BFS itself is O(V+E) and runs in time the adjacency list creation takes O(n2) time and therefore overall complexity becomes O(n2). is there any way i can decrease the time complexity of adjacency list creation? or is there a better and more efficient way of solving the problem? int main(){ vector<int> v; string str; vector<int> sets[10]; cin>>str; int in; for(int i=0;i<str.length();i++){ in=str[i]-'0'; v.push_back(in); sets[in].push_back(i); } int n=v.size(); if(n==1){ cout<<"0\n"; return 0; } if(v[0]==v[n-1]){ cout<<"1\n"; return 0; } vector<int> adj[100001]; for(int i=0;i<10;i++){ for(int j=0;j<sets[i].size();j++){ if(sets[i][j]>0) adj[sets[i][j]].push_back(sets[i][j]-1); if(sets[i][j]<n-1) adj[sets[i][j]].push_back(sets[i][j]+1); for(int k=j+1;k<sets[i].size();k++){ if(abs(sets[i][j]-sets[i][k])!=1){ adj[sets[i][j]].push_back(sets[i][k]); adj[sets[i][k]].push_back(sets[i][j]); } } } } queue<int> q; q.push(0); int dist[100001]; bool visited[100001]={false}; dist[0]=0; visited[0]=true; int c=0; while(!q.empty()){ int dq=q.front(); q.pop(); c++; for(int i=0;i<adj[dq].size();i++){ if(visited[adj[dq][i]]==false){ dist[adj[dq][i]]=dist[dq]+1; visited[adj[dq][i]]=true; q.push(adj[dq][i]); } } } cout<<dist[n-1]<<"\n"; return 0; }

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  • Finding an A* heuristic for a directed graph

    - by Janis Peisenieks
    In a previous question, I asked about finding a route (or path if you will) in a city. That is all dandy. The solution I chose was with the A* algorithm, which really seems to suit my needs. What I find puzzling is heuristic. How do I find one in an environment without constant distance between 2 nodes? Meaning, not every 2 nodes have the same distance between them. What I have is nodes (junctures), streets with weight (which may also be one-way), a start/finish node (since the start and end is always in the same place) and a goal node. In an ordinary case, I would just use the same way I got to goal to go back, but since one of the streets could have been a one-way, that may not be possible. The main question How do I find a heuristic in a directed graph?

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  • Finding the shortest path through a digraph that visits all nodes

    - by Boluc Papuccuoglu
    I am trying to find the shortest possible path that visits every node through a graph (a node may be visited more than once, the solution may pick any node as the starting node.). The graph is directed, meaning that being able to travel from node A to node B does not mean one can travel from node B to node A. All distances between nodes are equal. I was able to code a brute force search that found a path of only 27 nodes when I had 27 nodes and each node had a connection to 2 or 1 other node. However, the actual problem that I am trying to solve consists of 256 nodes, with each node connecting to either 4 or 3 other nodes. The brute force algorithm that solved the 27 node graph can produce a 415 node solution (not optimal) within a few seconds, but using the processing power I have at my disposal takes about 6 hours to arrive at a 402 node solution. What approach should I use to arrive at a solution that I can be certain is the optimal one? For example, use an optimizer algorithm to shorten a non-optimal solution? Or somehow adopt a brute force search that discards paths that are not optimal? EDIT: (Copying a comment to an answer here to better clarify the question) To clarify, I am not saying that there is a Hamiltonian path and I need to find it, I am trying to find the shortest path in the 256 node graph that visits each node AT LEAST once. With the 27 node run, I was able to find a Hamiltonian path, which assured me that it was an optimal solution. I want to find a solution for the 256 node graph which is the shortest.

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  • Looking for an algorithm to connect dots - shortest route

    - by e4ch
    I have written a program to solve a special puzzle, but now I'm kind of stuck at the following problem: I have about 3200 points/nodes/dots. Each of these points is connected to a few other points (usually 2-5, theoretical limit is 1-26). I have exactly one starting point and about 30 exit points (probably all of the exit points are connected to each other). Many of these 3200 points are probably not connected to neither start nor end point in any way, like a separate net, but all points are connected to at least one other point. I need to find the shortest number of hops to go from entry to exit. There is no distance between the points (unlike the road or train routing problem), just the number of hops counts. I need to find all solutions with the shortest number of hops, and not just one solution, but all. And potentially also solutions with one more hop etc. I expect to have a solution with about 30-50 hops to go from start to exit. I already tried: 1) randomly trying possibilities and just starting over when the count was bigger than a previous solution. I got first solution with 3500 hops, then it got down to about 97 after some minutes, but looking at the solutions I saw problems like unnecessary loops and stuff, so I tried to optimize a bit (like not going back where it came from etc.). More optimizations are possible, but this random thing doesn't find all best solutions or takes too long. 2) Recursively run through all ways from start (chess-problem-like) and breaking the try when it reached a previous point. This was looping at about a length of 120 nodes, so it tries chains that are (probably) by far too long. If we calculate 4 possibilities and 120 nodes, we're reaching 1.7E72 possibilities, which is not possible to calculate through. This is called Depth-first search (DFS) as I found out in the meantime. Maybe I should try Breadth-first search by adding some queue? The connections between the points are actually moves you can make in the game and the points are how the game looks like after you made the move. What would be the algorithm to use for this problem? I'm using C#.NET, but the language shouldn't matter.

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  • Approaches to create a nested tree structure of NSDictionaries?

    - by d11wtq
    I'm parsing some input which produces a tree structure containing NSDictionary instances on the branches and NSString instance at the nodes. After parsing, the whole structure should be immutable. I feel like I'm jumping through hoops to create the structure and then make sure it's immutable when it's returned from my method. We can probably all relate to the input I'm parsing, since it's a query string from a URL. In a string like this: a=foo&b=bar&a=zip We expect a structure like this: NSDictionary { "a" => NSDictionary { 0 => "foo", 1 => "zip" }, "b" => "bar" } I'm keeping it just two-dimensional in this example for brevity, though in the real-world we sometimes see var[key1][key2]=value&var[key1][key3]=value2 type structures. The code hasn't evolved that far just yet. Currently I do this: - (NSDictionary *)parseQuery:(NSString *)queryString { NSMutableDictionary *params = [NSMutableDictionary dictionary]; NSArray *pairs = [queryString componentsSeparatedByString:@"&"]; for (NSString *pair in pairs) { NSRange eqRange = [pair rangeOfString:@"="]; NSString *key; id value; // If the parameter is a key without a specified value if (eqRange.location == NSNotFound) { key = [pair stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; value = @""; } else { // Else determine both key and value key = [[pair substringToIndex:eqRange.location] stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; if ([pair length] > eqRange.location + 1) { value = [[pair substringFromIndex:eqRange.location + 1] stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; } else { value = @""; } } // Parameter already exists, it must be a dictionary if (nil != [params objectForKey:key]) { id existingValue = [params objectForKey:key]; if (![existingValue isKindOfClass:[NSDictionary class]]) { value = [NSDictionary dictionaryWithObjectsAndKeys:existingValue, [NSNumber numberWithInt:0], value, [NSNumber numberWithInt:1], nil]; } else { // FIXME: There must be a more elegant way to build a nested dictionary where the end result is immutable? NSMutableDictionary *newValue = [NSMutableDictionary dictionaryWithDictionary:existingValue]; [newValue setObject:value forKey:[NSNumber numberWithInt:[newValue count]]]; value = [NSDictionary dictionaryWithDictionary:newValue]; } } [params setObject:value forKey:key]; } return [NSDictionary dictionaryWithDictionary:params]; } If you look at the bit where I've added FIXME it feels awfully clumsy, pulling out the existing dictionary, creating an immutable version of it, adding the new value, then creating an immutable dictionary from that to set back in place. Expensive and unnecessary? I'm not sure if there are any Cocoa-specific design patterns I can follow here?

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  • How to structure classes in the filesystem?

    - by da_b0uncer
    I have a few (view) classes. Table, Tree, PagingColumn, SelectionColumn, SparkLineColumn, TimeColumn. currently they're flat under app/view like this: app/view/Table app/view/Tree app/view/PagingColumn ... I thought about restructuring it, because the Trees and Tables use the columns, but there are some columns, which only work in a tree, some who work in trees and tables and in the future there are probably some who only work in tables, I don't know. My first idea was like this: app/view/Table app/view/Tree app/view/column/PagingColumn app/view/column/SelectionColumn app/view/column/SparkLineColumn app/view/column/TimeColumn But since the SelectionColumn is explicitly for trees, I have the fear that future developers could get the idea of missuse them. But how to restructure it probably? Like this: app/view/table/panel/Table app/view/tree/panel/Tree app/view/tree/column/PagingColumn app/view/tree/column/SelectionColumn app/view/column/SparkLineColumn app/view/column/TimeColumn Or like this: app/view/Table app/view/Tree app/view/column/SparkLineColumn app/view/column/TimeColumn app/view/column/tree/PagingColumn app/view/column/tree/SelectionColumn

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  • Pseudo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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  • Parsing a given binary tree using python?

    - by kaushik
    Parse a binary tree,referring to given set of features,answering decision tree question at each node to decide left child or right child and find the path to leaf node according to answer given to the decision tree.. input wil be a set of feature which wil help in answering the question at each level to choose the left or right half and the output will be the leaf node.. i need help in implementing this can anyone suggest methods?? Please answer... thanks in advance..

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  • gunit syntax for tree walker with a flat list of nodes

    - by Kaleb Pederson
    Here's a simple gunit test for a portion of my tree grammar which generates a flat list of nodes: objectOption walks objectOption: <<one:"value">> -> (one "value") Although you define a tree in ANTLR's rewrite syntax using a caret (i.e. ^(ROOT child...)), gunit matches trees without the caret, so the above represents a tree and it's not surprising that it fails: it's a flat list of nodes and not a tree. This results in a test failure: 1 failures found: test2 (objectOption walks objectOption, line17) - expected: (one \"value\") actual: one \"value\" Another option which seems intuitive is to leave off the parenthesis, like this: objectOption walks objectOption: <<one:"value">> -> one "value" But gunit doesn't like this syntax. It seems to result in a parse failure in the gunit grammar: line 17:20 no viable alternative at input 'one' line 17:24 missing ':' at 'value' line 0:-1 no viable alternative at input '<EOF>' java.lang.NullPointerException at org.antlr.gunit.OutputTest.getExpected(OutputTest.java:65) at org.antlr.gunit.gUnitExecutor.executeTests(gUnitExecutor.java:245) ... What is the correct way to match a flat tree?

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  • Psuedo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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  • Output of gcc -fdump-tree-original

    - by Job
    If I dump the code generated by GCC for a virtual destructor (with -fdump-tree-original), I get something like this: ;; Function virtual Foo::~Foo() (null) ;; enabled by -tree-original { <<cleanup_point <<< Unknown tree: expr_stmt (void) (((struct Foo *) this)->_vptr.Foo = &_ZTV3Foo + 8) >>> >>; } <D.20148>:; if ((bool) (__in_chrg & 1)) { <<cleanup_point <<< Unknown tree: expr_stmt operator delete ((void *) this) >>> >>; } My question is: where is the code after "<D.20148>:;" located? It is outside of the destructor so when is this code executed?

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  • What is validating a binary search tree?

    - by dotnetdev
    I read on here of an exercise in interviews known as validating a binary search tree. How exactly does this work? What would one be looking for in validating a binary search tree? I have written a basic search tree, but never heard of this concept. Thanks

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  • Level-order in Haskell

    - by brain_damage
    I have a structure for a tree and I want to print the tree by levels. data Tree a = Nd a [Tree a] deriving Show type Nd = String tree = Nd "a" [Nd "b" [Nd "c" [], Nd "g" [Nd "h" [], Nd "i" [], Nd "j" [], Nd "k" []]], Nd "d" [Nd "f" []], Nd "e" [Nd "l" [Nd "n" [Nd "o" []]], Nd "m" []]] preorder (Nd x ts) = x : concatMap preorder ts postorder (Nd x ts) = (concatMap postorder ts) ++ [x] But how to do it by levels? "levels tree" should print ["a", "bde", "cgflm", "hijkn", "o"]. I think that "iterate" would be suitable function for the purpose, but I cannot come up with a solution how to use it. Would you help me, please?

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  • Design suggestion for expression tree evaluation with time-series data

    - by Lirik
    I have a (C#) genetic program that uses financial time-series data and it's currently working but I want to re-design the architecture to be more robust. My main goals are: sequentially present the time-series data to the expression trees. allow expression trees to access previous data rows when needed. to optimize performance of the data access while evaluating the expression trees. keep a common interface so various types of data can be used. Here are the possible approaches I've thought about: I can evaluate the expression tree by passing in a data row into the root node and let each child node use the same data row. I can evaluate the expression tree by passing in the data row index and letting each node get the data row from a shared DataSet (currently I'm passing the row index and going to multiple synchronized arrays to get the data). Hybrid: an immutable data set is accessible by all of the expression trees and each expression tree is evaluated by passing in a data row. The benefit of the first approach is that the data row is being passed into the expression tree and there is no further query done on the data set (which should increase performance in a multithreaded environment). The drawback is that the expression tree does not have access to the rest of the data (in case some of the functions need to do calculations using previous data rows). The benefit of the second approach is that the expression trees can access any data up to the latest data row, but unless I specify what that row is, I'll have to iterate through the rows and figure out which one is the last one. The benefit of the hybrid is that it should generally perform better and still provide access to the earlier data. It supports two basic "views" of data: the latest row and the previous rows. Do you guys know of any design patterns or do you have any tips that can help me build this type of system? Should I use a DataSet to hold and present the data, or are there more efficient ways to present rows of data while maintaining a simple interface? FYI: All of my code is written in C#.

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