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  • Solving the context menu problem with drag and drop in trees

    - by Frank Nimphius
    Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} The following drag-and-drop problem has been reported on OTN: An ADF Faces tree component is configured with a af:collectionDropTarget tag to handle drop events. The same tree component also has a context menu defined that is shown when users select the tree with the right mouse button. The problem now was - and I could reproduce this - that the context menu stopped working after the first time the tree handled a drop event. The drag and drop use case is to associate employees from a table to a department in the tree using drag and drop. The drop handler code in the managed bean looked up the tree node that received the drop event to determine the department ID to assign to the employee. For this code similar to the one shown below was used List dropRowKey = (List) dropEvent.getDropSite(); //if no dropsite then drop area was not a data area if(dropRowKey == null){    return DnDAction.NONE; }                tree.setRowKey(dropRowKey); JUCtrlHierNodeBinding dropNode = (JUCtrlHierNodeBinding) tree.getRowData(); Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} So what happens in this code? The drop event contains the dropSite reference, which is the row key of the tree node that received the drop event. The code then sets the key to the tree in a call to getRowDate() returns the node information for the drop target (the department). This however causes the tree state to go out of synch with its model (ADF tree binding), which is known to cause issues. In this use case the issue caused by this is that the context menu no longer shows up. To fix the problem, the code needs to be changes to read the current row key from the key, then perform the drop operation and at the end set the origin (or model) row key back //memorize current row key Object currentRowKey = tree.getRowKey();        List dropRowKey = (List) dropEvent.getDropSite(); //if no dropsite then drop area was not a data area if(dropRowKey == null){   return DnDAction.NONE;   }              tree.setRowKey(dropRowKey); JUCtrlHierNodeBinding dropNode = (JUCtrlHierNodeBinding) tree.getRowData(); ... do your stuff here .... //set current row key back tree.setRowKey(currentRowKey); AdfFacesContext.getCurrentInstance().addPartialTarget(tree); Node the code line that sets the row key back to its original value.

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  • How to generate and encode (for use in GA), random, strict, binary rooted trees with N leaves?

    - by Peter Simon
    First, I am an engineer, not a computer scientist, so I apologize in advance for any misuse of nomenclature and general ignorance of CS background. Here is the motivational background for my question: I am contemplating writing a genetic algorithm optimizer to aid in designing a power divider network (also called a beam forming network, or BFN for short). The BFN is intended to distribute power to each of N radiating elements in an array of antennas. The fraction of the total input power to be delivered to each radiating element has been specified. Topologically speaking, a BFN is a strictly binary, rooted tree. Each of the (N-1) interior nodes of the tree represents the input port of an unequal, binary power splitter. The N leaves of the tree are the power divider outputs. Given a particular power divider topology, one is still free to map the power divider outputs to the array inputs in an arbitrary order. There are N! such permutations of the outputs. There are several considerations in choosing the desired ordering: 1) The power ratio for each binary coupler should be within a specified range of values. 2) The ordering should be chosen to simplify the mechanical routing of the transmission lines connecting the power divider. The number of ouputs N of the BFN may range from, say, 6 to 22. I have already written a genetic algorithm optimizer that, given a particular BFN topology and desired array input power distribution, will search through the N! permutations of the BFN outputs to generate a design with compliant power ratios and good mechanical routing. I would now like to generalize my program to automatically generate and search through the space of possible BFN topologies. As I understand it, for N outputs (leaves of the binary tree), there are $C_{N-1}$ different topologies that can be constructed, where $C_N$ is the Catalan number. I would like to know how to encode an arbitrary tree having N leaves in a way that is consistent with a chromosomal description for use in a genetic algorithm. Also associated with this is the need to generate random instances for filling the initial population, and to implement crossover and mutations operators for this type of chromosome. Any suggestions will be welcome. Please minimize the amount of CS lingo in your reply, since I am not likely to be acquainted with it. Thanks in advance, Peter

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  • --log-slave-updates is OFF but updates received from master are still logged to slave binary log?

    - by quanta
    MySQL version 5.5.14 According to the document, by the default, slave does not log to its binary log any updates that are received from a master server. Here are my config. on the slave: # egrep 'bin|slave' /etc/my.cnf relay-log=mysqld-relay-bin log-bin = /var/log/mysql/mysql-bin binlog-format=MIXED sync_binlog = 1 log-bin-trust-function-creators = 1 mysql> show global variables like 'log_slave%'; +-------------------+-------+ | Variable_name | Value | +-------------------+-------+ | log_slave_updates | OFF | +-------------------+-------+ 1 row in set (0.01 sec) mysql> select @@log_slave_updates; +---------------------+ | @@log_slave_updates | +---------------------+ | 0 | +---------------------+ 1 row in set (0.00 sec) but slave still logs the updates that are received from a master to its binary logs, let's see the file size: -rw-rw---- 1 mysql mysql 37M Apr 1 01:00 /var/log/mysql/mysql-bin.001256 -rw-rw---- 1 mysql mysql 25M Apr 2 01:00 /var/log/mysql/mysql-bin.001257 -rw-rw---- 1 mysql mysql 46M Apr 3 01:00 /var/log/mysql/mysql-bin.001258 -rw-rw---- 1 mysql mysql 115M Apr 4 01:00 /var/log/mysql/mysql-bin.001259 -rw-rw---- 1 mysql mysql 105M Apr 4 18:54 /var/log/mysql/mysql-bin.001260 and the sample query when reading these binary files with mysqlbinlog utility: #120404 19:08:57 server id 3 end_log_pos 110324763 Query thread_id=382435 exec_time=0 error_code=0 SET TIMESTAMP=1333541337/*!*/; INSERT INTO norep_SplitValues VALUES ( NAME_CONST('cur_string',_utf8'118212' COLLATE 'utf8_general_ci')) /*!*/; # at 110324763 Did I miss something?

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  • Understanding Application binary interface (ABI)

    - by Tim
    I am trying to understand the concept of Application binary interface (ABI). From The Linux Kernel Primer: An ABI is a set of conventions that allows a linker to combine separately compiled modules into one unit without recompilation, such as calling conventions, machine interface, and operating-system interface. Among other things, an ABI defines the binary interface between these units. ... The benefits of conforming to an ABI are that it allows linking object files compiled by different compilers. From Wikipedia: an application binary interface (ABI) describes the low-level interface between an application (or any type of) program and the operating system or another application. ABIs cover details such as data type, size, and alignment; the calling convention, which controls how functions' arguments are passed and return values retrieved; the system call numbers and how an application should make system calls to the operating system; and in the case of a complete operating system ABI, the binary format of object files, program libraries and so on. I was wondering whether ABI depends on both the instruction set and the OS. Are the two all that ABI depends on? What kinds of role does ABI play in different stages of compilation: preprocessing, conversion of code from C to Assembly, conversion of code from Assembly to Machine code, and linking? From the first quote above, it seems to me that ABI is needed for only linking stage, not the other stages. Is it correct? When is ABI needed to be considered? Is ABI needed to be considered during programming in C, Assembly or other languages? If yes, how are ABI and API different? Or is it only for linker or compiler? Is ABI specified for/in machine code, Assembly language, and/or of C?

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  • What would be the time complexity of counting the number of all structurally different binary trees?

    - by ktslwy
    Using the method presented here: http://cslibrary.stanford.edu/110/BinaryTrees.html#java 12. countTrees() Solution (Java) /** For the key values 1...numKeys, how many structurally unique binary search trees are possible that store those keys? Strategy: consider that each value could be the root. Recursively find the size of the left and right subtrees. */ public static int countTrees(int numKeys) { if (numKeys <=1) { return(1); } else { // there will be one value at the root, with whatever remains // on the left and right each forming their own subtrees. // Iterate through all the values that could be the root... int sum = 0; int left, right, root; for (root=1; root<=numKeys; root++) { left = countTrees(root-1); right = countTrees(numKeys - root); // number of possible trees with this root == left*right sum += left*right; } return(sum); } } I have a sense that it might be n(n-1)(n-2)...1, i.e. n!

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  • Collaborative Whiteboard using WebSocket in GlassFish 4 - Text/JSON and Binary/ArrayBuffer Data Transfer (TOTD #189)

    - by arungupta
    This blog has published a few blogs on using JSR 356 Reference Implementation (Tyrus) as its integrated in GlassFish 4 promoted builds. TOTD #183: Getting Started with WebSocket in GlassFish TOTD #184: Logging WebSocket Frames using Chrome Developer Tools, Net-internals and Wireshark TOTD #185: Processing Text and Binary (Blob, ArrayBuffer, ArrayBufferView) Payload in WebSocket TOTD #186: Custom Text and Binary Payloads using WebSocket One of the typical usecase for WebSocket is online collaborative games. This Tip Of The Day (TOTD) explains a sample that can be used to build such games easily. The application is a collaborative whiteboard where different shapes can be drawn in multiple colors. The shapes drawn on one browser are automatically drawn on all other peer browsers that are connected to the same endpoint. The shape, color, and coordinates of the image are transfered using a JSON structure. A browser may opt-out of sharing the figures. Alternatively any browser can send a snapshot of their existing whiteboard to all other browsers. Take a look at this video to understand how the application work and the underlying code. The complete sample code can be downloaded here. The code behind the application is also explained below. The web page (index.jsp) has a HTML5 Canvas as shown: <canvas id="myCanvas" width="150" height="150" style="border:1px solid #000000;"></canvas> And some radio buttons to choose the color and shape. By default, the shape, color, and coordinates of any figure drawn on the canvas are put in a JSON structure and sent as a message to the WebSocket endpoint. The JSON structure looks like: { "shape": "square", "color": "#FF0000", "coords": { "x": 31.59999942779541, "y": 49.91999053955078 }} The endpoint definition looks like: @WebSocketEndpoint(value = "websocket",encoders = {FigureDecoderEncoder.class},decoders = {FigureDecoderEncoder.class})public class Whiteboard { As you can see, the endpoint has decoder and encoder registered that decodes JSON to a Figure (a POJO class) and vice versa respectively. The decode method looks like: public Figure decode(String string) throws DecodeException { try { JSONObject jsonObject = new JSONObject(string); return new Figure(jsonObject); } catch (JSONException ex) { throw new DecodeException("Error parsing JSON", ex.getMessage(), ex.fillInStackTrace()); }} And the encode method looks like: public String encode(Figure figure) throws EncodeException { return figure.getJson().toString();} FigureDecoderEncoder implements both decoder and encoder functionality but thats purely for convenience. But the recommended design pattern is to keep them in separate classes. In certain cases, you may even need only one of them. On the client-side, the Canvas is initialized as: var canvas = document.getElementById("myCanvas");var context = canvas.getContext("2d");canvas.addEventListener("click", defineImage, false); The defineImage method constructs the JSON structure as shown above and sends it to the endpoint using websocket.send(). An instant snapshot of the canvas is sent using binary transfer with WebSocket. The WebSocket is initialized as: var wsUri = "ws://localhost:8080/whiteboard/websocket";var websocket = new WebSocket(wsUri);websocket.binaryType = "arraybuffer"; The important part is to set the binaryType property of WebSocket to arraybuffer. This ensures that any binary transfers using WebSocket are done using ArrayBuffer as the default type seem to be blob. The actual binary data transfer is done using the following: var image = context.getImageData(0, 0, canvas.width, canvas.height);var buffer = new ArrayBuffer(image.data.length);var bytes = new Uint8Array(buffer);for (var i=0; i<bytes.length; i++) { bytes[i] = image.data[i];}websocket.send(bytes); This comprehensive sample shows the following features of JSR 356 API: Annotation-driven endpoints Send/receive text and binary payload in WebSocket Encoders/decoders for custom text payload In addition, it also shows how images can be captured and drawn using HTML5 Canvas in a JSP. How could this be turned in to an online game ? Imagine drawing a Tic-tac-toe board on the canvas with two players playing and others watching. Then you can build access rights and controls within the application itself. Instead of sending a snapshot of the canvas on demand, a new peer joining the game could be automatically transferred the current state as well. Do you want to build this game ? I built a similar game a few years ago. Do somebody want to rewrite the game using WebSocket APIs ? :-) Many thanks to Jitu and Akshay for helping through the WebSocket internals! Here are some references for you: JSR 356: Java API for WebSocket - Specification (Early Draft) and Implementation (already integrated in GlassFish 4 promoted builds) Subsequent blogs will discuss the following topics (not necessary in that order) ... Error handling Interface-driven WebSocket endpoint Java client API Client and Server configuration Security Subprotocols Extensions Other topics from the API

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  • [BST] Deletion procedure

    - by Metz
    Hi all. 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. Thank you. 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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  • How to find largest common sub-tree in the given two binary search trees?

    - by Bhushan
    Two BSTs (Binary Search Trees) are given. How to find largest common sub-tree in the given two binary trees? EDIT 1: Here is what I have thought: Let, r1 = current node of 1st tree r2 = current node of 2nd tree There are some of the cases I think we need to consider: Case 1 : r1.data < r2.data 2 subproblems to solve: first, check r1 and r2.left second, check r1.right and r2 Case 2 : r1.data > r2.data 2 subproblems to solve: - first, check r1.left and r2 - second, check r1 and r2.right Case 3 : r1.data == r2.data Again, 2 cases to consider here: (a) current node is part of largest common BST compute common subtree size rooted at r1 and r2 (b)current node is NOT part of largest common BST 2 subproblems to solve: first, solve r1.left and r2.left second, solve r1.right and r2.right I can think of the cases we need to check, but I am not able to code it, as of now. And it is NOT a homework problem. Does it look like?

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  • More localized, efficient Lowest Common Ancestor algorithm given multiple binary trees?

    - by mstksg
    I have multiple binary trees stored as an array. In each slot is either nil (or null; pick your language) or a fixed tuple storing two numbers: the indices of the two "children". No node will have only one child -- it's either none or two. Think of each slot as a binary node that only stores pointers to its children, and no inherent value. Take this system of binary trees: 0 1 / \ / \ 2 3 4 5 / \ / \ 6 7 8 9 / \ 10 11 The associated array would be: 0 1 2 3 4 5 6 7 8 9 10 11 [ [2,3] , [4,5] , [6,7] , nil , nil , [8,9] , nil , [10,11] , nil , nil , nil , nil ] I've already written simple functions to find direct parents of nodes (simply by searching from the front until there is a node that contains the child) Furthermore, let us say that at relevant times, both all trees are anywhere between a few to a few thousand levels deep. I'd like to find a function P(m,n) to find the lowest common ancestor of m and n -- to put more formally, the LCA is defined as the "lowest", or deepest node in which have m and n as descendants (children, or children of children, etc.). If there is none, a nil would be a valid return. Some examples, given our given tree: P( 6,11) # => 2 P( 3,10) # => 0 P( 8, 6) # => nil P( 2,11) # => 2 The main method I've been able to find is one that uses an Euler trace, which turns the given tree, with a node A to be the invisible parent of 0 and 1 with a depth of -1, into: A-0-2-6-2-7-10-7-11-7-2-0-3-0-A-1-4-1-5-8-5-9-5-1-A And from that, simply find the node between your given m and n that has the lowest number; For example, to find P(6,11), look for a 6 and an 11 on the trace. The number between them that is the lowest is 2, and that's your answer. If A is in between them, return nil. -- Calculating P(6,11) -- A-0-2-6-2-7-10-7-11-7-2-0-3-0-A-1-4-1-5-8-5-9-5-1-A ^ ^ ^ | | | m lowest n Unfortunately, I do believe that finding the Euler trace of a tree that can be several thousands of levels deep is a bit machine-taxing...and because my tree is constantly being changed throughout the course of the programming, every time I wanted to find the LCA, I'd have to re-calculate the Euler trace and hold it in memory every time. Is there a more memory efficient way, given the framework I'm using? One that maybe iterates upwards? One way I could think of would be the "count" the generation/depth of both nodes, and climb the lowest node until it matched the depth of the highest, and increment both until they find someone similar. But that'd involve climbing up from level, say, 3025, back to 0, twice, to count the generation, and using a terribly inefficient climbing-up algorithm in the first place, and then re-climbing back up. Are there any other better ways?

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  • New binary analysis tool finds FOSS in device firmware

    <b>ars Technica:</b> "Software development company Loohuis Consulting and process management consultancy OpenDawn have released a new binary analysis tool that is designed to detect Linux and BusyBox in binary firmware. The program, which is freely available for download, is intended to aid open source license compliance efforts."

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  • QotD: Maurizio Cimadamore on Project Lambda Binary Snapshots

    - by $utils.escapeXML($entry.author)
    I'm glad to announce that the first binary snapshots of the lambda repository are available at the following URL:http://jdk8.java.net/lambda/As you can imagine, as the implementation of the compiler/libraries is still under heavy development, there are still many rough corners that need to be polished. I'd like to thank you all for all the patience and the valuable feedback provided so far - please keep it coming!Maurizio Cimadamore announcing the Project Lambda binary snapshots on the lambda-dev OpenJDK mailing list.

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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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  • How to perform a binary search on IList<T>?

    - by Daniel Brückner
    Simple question - given an IList<T> how do you perform a binary search without writing the method yourself and without copying the data to a type with build-in binary search support. My current status is the following. List<T>.BinarySearch() is not a member of IList<T> There is no equivalent of the ArrayList.Adapter() method for List<T> IList<T> does not inherit from IList, hence using ArrayList.Adapter() is not possible I tend to believe that is not possible with build-in methods, but I cannot believe that such a basic method is missing from the BCL/FCL. If it is not possible, who can give the shortest, fastest, smartest, or most beatiful binary search implementation for IList<T>? UPDATE We all know that a list must be sorted before using binary search, hence you can assume that it is. But I assume (but did not verify) it is the same problem with sort - how do you sort IList<T>? CONCLUSION There seems to be no build-in binary search for IList<T>. One can use First() and OrderBy() LINQ methods to search and sort, but it will likly have a performance hit. Implementing it yourself (as an extension method) seems the best you can do.

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  • How to tell Subversion to display binary files using an external program?

    - by lamcro
    I have some code which, like java, is stored in a binary format, and I have the applications to display and modify this code setup in the Subversion's config file. But when I run svn diff for these file, Subversion prevents me =================================================================== Cannot display: file marked as a binary type. svn:mime-type = application/octet-stream I can still view them, but only with the --force argument Since all the files in the repository are of this binary code, how can I permanently force subversion to open the files for diff or edit mode?

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  • Transferring binary data through a SOAP webservice? C# / .NET

    - by Jason
    I have a webservice that returns the binary array of an object. Is there an easier way to transfer this with SOAP or does it need to be contained in XML? It's working, but I had to increase the send and receive buffer to a large value. How much is too much? Transferring binary in XML as an array seems really inefficient, but I can't see any way to add a binary attachment using .NET.

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  • Serialize a C# class to binary be used by C++. How to handle alignment?

    - by glenn.danthi
    I am currently serializing a C# class into a binary stream using BinaryWriter. I take each element of the class and write it out using BinaryWriter. This worked fine as the C++ application reading this binary file supported packed structs and hence the binary file could be loaded directly. Now I have got a request to handle alignment as a new application has popped up which cannot support packed structs. What's the best way to convert the C# class and exporting it out as a binary keeping both 2 byte as well as 4 byte alignment in mind? The user can choose the alignment.

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  • Detecting if a file is binary or plain text?

    - by dr. evil
    How can I detect if a file is binary or a plain text? Basically my .NET app is processing batch files and extracting data however I don't want to process binary files. As a solution I'm thinking about analysing first X bytes of the file and if there are more unprintable characters than printable characters it should be binary. Is this the right way to do it? Is there nay better implementation for this task?

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  • How to create a complete binary tree of height 'h' using Python?

    - by Jack
    Here is the node structure class Node: def __init__(self, data): # initializes the data members self.left = None self.right = None self.parent = None self.data = data complete binary tree Definition: A binary tree in which every level, except possibly the deepest, is completely filled. At depth n, the height of the tree, all nodes must be as far left as possible. -- http://www.itl.nist.gov/div897/sqg/dads/HTML/completeBinaryTree.html I am looking for an efficient algorithm.

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  • How could I represent 1.625 by 0 or a 1 (binary digit)?

    - by pepito
    This is an excerpt from wikipedia about 'full rate' speech coding standard. Full Rate or FR or GSM-FR or GSM 06.10 was the first digital speech coding standard used in the GSM digital mobile phone system. The bit rate of the codec is 13 kbit/s, or 1.625 bits/audio sample. And this one is an excerpt from wikipedia about bit. In computing parlance, bit is the abbreviation for a single binary digit, represented by a 0 or a 1. How could I represent 1.625 by 0 or a 1? Actually, that's my lecturer's question that I could not answer. Some links to papers are more than welcome. Thanks in advance.

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