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  • Recursion problem in C

    - by jake
    Hi there. I've been trying to solve this problem for a few days now but it seems I haven't grasped the concept of recursion,yet. I have to build a program in C (recursion is a must here but loops are allowed as well) which does the following: The user inputs 2 different strings.For example: String 1 - ABC String 2 - DE The program is supposed to print strings which are combined of the ones the user has entered. the rule is that the inner order of the letters in each string (1&2) must remain. That's the output for string1=ABC & string2=DE ": abcde abdce abdec adbce adbec adebc dabce dabec daebc deabc If anyone could give me a hand here, it would be great. Thanks guys.

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  • Help needed inorder to avoid recursion

    - by Srivatsa
    Hello All I have a method, which gives me the required number of Boxes based on number of devices it can hold.Currently i have implemented this logic using recursion private uint PerformRecursiveDivision(uint m_oTotalDevices,uint m_oDevicesPerBox, ref uint BoxesRequired) { if (m_oTotalDevices< m_oDevicesPerBox) { BoxesRequired = 1; } else if ((m_oTotalDevices- m_oDevicesPerBox>= 0) && (m_oTotalDevices- m_oDevicesPerBox) < m_oDevicesPerBox) { //Terminating condition BoxesRequired++; return BoxesRequired; } else { //Call recursive function BoxesRequired++; return PerformRecursiveDivision((m_oTotalDevices- m_oDevicesPerBox), m_oDevicesPerBox, ref BoxesRequired); } return BoxesRequired; } Is there any better method to implement the same logic without using recursion. Cos this method is making my application very slow for cases when number of devices exceeds 50000. Thanks in advance for your support Constant learner

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  • merge sort recursion tree height

    - by Tony
    Hello! I am learning about recursion tree's and trying to figure out how the height of the tree is log b of n where n = 2 and one has 10 elements as input size. I am working with Merge sort. The number of times the split is done is the height of the tree as far as I understood, and the number of levels in the tree is height + 1. But if you take (for merge sort) log2 of 10 you get 1, where if you draw the tree you get at least 2 times that the recursion occurs. Where have I gone wrong? (I hope I am making sense here) NOTE: I am doing a self study, this is not homework!

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  • python multithread "maximum recursion depth exceed"

    - by user293487
    I use Python multithread to realize Quicksort. Quicksort is implement in a function. It is a recursive function. Each thread calls Quicksort to sort the array it has. Each thread has its own array that stores the numbers needs to be sorted. If the array size is smaller (<10,000). It runs ok. However, if the array size is larger, it shows the "maximum recursion depth exceed". So, I use setrecursionlimit () function to reset the recursion depth to 1500. But the program crash directly...

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  • Chaining jQuery animations using recursion crashes browser

    - by Rob Sobers
    Here's the basic idea of what I'm trying to do: Set the innerHTML of a DIV to some value X Animate the DIV When the animation finishes, change the value of X and repeat N times If I do this in a loop, what ends up happening is, because the animations occur asynchronously, the loop finishes and the DIV is set to its final value before the animations have had a chance to run for each value of X. As this question notes, the best way to solve this problem is to make a recursive call to the function in the callback handler for the animation. This way the value of the DIV doesn't change until the animation of the previous value is complete. This works perfectly...to a point. If I animate a bunch of these DIVs at the same time, my browser gets overwhelmed and crashes. Too much recursion. Can anyone think of a way to do this without using recursion?

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  • How to return a string from char[] array using recursion loop.(java)

    - by Daniel
    I am very bed in recursion... I need to convert a char[] array by using recursion loop only, into string. Without using for(),while()... loops. For example if i have char array: a[0]='H', a[1]='e', a[2]='l',a[3]= 'l',a[4]= 'o', it returns H e l l o. What I doing wrong? public String toFormattedString(char[] a) { int temp =a.length; if (a == null) return "null"; if (a.length == 0) return "0"; if( a.length == 1 ) else if( a[0] == a[a.length] ) return toFormattedString (a[a.length -1])+a[a.length];

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  • "Too much recursion" error when loading the same page with a hash

    - by Elliott
    Hi, I have a site w/ an image gallery ("Portfolio") page. There is drop-down navigation that allows a user to view a specific image in the portfolio from any page on the site. The links in the navigation use a hash, and that hash is read and converted into a string of an image filename. The image src attribute on the /portfolio/ page is then swapped out with the new image filename. This works fine if I'm clicking the dropdown link from a page OTHER THAN the /portfolio/ page itself. However if I take the same action from the /portfolio/ page, I get a "too much recursion" error in Firefox. Here's the code: Snippet of the nav markup: <li>Portfolio Category A <ul> <li><a href="/portfolio/#dining-room-table">Dining Room Table</a></li> <li><a href="/portfolio/#bathroom-mirror">Bathroom Mirror</a></li> </ul> </li> JS that reads the hash, converts it to an image filename, and swaps out the image on the page: $(document).ready(function() { if(location.pathname.indexOf("/portfolio/") > -1) { var hash = location.hash; var new_image = hash.replace("#", "")+".jpg"; swapImage(new_image); } }); function swapImage(new_image) { setTimeout(function() { $("img#current-image").attr("src", "/images/portfolio/work/"+new_image); }, 100); } I'm using the setTimeout function because I'm fading out the old image before making the swap, then fading it back in. I initially thought this was the function that was causing the recursion error, but when I remove the setTimeout I still have this problem. Does this have to do with a closure I'm not aware of? I'm pretty green on closures. JS that listens for the click on the nav: $("nav.main li.dropdown li ul li").click(function() { $(this).find("a").click(); $("nav.main").find("ul ul").hide(); $("nav.main li.hover").removeClass("hover"); }); I haven't implemented the fade in/out functionality for the dropdown nav yet, but I have implemented it for Next and Previous arrows, which can also be used to swap out images using the same swapImage function. Here's that code: $("#scroll-arrows a").click(function() { $("#current-image").animate({ opacity: 0 }, 100); var current_image = $("#current-image").attr("src").split("/").pop(); var new_image; var positions = getPositions(current_image); if($(this).is(".right")) { new_image = positions.next_img; } else { new_image = positions.prev_img; } swapImage(new_image); $("#current-image").animate({ opacity: 1 }, 100); return false; }); Here's the error I'm getting in Firefox: too much recursion var ret = handleObj.handler.apply( this, arguments ); jquery.js (line 1936) Thanks for any advice.

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  • Managing lots of callback recursion in Nodejs

    - by Maciek
    In Nodejs, there are virtually no blocking I/O operations. This means that almost all nodejs IO code involves many callbacks. This applies to reading and writing to/from databases, files, processes, etc. A typical example of this is the following: var useFile = function(filename,callback){ posix.stat(filename).addCallback(function (stats) { posix.open(filename, process.O_RDONLY, 0666).addCallback(function (fd) { posix.read(fd, stats.size, 0).addCallback(function(contents){ callback(contents); }); }); }); }; ... useFile("test.data",function(data){ // use data.. }); I am anticipating writing code that will make many IO operations, so I expect to be writing many callbacks. I'm quite comfortable with using callbacks, but I'm worried about all the recursion. Am I in danger of running into too much recursion and blowing through a stack somewhere? If I make thousands of individual writes to my key-value store with thousands of callbacks, will my program eventually crash? Am I misunderstanding or underestimating the impact? If not, is there a way to get around this while still using Nodejs' callback coding style?

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  • PocketPC c++ windows message processing recursion problem

    - by user197350
    Hello, I am having a problem in a large scale application that seems related to windows messaging on the Pocket PC. What I have is a PocketPC application written in c++. It has only one standard message loop. while (GetMessage (&msg, NULL, 0, 0)) { { TranslateMessage (&msg); DispatchMessage (&msg); } } We also have standard dlgProc's. In the switch of the dlgProc, we will call a proprietary 3rd party API. This API uses a socket connection to communicate with another process. The problem I am seeing is this: whenever two of the same messages come in quickly (from the user clicking the screen twice too fast and shouldn't be) it seems as though recursion is created. Windows begins processing the first message, gets the api into a thread safe state, and then jumps to process the next (identical ui) message. Well since the second message also makes the API call, the call fails because it is locked. Because of the design of this legacy system, the API will be locked until the recursion comes back out (which also is triggered by the user; so it could be locked the entire working day). I am struggling to figure out exactly why this is happening and what I can do about it. Is this because windows recognizes the socket communication will take time and preempts it? Is there a way I can force this API call to complete before preemption? Is there a way I can slow down the message processing or re-queue the message to ensure the first will execute (capturing it and doing a PostMessage back to itself didnt work). We don't want to lock the ui down while the first call completes. Any insight is greatly appreciated! Thanks!!

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  • Invert a string: Recursion vs iteration in javascript

    - by steweb
    Hi all, one month ago I've been interviewed by some google PTO members. One of the questions was: Invert a string recursively in js and explain the running time by big O notation this was my solution: function invert(s){ return (s.length > 1) ? s.charAt(s.length-1)+invert(s.substring(0,s.length-1)) : s; } Pretty simple, I think. And, about the big-o notation, I quickly answered O(n) as the running time depends linearly on the input. - Silence - and then, he asked me, what are the differences in terms of running time if you implement it by iteration? I replied that sometimes the compiler "translate" the recursion into iteration (some programming language course memories) so there are no differences about iteration and recursion in this case. Btw since I had no feedback about this particular question, and the interviewer didn't answer "ok" or "nope", I'd like to know if you maybe agree with me or if you can explain me whether there could be differences about the 2 kind of implementations. Thanks a lot and Regards!

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  • Recursion - Ship Battle

    - by rgorrosini
    I'm trying to write a little ship battle game in java. It is 100% academic, I made it to practice recursion, so... I want to use it instead of iteration, even if it's simpler and more efficient in most some cases. Let's get down to business. These are the rules: Ships are 1, 2 or 3 cells wide and are placed horizontally only. Water is represented with 0, non-hit ship cells are 1, hit ship cells are 2 and sunken ships have all it's cells in 3. With those rules set, I'm using the following array for testing: int[][] board = new int[][] { {0, 1, 2, 0, 1, 0}, {0, 0, 1, 1, 1, 0}, {0, 3, 0, 0, 0, 0}, {0, 0, 2, 1, 2, 0}, {0, 0, 0, 1, 1, 1}, }; It works pretty good so far, and to make it more user-friendly I would like to add a couple of reports. these are the methods I need for them: Given the matrix, return the amount of ships in it. Same as a), but separating them by state (amount of non-hit ships, hit and sunken ones). I will need a hand with those reports, and I would like to get some ideas. Remember it must be done using recursion, I want to understand this, and the only way to go is practice! Thanks a lot for your time and patience :).

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  • Permutation algorithm without recursion? Java

    - by Andreas Hornig
    Hi, I would like to get all combination of a number without any repetation. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme but couldn't find so I drawed a graph/tree for it and this screams to use recursion. But I would like to do it without, if this is possible. So could anyone please help me how to do that? Thank you in advance, Andreas

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  • Permutatation algorithm without recursion? Java

    - by Andreas Hornig
    Hi, I would like to get all combination of a number without any repetation. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme but couldn't find so I drawed a graph/tree for it and this screams to use recursion. But I would like to do it without, if this is possible. So could anyone please help mw how to do that? Thank you in advance, Andreas

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  • XSLT 1.0 recursion

    - by DashaLuna
    I'm stuck with recursion, was wondering if anyone can help me out with it. I have <Receipts> and <Deposits> elements, that are not verbose, i.e. that a <Receipts> element doesn't have an attribute to show what <Deposit> it is towards. I need to figure out <Deposits> "still amount due" and when a last receipt towards it was paid if any. I'm trying to do it with the following code: The idea was to take 1st deposit and see if there are receipts. If the deposit isn't fully paid and there are more receipts - call that recorsive function with all the same params except now count in following receipt. If there aren't any more receipts or deposit is payed - process it correctly (add required attributes). Otherwise proceed with 2nd deposit. And so on. However, the XSLT crashes with error message that "a processor stack has overflowed - possible cause is infinite template recursion" I would really appreciate any help/teps... I'm not that great with recursion and can't understand why mine here doesn't work. Thanks! :) <!-- Accumulate all the deposits with @DueAmount attribute --> <xsl:variable name="depositsClassified"> <xsl:call-template name="classifyDeposits"> <!-- a node-list of all Deposits elements ordered by DueDate Acs --> <xsl:with-param name="depositsAll" select="$deposits"/> <xsl:with-param name="depositPrevAmount" select="'0'"/> <!-- a node-list of all Receipts elements ordered by ReceivedDate Acs --> <xsl:with-param name="receiptsAll" select="$receiptsAsc"/> <xsl:with-param name="receiptCount" select="'1'"/> </xsl:call-template> </xsl:variable> <xsl:template name="classifyDeposits"> <xsl:param name="depositsAll"/> <xsl:param name="depositPrevAmount" select="'0'"/> <xsl:param name="receiptsAll"/> <xsl:param name="receiptCount"/> <xsl:if test="$deposits"> <!-- Do required operations for the 1st deposit --> <xsl:variable name="depositFirst" select="$depositsAll[1]"/> <xsl:variable name="receiptSum"> <xsl:choose> <xsl:when test="$receiptsAll"> <xsl:value-of select="sum($receiptsAll[position() &lt;= $receiptCount]/@ActionAmount)"/> </xsl:when> <xsl:otherwise>0</xsl:otherwise> </xsl:choose> </xsl:variable> <xsl:variable name="diff" select="$depositPrevAmount + $depositFirst/@DepositTotalAmount - $receiptSum"/> <xsl:choose> <xsl:when test="$diff &gt; 0 and $receiptCount &lt; $receiptsQuantityOf"> <xsl:call-template name="classifyDeposits"> <xsl:with-param name="depositsAll" select="$depositsAll"/> <xsl:with-param name="depositPrevAmount" select="$depositPrevAmount"/> <xsl:with-param name="receiptsAll" select="$receiptsAll"/> <xsl:with-param name="receiptCount" select="$receiptCount + 1"/> </xsl:call-template> </xsl:when> <xsl:otherwise> <!-- Record changes to the deposit (@DueAmount and receipt ReceivedDate) --> <xsl:apply-templates select="$depositFirst" mode="defineDeposit"> <xsl:with-param name="diff" select="$diff"/> <xsl:with-param name="latestReceiptForDeposit" select="$receiptsAll[position() = $receiptCount]"/> </xsl:apply-templates> <!-- Recursive call to the next deposit --> <xsl:call-template name="classifyDeposits"> <xsl:with-param name="depositsAll" select="$depositsAll[position() &gt; 1]"/> <xsl:with-param name="depositPrevAmount" select="$depositPrevAmount + $depositFirst/@DepositTotalAmount"/> <xsl:with-param name="receiptsAll" select="$receiptsAll"/> <xsl:with-param name="receiptCount" select="'1'"/> </xsl:call-template> </xsl:otherwise> </xsl:choose> </xsl:if> </xsl:template> <!-- Determine deposit's status, due amount and payment received date if any --> <xsl:template match="Deposits" mode="defineDeposit"> <xsl:param name="diff"/> <xsl:param name="latestReceiptForDeposit"/> <xsl:choose> <xsl:when test="$diff &lt;= 0"> <xsl:apply-templates select="." mode="addAttrs"> <xsl:with-param name="status" select="'paid'"/> <xsl:with-param name="dueAmount" select="'0'"/> <xsl:with-param name="receipt" select="$latestReceiptForDeposit"/> </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="$diff &lt; ./@DepositTotalAmount"> <xsl:apply-templates select="." mode="addAttrs"> <xsl:with-param name="status" select="'outstanding'"/> <xsl:with-param name="dueAmount" select="$diff"/> <xsl:with-param name="receipt" select="$latestReceiptForDeposit"/> </xsl:apply-templates> </xsl:when> <xsl:otherwise/> </xsl:choose> </xsl:template> <xsl:template match="Deposits" mode="addAttrs"> <xsl:param name="status"/> <xsl:param name="dueAmount"/> <xsl:param name="receipt" select="''"/> <!-- Constract a new MultiDeposits element with required info --> <xsl:copy> <xsl:copy-of select="./@*"/> <xsl:attribute name="Status"><xsl:value-of select="$status"/></xsl:attribute> <xsl:attribute name="DueAmount"><xsl:value-of select="$dueAmount"/></xsl:attribute> <xsl:if test="$receipt"> <xsl:attribute name="latestReceiptDate"> <xsl:value-of select="$receipt/@ActionDate"/> </xsl:attribute> </xsl:if> <xsl:copy-of select="./*"/> </xsl:copy> </xsl:template>

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  • Good use of recursion in chess programming?

    - by JDelage
    Hi, As part of a homework assignment, I have to program a simple chess game in Java. I was thinking of taking the opportunity to experiment with recursion, and I was wondering if there's an obvious candidate in chess for recursive code? Thank you, JDelage

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  • Print Alternating Characters From Two Strings (Interleaving) Using Recursion Java

    - by babi2balla
    I'm trying to write a method that uses recursion to print the string formed by "interleaving" the strings str1 and str2. In other words, it should alternate characters from the two strings: the first character from str1, followed by the first character from str2, followed by the second character from str1, followed by the second character from str2, etc. How would I go about this? Any ideas are greatly appreciated!! Thanks

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  • Recursion problem; completely lost

    - by timeNomad
    So I've been trying to solve this assignment whole day, just can't get it. The following function accepts 2 strings, the 2nd (not 1st) possibly containing *'s (asterisks). An * is a replacement for a string (empty, 1 char or more), it can appear appear (only in s2) once, twice, more or not at all, it cannot be adjacent to another * (ab**c), no need to check that. public static boolean samePattern(String s1, String s2) It returns true if strings are of the same pattern. It must be recursive, not use any loops, static & global variables. Can use local variables & method overloading. Can use only these methods: charAt(i), substring(i), substring(i, j), length(). Examples: 1: TheExamIsEasy; 2: "The*xamIs*y" --- true 1: TheExamIsEasy; 2: "Th*mIsEasy*" --- true 1: TheExamIsEasy; 2: "*" --- true 1: TheExamIsEasy; 2: "TheExamIsEasy" --- true 1: TheExamIsEasy; 2: "The*IsHard" --- FALSE I tried comparing the the chars one by one using charAt until an asterisk is encountered, then check if the asterisk is an empty one by comparing is successive char (i+1) with the char of s1 at position i, if true -- continue recursion with i+1 as counter for s2 & i as counter for s1; if false -- continue recursion with i+1 as counters for both. Continue this until another asterisk is found or end of string. I dunno, my brain loses track of things, can't concentrate, any pointers / hints? Am I in the right direction? Also, it's been told that a backtracking technique is to be used to solve this. My code so far (doesn't do the job, even theoretically): public static boolean samePattern(String s1, String s2) { if (s1.equals(s2) || s2 == "*") { return true; } return samePattern(s1, s2, 1); } public static boolean samePattern(String s1, String s2, int i) { if (s1.equals(s2)) return true; if (i == s2.length() - 1) // No *'s found -- not same pattern. return false; if (s1.substring(0, i).equals(s2.substring(0, i))) samePattern(s1, s2, i+1); else if (s2.charAt(i-1) == '*') samePattern(s1.substring(0, i-1), s2.substring(0, i), 1); // new smaller strings. else samePattern(s1.substring(1), s2, i); }

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  • Pyjamas & JavaScript: Too much recursion

    - by Wraith
    I'm doing a Pyjamas example and get this error: TodoApp InternalError: too much recursion Has anyone else encountered this? Some articles around the web recommend adjusting the C++ code of your browser to fix it, but that doesn't seem safe to me.

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  • C++ Recursion Issue

    - by stupidmonkey
    Hi guys, I feel a little dumb asking this, but here we go... When trying to follow the Recursion example at the following website http://www.cplusplus.com/doc/tutorial/functions2/, I ran into a road bump that has me perplexed. I altered the code slightly just to get my head around the code in the recursion example and I pretty much have my head around it, but I can't figure out why the variable 'n' increments in 'Pass B' when I have not told the program to increment 'n'. Could you please help explain this? #include <stdlib.h> #include <iostream> using namespace std; long factorial (long n) { if (n > 1) { long r(0); cout << "Pass A" << endl; cout << "n = " << n << endl; cout << "r = " << r << endl; r = n * factorial (n-1); cout << "Pass B" << endl; cout << "n = " << n << endl; cout << "r = " << r << endl; return (r); } else return (1); } int main () { long number; cout << "Please type a number: "; cin >> number; cout << number << "! = " << factorial (number) << endl; system ("pause"); return 0; }

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  • c++ recursion with arrays

    - by Sam
    I have this project that I'm working on for a class, and I'm not looking for the answer, but maybe just a few tips since I don't feel like I'm using true recursion. The problem involves solving the game of Hi Q or "peg solitaire" where you want the end result to be one peg remaining (it's played with a triangular board and one space is open at the start.) I've represented the board with a simple array, each index being a spot and having a value of 1 if it has a peg, and 0 if it doesn't and also the set of valid moves with a 2 dimensional array that is 36, 3 (each move set contains 3 numbers; the peg you're moving, the peg it hops over, and the destination index.) So my problem is that in my recursive function, I'm using a lot of iteration to determine things like which space is open (or has a value of 0) and which move to use based on which space is open by looping through the arrays. Secondly, I don't understand how you would then backtrack with recursion, in the event that an incorrect move was made and you wanted to take that move back in order to choose a different one. This is what I have so far; the "a" array represents the board, the "b" array represents the moves, and the "c" array was my idea of a reminder as to which moves I used already. The functions used are helper functions that basically just loop through the arrays to find an empty space and corresponding move. : void solveGame(int a[15], int b[36][3], int c[15][3]){ int empSpace; int moveIndex; int count = 0; if(pegCount(a) < 2){ return; } else{ empSpace = findEmpty(a); moveIndex = chooseMove( a, b, empSpace); a[b[moveIndex][0]] = 0; a[b[moveIndex][1]] = 0; a[b[moveIndex][2]] = 1; c[count][0] = b[moveIndex][0]; c[count][1] = b[moveIndex][1]; c[count][2] = b[moveIndex][2]; solveGame(a,b,c); } }

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