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What is recursion? -- In plain english.

- by Christopher Altman

I hear this word everyday, but cannot give a meaningful, concise, or plain-english answer to what it is. Recursion is defined by the bastian of knowledge as: Recursion in computer science is a method where the solution to a problem depends on solutions to smaller instances of the same problem.1 The approach can be applied to many types of problems, and is one of the central ideas of computer science.[2] Source Is recursion simply something that repeats itself to get a solution? I am looking for a "Recursion for Dummies" definition, and maybe simple examples. My goal is to be able to understand and explain recursion in my own words. I do not like simply thinking I know the meaning of something because I hear it referenced daily, but have not paused to form my own understanding.

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Is recursion preferred compare to iteration in multicore era?

- by prM

Or say, do multicore CPUs process recursion faster than iteration? Or it simply depends on how one language runs on the machine? like c executes function calls with large cost, comparing to doing simple iterations. I had this question because one day I told one of my friend that recursion isn't any amazing magic that can speed up programs, and he told me that with multicore CPUs recursion can be faster than iteration. EDIT: If we consider the most recursion-loved situation (data structure, function call), is it even possible for recursion to be faster?

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Recursion in Java enums?

- by davidrobles

I've been trying for 3 hours and I just can't understand what is happening here. I have an enum 'Maze'. For some reason, when the method 'search' is called on this enum it's EXTREMELY slow (3 minutes to run). However, if I copy the same method to another class as a static method, and I call it from the enum 'Maze' it runs in one second! I don't understand why? is there any problem with recursive methods in Java enums?? What am I doing wrong? public enum Maze { A("A.txt"), B("B.txt"); // variables here... Maze(String fileName) { loadMap(fileName); nodeDistances = new int[nodes.size()][nodes.size()]; setNeighbors(); setDistances(); } ... more methods here ... private void setDistances() { nodeDistances = new int[nodes.size()][nodes.size()]; for (int i = 0; i < nodes.size(); i++) { setMax(nodeDistances[i]); // This works!!! TestMaze.search(nodes, nodeDistances[i], i, 0); // This DOESN'T WORK //search(nodes, nodeDistances[i], i, 0); } } public void setMax(int[] a) { for (int i=0; i<a.length; i++) { a[i] = Integer.MAX_VALUE; } } public void search(List<Node> allNodes, int[] distances, int curNodeIndex, int curDist) { if (curDist < distances[curNodeIndex]) { distances[curNodeIndex] = curDist; for (Node n : allNodes.get(curNodeIndex).getNeighbors()) { search(allNodes, distances, n.getNodeIndex(), curDist + 1); } } } } public class TestMaze { public static void search(List<Node> allNodes, int[] distances, int curNodeIndex, int curDist) { if (curDist < distances[curNodeIndex]) { distances[curNodeIndex] = curDist; for (Node n : allNodes.get(curNodeIndex).getNeighbors()) { search(allNodes, distances, n.getNodeIndex(), curDist + 1); } } } }

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Dynamic Programming Recursion and a sprinkle of Memoization

- by Auburnate

I have this massive array of ints from 0-4 in this triangle. I am trying to learn dynamic programming with Ruby and would like some assistance in calculating the number of paths in the triangle that meet three criterion: You must start at one of the zero points in the row with 70 elements. Your path can be directly above you one row (if there is a number directly above) or one row up heading diagonal to the left. One of these options is always available The sum of the path you take to get to the zero on the first row must add up to 140. Example, start at the second zero in the bottom row. You can move directly up to the one or diagonal left to the 4. In either case, the number you arrive at must be added to the running count of all the numbers you have visited. From the 1 you can travel to a 2 (running sum = 3) directly above or to the 0 (running sum = 1) diagonal to the left. 0 41 302 2413 13024 024130 4130241 30241302 241302413 1302413024 02413024130 413024130241 3024130241302 24130241302413 130241302413024 0241302413024130 41302413024130241 302413024130241302 2413024130241302413 13024130241302413024 024130241302413024130 4130241302413024130241 30241302413024130241302 241302413024130241302413 1302413024130241302413024 02413024130241302413024130 413024130241302413024130241 3024130241302413024130241302 24130241302413024130241302413 130241302413024130241302413024 0241302413024130241302413024130 41302413024130241302413024130241 302413024130241302413024130241302 2413024130241302413024130241302413 13024130241302413024130241302413024 024130241302413024130241302413024130 4130241302413024130241302413024130241 30241302413024130241302413024130241302 241302413024130241302413024130241302413 1302413024130241302413024130241302413024 02413024130241302413024130241302413024130 413024130241302413024130241302413024130241 3024130241302413024130241302413024130241302 24130241302413024130241302413024130241302413 130241302413024130241302413024130241302413024 0241302413024130241302413024130241302413024130 41302413024130241302413024130241302413024130241 302413024130241302413024130241302413024130241302 2413024130241302413024130241302413024130241302413 13024130241302413024130241302413024130241302413024 024130241302413024130241302413024130241302413024130 4130241302413024130241302413024130241302413024130241 30241302413024130241302413024130241302413024130241302 241302413024130241302413024130241302413024130241302413 1302413024130241302413024130241302413024130241302413024 02413024130241302413024130241302413024130241302413024130 413024130241302413024130241302413024130241302413024130241 3024130241302413024130241302413024130241302413024130241302 24130241302413024130241302413024130241302413024130241302413 130241302413024130241302413024130241302413024130241302413024 0241302413024130241302413024130241302413024130241302413024130 41302413024130241302413024130241302413024130241302413024130241 302413024130241302413024130241302413024130241302413024130241302 2413024130241302413024130241302413024130241302413024130241302413 13024130241302413024130241302413024130241302413024130241302413024 024130241302413024130241302413024130241302413024130241302413024130 4130241302413024130241302413024130241302413024130241302413024130241 30241302413024130241302413024130241302413024130241302413024130241302 241302413024130241302413024130241302413024130241302413024130241302413 1302413024130241302413024130241302413024130241302413024130241302413024 02413024130241302413024130241302413024130241302413024130241302413024130

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Python recursion , Sierpinski triangle with color at each depth

- by ???? ???

import turtle w=turtle.Screen() def Tri(t, order, size): if order==0: t.forward(size) t.left(120) t.forward(size) t.left(120) t.forward(size) t.left(120) else: t.pencolor('red') Tri(t, order-1, size/2, color-1) t.fd(size/2) t.pencolor('blue') Tri(t, order-1, size/2, color-1) t.fd(size/2) t.lt(120) t.fd(size) t.lt(120) t.fd(size/2) t.lt(120) t.pencolor('green') Tri(t, order-1, size/2,color-1) t.rt(120) t.fd(size/2) t.lt(120) can anyone help with this problem ? i want to a sierpinski triangle that have color at specific depth like this http://openbookproject.net/thinkcs/python/english3e/_images/sierpinski_color.png i dont know how to make the the triangle color change at specific depth

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JavaScript recursion does not work properly

- by misha-moroshko

Hi, Could anyone say why the following recursive function does not work for me ? It should collect recursively all radio buttons in a given element. But, it does not found any for some reason !? Thanks !! <?xml version="1.0" encoding="Windows-1255"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <script type="text/javascript"> function AllInputs(radioElement) { this.radioInputs = ((arguments.length == 1) ? [radioElement] : []); } AllInputs.prototype.toString = function() { return "[object AllInputs: radioInputs: " + this.radioInputs.length + "]"; } AllInputs.prototype.add = function(otherAllInputs) { this.radioInputs = this.radioInputs.concat(otherAllInputs.radioInputs); } function getAllInputsOfElement(element) { if (element.tagName.toLowerCase() == "input") { if (element.getAttribute("type").toLowerCase() == "radio") { return new AllInputs(element); } else { return new AllInputs(); } } else { var result = new AllInputs(); for (i = 0; i < element.children.length; i++) { result.add(getAllInputsOfElement(element.children[i])); } return result; } } function main() { alert(getAllInputsOfElement(document.getElementById("MyTable"))); } </script> </head> <body onload="main()"> <table id="MyTable"> <tr><td>Day</td></tr> <tr><td> <input type="radio" name="DayOfTheWeek" value="1" /><label>Monday</label> <input type="radio" name="DayOfTheWeek" value="2" /><label>Tuesday</label> <input type="radio" name="DayOfTheWeek" value="3" /><label>Wednesday</label> </td></tr> </table> </body> </html>

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Using recursion to find paths in a 2D array

- by rikkit

Hi, I'm working on a project for my A level. It involves finding the maximum flow of a network, and I'm using javascript. I have a 2D array, with values in the array representing a distance between the two points. An example of the array: 0 2 2 0 0 0 1 2 0 0 0 2 0 0 0 0 I think I need to use a recursive technique to find a path; below is some pseudocode, assuming that the array is 4x4. a is (0,0), b is (3,3). function search(a,b) from a to b if element(i,j) != 0 then store value of element search(j,3) I was wondering if that was the right construction for a depth first search. Thanks for any help.

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XSLT 1.0 help with recursion logic

- by DashaLuna

Hello guys, I'm having troubles with the logic and would apprecite any help/tips. I have <Deposits> elements and <Receipts> elements. However there isn't any identification what receipt was paid toward what deposit. I am trying to update the <Deposits> elements with the following attributes: @DueAmont - the amount that is still due to pay @Status - whether it's paid, outstanding (partly paid) or due @ReceiptDate - the latest receipt's date that was paid towards this deposit Every deposit could be paid with one or more receipts. It also could happen, that 1 receipt could cover one or more deposits. For example. If there are 3 deposits: 500 100 450 That are paid with the following receipts: 200 100 250 I want to get the following info: Deposit 1 is fully paid (status=paid, dueAmount=0, receiptNum=3. Deposit 2 is partly paid (status=outstanding, dueAmount=50, receiptNum=3. Deposit 3 is not paid (status=due, dueAmount=450, receiptNum=NAN. I've added comments in the code explaining what I'm trying to do. I am staring at this code for the 3rd day now non stop - can't see what I'm doing wrong. Please could anyone help me with it? :) Thanks! Set up: $deposits - All the available deposits $receiptsAsc - All the available receipts sorted by their @ActionDate Code:  <xsl:variable name="depositsClassified"> <xsl:call-template name="classifyDeposits"> <xsl:with-param name="depositsAll" select="$deposits"/> <xsl:with-param name="receiptsAll" select="$receiptsAsc"/> <xsl:with-param name="receiptCount" select="'1'"/> </xsl:call-template> </xsl:variable>  <xsl:template name="classifyDeposits"> <xsl:param name="depositsAll"/> <xsl:param name="receiptsAll"/> <xsl:param name="receiptCount"/>  <xsl:if test="$depositsAll">  <xsl:variable name="deposit" select="$depositsAll[1]"/>  <xsl:variable name="receiptSum"> <xsl:choose> <xsl:when test="$receiptsAll"> <xsl:value-of select="sum($receiptsAll[position() <= $receiptCount]/@ReceiptAmount)"/> </xsl:when> <xsl:otherwise>0</xsl:otherwise> </xsl:choose> </xsl:variable>  <xsl:variable name="diff" select="$deposit/@DepositTotalAmount - $receiptSum"/> <xsl:choose>  <xsl:when test="($diff > 0) and ($receiptCount < count($receiptsAll))"> <xsl:call-template name="classifyDeposits"> <xsl:with-param name="depositsAll" select="$depositsAll"/> <xsl:with-param name="receiptsAll" select="$receiptsAll"/> <xsl:with-param name="receiptCount" select="$receiptCount + 1"/> </xsl:call-template> </xsl:when>  <xsl:otherwise>  <xsl:apply-templates select="$deposit" mode="defineDeposit"> <xsl:with-param name="diff" select="$diff"/> <xsl:with-param name="receiptNum" select="$receiptCount"/> </xsl:apply-templates>  <xsl:variable name="receiptsUpdatedRTF"> <xsl:for-each select="$receiptsAll"> <xsl:choose>  <xsl:when test="position() < $receiptCount"> <xsl:copy> <xsl:copy-of select="./@*"/> <xsl:attribute name="ReceiptAmount">0</xsl:attribute> </xsl:copy> </xsl:when>  <xsl:when test="position() = $receiptCount"> <xsl:copy> <xsl:copy-of select="./@*"/> <xsl:attribute name="ReceiptAmount"> <xsl:value-of select="format-number($diff, '#.00;#.00')"/> </xsl:attribute> </xsl:copy> </xsl:when>  <xsl:otherwise> <xsl:copy-of select="."/> </xsl:otherwise> </xsl:choose> </xsl:for-each> </xsl:variable> <xsl:variable name="receiptsUpdated" select="msxsl:node-set($receiptsUpdatedRTF)/Receipts"/>  <xsl:call-template name="classifyDeposits"> <xsl:with-param name="depositsAll" select="$deposits[position() != 1]"/> <xsl:with-param name="receiptsAll" select="$receiptsUpdated"/> <xsl:with-param name="receiptCount" select="$receiptCount"/> </xsl:call-template> </xsl:otherwise> </xsl:choose> </xsl:if> </xsl:template>  <xsl:template match="MultiDeposits" mode="defineDeposit"> <xsl:param name="diff"/> <xsl:param name="receiptNum"/> <xsl:choose> <xsl:when test="$diff <= 0"> <xsl:apply-templates select="." mode="addAttrs"> <xsl:with-param name="status" select="'paid'"/> <xsl:with-param name="dueAmount" select="'0'"/> <xsl:with-param name="receiptNum" select="$receiptNum"/> </xsl:apply-templates> </xsl:when> <xsl:when test="$diff = ./@DepositTotalAmount"> <xsl:apply-templates select="." mode="addAttrs"> <xsl:with-param name="status" select="'due'"/> <xsl:with-param name="dueAmount" select="$diff"/> </xsl:apply-templates> </xsl:when> <xsl:when test="$diff < ./@DepositTotalAmount"> <xsl:apply-templates select="." mode="addAttrs"> <xsl:with-param name="status" select="'outstanding'"/> <xsl:with-param name="dueAmount" select="$diff"/> <xsl:with-param name="receiptNum" select="$receiptNum"/> </xsl:apply-templates> </xsl:when> <xsl:otherwise/> </xsl:choose> </xsl:template>  <xsl:template match="MultiDeposits" mode="addAttrs"> <xsl:param name="status"/> <xsl:param name="dueAmount"/> <xsl:param name="receiptNum" select="''"/> <xsl:copy> <xsl:copy-of select="./@*"/> <xsl:attribute name="Status"><xsl:value-of select="$status"/></xsl:attribute> <xsl:attribute name="DueAmount"><xsl:value-of select="$dueAmount"/></xsl:attribute> <xsl:if test="$receiptNum != ''"> <xsl:attribute name="ReceiptDate"> <xsl:value-of select="$receiptsAsc[position() = $receiptNum]/@ActionDate"/> </xsl:attribute> </xsl:if> <xsl:copy-of select="./*"/> </xsl:copy> </xsl:template>

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HQL recursion, how do I do this?

- by niklassaers

Hi guys, I have a tree structure where each Node has a parent and a Set<Node> children. Each Node has a String title, and I want to make a query where I select Set<String> titles, being the title of this node and of all parent nodes. How do I write this query? The query for a single title is this, but like I said, I'd like it expanded for the entire branch of parents. SELECT node.title FROM Node node WHERE node.id = :id Cheers Nik

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Recursion causes exit to exit all JFrames (terminates app)

- by Trizicus

I have made an application that gives the user the option to open up a new spawn of the application entirely. When the user does so and closes the application the entire application terminates; not just the window. How should I go about recursively spawning an application and then when the user exits the JFrame spawn; killing just that JFrame and not the entire instance? Here is the relevant code: [...] JMenuItem newMenuItem = new JMenuItem ("New"); newMenuItem.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { new MainWindow(); } }); fileMenu.add(newMenuItem); [....] JMenuItem exit = new JMenuItem("Exit"); exit.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); } }); fileMenu.add(exit); [...]

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Fortran recursion segmentation faults

- by ConnorG

Hey all - I have to design and implement a Fortran routine to determine the size of clusters on a square lattice, and it seemed extremely convenient to code the subroutine recursively. However, whenever my lattice size grows beyond a certain value (around 200/side), the subroutine consistently segfaults. Here's my cluster-detection routine: RECURSIVE SUBROUTINE growCluster(lattice, adj, idx, area) INTEGER, INTENT(INOUT) :: lattice(:), area INTEGER, INTENT(IN) :: adj(:,:), idx lattice(idx) = -1 area = area + 1 IF (lattice(adj(1,idx)).GT.0) & CALL growCluster(lattice,adj,adj(1,idx),area) IF (lattice(adj(2,idx)).GT.0) & CALL growCluster(lattice,adj,adj(2,idx),area) IF (lattice(adj(3,idx)).GT.0) & CALL growCluster(lattice,adj,adj(3,idx),area) IF (lattice(adj(4,idx)).GT.0) & CALL growCluster(lattice,adj,adj(4,idx),area) END SUBROUTINE growCluster where adj(1,n) represents the north neighbor of site n, adj(2,n) represents the west and so on. What would cause the erratic segfault behavior? Is the cluster just "too huge" for large lattice sizes?

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condition in recursion - best practise

- by mo

hi there! what's the best practise to break a loop? my ideas were: Child Find(Parent parent, object criteria) { Child child = null; foreach(Child wannabe in parent.Childs) { if (wannabe.Match(criteria)) { child = wannabe; break; } else { child = Find(wannabe, criteria); } } return child; } or Child Find(Parent parent, object criteria) { Child child = null; var conditionator = from c parent.Childs where child != null select c; foreach(Child wannabe in conditionator) { if (wannabe.Match(criteria)) { child = wannabe; } else { child = Find(wannabe, criteria); } } return child; } or Child Find(Parent parent, object criteria) { Child child = null; var enumerator = parent.Childs.GetEnumerator(); while(child != null && enumerator.MoveNext()) { if (enumerator.Current.Match(criteria)) { child = wannabe; } else { child = Find(wannabe, criteria); } } return child; } what do u think, any better ideas? i'm looking for the niciest solution :D mo

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Recursion and Iteration

- by Doug

What is the difference? Are these the same? If not, can someone please give me an example? MW: Iteration - 1 : the action or a process of iterating or repeating: as a : a procedure in which repetition of a sequence of operations yields results successively closer to a desired result b : the repetition of a sequence of computer instructions a specified number of times or until a condition is met Recusion - 3 : a computer programming technique involving the use of a procedure, subroutine, function, or algorithm that calls itself one or more times until a specified condition is met at which time the rest of each repetition is processed from the last one called to the first

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build a bst as an array using recursion?

- by Jack B

String[] dictionary = new String[dictSize]; //arrray of strings from dictionary String[] tree = new String[3*dictSize]; //array of tree void makeBST() { recMakeBST(0, dictionary.length-1); }//makeBST() int a=0; void recMakeBST(int low, int high) { if(high-low==0){ return; } else{ int mid=(high-low)/2; tree[a]=dictionary[mid]; a=a+1; recMakeBST(low, mid-1); a=a+1; recMakeBST(mid+1, high); } }

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Basic Recursion, Check Balanced Parenthesis

- by pws5068

Greetings all, I've written software in the past that uses a stack to check for balanced equations, but now I'm asked to write a similar algorithm recursively to check for properly nested brackets and parenthesis. Good examples: () [] () ([]()[]) Bad examples: ( (] ([)] Suppose my function is called: isBalanced. Should each pass evaluate a smaller substring (until reaching a base case of 2 left)? Or, should I always evaluate the full string and move indices inward?

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Using recursion an append in prolog

- by Adrian

Lets say that I would like to construct a list (L2) by appending elements of another list (L) one by one. The result should be exactly the same as the input. This task is silly, but it'll help me understand how to recurse through a list and remove certain elements. I have put together the following code: create(L, L2) :- (\+ (L == []) -> L=[H|T], append([H], create(T, L2), L2);[]). calling it by create([1,2,3,4], L2) returns L2 = [1|create([2,3,4], **)\. which is not a desired result.

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sorting using recursion

- by user310587

I have the following function to sort an array with even numbers in the front and odd numbers in the back. Is there a way to get it done without using any loops? //front is 0, back =array.length-1; arrangeArray (front, back); public static void arrangeArray (int front, int back) { if (front != back || front<back) { while (numbers [front]%2 == 0) front++; while (numbers[back]%2!=0) back--; if (front < back) { int oddnum = numbers [front]; numbers[front]= numbers[back]; numbers[back]=oddnum; arrangeArray (front+1, back-1); } } }

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Recursion Problems in Prolog

- by Humble_Student

I'm having some difficulties in prolog, I'm trying to write a predicate that will return all paths between two cities, although at the moment it returns the first path it finds on an infinite loop. Not sure where I'm going wrong but I've been trying to figure this out all day and I'm getting nowhere. Any help that could be offered would be appreciated. go:- repeat, f([],0,lon,spa,OP,OD), write(OP), write(OD), fail. city(lon). city(ath). city(spa). city(kol). path(lon,1,ath). path(ath,3,spa). path(spa,2,kol). path(lon,1,kol). joined(X,Y,D):- path(X,D,Y);path(Y,D,X). f(Ci_Vi,Di,De,De,PaO,Di):- append([De],Ci_Vi,PaO), !. f(Cities_Visited,Distance,Start,Destination,Output_Path,Output_Distance):- repeat, city(X), joined(Start,X,D), not_member(X,Cities_Visited), New_Distance is Distance + D, f([Start|Cities_Visited],New_Distance,X,Destination,Output_Path,Output_Distance). not_member(X,List):- member(X,List), !, fail. not_member(X,List). The output I'm expecting here is [spa,ath,lon]4 [spa,kol,lon]3. Once again, any help would be appreciated. Many thanks in advance.

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Recursion Question : Revision

- by stan

My slides say that: A recursive call should always be on a smaller data structure than the current one There must be a non recursive option if the data structure is too small You need a wrapper method to make the recursive method accessible Just reading this from the slides makes no sense, especially seeing as it was a topic from before christmas! Could anyone try and clear up what it means please? Thank you

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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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Recursion in prepared statements

- by Rob

I've been using PDO and preparing all my statements primarily for security reasons. However, I have a part of my code that does execute the same statement many times with different parameters, and I thought this would be where the prepared statements really shine. But they actually break the code... The basic logic of the code is this. function someFunction($something) { global $pdo; $array = array(); static $handle = null; if (!$handle) { $handle = $pdo->prepare("A STATEMENT WITH :a_param"); } $handle->bindValue(":a_param", $something); if ($handle->execute()) { while ($row = $handle->fetch()) { $array[] = someFunction($row['blah']); } } return $array; } It looked fine to me, but it was missing out a lot of rows. Eventually I realised that the statement handle was being changed (executed with different param), which means the call to fetch in the while loop will only ever work once, then the function calls itself again, and the result set is changed. So I am wondering what's the best way of using PDO prepared statements in a recursive way. One way could be to use fetchAll(), but it says in the manual that has a substantial overhead. The whole point of this is to make it more efficient. The other thing I could do is not reuse a static handle, and instead make a new one every time. I believe that since the query string is the same, internally the MySQL driver will be using a prepared statement anyway, so there is just the small overhead of creating a new handle on each recursive call. Personally I think that defeats the point. Or is there some way of rewriting this?

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Recursion inside while loop, How does it work ?

- by M.H

Can you please tell me how does this java code work? : public class Main { public static void main (String[] args) { Strangemethod(5); } public static void Strangemethod(int len) { while(len > 1){ System.out.println(len-1); Strangemethod(len - 1); } } } I tried to debug it and follow the code step by step but I didn't understand it.

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Create a triangle out of stars using only recursion

- by Ramblingwood

I need to to write a method that is called like printTriangle(5);. We need to create an iterative method and a recursive method (without ANY iteration). The output needs to look like this: * ** *** **** ***** This code works with the iterative but I can't adapt it to be recursive. public void printTriangle (int count) { int line = 1; while(line <= count) { for(int x = 1; x <= line; x++) { System.out.print("*"); } System.out.print("\n"); line++; } } I should not that you cannot use any class level variables or any external methods. Thanks.

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Sudoku Recursion Issue (Java)

- by SkylineAddict

I'm having an issue with creating a random Sudoku grid. I tried modifying a recursive pattern that I used to solve the puzzle. The puzzle itself is a two dimensional integer array. This is what I have (By the way, the method doesn't only randomize the first row. I had an idea to randomize the first row, then just decided to do the whole grid): public boolean randomizeFirstRow(int row, int col){ Random rGen = new Random(); if(row == 9){ return true; } else{ boolean res; for(int ndx = rGen.nextInt() + 1; ndx <= 9;){ //Input values into the boxes sGrid[row][col] = ndx; //Then test to see if the value is valid if(this.isRowValid(row, sGrid) && this.isColumnValid(col, sGrid) && this.isQuadrantValid(row, col, sGrid)){ // grid valid, move to the next cell if(col + 1 < 9){ res = randomizeFirstRow(row, col+1); } else{ res = randomizeFirstRow( row+1, 0); } //If the value inputed is valid, restart loop if(res == true){ return true; } } } } //If no value can be put in, set value to 0 to prevent program counting to 9 setGridValue(row, col, 0); //Return to previous method in stack return false; } This results in an ArrayIndexOutOfBoundsException with a ridiculously high or low number (+- 100,000). I've tried to see how far it goes into the method, and it never goes beyond this line: if(this.isRowValid(row, sGrid) && this.isColumnValid(col, sGrid) && this.isQuadrantValid(row, col, sGrid)) I don't understand how the array index goes so high. Can anyone help me out?

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LL(8) and left-recursion

- by Peregring-lk

I want to understand the relation between LL/LR grammars and the left-recursion problem (for any question I know parcially the answer, but I ask them as I don't know nothing, because I am a little confused now, and prefer complete answers) I'm happy with sintetized or short and direct answers (or just links solving it unambiguously): What type of language isn't LL(8) languages? LL(K) and LL(8) have problems with left-recursion? Or only LL(k) parsers? LALR(1) parser have troubles with left or right recursion? What type of troubles? Only in terms of the LL/LALR comparision. What is better, Bison (LALR(1)) or Boost.Spirit (LL(8))? (Let's suppose other features of them are irrelevant in this question) Why GCC use a (hand-made) LL(8) parser? Only for the "handling-error" problem?

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