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• Problems in Binary Search Tree

  - by user2782324

  This is my first ever trial at implementing the BST, and I am unable to get it done. Please help The problem is that When I delete the node if the node is in the right subtree from the root or if its a right child in the left subtree, then it works fine. But if the node is in the left subtree from root and its any left child, then it does not get deleted. Can someone show me what mistake am I doing?? the markedNode here gets allocated to the parent node of the node to be deleted. the minValueNode here gets allocated to a node whose left value child is the smallest value and it will be used to replace the value to be deleted. package DataStructures; class Node { int value; Node rightNode; Node leftNode; } class BST { Node rootOfTree = null; public void insertintoBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { Node newNode = new Node(); newNode.value = value; rootOfTree = newNode; newNode.rightNode = null; newNode.leftNode = null; } else { while (true) { if (value >= markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.rightNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.leftNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } } } } public void searchBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { System.out.println("Element Not Found"); } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { System.out.println("Element Not Found"); break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { System.out.println("Element Not Found"); break; } } if (value == markedNode.value) { System.out.println("Element Found"); break; } } } } public void deleteFromBST(int value) { Node markedNode = rootOfTree; Node minValueNode = null; if (rootOfTree == null) { System.out.println("Element Not Found"); return; } if (rootOfTree.value == value) { if (rootOfTree.leftNode == null && rootOfTree.rightNode == null) { rootOfTree = null; return; } else if (rootOfTree.leftNode == null ^ rootOfTree.rightNode == null) { if (rootOfTree.rightNode != null) { rootOfTree = rootOfTree.rightNode; return; } else { rootOfTree = rootOfTree.leftNode; return; } } else { minValueNode = rootOfTree.rightNode; if (minValueNode.leftNode == null) { rootOfTree.rightNode.leftNode = rootOfTree.leftNode; rootOfTree = rootOfTree.rightNode; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node rootOfTree.value = minValueNode.leftNode.value; // The value has been swapped if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { break; } else { markedNode = markedNode.rightNode; } } else { System.out.println("Element Not Found"); return; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { break; } else { markedNode = markedNode.leftNode; } } else { System.out.println("Element Not Found"); return; } } } // Parent of the required element found // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { if (markedNode.rightNode.rightNode == null && markedNode.rightNode.leftNode == null) { markedNode.rightNode = null; return; } else if (markedNode.rightNode.rightNode == null ^ markedNode.rightNode.leftNode == null) { if (markedNode.rightNode.rightNode != null) { markedNode.rightNode = markedNode.rightNode.rightNode; return; } else { markedNode.rightNode = markedNode.rightNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.rightNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { if (markedNode.leftNode.rightNode == null && markedNode.leftNode.leftNode == null) { markedNode.leftNode = null; return; } else if (markedNode.leftNode.rightNode == null ^ markedNode.leftNode.leftNode == null) { if (markedNode.leftNode.rightNode != null) { markedNode.leftNode = markedNode.leftNode.rightNode; return; } else { markedNode.leftNode = markedNode.leftNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.leftNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// } } } } } } public class BSTImplementation { public static void main(String[] args) { BST newBst = new BST(); newBst.insertintoBST(19); newBst.insertintoBST(13); newBst.insertintoBST(10); newBst.insertintoBST(20); newBst.insertintoBST(5); newBst.insertintoBST(23); newBst.insertintoBST(28); newBst.insertintoBST(16); newBst.insertintoBST(27); newBst.insertintoBST(9); newBst.insertintoBST(4); newBst.insertintoBST(22); newBst.insertintoBST(17); newBst.insertintoBST(30); newBst.insertintoBST(40); newBst.deleteFromBST(5); newBst.deleteFromBST(4); newBst.deleteFromBST(9); newBst.deleteFromBST(10); newBst.deleteFromBST(13); newBst.deleteFromBST(16); newBst.deleteFromBST(17); newBst.searchBST(5); newBst.searchBST(4); newBst.searchBST(9); newBst.searchBST(10); newBst.searchBST(13); newBst.searchBST(16); newBst.searchBST(17); System.out.println(); newBst.deleteFromBST(20); newBst.deleteFromBST(23); newBst.deleteFromBST(27); newBst.deleteFromBST(28); newBst.deleteFromBST(30); newBst.deleteFromBST(40); newBst.searchBST(20); newBst.searchBST(23); newBst.searchBST(27); newBst.searchBST(28); newBst.searchBST(30); newBst.searchBST(40); } }

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• Command to find the source package of a binary?

  - by Delan Azabani

  I know there's a which command, that echoes the full name of a binary (e.g. which sh). However, I'm fairly sure there's a command that echoes the package that provides a particular binary. Is there such a command? If so, what is it? I'd like to be able to run this: commandName ls and get coreutils for example.

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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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• Using MSpec runner in Visual Studio 2010 and .NET 4

  - by spapaseit

  Hi Everyone, I'm a big fan of MSpec so naturally I wanted to use is right away with VS2010 as well. I have the MSpec runner defined as an external tool in Visual Studio to be able to have it always visible as a toolbar item. Anyway, whenever I try to use the MSpec runner (mspec.exe) with a .NET 4.0 solution I get the following error: Could not load file or assembly 'file:///C:\Users\[SOMEUSER]\[SOME_FOLDERS}\bin\Debug\[PROJECT].Specs.dll' or one of its dependencies. This assembly is built by a runtime newer than the currently loaded runtime and cannot be loaded. I can still run my specs with the Resharper 5 runner so it's no big drama, but I bothers me to no end :þ Do you guys have any idea what the problem could be? Is there any solution other than recompiling the whole Mspec source code as a .NET 4.0 solution, which I really, really don't want to do? Thanks in advance. Sergi

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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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• 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 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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• Is it possible to compare a binary file in c#?

  - by Blankman

  I want to replace a binary file if the contents are different. So I need to be able to compare the binary file (without having to deserialize it). Is this possible? I used binary formatter to save the file.

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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 do a binary diff on two identically sized files under linux?

  - by Jon Cage

  I have two identically sized files and I need to do a binary diff to check whether they're the same. I'm currently runnnig diff file1.img file2.img but it's taking quite a while to process my 4GB files. Is this the most efficient way to do this?

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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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• How to implement dynamic binary search for search and insert operations of n element (C or C++)

  - by iecut

  The idea is to use multiple arrays, each of length 2^k, to store n elements, according to binary representation of n.Each array is sorted and different arrays are not ordered in any way. In the above mentioned data structure, SEARCH is carried out by a sequence of binary search on each array. INSERT is carried out by a sequence of merge of arrays of the same length until an empty array is reached. More Detail: Lets suppose we have a vertical array of length 2^k and to each node of that array there attached horizontal array of length 2^k. That is, to the first node of vertical array, a horizontal array of length 2^0=1 is connected,to the second node of vertical array, a horizontal array of length 2^1= 2 is connected and so on. So the insert is first carried out in the first horizontal array, for the second insert the first array becomes empty and second horizontal array is full with 2 elements, for the third insert 1st and 2nd array horiz. array are filled and so on. I implemented the normal binary search for search and insert as follows: int main() { int a[20]= {0}; int n, i, j, temp; int *beg, *end, *mid, target; printf(" enter the total integers you want to enter (make it less then 20):\n"); scanf("%d", &n); if (n = 20) return 0; printf(" enter the integer array elements:\n" ); for(i = 0; i < n; i++) { scanf("%d", &a[i]); } // sort the loaded array, binary search! for(i = 0; i < n-1; i++) { for(j = 0; j < n-i-1; j++) { if (a[j+1] < a[j]) { temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; } } } printf(" the sorted numbers are:"); for(i = 0; i < n; i++) { printf("%d ", a[i]); } // point to beginning and end of the array beg = &a[0]; end = &a[n]; // use n = one element past the loaded array! // mid should point somewhere in the middle of these addresses mid = beg += n/2; printf("\n enter the number to be searched:"); scanf("%d",&target); // binary search, there is an AND in the middle of while()!!! while((beg <= end) && (*mid != target)) { // is the target in lower or upper half? if (target < *mid) { end = mid - 1; // new end n = n/2; mid = beg += n/2; // new middle } else { beg = mid + 1; // new beginning n = n/2; mid = beg += n/2; // new middle } } // find the target? if (*mid == target) { printf("\n %d found!", target); } else { printf("\n %d not found!", target); } getchar(); // trap enter getchar(); // wait return 0; } Could anyone please suggest how to modify this program or a new program to implement dynamic binary search that works as explained above!!

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• ubuntu 13.10 kvm binary is deprecated, please use qemu-system-x86_64

  - by ??1986

  I just upgrade from 13.04 to 13.10 and I have this issue when I run my KVM Unable to complete install: 'internal error: process exited while connecting to monitor: W: kvm binary is deprecated, please use qemu-system-x86_64 instead char device redirected to /dev/pts/10 (label charserial0) failed to initialize KVM: Device or resource busy Detail Error: Traceback (most recent call last): File "/usr/share/virt-manager/virtManager/asyncjob.py", line 96, in cb_wrapper callback(asyncjob, *args, **kwargs) File "/usr/share/virt-manager/virtManager/create.py", line 1983, in do_install guest.start_install(False, meter=meter) File "/usr/lib/python2.7/dist-packages/virtinst/Guest.py", line 1246, in start_install noboot) File "/usr/lib/python2.7/dist-packages/virtinst/Guest.py", line 1314, in _create_guest dom = self.conn.createLinux(start_xml or final_xml, 0) File "/usr/lib/python2.7/dist-packages/libvirt.py", line 2892, in createLinux if ret is None:raise libvirtError('virDomainCreateLinux() failed', conn=self) libvirtError: internal error: process exited while connecting to monitor: W: kvm binary is deprecated, please use qemu-system-x86_64 instead char device redirected to /dev/pts/8 (label charserial0) failed to initialize KVM: Device or resource busy

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• Is it safe to delete rotated MySQL binary logs?

  - by Milan Babuškov

  I have a MySQL server with binary logging active. Once a days logs file is "rotated", i.e. MySQL seems to stop writing to it and creates and new log file. For example, I currently have these files in /var/lib/mysql -rw-rw---- 1 mysql mysql 10485760 Jun 7 09:26 ibdata1 -rw-rw---- 1 mysql mysql 5242880 Jun 7 09:26 ib_logfile0 -rw-rw---- 1 mysql mysql 5242880 Jun 2 15:20 ib_logfile1 -rw-rw---- 1 mysql mysql 1916844 Jun 6 09:20 mybinlog.000004 -rw-rw---- 1 mysql mysql 61112500 Jun 7 09:26 mybinlog.000005 -rw-rw---- 1 mysql mysql 15609789 Jun 7 13:57 mybinlog.000006 -rw-rw---- 1 mysql mysql 54 Jun 7 09:26 mybinlog.index and mybinlog.000006 is growing. Can I simply take mybinlog.000004 and mybinlog.000005, zip them up and transfer to another server, or I need to do something else before? What info is stored in mybinlog.index? Only the info about the latest binary log? UPDATE: I understand I can delete the logs with PURGE BINARY LOGS which updates mybinlog.index file. However, I need to transfer logs to another computer before deleting them (I test if backup is valid on another machine). To reduce the transfer size, I wish to bzip2 the files. What will PURGE BINARY LOGS do if log files are not "there" anymore?

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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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• BackgroundWorker not working with TeamCity NUnit runner

  - by Catalin DICU

  I'm using NUnit to test View Models in a WPF 3.5 application and I'm using the BackgroundWorker class to execute asynchronous commands.The unit test are running fine with the NUnit runner or ReSharper runner but fail on TeamCity 5.1 server. How is it implemented : I'm using a ViewModel property named IsBusy and set it to false on BackgroundWorker.RunWorkerCompleted event. In my test I'm using this method to wait for the BackgroundWorker to finish : protected void WaitForBackgroundOperation(ViewModel viewModel) { int count = 0; while (viewModel.IsBusy) { RunBackgroundWorker(); if (count++ >= 100) { throw new Exception("Background operation too long"); } Thread.Sleep(100); } } private static void RunBackgroundWorker() { Dispatcher.CurrentDispatcher.Invoke(DispatcherPriority.Background, new ThreadStart(delegate { })); System.Windows.Forms.Application.DoEvents(); } Well, sometimes it works and sometimes it hangs the build. I suppose it's the Application.DoEvents() but I don't know why...

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• ReSharper Unit Test Runner: Support for Deployment Items

  - by driis

  I like the Unit test runner in ReSharper 4.5, and would like to use it with my MSTest tests, but one thing annoys me: In some of our solutions, we have set up some Deployment Items in the .testrunconfig file. The ReSharper Unit Test runner does not seem to respect this, so I get errors when trying to run the unit tests from ReSharper. Is there any workraound for this ? Update: citizenmatt's answer was correct, the option to use a .testrunconfig with ReSharper exists in the Options dialog of ReSharper. You have to select the unit test provider on the list, then the controls to do that appears. (That was not obvious or discoverable, at least not for me ;-)

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• Java: confirm method Binary division and find remainder is correct?

  - by cadwag

  I am parsing binary files and have to implement a CRC algorithm to ensure the file is not corrupted. Problem is that I can't seem to get the binary math working when using larger numbers. The example I'm trying to get working: BigInteger G = new BigInteger("11001", 2); BigInteger M = new BigInteger("1110010000", 2); BigInteger R = M.remainder(G); I am expecting: R = "0101" But I am getting: R = "1100" I am assuming the remainder of 0101 is correct since it is given to me in this book I am using as a reference for the CRC algorithm (it's not based in Java), but I can't seem to get it working. I can get small binary calculations to work that I have solved by hand, but not the larger ones. I'll admit that I haven't worked the larger ones by hand yet, that is my next step, but I wanted to see if someone could point out a glaring flaw I have in my code. Can anyone confirm or deny that my methodology is correct? Thanks

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• what is the format of a binary image & how is it different from jpg, png images?

  - by Rahulsingh190

  I searched the internet for the basic formats of image files (e.g. .jpg, .png, .gif) as there is a specific format for .doc, .pdf etc. But didn't got anything relevant. And today I also came with an .bin image format. BIN signifies that the image is in the Binary format. So, what is the Internal format of .jpg image file. And How is it different from .bin (Binary) format. Because everything is Basically saved in Binary Form. And How is BITMAP Image different from .jpg format.

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