Search Results

Search found 4504 results on 181 pages for 'tree grammar'.

Page 30/181 | < Previous Page | 26 27 28 29 30 31 32 33 34 35 36 37  | Next Page >

  • B-trees that use redistribution on insertion

    - by Phenom
    If I insert the following keys into a B-tree of order 4 (meaning 4 pointers and 3 elements in each node), I get the following B-tree. G / \ A IY Would it look any different if redistribution on insertion were used? How does redistribution on insertion work?

    Read the article

  • C++ vector and segmentation faults

    - by Headspin
    I am working on a simple mathematical parser. Something that just reads number = 1 + 2; I have a vector containing these tokens. They store a type and string value of the character. I am trying to step through the vector to build an AST of these tokens, and I keep getting segmentation faults, even when I am under the impression my code should prevent this from happening. Here is the bit of code that builds the AST: struct ASTGen { const vector<Token> &Tokens; unsigned int size, pointer; ASTGen(const vector<Token> &t) : Tokens(t), pointer(0) { size = Tokens.size() - 1; } unsigned int next() { return pointer + 1; } Node* Statement() { if(next() <= size) { switch(Tokens[next()].type) { case EQUALS : Node* n = Assignment_Expr(); return n; } } advance(); } void advance() { if(next() <= size) ++pointer; } Node* Assignment_Expr() { Node* lnode = new Node(Tokens[pointer], NULL, NULL); advance(); Node* n = new Node(Tokens[pointer], lnode, Expression()); return n; } Node* Expression() { if(next() <= size) { advance(); if(Tokens[next()].type == SEMICOLON) { Node* n = new Node(Tokens[pointer], NULL, NULL); return n; } if(Tokens[next()].type == PLUS) { Node* lnode = new Node(Tokens[pointer], NULL, NULL); advance(); Node* n = new Node(Tokens[pointer], lnode, Expression()); return n; } } } }; ... ASTGen AST(Tokens); Node* Tree = AST.Statement(); cout << Tree->Right->Data.svalue << endl; I can access Tree->Data.svalue and get the = Node's token info, so I know that node is getting spawned, and I can also get Tree->Left->Data.svalue and get the variable to the left of the = I have re-written it many times trying out different methods for stepping through the vector, but I always get a segmentation fault when I try to access the = right node (which should be the + node) Any help would be greatly appreciated.

    Read the article

  • Binary Trees in Scheme

    - by Javier
    Consider the following BNF defining trees of numbers. Notice that a tree can either be a leaf, a node-1 with one subtrees, or a node-2 with two subtrees. tree ::= (’leaf number) | (’node-1 tree) | (’node-2 tree tree) a. Write a template for recursive procedures on these trees. b. Define the procedure (leaf-count t) that returns the number of leaves in t > (leaf-count ’(leaf 5)) 1 > (leaf-count ’(node-2 (leaf 25) (leaf 17))) 2 > (leaf-count ’(node-1 (node-2 (leaf 4) (node-2 (leaf 2) (leaf 3))))) 3 Here's what I have so far: ;define what a leaf, node-1, and node-2 is (define leaf list) (define node-1 list) (define node-2 list) ;procedure to decide if a list is a leaf or a node (define (leaf? tree) (number? (car tree))) (define (node? tree) (pair? (car tree))) (define (leaf-count tree) (cond ((null? tree) 0) ((number? tree) 0) ((leaf? tree) 1) (else (+ (leaf-count (car tree)) (leaf-count (cdr tree)))))) It looks like it should run just fine, but when I try to run it using a simple test case like (leaf-count '(leaf 5)) I get the following error message: car: expects argument of type pair; given leaf What does this error message mean? I am defining a leaf as a list. But for some reason, it's not seeing that and gives me that error message.

    Read the article

  • How to make a tree view from MySQL and PHP and jquery

    - by Mac Taylor
    hey guys i need to show a treeview of my categories , saved in my mysql database . Database table : table : cats : columns: id,name,parent Here is a sample of what I want the markup to be like: <ul id="browser" class="filetree"> <li><span class="folder">Folder 1</span> <ul> <li><span class="file">Item 1.1</span></li> </ul> </li> <li><span class="folder">Folder 2</span> <ul> <li><span class="folder">Subfolder 2.1</span> <ul id="folder21"> <li><span class="file">File 2.1.1</span></li> <li><span class="file">File 2.1.2</span></li> </ul> </li> <li><span class="file">File 2.2</span></li> </ul> </li> <li><span class="file">File 4</span></li> </ul> i used this script to show treeview : http://www.dynamicdrive.com/dynamicindex1/treeview now problem is in php part : //function to build tree menu from db table test1 function tree_set($index) { global $menu; $q=mysql_query("select * from cats where parent='$index'"); if(!mysql_num_rows($q)) return; $menu .= '<ul>'."\n"; while($arr=mysql_fetch_assoc($q)) { $menu .= '<li>'; $menu .= '<span class="file">'.$arr['name'].'</span>';//you can add another output there $menu .=tree_set("".$arr['id'].""); $menu .= '</li>'."\n"; } $menu.= '</ul>'."\n"; return $menu; } //variable $menu must be defined before the function call $menu = ' <link rel="stylesheet" href="modules/Topics/includes/jquery.treeview.css" /> <script src="modules/Topics/includes/lib/jquery.cookie.js" type="text/javascript"></script> <script src="modules/Topics/includes/jquery.treeview.js" type="text/javascript"></script> <script type="text/javascript" src="modules/Topics/includes/demo/demo.js"></script> <ul id="browser" class="filetree">'."\n"; $menu .= tree_set(0); $menu .= '</ul>'; echo $menu; i even asked in this forum : http://forums.tizag.com/showthread.php?p=60649 problem is in php part of my codes that i mentioned . i cant show sub menus , i mean , really i dont know how to show sub menus is there any chance of a pro php coder helping me here ?

    Read the article

  • Lexer antlr3 token problem

    - by nioo
    Can I construct a token ENDPLUS: '+' (options (greedy = false;):.) * '+' ; being considered by the lexer only if it is preceded by a token PREwithout including in ENDPLUS? PRE: '<<' ; Thanks.

    Read the article

  • GNU Flex, multiline rule

    - by Simone Margaritelli
    Hi there i have a flex rule inside my lexer definition : operators "[]"|"[]="|"[]<"|".."|"."|".="|"+"|"+="|"-"|"-="|"/"|"/="|"*"|"*="|"%"|"%="|"++"|"--"|"^"|"^="|"~"|"&"|"&="|"|"|"|="|"<<"|"<<="|">>"|"!"|"<"|">"|">="|"<="|"=="|"!="|"&&"|"||"|"~=" Is there any way to split this ruole on more lines to keep it clearer? I tried with \ just like macros but it does not seem to be accepted by flex :( PS: I don't want to split the rule in more sub-rules, but only split its regex in more lines to keep the code clearer.

    Read the article

  • What makes Ometa special?

    - by Brian
    Ometa is "a new object-oriented language for pattern matching." I've encountered pattern matching in languages like Oz tools to parse grammars like Lexx/Yacc or Pyparsing before. Despite looking at example code, reading discussions, and talking to a friend, I still am not able to get a real understanding of what makes Ometa special (or at least, why some people think it is). Any explanation?

    Read the article

  • Dependency parsing

    - by C.
    Hi I particularly like the transduce feature offered by agfl in their EP4IR http://www.agfl.cs.ru.nl/EP4IR/english.html The download page is here: http://www.agfl.cs.ru.nl/download.html Is there any way i can make use of this in a c# program? Do I need to convert classes to c#? Thanks :)

    Read the article

  • Theory of computation - Using the pumping lemma for context free languages

    - by Tony
    I'm reviewing my notes for my course on theory of computation and I'm having trouble understanding how to complete a certain proof. Here is the question: A = {0^n 1^m 0^n | n>=1, m>=1} Prove that A is not regular. It's pretty obvious that the pumping lemma has to be used for this. So, we have |vy| = 1 |vxy| <= p (p being the pumping length, = 1) uv^ixy^iz exists in A for all i = 0 Trying to think of the correct string to choose seems a bit iffy for this. I was thinking 0^p 1^q 0^p, but I don't know if I can obscurely make a q, and since there is no bound on u, this could make things unruly.. So, how would one go about this?

    Read the article

  • Why is this c# snippet legal?

    - by Sir Psycho
    Silly question, but why does the following line compile? int[] i = new int[] {1,}; As you can see, I haven't entered in the second element and left a comma there. Still compiles even though you would expect it not to.

    Read the article

  • Data Quality Check - SQL Server

    - by user319384
    I am trying to find a good mechanism where I can check whether the data being entered by a group of people is grammatically correct, has correct spellings, etc, etc. I also would like to compute words per minute and accuracy. Is there any process to do this so that I do not have to re-invent the wheel? Thanks in advance.

    Read the article

  • The Definition of Regular Languages

    - by AraK
    Good Day, I have tried, and burned my brain to understand the definition of Regular Languages in Discrete Mathematics and its Applications(Rosen) without reaching the goal of understanding why the definition is like that in this book. On page(789), I am rephrasing the definition: Type 3 grammars are defined as: w1 --> w2 Where w1 is a non-terminal, and w2 is of the form: w2 = aB w2 = a Where B is a non-terminal, and a is a terminal. A special case is when w1 is the starting symbol and w2 is lambda(the empty string): w1 = S S --> lambda Two questions I couldn't find an answer for. First, Why can't w2 be of the form Ba. Second, Why lambda is only allowed for the starting symbol only. The book states that, regular languages are equivalent to Finite State Automaton, and we can easily see that a we can build FSA for both cases. I took a look at other resources, and these restrictions don't exist in these resources. Thanks,

    Read the article

  • Theory of computation - Using the pumping lemma for CFLs

    - by Tony
    I'm reviewing my notes for my course on theory of computation and I'm having trouble understanding how to complete a certain proof. Here is the question: A = {0^n 1^m 0^n | n>=1, m>=1} Prove that A is not regular. It's pretty obvious that the pumping lemma has to be used for this. So, we have |vy| = 1 |vxy| <= p (p being the pumping length, = 1) uv^ixy^iz exists in A for all i = 0 Trying to think of the correct string to choose seems a bit iffy for this. I was thinking 0^p 1^q 0^p, but I don't know if I can obscurely make a q, and since there is no bound on u, this could make things unruly.. So, how would one go about this?

    Read the article

  • How do I best do balanced quoting with Perl's Regexp::Grammars?

    - by Evan Carroll
    Using Damian Conway's Regexp::Grammars, I'm trying to match different balanced quoting ('foo', "foo", but not 'foo") mechanisms -- such as parens, quotes, double quotes, and double dollars. This is the code I'm currently using. <token: pair> \'<literal>\'|\"<literal>\"|\$\$<literal>\$\$ <token: literal> [\S]+ This generally works fine and allows me to say something like: <rule: quote> QUOTE <.as>? <pair> My question is how do I reform the output, to exclude the needles notation for the pair token? { '' => 'QUOTE AS \',\'', 'quote' => { '' => 'QUOTE AS \',\'', 'pair' => { 'literal' => ',', '' => '\',\'' } } }, Here, there is obviously no desire to have pair in between, quote, and the literal value of it. Is there a better way to match 'foo', "foo", and $$foo$$, and maybe sometimes ( foo ) without each time creating a needless pair token? Can I preprocess-out that token or fold it into the above? Or, write a better construct entirely that eliminates the need for it?

    Read the article

  • How do LL(*) parsers work?

    - by freezer878
    I cannot find any complete description about LL(*) parser, such as ANTLR, on Internet. I'm wondering what is the difference between an LL(k) parser and an LL(*) one and why they can't support left-recusrive grammars despite their flexibility.

    Read the article

  • Regular Expression

    - by equilibrium
    Ohh! this regular expression thing is eating my brain up. I have been reading it from Introduction to Automata Theory, Languages and Computer by Hopcroft, Motwani and Ullman. I have solved a few exercises too but could not solve the following even after trying for almost one hr. The problem is to write a regular expression that defines a language consisting of all strings of 0s and 1s except the substring 011. Is the answer (0+1)* - 011 correct ? If not what should be the correct answer for this?

    Read the article

  • regular expression for bit strings with even number of 1s

    - by equilibrium
    Let L= { w in (0+1)* | w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expressions below represents L? A) (0*10*1)* B) 0*(10*10*)* C) 0*(10*1)* 0* D) 0*1(10*1)* 10* According to me option D is never correct because it does not represent the bit string with zero 1s. But what about the other options? We are concerned about the number of 1s(even or not) not the number of zeros doesn't matter. Then which is the correct option and why?

    Read the article

  • Demantra 7.3.1.3 Controlling MDP_MATRIX Combinations Assigned to Forecasting Tasks Using TargetTaskSize

    - by user702295
    New 7.3.1.3 parameter: TargetTaskSize Old parameter: BranchID  Multiple, deprecated  7.3.1.3 onwards Parameter Location: Parameters > System Parameters > Engine > Proport   Default: 0   Engine Mode: Both   Details: Specifies how many MDP_MATRIX combinations the analytical engine attempts to assign to each forecasting task.  Allocation will be affected by forecsat tree branch size.  TaskTargetSize is automcatically calculated.  It holds the perferred branch size, in number of combinations in the lowest level. This parameter is adjusted to a lower value for smaller schemas, depending on the number of available engines.   - As the forecast is generated the engine goes up the tree using max_fore_level and not top_level -1.  Max_fore_level has     to be less than or equal to top_level -1.  Due to this requirement, combinations falling under the same top level -1     member must be in the same task.  A member of the top level -1 of the forecast tree is known as a branch.  An engine     task is therefore comprised of one or more branches.     - Reveal current task size       go to Engine Administrator --> View --> Branch Information and run the application on your Demantra schema.  This will be deprecated in 7.3.1.3 since there is no longer a means of adjusting the brach size directly.  The focus is now on proper hierarchy / forecast design.     - Control of tasks       The number of tasks created is the lowest of number of branches, as defined by top level -1 members in forecast       tree, and engine sessions and the value of TargetTaskSize.  You are used to using the branch multiplier in this       calculation.  As of 7.3.1.3, the branch ID multiple is deprecated.     - Discovery of current branch size       To resolve this you must review the 2nd highest level in the forecast tree (below highest/highest) as this is the       level which determines the size of the branches.  If a few resulting tasks are too large it is recommended that       the forecast tree level driving branches be revised or at times completely removed from the forecast tree.     - Control of foreacast tree branch size         - Run the following sql to determine how even the branches are being split by the engine:             select count(*),branch_id from mdp_matrix where prediction_status = 1 and do_fore = 1 group by branch_id;             This will give you an understanding if some of the individual branches have an unusually large number of           rows and thus might indicate that the engine is not efficiently dividing up the parallel tasks.         - Based on the results of this sql, we may want to adjust the branch id multiplier and/or the number of engines           (both of these settings are found in the Engine Administrator)           select count(*), level_id from mdp_matrix where prediction_status = 1 and do_fore = 1 group by level_id;           This will give us an understanding at which level of the Forecast tree where the forecast is being generated.            Having a majority of combinations higher on the forecast tree might indicate either a poorly designed forecast           tree and/or engine parameters that are too strict           Based on the results of this we would adjust the Forecast Tree to see if choosing a different hierarchy might           produce a forecast, with more combinations, at a lower level.           For example:             - Review the 2nd highest level in the forecast tree, below highest/highest, as this is the level which               determines the size of the branches.             - If a few resulting tasks are too large it is recommended that the forecast tree level driving branches               be revised or at times completely removed from the forecast tree.               - For example, if the highest level of the forecast tree is set to Brand/All Locations.             - You have 10 brands but 2 of the brands account for 67% and 29% of all combinations.             - There is a distinct possibility that the tasks resulting from these 2 branches will be too large for               a single engine to process.  Some possible solutions could be to remove the Brand level and instead               use a different product grouping which has a more even distribution, possibly Product Group.               - It is also possible to add a location dimension to this forecast tree level, for example Customer.                This will also reduce forecast tree branch size and will deliver a balanced task allocation.             - A correctly configured Forecast Tree is something that is done by the Implementation team and is               not the responsibility of Oracle Support.  Allocation will be affected by forecast tree branch size.  When TargetTaskSize is set to 0, the default value, the system automatically calculates a value for 'TargetTaskSize' depending on the number of engines.   - QUESTION:  Does this mean that if TargetTaskSize is 1, we use tree branch size to allocate branches to tasks instead                of automatically calculating the size?     ANSWER: DEV Strongly recommends that the setting of TargetTaskSize remain at the DEFAULT of ZERO (0).   - How to control the number of engines?     Determine how many CPUs are on the machine(s) that is (are) running the engine.  As mentioned earlier, the general     rule is that you should designate 2 engines per each CPU that is available.  So for example, if you are running the     engine on a machine that has 4 CPU then you can have up to 8 engines designated in the Engine Administrator.  In this     type of architecture then instead of having one 'localhost' in your Engine Settings Screen, you would have 'localhost'     repeated eight times in this field.     Where do I set the number of engines?                 To add multiples computers where engine will run, please do a back-up of Settings.xml file under         Analytical Engines\bin\ folder, then edit it and add there the selected machines.                 Example, this will allow 3 engines to start:         - <Entry>           <Key argument="ComputerNames" />           <Value type="string" argument="localhost,localhost,localhost" />           </Entry Otherwise, if there are no additional engines defined, the calculated value of 'TargetTaskSize' is used. (Oracle does not recommend changing the default value.) The TargetTaskSize holds the engines prefered branch size, in number of level 1 combinations.   - Level 1 combinations, known as group size The engine manager will use this parameter to attempt creating branches with similar size.   * The engine manager will not create engines that do not have a branch. The engine divider algorithm uses the value of 'TargetTaskSize' as a system-preferred branch size to create branches that are more equal in size which improves engine performance.  The engine divider will try to add as many tasks as possible to an existing branch, up to the limit of 'TargetTaskSize' level 1 combinations, before adding new branches. Coming up next: - The engine divider - Group size - Level 1 combinations - MAX_FORE_LEVEL - Engine Parameters  

    Read the article

< Previous Page | 26 27 28 29 30 31 32 33 34 35 36 37  | Next Page >