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  • Getting Image size of JPEG from its binary

    - by rajeshsr
    Hi I have a lot of jpeg files with varying image size. For instance, here is the first 640 bytes as given by hexdump of an image of size 256*384(pixels): 0000000: ffd8 ffe0 0010 4a46 4946 0001 0101 0048 ......JFIF.....H 0000010: 0048 0000 ffdb 0043 0003 0202 0302 0203 .H.....C........ 0000020: 0303 0304 0303 0405 0805 0504 0405 0a07 ................ 0000030: 0706 080c 0a0c 0c0b 0a0b 0b0d 0e12 100d ................ I guess the size information mus be within these lines. But am unable to see which bytes give the sizes correctly. Can anyone help me find the fields that contains the size information?

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  • C++ string array binary search

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

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  • Need algorithm to add Node in binary tree

    - by m.qayyum
    •if your new element is less or equal than the current node, you go to the left subtree, otherwise to the right subtree and continue traversing •if you arrived at a node, where you can not go any deeper, because there is no subtree, this is the place to insert your new element (5)Root (3)-------^--------(7) (2)---^----(5) ^-----(8) (5)--^ i want to add this last node with data 5...but i can't figure it out...I need a algorithm to do that or in java language

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  • Recursive Binary Search Tree Insert

    - by Nick Sinklier
    So this is my first java program, but I've done c++ for a few years. I wrote what I think should work, but in fact it does not. So I had a stipulation of having to write a method for this call: tree.insertNode(value); where value is an int. I wanted to write it recursively, for obvious reasons, so I had to do a work around: public void insertNode(int key) { Node temp = new Node(key); if(root == null) root = temp; else insertNode(temp); } public void insertNode(Node temp) { if(root == null) root = temp; else if(temp.getKey() <= root.getKey()) insertNode(root.getLeft()); else insertNode(root.getRight()); } Thanks for any advice.

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  • Nicely printing/showing a binary tree in Haskell

    - by nicole
    I have a tree data type: data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a ...and I need to make it an instance of Show, without using deriving. I have found that nicely displaying a little branch with two leaves is easy: instance (Show a, Show b) => Show (Tree a b) where show (Leaf x) = show x show (Branch val l r) = " " ++ show val ++ "\n" ++ show l ++ " " ++ show r But how can I extend a nice structure to a tree of arbitrary size? It seems like determining the spacing would require me to know just how many leaves will be at the very bottom (or maybe just how many leaves there are in total) so that I can allocate all the space I need there and just work 'up.' I would probably need to call a size function. I can see this being workable, but is that making it harder than it is?

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  • A balanced binary search tree which is also a heap

    - by saeedn
    I'm looking for a data structure where each element in it has two keys. With one of them the structure is a BST and looking at the other one, data structure is a heap. With a little search, I found a structure called Treap. It uses the heap property with a random distribution on heap keys to make the BST balanced! What I want is a Balanced BST, which can be also a heap. The BST in Treap could be unbalanced if I insert elements with heap Key in the order of my choice. Is there such a data structure?

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  • Reverse reading WORD from a binary file?

    - by Angel
    Hi, I have a structure: struct JFIF_HEADER { WORD marker[2]; // = 0xFFD8FFE0 WORD length; // = 0x0010 BYTE signature[5]; // = "JFIF\0" BYTE versionhi; // = 1 BYTE versionlo; // = 1 BYTE xyunits; // = 0 WORD xdensity; // = 1 WORD ydensity; // = 1 BYTE thumbnwidth; // = 0 BYTE thumbnheight; // = 0 }; This is how I read it from the file: HANDLE file = CreateFile(filename, GENERIC_READ, FILE_SHARE_READ, NULL, OPEN_EXISTING, FILE_ATTRIBUTE_NORMAL, 0); DWORD tmp = 0; DWORD size = GetFileSize(file, &tmp); BYTE *DATA = new BYTE[size]; ReadFile(file, DATA, size, &tmp, 0); JFIF_HEADER header; memcpy(&header, DATA, sizeof(JFIF_HEADER)); This is how the beginning of my file looks in hex editor: 0xFF 0xD8 0xFF 0xE0 0x00 0x10 0x4A 0x46 0x49 0x46 0x00 0x01 0x01 0x00 0x00 0x01 When I print header.marker, it shows exactly what it should (0xFFD8FFE0). But when I print header.length, it shows 0x1000 instead of 0x0010. The same thing is with xdensity and ydensity. Why do I get wrong data when reading a WORD? Thank you.

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  • Return parent of node in Binary Tree

    - by user188995
    I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs. //Assume that nodes are represented by numbers from 1...n where 1=root and even //nos.=left child and odd nos=right child. public int parent(Node node){ if (node % 2 == 0){ if (root.left==node) return root; else return parent(root.left); } //same case for right } But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right. If it's the child or if the node that was queried else, recurses it with the child.

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  • Write to a binary file?

    - by rick irby
    Here is data structure w/ variables: struct Part_record { char id_no[3]; int qoh; string desc; double price: }; --- (Using "cin" to input data) --- Part_record null_part = {" ", 0," ",0.0}; --- --- file.seekg( -(long)sizeof(Part_record), ios::cur); file.write( ( char *)&part, sizeof(Part_record) ); The three variables, qoh, Id_no & price, write out correctly, but the "desc" variable is not right. Do I need to initialize Part_record some other way? It should be 20 characters in length. If you have enough info here, pls share your advice,thanks.

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  • findNode in binary search tree

    - by Weadadada Awda
    Does this look right? I mean I am trying to implement the delete function. Node* BST::findNode(int tofind) { Node* node = new Node; node = root; while (node != NULL) { if (node->val == tofind) { return node; } else if (tofind < node->val) { node = node->left; } else { node = node->right; } } } Here is the delete, it's not even close to done but, void BST::Delete(int todelete) { // bool found = false; Node* toDelete = new Node(); toDelete=findNode(todelete); if(toDelete->val!=NULL) { cout << toDelete->val << endl; } } This causes a segmentation fault just running that, any ideas?

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  • .NET binary serialization conditionally without ISerializable

    - by SillyWhy
    I got 2 classes, for example: public class A { private B b; ... } public class B { ... } I need to serialize an object A using BinaryFormatter. When remoting it shall include the field b, but not when serialize to file. Here is what I added: [Serializable] public class A : MarshalByRefObject { private B b; [OnSerializing] private void OnSerializing(StreamingContext context) { if (context.State == StreamingContextStates.File) { this.b = null; } } ... } [Serializable] public class B : MarshalByRefObject { ... } I think this is a bad design because if another class C also contains B, in class C we must add the duplicate OnSerializing() logic as in A. Class B should decide what to do, not class A or C. I don't want to use ISerializable interface because there are too many variables in class B have to be added to SerializationInfo. I can create a SerializationSurrogate for class B, which perform nothing in GetObjectData() & SetObjectData(), then use it when serializing to file. However the same maintenance issue because whoever modify class B can't notice what going to happen during serialization & the existence of SerializationSurrogate. Is there a better alternative?

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  • Binary Search Tree can't delete the root

    - by Ali Zahr
    Everything is working fine in this function, but the problem is that I can't delete the root, I couldn't figure out what's the bug here.I've traced the "else part" it works fine until the return, it returns the old value I don't know why. Plz Help! node *removeNode(node *Root, int key) { node *tmp = new node; if(key > Root->value) Root->right = removeNode(Root->right,key); else if(key < Root->value) Root->left = removeNode(Root->left, key); else if(Root->left != NULL && Root->right != NULL) { node *minNode = findNode(Root->right); Root->value = minNode->value; Root->right = removeNode(Root->right,Root->value); } else { tmp = Root; if(Root->left == NULL) Root = Root->right; else if(Root->right == NULL) Root = Root->left; delete tmp; } return Root; }

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  • Error inserting data in binary tree

    - by chepe263
    I copied this code (in spanish) http://www.elrincondelc.com/nuevorincon/index.php?pag=codigos&id=4 and wrote a new one. This is my code: #include <cstdlib> #include <conio.h> #include <iostream> using namespace std; struct nodoarbol { int dato; struct nodoarbol *izq; struct nodoarbol *der; }; typedef nodoarbol Nodo; typedef Nodo *Arbol; void insertar(Arbol *, int); void inorden(Arbol); void postorden(Arbol); void preorden(Arbol); void insertar(Arbol *raiz, int nuevo){ if (*raiz==NULL){ *raiz = (Nodo *)malloc(sizeof(Nodo)); if (*raiz != NULL){ (*raiz)->dato=nuevo; (*raiz)->der=NULL; (*raiz)->izq=NULL; } else{ cout<<"No hay memoria suficiente u ocurrio un error"; } } else{ if (nuevo < (*raiz)->dato) insertar( &((*raiz)->izq), nuevo ); else if (nuevo > (*raiz)->dato) insertar(&((*raiz)->der), nuevo); } }//inseertar void inorden(Arbol raiz){ if (raiz != NULL){ inorden(raiz->izq); cout << raiz->dato << " "; inorden(raiz->der); } } void preorden(Arbol raiz){ if (raiz != NULL){ cout<< raiz->dato << " "; preorden(raiz->izq); preorden(raiz->der); } } void postorden(Arbol raiz){ if (raiz!=NULL){ postorden(raiz->izq); postorden(raiz->der); cout<<raiz->dato<<" "; } } int main() { int i; i=0; int val; Arbol raiz = NULL; for (i=0; i<10; i++){ cout<<"Inserte un numero"; cin>>val; insertar( (raiz), val); } cout<<"\nPreorden\n"; preorden(raiz); cout<<"\nIneorden\n"; inorden(raiz); cout<<"\nPostorden\n"; postorden(raiz); return 0; } I'm using netbeans 7.1.1, mingw32 compiler This is the output: make[2]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' make[1]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' main.cpp: In function 'int main()': main.cpp:110:30: error: cannot convert 'Arbol {aka nodoarbol*}' to 'Nodo** {aka nodoarbol**}' for argument '1' to 'void insertar(Nodo**, int)' make[2]: *** [build/Release/MinGW-Windows/main.o] Error 1 make[1]: *** [.build-conf] Error 2 make: *** [.build-impl] Error 2 BUILD FAILED (exit value 2, total time: 11s) I don't understand what's wrong since i just copied the code (and rewrite it to my own code). I'm really good in php, asp.net (vb) and other languages but c is a headche for me. I've been struggling with this problem for about an hour. Could somebody tell me what could it be?

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  • Better way to download a binary file?

    - by geoff
    I have a site where a user can download a file. Some files are extremely large (the largest being 323 MB). When I test it to try and download this file I get an out of memory exception. The only way I know to download the file is below. The reason I'm using the code below is because the URL is encoded and I can't let the user link directly to the file. Is there another way to download this file without having to read the whole thing into a byte array? FileStream fs = new FileStream(context.Server.MapPath(url), FileMode.Open, FileAccess.Read); BinaryReader br = new BinaryReader(fs); long numBytes = new FileInfo(context.Server.MapPath(url)).Length; byte[] bytes = br.ReadBytes((int) numBytes); string filename = Path.GetFileName(url); context.Response.Buffer = true; context.Response.Charset = ""; context.Response.Cache.SetCacheability(HttpCacheability.NoCache); context.Response.ContentType = "application/x-rar-compressed"; context.Response.AddHeader("content-disposition", "attachment;filename=" + filename); context.Response.BinaryWrite(bytes); context.Response.Flush(); context.Response.End();

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  • efficient video format/codec for sparse & binary blob tracking

    - by user391339
    I am working on a blob tracking project and have many high-definition videos that I would like to reduce in size for storage and downstream tracking/shape-analysis. I want to use a lossless method that takes advantage of the black and white nature of the video as well as the fact that not much is moving between individual frames. The videos are quite sparse, with 5 to 10 b&w blobs per frame occupying <30% of the space in total, with each blob moving <5-10% of the field of view between frames and not changing shape too much between 2-3 frames. I will work in Python, Matlab, or LabView for this project, and could use a batch utility if available. It may be worthwhile to export the files as compressed image stacks if a proper video format can't be found. What are the pros and cons of this? A video codec uses correlations between neighboring frames, so it should be more efficient, but not if the wrong one is chosen or if it is improperly configured.

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  • How many posibilities on a binary ?

    - by Val
    in hexadecimal "10 10 10 10" system you have 0-255 posibilities right? in total 256 different posibilities as there are 8 1s and 0s. how many different posibilities would i get? if i had 10 digits. instead of 8? or how would i calculate that in php ?

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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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  • What library for octrees or kd-trees?

    - by Will
    Are there any robust performant libraries for indexing objects? It would need frustum culling and visiting objects hit by a ray as well as neighbourhood searches. I can find lots of articles showing the math for the component parts, often as algebra rather than simple C, but nothing that puts it all together (apart from perhaps Ogre, which has rather more involved and isn't so stand-alone). Surely hobby game makers don't all have to make their own octrees? (Python or C/C++ w/bindings preferred)

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  • Behavior Trees and Animations

    - by Tom
    I have started working on the AI for a game, but am confused how I should handle animations. I will be using a Behavior Tree for AI behavior and Cocos2D for my game engine. Should my "PlayAnimationWalk" just be another node in the tree? Something similar to this: [Approach Player] - Play Walk animation - Move Towards player - Stop Walk animation Or should the node just update an AnimationState in the blackboard and have some type of animation handler/component reference this for which animation should be playing? This has been driving me nuts :)

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