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• MATLAB - Delete elements of binary files without loading entire file

  - by Doresoom

  This may be a stupid question, but Google and MATLAB documentation have failed me. I have a rather large binary file (10 GB) that I need to open and delete the last forty million bytes or so. Is there a way to do this without reading the entire file to memory in chunks and printing it out to a new file? It took 6 hours to generate the file, so I'm cringing at the thought of re-reading the whole thing. EDIT: The file is 14,440,000,000 bytes in size. I need to chop it to 14,400,000,000.

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• In a binary search Tree

  - by user1044800

  In a binary search tree that takes a simple object.....when creating the getter and setter methods for the left, right, and parent. do I a do a null pointer? as in this=this or do I create the object in each method? Code bellow... This is my code: public void setParent(Person parent) { parent = new Person( parent.getName(), parent.getWeight()); //or is the parent supposed to be a null pointer ???? This is the code it came from: public void setParent(Node parent) { this.parent = parent; } Their code takes a node from the node class...my set parent is taking a person object from my person class.....

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• Optimize Binary Search Algorithm

  - by Ganesh M

  In a binary search, we have two comparisons one for greater than and other for less than, otherwise its the mid value. How would you optimize so that we need to check only once? bool binSearch(int array[], int key, int left, int right) { mid = left + (right-left)/2; if (key < array[mid]) return binSearch(array, key, left, mid-1); else if (key > array[mid]) return binSearch(array, key, mid+1, right); else if (key == array[mid]) return TRUE; // Found return FALSE; // Not Found }

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• printing double in binary

  - by Happy Mittal

  In Thinking in C++ by Bruce eckel, there is a program given to print a double value in binary.(Chapter 3, page no. 189) int main(int argc, char* argv[]) { if(argc != 2) { cout << "Must provide a number" << endl; exit(1); } double d = atof(argv[1]); unsigned char* cp = reinterpret_cast<unsigned char*>(&d); for(int i = sizeof(double); i > 0 ; i -= 2) { printBinary(cp[i-1]); printBinary(cp[i]); } } Here while printing cp[i] when i=8(assuming double is of 8 bytes), wouldn't it be undefined behaviour? I mean this code doesn't work as it doesn't print cp[0].

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• Dynamic binary from file

  - by Aurel300

  This is a little bit of weird problem here. Say I have a C++ code, running on a specific platform. The only purpose of this code is to run files containing binary, NATIVE to that platform. Now - my question is - HOW would I get the data from these files (could even be by bits, like 1024 bits a cycle) to the memory of machine running my code so that this data would be in the execution part? In other words, can I get the data to somewhere where I can point the instruction pointer? If yes, how? I don't mind if I have to use assembler for this - just so it would work.

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• JavaScript pack("d") - binary strings

  - by Tim Whitlock

  I'm trying to replicate the Perl and PHP style pack and unpack functions in JavaScript. Unsigned integers were easy enough, so my pack('n') and pack('N') are ok. But my lack of a computer science background is a hurdle now and I don't know where to start with pack('d') for packing JavaScript's standard floating point. Is there a JavaScript library for this out there? If not, is there a good resource where I can learn how to do this? I am fine with bitwise and binary level operations in JS, I just don't know where to start with the logic. Thanks.

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• Preferred way of filling up a C++ vector of structs

  - by henle

  Alternative 1, reusing a temporary variable: Sticker sticker; sticker.x = x + foreground.x; sticker.y = foreground.y; sticker.width = foreground.width; sticker.height = foreground.height; board.push_back(sticker); sticker.x = x + outline.x; sticker.y = outline.y; sticker.width = outline.width; sticker.height = outline.height; board.push_back(sticker); Alternative 2, scoping the temporary variable: { Sticker sticker; sticker.x = x + foreground.x; sticker.y = foreground.y; sticker.width = foreground.width; sticker.height = foreground.height; board.push_back(sticker); } { Sticker sticker; sticker.x = x + outline.x; sticker.y = outline.y; sticker.width = outline.width; sticker.height = outline.height; board.push_back(sticker); } Alternative 3, writing straight to the vector memory: { board.push_back(Sticker()); Sticker &sticker = board.back(); sticker.x = x + foreground.x; sticker.y = foreground.y; sticker.width = foreground.width; sticker.height = foreground.height; } { board.push_back(Sticker()); Sticker &sticker = board.back(); sticker.x = x + outline.x; sticker.y = outline.y; sticker.width = outline.width; sticker.height = outline.height; } Which approach do you prefer?

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• Points on lines where the two lines are the closest together

  - by James Bedford

  Hey guys, I'm trying to find the points on two lines where the two lines are the closest. I've implemented the following method (Points and Vectors are as you'd expect, and a Line consists of a Point on the line and a non-normalized direction Vector from that point): void CDClosestPointsOnTwoLines(Line line1, Line line2, Point* closestPoints) { closestPoints[0] = line1.pointOnLine; closestPoints[1] = line2.pointOnLine; Vector d1 = line1.direction; Vector d2 = line2.direction; float a = d1.dot(d1); float b = d1.dot(d2); float e = d2.dot(d2); float d = a*e - b*b; if (d != 0) // If the two lines are not parallel. { Vector r = Vector(line1.pointOnLine) - Vector(line2.pointOnLine); float c = d1.dot(r); float f = d2.dot(r); float s = (b*f - c*e) / d; float t = (a*f - b*c) / d; closestPoints[0] = line1.positionOnLine(s); closestPoints[1] = line2.positionOnLine(t); } else { printf("Lines were parallel.\n"); } } I'm using OpenGL to draw three lines that move around the world, the third of which should be the line that most closely connects the other two lines, the two end points of which are calculated using this function. The problem is that the first point of closestPoints after this function is called will lie on line1, but the second point won't lie on line2, let alone at the closest point on line2! I've checked over the function many times but I can't see where the mistake in my implementation is. I've checked my dot product function, scalar multiplication, subtraction, positionOnLine() etc. etc. So my assumption is that the problem is within this method implementation. If it helps to find the answer, this is function supposed to be an implementation of section 5.1.8 from 'Real-Time Collision Detection' by Christer Ericson. Many thanks for any help!

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• Python: fetching SVG file using urllib is returning binary when I need ASCII

  - by Drew Dara-Abrams

  I'm using urllib (in Python) to fetch an SVG file: import urllib urllib.urlopen('http://alpha.vectors.cloudmade.com/BC9A493B41014CAABB98F0471D759707/-122.2487,37.87588,-122.265823,37.868054?styleid=1&viewport=400x231').read() which produces output of the sort: xb6\xf6\x00\xb3\xfb2\xff\xda\xc5\xf2\xc2\x14\xef\xcd\x82\x0b\xdbU\xb0\x81\xcaF\xd8\x1a\xf6\xdf[i)\xba\xcf\x80\xab\xd6\x8c\xe3l_\xe7\n\xed2,\xbdm\xa0_|\xbb\x12\xff\xb6\xf8\xda\xd9\xc3\xd9\t\xde\x9a\xf8\xae\xb3T\xa3\r`\x8a\x08!T\xfb8\x92\x95\x0c\xdd\x8b!\x02P\xea@\x98\x1c^\xc7\xda\\\xec\xe3\xe1\xbe,0\xcd\xbeZ~\x92\xb3\xfa\xdd\xfcbyu\xb8\x83\xbb\xbdS\x0f\x82\x0b\xfe\xf5_\xdawn\xff\xef_\xff\xe5\xfa\x1f?\xbf\xffoZ\x0f\x8b\xbfV\xf4\x04\x00' when I was expecting more like this: <?xml version='1.0' encoding='UTF-8'?> <svg xmlns="http://www.w3.org/2000/svg" xmlns:cm="http://cloudmade.com/" width="400" height="231"> <rect width="100%" height="100%" fill="#eae8dd" opacity="1"/> <g transform="scale(0.209849975856)"> <g transform="translate(13610569, 4561906)" flood-opacity="0.1" flood-color="grey"> <path d="M -13610027.720000000670552 -4562403.660000000149012 I guess this is an issue of binary vs. ASCII. Can anyone help me (a Python newbie) with the appropriate conversion so that I can get on with parsing and manipulating the SVG code?

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• Binary file email attachment problem

  - by Alan Harris-Reid

  Hi there, Using Python 3.1.2 I am having a problem sending binary attachment files (jpeg, pdf, etc.) - MIMEText attachments work fine. The code in question is as follows... for file in self.attachments: part = MIMEBase('application', "octet-stream") part.set_payload(open(file,"rb").read()) encoders.encode_base64(part) part.add_header('Content-Disposition', 'attachment; filename="%s"' % file) msg.attach(part) # msg is an instance of MIMEMultipart() server = smtplib.SMTP(host, port) server.login(username, password) server.sendmail(from_addr, all_recipients, msg.as_string()) However, way down in the calling-stack (see traceback below), it looks as though msg.as_string() has received an attachment which creates a payload of 'bytes' type instead of string. Has anyone any idea what might be causing the problem? Any help would be appreciated. Alan builtins.TypeError: string payload expected: File "c:\Dev\CommonPY\Scripts\email_send.py", line 147, in send server.sendmail(self.from_addr, all_recipients, msg.as_string()) File "c:\Program Files\Python31\Lib\email\message.py", line 136, in as_string g.flatten(self, unixfrom=unixfrom) File "c:\Program Files\Python31\Lib\email\generator.py", line 76, in flatten self._write(msg) File "c:\Program Files\Python31\Lib\email\generator.py", line 101, in _write self._dispatch(msg) File "c:\Program Files\Python31\Lib\email\generator.py", line 127, in _dispatch meth(msg) File "c:\Program Files\Python31\Lib\email\generator.py", line 181, in _handle_multipart g.flatten(part, unixfrom=False) File "c:\Program Files\Python31\Lib\email\generator.py", line 76, in flatten self._write(msg) File "c:\Program Files\Python31\Lib\email\generator.py", line 101, in _write self._dispatch(msg) File "c:\Program Files\Python31\Lib\email\generator.py", line 127, in _dispatch meth(msg) File "c:\Program Files\Python31\Lib\email\generator.py", line 155, in _handle_text raise TypeError('string payload expected: %s' % type(payload))

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• C# - parse content away from structure in a binary file

  - by Jeff Godfrey

  Using C#, I need to read a packed binary file created using FORTRAN. The file is stored in an "Unformatted Sequential" format as described here (about half-way down the page in the "Unformatted Sequential Files" section): http://www.tacc.utexas.edu/services/userguides/intel8/fc/f_ug1/pggfmsp.htm As you can see from the URL, the file is organized into "chunks" of 130 bytes or less and includes 2 length bytes (inserted by the FORTRAN compiler) surrounding each chunk. So, I need to find an efficient way to parse the actual file payload away from the compiler-inserted formatting. Once I've extracted the actual payload from the file, I'll then need to parse it up into its varying data types. That'll be the next exercise. My first thoughts are to slurp up the entire file into a byte array using File.ReadAllBytes. Then, just iterate through the bytes, skipping the formatting and transferring the actual data to a second byte array. In the end, that second byte array should contain the actual file contents minus all the formatting, which I'd then need to go back through to get what I need. As I'm fairly new to C#, I thought there might be a better, more accepted way of tackling this. Also, in case it's helpful, these files could be fairly large (say 30MB), though most will be much smaller...

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• compile Boost as static Universal binary lib

  - by Albert

  I want to have a static Universal binary lib of Boost. (Preferable the latest stable version, that is 1.43.0, or newer.) I found many Google hits with similar problems and possible solutions. However, most of them seems outdated. Also none of them really worked. Right now, I am trying sudo ./bjam --toolset=darwin --link=static --threading=multi \ --architecture=combined --address-model=32_64 \ --macosx-version=10.4 --macosx-version-min=10.4 \ install That compiles and install fine. However, the produced binaries seems broken. az@ip245 47 (openlierox) %file /usr/local/lib/libboost_signals.a /usr/local/lib/libboost_signals.a: current ar archive random library az@ip245 49 (openlierox) %lipo -info /usr/local/lib/libboost_signals.a input file /usr/local/lib/libboost_signals.a is not a fat file Non-fat file: /usr/local/lib/libboost_signals.a is architecture: x86_64 az@ip245 48 (openlierox) %otool -hv /usr/local/lib/libboost_signals.a Archive : /usr/local/lib/libboost_signals.a /usr/local/lib/libboost_signals.a(trackable.o): Mach header magic cputype cpusubtype caps filetype ncmds sizeofcmds flags MH_MAGIC_64 X86_64 ALL 0x00 OBJECT 3 1536 SUBSECTIONS_VIA_SYMBOLS /usr/local/lib/libboost_signals.a(connection.o): Mach header magic cputype cpusubtype caps filetype ncmds sizeofcmds flags MH_MAGIC_64 X86_64 ALL 0x00 OBJECT 3 1776 SUBSECTIONS_VIA_SYMBOLS /usr/local/lib/libboost_signals.a(named_slot_map.o): Mach header magic cputype cpusubtype caps filetype ncmds sizeofcmds flags MH_MAGIC_64 X86_64 ALL 0x00 OBJECT 3 1856 SUBSECTIONS_VIA_SYMBOLS /usr/local/lib/libboost_signals.a(signal_base.o): Mach header magic cputype cpusubtype caps filetype ncmds sizeofcmds flags MH_MAGIC_64 X86_64 ALL 0x00 OBJECT 3 1776 SUBSECTIONS_VIA_SYMBOLS /usr/local/lib/libboost_signals.a(slot.o): Mach header magic cputype cpusubtype caps filetype ncmds sizeofcmds flags MH_MAGIC_64 X86_64 ALL 0x00 OBJECT 3 1616 SUBSECTIONS_VIA_SYMBOLS Any suggestion how to get that correct?

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• Ruby GraphViz Binary Tree Record

  - by Jason M

  I'm using the ruby-graphviz gem and I'm trying to draw binary trees. I'd like to use the record shape so that each node can have a left, middle, and right field and, thus, if there are two edges leaving a node, the left and right edges can be distinguished. I tried specifying the field by concatenating the field name like this: @node1.name + ":left" But that did not work. What is the correct way of specifying the field? require 'rubygems' require 'graphviz' @graph = GraphViz.new( :G, :type => :digraph ) @node1 = @graph.add_node("1", "shape" => "record", "label" => "<left>|<f1> 1|<right>" ) @node2 = @graph.add_node("2", "shape" => "record", "label" => "<left>|<f1> 2|<right>" ) @graph.add_edge(@node1.name + ":left", @node2) # generate a random filename filename = "/tmp/#{(0...8).map{65.+(rand(25)).chr}.join}.png" @graph.output( :png => filename ) exec "open #{filename}"

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• Passing Binary Data to a Stored Procedure in SQL Server 2008

  - by Joe Majewski

  I'm trying to figure out a way to store files in a database. I know it's recommended to store files on the file system rather than the database, but the job I'm working on would highly prefer using the database to store these images (files). There are also some constraints. I'm not an admin user, and I have to make stored procedures to execute all the commands. This hasn't been of much difficulty so far, but I cannot for the life of me establish a way to store a file (image) in the database. When I try to use the BULK command, I get an error saying "You do not have permission to use the bulk load statement." The bulk utility seemed like the easy way to upload files to the database, but without permissions I have to figure a work-a-round. I decided to use an HTML form with a file upload input type and handle it with PHP. The PHP calls the stored procedure and passes in the contents of the file. The problem is that now it's saying that the max length of a parameter can only be 128 characters. Now I'm completely stuck. I don't have permissions to use the bulk command and it appears that the max length of a parameter that I can pass to the SP is 128 characters. I expected to run into problems because binary characters and ascii characters don't mix well together, but I'm at a dead end... Thanks

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• parse content away from structure in a binary file

  - by Jeff Godfrey

  Using C#, I need to read a packed binary file created using FORTRAN. The file is stored in an "Unformatted Sequential" format as described here (about half-way down the page in the "Unformatted Sequential Files" section): http://www.tacc.utexas.edu/services/userguides/intel8/fc/f_ug1/pggfmsp.htm As you can see from the URL, the file is organized into "chunks" of 130 bytes or less and includes 2 length bytes (inserted by the FORTRAN compiler) surrounding each chunk. So, I need to find an efficient way to parse the actual file payload away from the compiler-inserted formatting. Once I've extracted the actual payload from the file, I'll then need to parse it up into its varying data types. That'll be the next exercise. My first thoughts are to slurp up the entire file into a byte array using File.ReadAllBytes. Then, just iterate through the bytes, skipping the formatting and transferring the actual data to a second byte array. In the end, that second byte array should contain the actual file contents minus all the formatting, which I'd then need to go back through to get what I need. As I'm fairly new to C#, I thought there might be a better, more accepted way of tackling this. Also, in case it's helpful, these files could be fairly large (say 30MB), though most will be much smaller...

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• Binary Search Tree for specific intent

  - by Luís Guilherme

  We all know there are plenty of self-balancing binary search trees (BST), being the most famous the Red-Black and the AVL. It might be useful to take a look at AA-trees and scapegoat trees too. I want to do deletions insertions and searches, like any other BST. However, it will be common to delete all values in a given range, or deleting whole subtrees. So: I want to insert, search, remove values in O(log n) (balanced tree). I would like to delete a subtree, keeping the whole tree balanced, in O(log n) (worst-case or amortized) It might be useful to delete several values in a row, before balancing the tree I will most often insert 2 values at once, however this is not a rule (just a tip in case there is a tree data structure that takes this into account) Is there a variant of AVL or RB that helps me on this? Scapegoat-trees look more like this, but would also need some changes, anyone who has got experience on them can share some thougts? More precisely, which balancing procedure and/or removal procedure would help me keep this actions time-efficient?

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• Using recursion to to trim a binary tree based on a given min and max value

  - by Justin

  As the title says, I have to trim a binary tree based on a given min and max value. Each node stores a value, and a left/right node. I may define private helper methods to solve this problem, but otherwise I may not call any other methods of the class nor create any data structures such as arrays, lists, etc. An example would look like this: overallRoot _____[50]____________________ / \ __________[38] _______________[90] / \ / _[14] [42] [54]_____ / \ \ [8] [20] [72] \ / \ [26] [61] [83] trim(52, 65); should return: overallRoot [54] \ [61] My attempted solution has three methods: public void trim(int min, int max) { rootFinder(overallRoot, min, max); } First recursive method finds the new root perfectly. private void rootFinder(IntTreeNode node, int min, int max) { if (node == null) return; if (overallRoot.data < min) { node = overallRoot = node.right; rootFinder(node, min, max); } else if (overallRoot.data > max) { node = overallRoot = node.left; rootFinder(node, min, max); } else cutter(overallRoot, min, max); } This second method should eliminate any further nodes not within the min/max, but it doesn't work as I would hope. private void cutter(IntTreeNode node, int min, int max) { if (node == null) return; if (node.data <= min) { node.left = null; } if (node.data >= max) { node.right = null; } if (node.data < min) { node = node.right; } if (node.data > max) { node = node.left; } cutter(node.left, min, max); cutter(node.right, min, max); } This returns: overallRoot [54]_____ \ [72] / [61] Any help is appreciated. Feel free to ask for further explanation as needed.

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• Java Binary Tree. Priting InOrder traversal

  - by user69514

  I am having some problems printing an inOrder traversal of my binary tree. Even after inserting many items into the tree it's only printing 3 items. public class BinaryTree { private TreeNode root; private int size; public BinaryTree(){ this.size = 0; } public boolean insert(TreeNode node){ if( root == null) root = node; else{ TreeNode parent = null; TreeNode current = root; while( current != null){ if( node.getData().getValue().compareTo(current.getData().getValue()) <0){ parent = current; current = current.getLeft(); } else if( node.getData().getValue().compareTo(current.getData().getValue()) >0){ parent = current; current = current.getRight(); } else return false; if(node.getData().getValue().compareTo(parent.getData().getValue()) < 0) parent.setLeft(node); else parent.setRight(node); } } size++; return true; } /** * */ public void inOrder(){ inOrder(root); } private void inOrder(TreeNode root){ if( root.getLeft() !=null) this.inOrder(root.getLeft()); System.out.println(root.getData().getValue()); if( root.getRight() != null) this.inOrder(root.getRight()); } }

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• Permuting a binary tree without the use of lists

  - by Banang

  I need to find an algorithm for generating every possible permutation of a binary tree, and need to do so without using lists (this is because the tree itself carries semantics and restraints that cannot be translated into lists). I've found an algorithm that works for trees with the height of three or less, but whenever I get to greater hights, I loose one set of possible permutations per height added. Each node carries information about its original state, so that one node can determine if all possible permutations have been tried for that node. Also, the node carries information on weather or not it's been 'swapped', i.e. if it has seen all possible permutations of it's subtree. The tree is left-centered, meaning that the right node should always (except in some cases that I don't need to cover for this algorithm) be a leaf node, while the left node is always either a leaf or a branch. The algorithm I'm using at the moment can be described sort of like this: if the left child node has been swapped swap my right node with the left child nodes right node set the left child node as 'unswapped' if the current node is back to its original state swap my right node with the lowest left nodes' right node swap the lowest left nodes two childnodes set my left node as 'unswapped' set my left chilnode to use this as it's original state set this node as swapped return null return this; else if the left child has not been swapped if the result of trying to permute left child is null return the permutation of this node else return the permutation of the left child node if this node has a left node and a right node that are both leaves swap them set this node to be 'swapped' The desired behaviour of the algoritm would be something like this: branch / | branch 3 / | branch 2 / | 0 1 branch / | branch 3 / | branch 2 / | 1 0 <-- first swap branch / | branch 3 / | branch 1 <-- second swap / | 2 0 branch / | branch 3 / | branch 1 / | 0 2 <-- third swap branch / | branch 3 / | branch 0 <-- fourth swap / | 1 2 and so on... Sorry for the ridiculisly long and waddly explanation, would really, really apreciate any sort of help you guys could offer me. Thanks a bunch!

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• Line Intersection from parametric equation

  - by Sidar

  I'm sure this question has been asked before. However, I'm trying to connect the dots by translating an equation on paper into an actual function. I thought It would be interesting to ask here instead on the Math sites (since it's going to be used for games anyway ). Let's say we have our vector equation : x = s + Lr; where x is the resulting vector, s our starting point/vector. L our parameter and r our direction vector. The ( not sure it's called like this, please correct me ) normal equation is : x.n = c; If we substitute our vector equation we get: (s+Lr).n = c. We now need to isolate L which results in L = (c - s.n) / (r.n); L needs to be 0 < L < 1. Meaning it needs to be between 0 and 1. My question: I want to know what L is so if I were to substitute L for both vector equation (or two lines) they should give me the same intersection coordinates. That is if they intersect. But I can't wrap my head around on how to use this for two lines and find the parameter that fits the intersection point. Could someone with a simple example show how I could translate this to a function/method?

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• Inorder tree traversal in binary tree in C

  - by srk

  In the below code, I'am creating a binary tree using insert function and trying to display the inserted elements using inorder function which follows the logic of In-order traversal.When I run it, numbers are getting inserted but when I try the inorder function( input 3), the program continues for next input without displaying anything. I guess there might be a logical error.Please help me clear it. Thanks in advance... #include<stdio.h> #include<stdlib.h> int i; typedef struct ll { int data; struct ll *left; struct ll *right; } node; node *root1=NULL; // the root node void insert(node *root,int n) { if(root==NULL) //for the first(root) node { root=(node *)malloc(sizeof(node)); root->data=n; root->right=NULL; root->left=NULL; } else { if(n<(root->data)) { root->left=(node *)malloc(sizeof(node)); insert(root->left,n); } else if(n>(root->data)) { root->right=(node *)malloc(sizeof(node)); insert(root->right,n); } else { root->data=n; } } } void inorder(node *root) { if(root!=NULL) { inorder(root->left); printf("%d ",root->data); inorder(root->right); } } main() { int n,choice=1; while(choice!=0) { printf("Enter choice--- 1 for insert, 3 for inorder and 0 for exit\n"); scanf("%d",&choice); switch(choice) { case 1: printf("Enter number to be inserted\n"); scanf("%d",&n); insert(root1,n); break; case 3: inorder(root1); break; default: break; } } }

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• Invoking a function (main()) from a binary file in C

  - by Dhara Darji

  I have simple c program like, my_bin.c: #include <stdio.h> int main() { printf("Success!\n"); return 0; } I compile it with gcc and got executable: my_bin. Now I want to invoke main (or run this my_bin) using another C program. That I did with mmap and function pointer like this: #include <stdio.h> #include <fcntl.h> #include <sys/mman.h> int main() { void (*fun)(); int fd; int *map; fd = open("./my_bin", O_RDONLY); map = mmap(0, 8378, PROT_READ, MAP_SHARED, fd, 0); fun = map; fun(); return 0; } PS: I went through some tutorial, for how to read binary file and execute. But this gives Seg fault, any help appreciated! Thanks!

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• SQL Binary Microsoft Access - Combining two tables if specific field values are equal

  - by Jordan

  I am new to Microsoft Access and SQL but have a decent programming background and I believe this problem should be relatively simple. I have two tables that I have imported into Access. I will give you a little context. One table is huge and contains generic, global data. The other table is still big but contains specific, regional data. There is only one common field (or column) between the two tables. Let’s call this common field CF. The other fields in both tables are different. I’ll take you through one iteration of what I need to do. I need to take each CF value in the regional, smaller table and find the common CF value in the larger, global table. After finding the match, I need to take the whole “record” or “row” from the global data and copy it over to the corresponding record in the smaller regional table (This should involve creating the new fields). I need to do this for all CF values in the regional, smaller table. I was recommended to use SQL and a binary search, but I am unfamiliar. Let me know if you have any questions. I appreciate the help!

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• Practicing inserting data into an array by using binary search, few problems

  - by HelpNeeder

  I'm trying to create a method which inserts and then sort elements in form of binary form. The problem I am experiencing that my code doesn't insert data correctly which means that output does not appear to be in order at all. The list is not organized, and data is added in order that is being inserted. Now, 2 questions, what am I doing wrong here? And how to fix this? public void insertBinarySearch(long value) // put element into array { int j = 0; int lower = 0; int upper = elems-1; int cur = 0; while (cur < elems) { curIn = (lower + upper ) / 2; if(a[cur] < value) { j = cur + 1; break; } else if(a[cur] > value) { j = cur; break; } else { if(a[cur] < value) lower = cur + 1; else upper = cur - 1; } } for(int k = elems; k > j; k--) a[k] = a[k-1]; a[j] = value; elems++; }

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• Convert a binary tree to linked list, breadth first, constant storage/destructive

  - by Merlyn Morgan-Graham

  This is not homework, and I don't need to answer it, but now I have become obsessed :) The problem is: Design an algorithm to destructively flatten a binary tree to a linked list, breadth-first. Okay, easy enough. Just build a queue, and do what you have to. That was the warm-up. Now, implement it with constant storage (recursion, if you can figure out an answer using it, is logarithmic storage, not constant). I found a solution to this problem on the Internet about a year back, but now I've forgotten it, and I want to know :) The trick, as far as I remember, involved using the tree to implement the queue, taking advantage of the destructive nature of the algorithm. When you are linking the list, you are also pushing an item into the queue. Each time I try to solve this, I lose nodes (such as each time I link the next node/add to the queue), I require extra storage, or I can't figure out the convoluted method I need to get back to a node that has the pointer I need. Even the link to that original article/post would be useful to me :) Google is giving me no joy. Edit: Jérémie pointed out that there is a fairly simple (and well known answer) if you have a parent pointer. While I now think he is correct about the original solution containing a parent pointer, I really wanted to solve the problem without it :) The refined requirements use this definition for the node: struct tree_node { int value; tree_node* left; tree_node* right; };

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