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  • Few Steps For Making Your Website SEO Friendly

    Currently going by the dynamics of the internet every web designer of class is expected to be acquainted with or have a basic working knowledge of search engine optimization tools. Web promotion as an optimization tool for enhancing increased traffic to a given site is one skill web designers just like online marketers require in their respective fields to conquer the market.

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  • How Local SEO Can Improve Your Business

    Local search engine optimization is a good first step to conquer the search engines and present your business. This article summarizes some of the positive aspects of Local SEO, and why businesses should not scared to embrace the internet as a new marketing medium.

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  • Converting Openfire IM datetime values in SQL Server to / from VARCHAR(15) and DATETIME data types

    - by Brian Biales
    A client is using Openfire IM for their users, and would like some custom queries to audit user conversations (which are stored by Openfire in tables in the SQL Server database). Because Openfire supports multiple database servers and multiple platforms, the designers chose to store all date/time stamps in the database as 15 character strings, which get converted to Java Date objects in their code (Openfire is written in Java).  I did some digging around, and, so I don't forget and in case someone else will find this useful, I will put the simple algorithms here for converting back and forth between SQL DATETIME and the Java string representation. The Java string representation is the number of milliseconds since 1/1/1970.  SQL Server's DATETIME is actually represented as a float, the value being the number of days since 1/1/1900, the portion after the decimal point representing the hours/minutes/seconds/milliseconds... as a fractional part of a day.  Try this and you will see this is true:     SELECT CAST(0 AS DATETIME) and you will see it returns the date 1/1/1900. The difference in days between SQL Server's 0 date of 1/1/1900 and the Java representation's 0 date of 1/1/1970 is found easily using the following SQL:   SELECT DATEDIFF(D, '1900-01-01', '1970-01-01') which returns 25567.  There are 25567 days between these dates. So to convert from the Java string to SQL Server's date time, we need to convert the number of milliseconds to a floating point representation of the number of days since 1/1/1970, then add the 25567 to change this to the number of days since 1/1/1900.  To convert to days, you need to divide the number by 1000 ms/s, then by  60 seconds/minute, then by 60 minutes/hour, then by 24 hours/day.  Or simply divide by 1000*60*60*24, or 86400000.   So, to summarize, we need to cast this string as a float, divide by 86400000 milliseconds/day, then add 25567 days, and cast the resulting value to a DateTime.  Here is an example:   DECLARE @tmp as VARCHAR(15)   SET @tmp = '1268231722123'   SELECT @tmp as JavaTime, CAST((CAST(@tmp AS FLOAT) / 86400000) + 25567 AS DATETIME) as SQLTime   To convert from SQL datetime back to the Java time format is not quite as simple, I found, because floats of that size do not convert nicely to strings, they end up in scientific notation using the CONVERT function or CAST function.  But I found a couple ways around that problem. You can convert a date to the number of  seconds since 1/1/1970 very easily using the DATEDIFF function, as this value fits in an Int.  If you don't need to worry about the milliseconds, simply cast this integer as a string, and then concatenate '000' at the end, essentially multiplying this number by 1000, and making it milliseconds since 1/1/1970.  If, however, you do care about the milliseconds, you will need to use DATEPART to get the milliseconds part of the date, cast this integer to a string, and then pad zeros on the left to make sure this is three digits, and concatenate these three digits to the number of seconds string above.  And finally, I discovered by casting to DECIMAL(15,0) then to VARCHAR(15), I avoid the scientific notation issue.  So here are all my examples, pick the one you like best... First, here is the simple approach if you don't care about the milliseconds:   DECLARE @tmp as VARCHAR(15)   DECLARE @dt as DATETIME   SET @dt = '2010-03-10 14:35:22.123'   SET @tmp = CAST(DATEDIFF(s, '1970-01-01 00:00:00' , @dt) AS VARCHAR(15)) + '000'   SELECT @tmp as JavaTime, @dt as SQLTime If you want to keep the milliseconds:   DECLARE @tmp as VARCHAR(15)   DECLARE @dt as DATETIME   DECLARE @ms as int   SET @dt = '2010-03-10 14:35:22.123'   SET @ms as DATEPART(ms, @dt)   SET @tmp = CAST(DATEDIFF(s, '1970-01-01 00:00:00' , @dt) AS VARCHAR(15))           + RIGHT('000' + CAST(@ms AS VARCHAR(3)), 3)   SELECT @tmp as JavaTime, @dt as SQLTime Or, in one fell swoop:   DECLARE @dt as DATETIME   SET @dt = '2010-03-10 14:35:22.123'   SELECT @dt as SQLTime     , CAST(DATEDIFF(s, '1970-01-01 00:00:00' , @dt) AS VARCHAR(15))           + RIGHT('000' + CAST( DATEPART(ms, @dt) AS VARCHAR(3)), 3) as JavaTime   And finally, a way to simply reverse the math used converting from Java date to SQL date. Note the parenthesis - watch out for operator precedence, you want to subtract, then multiply:   DECLARE @dt as DATETIME   SET @dt = '2010-03-10 14:35:22.123'   SELECT @dt as SQLTime     , CAST(CAST((CAST(@dt as Float) - 25567.0) * 86400000.0 as DECIMAL(15,0)) as VARCHAR(15)) as JavaTime Interestingly, I found that converting to SQL Date time can lose some accuracy, when I converted the time above to Java time then converted  that back to DateTime, the number of milliseconds is 120, not 123.  As I am not interested in the milliseconds, this is ok for me.  But you may want to look into using DateTime2 in SQL Server 2008 for more accuracy.

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  • Sublinear Extra Space MergeSort

    - by hulkmeister
    I am reviewing basic algorithms from a book called Algorithms by Robert Sedgewick, and I came across a problem in MergeSort that I am, sad to say, having difficulty solving. The problem is below: Sublinear Extra Space. Develop a merge implementation that reduces that extra space requirement to max(M, N/M), based on the following idea: Divide the array into N/M blocks of size M (for simplicity in this description, assume that N is a multiple of M). Then, (i) considering the blocks as items with their first key as the sort key, sort them using selection sort; and (ii) run through the array merging the first block with the second, then the second block with the third, and so forth. The problem I have with the problem is that based on the idea Sedgewick recommends, the following set of arrays will not be sorted: {0, 10, 12}, {3, 9, 11}, {5, 8, 13}. The algorithm I use is the following: Divide the full array into subarrays of size M. Run Selection Sort on each of the subarrays. Merge each of the subarrays using the method Sedgwick recommends in (ii). (This is where I encounter the problem of where to store the results after the merge.) This leads to wanting to increase the size of the auxiliary space needed to handle at least two subarrays at a time (for merging), but based on the specifications of the problem, that is not allowed. I have also considered using the original array as space for one subarray and using the auxiliary space for the second subarray. However, I can't envision a solution that does not end up overwriting the entries of the first subarray. Any ideas on other ways this can be done? NOTE: If this is suppose to be on StackOverflow.com, please let me know how I can move it. I posted here because the question was academic.

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  • MS-DOSdivide overflow

    - by repozitor
    when i want to install my dear application on MS-DOS os, i see error "Divide Overflow" what is the meaning of this error and how to fix it? the procedure of installing is: 1-partitioing my HDD disk 2-format C drive 3-installing MS-DOS 4-add the flowing lines to config.sys "DEVICE=C:\DOS\HIMEM.SYS DEVICE=C:\DOS\EMM386.EXE RAM DEVICE=C:\DOS\RAMDRIVE.SYS 6000 512 64 /e" 5-insert my floppy application and then restart 6-when boot process completed, all of the thing appear correctly and i now how to do the next steps note:for installing my application i need to boot from floppy app when i have ms-dos on my hdd disk i do it successfully on the Virtual machine, Q emulator but i can't do it on the real machine, Vectra HP PC, because of divide overflow

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  • Non-Profit Technololgy for Non-Profits?

    - by TomJ
    I've been looking around for a way to give back to the community, but I haven't found my right fit yet, so an idea came to mind: A non-profit technology "company" that targets non-profits. Do these exist? I've been doing some google searches and can only find software that is targeted for non-profits that is created by for-profit companies or that charges what I believe to be an outrages amount, conferences directed towards non-profits and technology they may use -- or articles complaining about the digital divide and how non-profits view technology as key but dont have the funds or the knowledge to employ it. Pseudo "Business Model" An open source 501(3)(c) organization that targets directly targets non-profits to fill the "digital divide." Most services would be free and consulting fees would be charged for customization. Donations would be accepted and government grants would be sought after. This would enable non-profits to keep pace with the for-profits in the technology sector, but at little to no cost. Perhaps the first "industry" to be targeted would be those that fill key social needs like unemployment, or food banks.

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  • Dividing up spritesheet in Javascript

    - by hustlerinc
    I would like to implement an object for my spritesheets in Javascript. I'm very new to this language and game-developement so I dont really know how to do it. My guess is I set spritesize to 16, use that to divide as many times as it fits on the spritesheet and store this value as "spritesheet". Then a for(i=0;i<spritesheet.length;i++) loop running over the coordinates. Then tile = new Image(); and tile.src = spritesheet[i] to store the individual sprites based on their coordinates on the spritesheet. My problem is how could I loop trough the spritesheet and make an array of that? The result should be similar to: var tile = Array( "img/ground.png", "img/crate.png" ); If possible this would be done with one single object that i only access once, and the tile array would be stored for later reference. I couldn't find anything similar searching for "javascript spritesheet". Edit: I made a small prototype of what I'm after: function Sprite(){ this.size = 16; this.spritesheet = new Image(); this.spritesheet.src = 'img/spritesheet.png'; this.countX = this.spritesheet.width / 16; this.countY = this.spritesheet.height / 16; this.spriteCount = this.countX * this.countY; this.divide = function(){ for(i=0;i<this.spriteCount;i++){ // define spritesheet coordinates and store as tile[i] } } } Am I on the right track?

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  • Inspiring the method of teaching. Example- C++ :)

    - by Ashwin
    A year ago I graduated with a degree in Computer Science and Engineering. Considering C++ as the first choice of programming language I have been in the process of learning C++ in many ways. At first - five years back - I had many conceptions, most of which were so abstract to me. It started when I knew almost everything about Structs in C and nothing about Classes in C++. I went through a great time experimenting them all and learning a lot. I had a hard time evaluating Procedural programming vs Object-Oriented Programming. Deciding when to choose Procedural or Object-Oriented Programming took a great deal of patience for me. I knew that I cannot underestimate any of these Programming styles... Though Procedural programming is often a better choice than simple sequential unstructured programming, when solving problems with procedural programming, we usually divide one problem into several steps in order regarded as functions. Then we call these functions one by one to get the result of the problem. When solving problems with Object Oriented Priciples we divide one problem into several classes and form the interaction between them. Evaluating these two at the beginning (as a learner) required a lot of inspiration and thoughts. Instructing to think step by step. Relative concepts to understand deeply. Intensive interests to contrast both solving in both POP and OOP. If you were ever a mentor: What ideas/methods would you teach to students in which it will Inspire them to learn a programming language (in general, computer sciences)?

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  • replace specefique codes in a script using through a tool

    - by Moudiz
    I have a script that contain random codes but I am searching for a way in notepad ++ or for a batch-file or any tool that can replace sepcifque codes, here is an example: Random If this equal that then you sould do this and do that therefore.. the code should be executed immediatly --stackb select * from user_error where object_name = name select * from user_error where table= randomly case 1 a = b else c=a --stacke Begin with the structure of the data and divide the codes end with what you know I want to replace the words between the comments stack b and stack a so the result will be like below Random If this equal that then you sould do this and do that therefore.. the code should be executed immediatly --stackb The codes here has been replaced, can you do that ? case 1 a = b else c=a --stacke Begin with the structure of the data and divide the codes end with what you know Is there a code in batch file or note pad ++ where I can acheive my result?

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  • Octrees and Vertex Buffer Objects

    - by sharethis
    As many others I want to code a game with a voxel based terrain. The data is represented by voxels which are rendered using triangles. I head of two different approaches and want to combine them. First, I could once divide the space in chunks of a fixed size like many games do. What I could do now is to generate a polygon shape for each chunk and store that in a vertex buffer object (vbo). Each time a voxel changes, the polygon and vbo of its chunk is recreated. Additionally it is easy to dynamically load and reload parts of the terrain. Another approach would be to use octrees and divide the space in eight cubes which are divided again and again. So I could efficiently render the terrain because I don't have to go deeper in a solid cube and can draw that as a single one (with a repeated texture). What I like to use for my game is an octree datastructure. But I can't imagine how to use vbos with that. How is that done, or is this impossible?

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  • nginx hashing on GET parameter

    - by Sparsh Gupta
    I have two Varnish servers and I plan to add more varnish servers. I am using a nginx load balancer to divide traffic to these varnish servers. To utilize maximum RAM of each varnish server, I need that same request reaches same varnish server. Same request can be identified by one GET parameter in the request URL say 'a' In a normal code, I would do something like- (if I need to divide all traffic between 2 Varnish servers) if($arg_a % 2 == 0) { proxy_pass varnish1; } if($arg_a % 2 == 1) { proxy_pass varnish2; } This is basically doing a even / odd check on GET parameter a and then deciding which upstream pool to send the request. My question are- What is the nginx equivalent of such a code. I dont know if nginx accepts modulas Is there a better/ efficient hashing function built in with nginx (0.8.54) which I can possibly use. In future I want to add more upstream pools so I need not to change %2 to %3 %4 and so on Any other alternate way to solve this problem

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  • How to reference or vlookup a list of values based on a comma separated list of column references within a cell in excel?

    - by glallen
    I want to do a vlookup (or similar) against a column which is a list of values. This works fine for looking up a value from a single row, but I want to be able to look up multiple rows, sum the results, and divide by the number of rows referenced. For example: A B C D E F G [----given values----------------] [Work/Auth] [sum(vlookup(each(G),table,5)) /count(G)] [given vals] 1 Item Authorized OnHand Working Operational% DependencyOR% Dependencies 2 A 1 1 1 1 .55 B 3 B 10 5 5 .50 .55 C,D 4 C 100 75 50 .50 .60 D 5 D 10 10 6 .60 1 I want to be able to show an Operational Rate, and an operational rate of the systems each system depends on (F). In order to get a value for F, I want to sum over each value in column-E that was referenced by a dependency in column-G then divide by the number of dependencies in G. Column-G can have varying lengths, and will be a comma separated list of values from column-A. Is there any way to do this in excel?

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  • Can i use a switch to hold a function?

    - by TIMOTHY
    I have a 3 file program, basically teaching myself c++. I have an issue. I made a switch to use the math function. I need and put it in a variable, but for some reason I get a zero as a result. Also another issue, when I select 4 (divide) it crashes... Is there a reason? Main file: #include <iostream> #include "math.h" #include <string> using namespace std; int opersel; int c; int a; int b; string test; int main(){ cout << "Welcome to Math-matrix v.34"<< endl; cout << "Shall we begin?" <<endl; //ASK USER IF THEY ARE READY TO BEGIN string answer; cin >> answer; if(answer == "yes" || answer == "YES" || answer == "Yes") { cout << "excellent lets begin..." << endl; cout << "please select a operator..." << endl << endl; cout << "(1) + " << endl; cout << "(2) - " << endl; cout << "(3) * " << endl; cout << "(4) / " << endl; cin >> opersel; switch(opersel){ case 1: c = add(a,b); break; case 2: c = sub(a,b); break; case 3: c = multi(a,b); break; case 4: c = divide(a,b); break; default: cout << "error... retry" << endl; }// end retry cout << "alright, how please select first digit?" << endl; cin >> a; cout << "excellent... and your second?" << endl; cin >> b; cout << c; cin >> test; }else if (answer == "no" || answer == "NO" || answer == "No"){ }//GAME ENDS }// end of int main Here is my math.h file #ifndef MATH_H #define MATH_H int add(int a, int b); int sub(int a, int b); int multi(int a, int b); int divide(int a, int b); #endif Here is my math.cpp: int add(int a, int b) { return a + b; } int sub(int a, int b) { return a - b; } int multi(int a, int b) { return a * b; } int divide(int a, int b) { return a / b; } }// end of int main

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  • Scala parser combinator runs out of memory

    - by user3217013
    I wrote the following parser in Scala using the parser combinators: import scala.util.parsing.combinator._ import scala.collection.Map import scala.io.StdIn object Keywords { val Define = "define" val True = "true" val False = "false" val If = "if" val Then = "then" val Else = "else" val Return = "return" val Pass = "pass" val Conj = ";" val OpenParen = "(" val CloseParen = ")" val OpenBrack = "{" val CloseBrack = "}" val Comma = "," val Plus = "+" val Minus = "-" val Times = "*" val Divide = "/" val Pow = "**" val And = "&&" val Or = "||" val Xor = "^^" val Not = "!" val Equals = "==" val NotEquals = "!=" val Assignment = "=" } //--------------------------------------------------------------------------------- sealed abstract class Op case object Plus extends Op case object Minus extends Op case object Times extends Op case object Divide extends Op case object Pow extends Op case object And extends Op case object Or extends Op case object Xor extends Op case object Not extends Op case object Equals extends Op case object NotEquals extends Op case object Assignment extends Op //--------------------------------------------------------------------------------- sealed abstract class Term case object TrueTerm extends Term case object FalseTerm extends Term case class FloatTerm(value : Float) extends Term case class StringTerm(value : String) extends Term case class Identifier(name : String) extends Term //--------------------------------------------------------------------------------- sealed abstract class Expression case class TermExp(term : Term) extends Expression case class UnaryOp(op : Op, exp : Expression) extends Expression case class BinaryOp(op : Op, left : Expression, right : Expression) extends Expression case class FuncApp(funcName : Term, args : List[Expression]) extends Expression //--------------------------------------------------------------------------------- sealed abstract class Statement case class ExpressionStatement(exp : Expression) extends Statement case class Pass() extends Statement case class Return(value : Expression) extends Statement case class AssignmentVar(variable : Term, exp : Expression) extends Statement case class IfThenElse(testBody : Expression, thenBody : Statement, elseBody : Statement) extends Statement case class Conjunction(left : Statement, right : Statement) extends Statement case class AssignmentFunc(functionName : Term, args : List[Term], body : Statement) extends Statement //--------------------------------------------------------------------------------- class myParser extends JavaTokenParsers { val keywordMap : Map[String, Op] = Map( Keywords.Plus -> Plus, Keywords.Minus -> Minus, Keywords.Times -> Times, Keywords.Divide -> Divide, Keywords.Pow -> Pow, Keywords.And -> And, Keywords.Or -> Or, Keywords.Xor -> Xor, Keywords.Not -> Not, Keywords.Equals -> Equals, Keywords.NotEquals -> NotEquals, Keywords.Assignment -> Assignment ) def floatTerm : Parser[Term] = decimalNumber ^^ { case x => FloatTerm( x.toFloat ) } def stringTerm : Parser[Term] = stringLiteral ^^ { case str => StringTerm(str) } def identifier : Parser[Term] = ident ^^ { case value => Identifier(value) } def boolTerm : Parser[Term] = (Keywords.True | Keywords.False) ^^ { case Keywords.True => TrueTerm case Keywords.False => FalseTerm } def simpleTerm : Parser[Expression] = (boolTerm | floatTerm | stringTerm) ^^ { case term => TermExp(term) } def argument = expression def arguments_aux : Parser[List[Expression]] = (argument <~ Keywords.Comma) ~ arguments ^^ { case arg ~ argList => arg :: argList } def arguments = arguments_aux | { argument ^^ { case arg => List(arg) } } def funcAppArgs : Parser[List[Expression]] = funcEmptyArgs | ( Keywords.OpenParen ~> arguments <~ Keywords.CloseParen ^^ { case args => args.foldRight(List[Expression]()) ( (a,b) => a :: b ) } ) def funcApp = identifier ~ funcAppArgs ^^ { case funcName ~ argList => FuncApp(funcName, argList) } def variableTerm : Parser[Expression] = identifier ^^ { case name => TermExp(name) } def atomic_expression = simpleTerm | funcApp | variableTerm def paren_expression : Parser[Expression] = Keywords.OpenParen ~> expression <~ Keywords.CloseParen def unary_operation : Parser[String] = Keywords.Not def unary_expression : Parser[Expression] = operation(0) ~ expression(0) ^^ { case op ~ exp => UnaryOp(keywordMap(op), exp) } def operation(precedence : Int) : Parser[String] = precedence match { case 0 => Keywords.Not case 1 => Keywords.Pow case 2 => Keywords.Times | Keywords.Divide | Keywords.And case 3 => Keywords.Plus | Keywords.Minus | Keywords.Or | Keywords.Xor case 4 => Keywords.Equals | Keywords.NotEquals case _ => throw new Exception("No operations with this precedence.") } def binary_expression(precedence : Int) : Parser[Expression] = precedence match { case 0 => throw new Exception("No operation with zero precedence.") case n => (expression (n-1)) ~ operation(n) ~ (expression (n)) ^^ { case left ~ op ~ right => BinaryOp(keywordMap(op), left, right) } } def expression(precedence : Int) : Parser[Expression] = precedence match { case 0 => unary_expression | paren_expression | atomic_expression case n => binary_expression(n) | expression(n-1) } def expression : Parser[Expression] = expression(4) def expressionStmt : Parser[Statement] = expression ^^ { case exp => ExpressionStatement(exp) } def assignment : Parser[Statement] = (identifier <~ Keywords.Assignment) ~ expression ^^ { case varName ~ exp => AssignmentVar(varName, exp) } def ifthen : Parser[Statement] = ((Keywords.If ~ Keywords.OpenParen) ~> expression <~ Keywords.CloseParen) ~ ((Keywords.Then ~ Keywords.OpenBrack) ~> statements <~ Keywords.CloseBrack) ^^ { case ifBody ~ thenBody => IfThenElse(ifBody, thenBody, Pass()) } def ifthenelse : Parser[Statement] = ((Keywords.If ~ Keywords.OpenParen) ~> expression <~ Keywords.CloseParen) ~ ((Keywords.Then ~ Keywords.OpenBrack) ~> statements <~ Keywords.CloseBrack) ~ ((Keywords.Else ~ Keywords.OpenBrack) ~> statements <~ Keywords.CloseBrack) ^^ { case ifBody ~ thenBody ~ elseBody => IfThenElse(ifBody, thenBody, elseBody) } def pass : Parser[Statement] = Keywords.Pass ^^^ { Pass() } def returnStmt : Parser[Statement] = Keywords.Return ~> expression ^^ { case exp => Return(exp) } def statement : Parser[Statement] = ((pass | returnStmt | assignment | expressionStmt) <~ Keywords.Conj) | ifthenelse | ifthen def statements_aux : Parser[Statement] = statement ~ statements ^^ { case st ~ sts => Conjunction(st, sts) } def statements : Parser[Statement] = statements_aux | statement def funcDefBody : Parser[Statement] = Keywords.OpenBrack ~> statements <~ Keywords.CloseBrack def funcEmptyArgs = Keywords.OpenParen ~ Keywords.CloseParen ^^^ { List() } def funcDefArgs : Parser[List[Term]] = funcEmptyArgs | Keywords.OpenParen ~> repsep(identifier, Keywords.Comma) <~ Keywords.CloseParen ^^ { case args => args.foldRight(List[Term]()) ( (a,b) => a :: b ) } def funcDef : Parser[Statement] = (Keywords.Define ~> identifier) ~ funcDefArgs ~ funcDefBody ^^ { case funcName ~ funcArgs ~ body => AssignmentFunc(funcName, funcArgs, body) } def funcDefAndStatement : Parser[Statement] = funcDef | statement def funcDefAndStatements_aux : Parser[Statement] = funcDefAndStatement ~ funcDefAndStatements ^^ { case stmt ~ stmts => Conjunction(stmt, stmts) } def funcDefAndStatements : Parser[Statement] = funcDefAndStatements_aux | funcDefAndStatement def parseProgram : Parser[Statement] = funcDefAndStatements def eval(input : String) = { parseAll(parseProgram, input) match { case Success(result, _) => result case Failure(m, _) => println(m) case _ => println("") } } } object Parser { def main(args : Array[String]) { val x : myParser = new myParser() println(args(0)) val lines = scala.io.Source.fromFile(args(0)).mkString println(x.eval(lines)) } } The problem is, when I run the parser on the following example it works fine: define foo(a) { if (!h(IM) && a) then { return 0; } if (a() && !h()) then { return 0; } } But when I add threes characters in the first if statement, it runs out of memory. This is absolutely blowing my mind. Can anyone help? (I suspect it has to do with repsep, but I am not sure.) define foo(a) { if (!h(IM) && a(1)) then { return 0; } if (a() && !h()) then { return 0; } } EDIT: Any constructive comments about my Scala style is also appreciated.

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  • Sorting Algorithms

    - by MarkPearl
    General Every time I go back to university I find myself wading through sorting algorithms and their implementation in C++. Up to now I haven’t really appreciated their true value. However as I discovered this last week with Dictionaries in C# – having a knowledge of some basic programming principles can greatly improve the performance of a system and make one think twice about how to tackle a problem. I’m going to cover briefly in this post the following: Selection Sort Insertion Sort Shellsort Quicksort Mergesort Heapsort (not complete) Selection Sort Array based selection sort is a simple approach to sorting an unsorted array. Simply put, it repeats two basic steps to achieve a sorted collection. It starts with a collection of data and repeatedly parses it, each time sorting out one element and reducing the size of the next iteration of parsed data by one. So the first iteration would go something like this… Go through the entire array of data and find the lowest value Place the value at the front of the array The second iteration would go something like this… Go through the array from position two (position one has already been sorted with the smallest value) and find the next lowest value in the array. Place the value at the second position in the array This process would be completed until the entire array had been sorted. A positive about selection sort is that it does not make many item movements. In fact, in a worst case scenario every items is only moved once. Selection sort is however a comparison intensive sort. If you had 10 items in a collection, just to parse the collection you would have 10+9+8+7+6+5+4+3+2=54 comparisons to sort regardless of how sorted the collection was to start with. If you think about it, if you applied selection sort to a collection already sorted, you would still perform relatively the same number of iterations as if it was not sorted at all. Many of the following algorithms try and reduce the number of comparisons if the list is already sorted – leaving one with a best case and worst case scenario for comparisons. Likewise different approaches have different levels of item movement. Depending on what is more expensive, one may give priority to one approach compared to another based on what is more expensive, a comparison or a item move. Insertion Sort Insertion sort tries to reduce the number of key comparisons it performs compared to selection sort by not “doing anything” if things are sorted. Assume you had an collection of numbers in the following order… 10 18 25 30 23 17 45 35 There are 8 elements in the list. If we were to start at the front of the list – 10 18 25 & 30 are already sorted. Element 5 (23) however is smaller than element 4 (30) and so needs to be repositioned. We do this by copying the value at element 5 to a temporary holder, and then begin shifting the elements before it up one. So… Element 5 would be copied to a temporary holder 10 18 25 30 23 17 45 35 – T 23 Element 4 would shift to Element 5 10 18 25 30 30 17 45 35 – T 23 Element 3 would shift to Element 4 10 18 25 25 30 17 45 35 – T 23 Element 2 (18) is smaller than the temporary holder so we put the temporary holder value into Element 3. 10 18 23 25 30 17 45 35 – T 23   We now have a sorted list up to element 6. And so we would repeat the same process by moving element 6 to a temporary value and then shifting everything up by one from element 2 to element 5. As you can see, one major setback for this technique is the shifting values up one – this is because up to now we have been considering the collection to be an array. If however the collection was a linked list, we would not need to shift values up, but merely remove the link from the unsorted value and “reinsert” it in a sorted position. Which would reduce the number of transactions performed on the collection. So.. Insertion sort seems to perform better than selection sort – however an implementation is slightly more complicated. This is typical with most sorting algorithms – generally, greater performance leads to greater complexity. Also, insertion sort performs better if a collection of data is already sorted. If for instance you were handed a sorted collection of size n, then only n number of comparisons would need to be performed to verify that it is sorted. It’s important to note that insertion sort (array based) performs a number item moves – every time an item is “out of place” several items before it get shifted up. Shellsort – Diminishing Increment Sort So up to now we have covered Selection Sort & Insertion Sort. Selection Sort makes many comparisons and insertion sort (with an array) has the potential of making many item movements. Shellsort is an approach that takes the normal insertion sort and tries to reduce the number of item movements. In Shellsort, elements in a collection are viewed as sub-collections of a particular size. Each sub-collection is sorted so that the elements that are far apart move closer to their final position. Suppose we had a collection of 15 elements… 10 20 15 45 36 48 7 60 18 50 2 19 43 30 55 First we may view the collection as 7 sub-collections and sort each sublist, lets say at intervals of 7 10 60 55 – 20 18 – 15 50 – 45 2 – 36 19 – 48 43 – 7 30 10 55 60 – 18 20 – 15 50 – 2 45 – 19 36 – 43 48 – 7 30 (Sorted) We then sort each sublist at a smaller inter – lets say 4 10 55 60 18 – 20 15 50 2 – 45 19 36 43 – 48 7 30 10 18 55 60 – 2 15 20 50 – 19 36 43 45 – 7 30 48 (Sorted) We then sort elements at a distance of 1 (i.e. we apply a normal insertion sort) 10 18 55 60 2 15 20 50 19 36 43 45 7 30 48 2 7 10 15 18 19 20 30 36 43 45 48 50 55 (Sorted) The important thing with shellsort is deciding on the increment sequence of each sub-collection. From what I can tell, there isn’t any definitive method and depending on the order of your elements, different increment sequences may perform better than others. There are however certain increment sequences that you may want to avoid. An even based increment sequence (e.g. 2 4 8 16 32 …) should typically be avoided because it does not allow for even elements to be compared with odd elements until the final sort phase – which in a way would negate many of the benefits of using sub-collections. The performance on the number of comparisons and item movements of Shellsort is hard to determine, however it is considered to be considerably better than the normal insertion sort. Quicksort Quicksort uses a divide and conquer approach to sort a collection of items. The collection is divided into two sub-collections – and the two sub-collections are sorted and combined into one list in such a way that the combined list is sorted. The algorithm is in general pseudo code below… Divide the collection into two sub-collections Quicksort the lower sub-collection Quicksort the upper sub-collection Combine the lower & upper sub-collection together As hinted at above, quicksort uses recursion in its implementation. The real trick with quicksort is to get the lower and upper sub-collections to be of equal size. The size of a sub-collection is determined by what value the pivot is. Once a pivot is determined, one would partition to sub-collections and then repeat the process on each sub collection until you reach the base case. With quicksort, the work is done when dividing the sub-collections into lower & upper collections. The actual combining of the lower & upper sub-collections at the end is relatively simple since every element in the lower sub-collection is smaller than the smallest element in the upper sub-collection. Mergesort With quicksort, the average-case complexity was O(nlog2n) however the worst case complexity was still O(N*N). Mergesort improves on quicksort by always having a complexity of O(nlog2n) regardless of the best or worst case. So how does it do this? Mergesort makes use of the divide and conquer approach to partition a collection into two sub-collections. It then sorts each sub-collection and combines the sorted sub-collections into one sorted collection. The general algorithm for mergesort is as follows… Divide the collection into two sub-collections Mergesort the first sub-collection Mergesort the second sub-collection Merge the first sub-collection and the second sub-collection As you can see.. it still pretty much looks like quicksort – so lets see where it differs… Firstly, mergesort differs from quicksort in how it partitions the sub-collections. Instead of having a pivot – merge sort partitions each sub-collection based on size so that the first and second sub-collection of relatively the same size. This dividing keeps getting repeated until the sub-collections are the size of a single element. If a sub-collection is one element in size – it is now sorted! So the trick is how do we put all these sub-collections together so that they maintain their sorted order. Sorted sub-collections are merged into a sorted collection by comparing the elements of the sub-collection and then adjusting the sorted collection. Lets have a look at a few examples… Assume 2 sub-collections with 1 element each 10 & 20 Compare the first element of the first sub-collection with the first element of the second sub-collection. Take the smallest of the two and place it as the first element in the sorted collection. In this scenario 10 is smaller than 20 so 10 is taken from sub-collection 1 leaving that sub-collection empty, which means by default the next smallest element is in sub-collection 2 (20). So the sorted collection would be 10 20 Lets assume 2 sub-collections with 2 elements each 10 20 & 15 19 So… again we would Compare 10 with 15 – 10 is the winner so we add it to our sorted collection (10) leaving us with 20 & 15 19 Compare 20 with 15 – 15 is the winner so we add it to our sorted collection (10 15) leaving us with 20 & 19 Compare 20 with 19 – 19 is the winner so we add it to our sorted collection (10 15 19) leaving us with 20 & _ 20 is by default the winner so our sorted collection is 10 15 19 20. Make sense? Heapsort (still needs to be completed) So by now I am tired of sorting algorithms and trying to remember why they were so important. I think every year I go through this stuff I wonder to myself why are we made to learn about selection sort and insertion sort if they are so bad – why didn’t we just skip to Mergesort & Quicksort. I guess the only explanation I have for this is that sometimes you learn things so that you can implement them in future – and other times you learn things so that you know it isn’t the best way of implementing things and that you don’t need to implement it in future. Anyhow… luckily this is going to be the last one of my sorts for today. The first step in heapsort is to convert a collection of data into a heap. After the data is converted into a heap, sorting begins… So what is the definition of a heap? If we have to convert a collection of data into a heap, how do we know when it is a heap and when it is not? The definition of a heap is as follows: A heap is a list in which each element contains a key, such that the key in the element at position k in the list is at least as large as the key in the element at position 2k +1 (if it exists) and 2k + 2 (if it exists). Does that make sense? At first glance I’m thinking what the heck??? But then after re-reading my notes I see that we are doing something different – up to now we have really looked at data as an array or sequential collection of data that we need to sort – a heap represents data in a slightly different way – although the data is stored in a sequential collection, for a sequential collection of data to be in a valid heap – it is “semi sorted”. Let me try and explain a bit further with an example… Example 1 of Potential Heap Data Assume we had a collection of numbers as follows 1[1] 2[2] 3[3] 4[4] 5[5] 6[6] For this to be a valid heap element with value of 1 at position [1] needs to be greater or equal to the element at position [3] (2k +1) and position [4] (2k +2). So in the above example, the collection of numbers is not in a valid heap. Example 2 of Potential Heap Data Lets look at another collection of numbers as follows 6[1] 5[2] 4[3] 3[4] 2[5] 1[6] Is this a valid heap? Well… element with the value 6 at position 1 must be greater or equal to the element at position [3] and position [4]. Is 6 > 4 and 6 > 3? Yes it is. Lets look at element 5 as position 2. It must be greater than the values at [4] & [5]. Is 5 > 3 and 5 > 2? Yes it is. If you continued to examine this second collection of data you would find that it is in a valid heap based on the definition of a heap.

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  • linear interpolation on 8bit microcontroller

    - by JB
    I need to do a linear interpolation over time between two values on an 8 bit PIC microcontroller (Specifically 16F627A but that shouldn't matter) using PIC assembly language. Although I'm looking for an algorithm here as much as actual code. I need to take an 8 bit starting value, an 8 bit ending value and a position between the two (Currently represented as an 8 bit number 0-255 where 0 means the output should be the starting value and 255 means it should be the final value but that can change if there is a better way to represent this) and calculate the interpolated value. Now PIC doesn't have a divide instruction so I could code up a general purpose divide routine and effectivly calculate (B-A)/(x/255)+A at each step but I feel there is probably a much better way to do this on a microcontroller than the way I'd do it on a PC in c++ Has anyone got any suggestions for implementing this efficiently on this hardware?

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  • How is schoolbook long division an O(n^2) algorithm?

    - by eSKay
    Premise: This Wikipedia page suggests that the computational complexity of Schoolbook long division is O(n^2). Deduction: Instead of taking "Two n-digit numbers", if I take one n-digit number and one m-digit number, then the complexity would be O(n*m). Contradiction: Suppose you divide 100000000 (n digits) by 1000 (m digits), you get 100000, which takes six steps to arrive at. Now, if you divide 100000000 (n digits) by 10000 (m digits), you get 10000 . Now this takes only five steps. Conclusion: So, it seems that the order of computation should be something like O(n/m). Question: Who is wrong, me or Wikipedia, and where?

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  • assembly language programming (prime number)

    - by chris
    Prompt the user for a positive three digit number, then read it. Let's call it N. Divide into N all integer values from 2 to (N/2)+1 and test to see if the division was even, in which case N is instantly shown to be non-prime. Output a message printing N and saying that it is not prime. If none of those integer values divide evenly (remainder never is zero), then N is shown to be prime. Output a message printing N and saying that it is prime. Ask the user if he or she wants to test another number; if the user types "n" or "N", quit. If "y" or "Y", jump back and repeat. Comments in your code are essential. Hi. I am kinda in rush to do this.. please help me doing it. I'll be much appreciated. thank you

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  • Clojure: Avoiding stack overflow in Sieve of Erathosthene?

    - by nixx
    Here's my implementation of Sieve of Erathosthene in Clojure (based on SICP lesson on streams): (defn nats-from [n] (iterate inc n)) (defn divide? [p q] (zero? (rem q p))) (defn sieve [stream] (lazy-seq (cons (first stream) (sieve (remove #(divide? (first stream) %) (rest stream)))))) (def primes (sieve (nats-from 2))) Now, it's all OK when i take first 100 primes: (take 100 primes) But, if i try to take first 1000 primes, program breaks because of stack overflow. I'm wondering if is it possible to change somehow function sieve to become tail-recursive and, still, to preserve "streamnes" of algorithm? Any help???

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  • Changing floating point behavior in Python to Numpy style.

    - by Tristan
    Is there a way to make Python floating point numbers follow numpy's rules regarding +/- Inf and NaN? For instance, making 1.0/0.0 = Inf. >>> from numpy import * >>> ones(1)/0 array([ Inf]) >>> 1.0/0.0 Traceback (most recent call last): File "<stdin>", line 1, in <module> ZeroDivisionError: float division Numpy's divide function divide(1.0,0.0)=Inf however it is not clear if it can be used similar to from __future__ import division.

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  • convert function from Access SQL to T-SQL 2005

    - by Pace
    Can someone please convert this access sql function for me to work in t-sql 2005. I am tring to take the selling price minus the cost as one number. And divide that by the original selling price to produce a second number Thanks :) =IIf([Selling Price]=0,0,([Selling Price]-Nz([Cost]))/[Selling Price]) IIRC it should be something along the lines of; ISNULL((ISNULL([Selling Price],0) - ISNULL(Cost,0)),0) / ISNULL([Selling Price],0) AS Margin But here I am getting a divide by Zero error. any suggestions?

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  • Second largest number in list python

    - by Manu Lakaster
    So I have to find THE SECOND LARGEST NUMBER IN A LIST. I am doing it through simple loops.My approach is I am going to divide a list into two parts and then find the largest number into two parts and then compare two nuumbers. I will choose the smaller number from two of them. I can not use ready functions or different approaches. Basically, this is my code....But it does not run correctly....Help me please to fix it because I spent a lot of time on it :( Thanks....P.S. Can we use indices to "divide" a list ??? #!/usr/local/bin/python2.7 alist=[-45,0,3,10,90,5,-2,4,18,45,100,1,-266,706] largest=alist[0] h=len(alist)/2 m=len(alist)-h print(alist) for i in alist: if alist[h]>largest: largest=alist[h] i=i+1 print(largest)

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