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  • What are the pros and cons of using manual list iteration vs recursion through fail

    - by magus
    I come up against this all the time, and I'm never sure which way to attack it. Below are two methods for processing some season facts. What I'm trying to work out is whether to use method 1 or 2, and what are the pros and cons of each, especially large amounts of facts. methodone seems wasteful since the facts are available, why bother building a list of them (especially a large list). This must have memory implications too if the list is large enough ? And it doesn't take advantage of Prolog's natural backtracking feature. methodtwo takes advantage of backtracking to do the recursion for me, and I would guess would be much more memory efficient, but is it good programming practice generally to do this? It's arguably uglier to follow, and might there be any other side effects? One problem I can see is that each time fail is called, we lose the ability to pass anything back to the calling predicate, eg. if it was methodtwo(SeasonResults), since we continually fail the predicate on purpose. So methodtwo would need to assert facts to store state. Presumably(?) method 2 would be faster as it has no (large) list processing to do? I could imagine that if I had a list, then methodone would be the way to go.. or is that always true? Might it make sense in any conditions to assert the list to facts using methodone then process them using method two? Complete madness? But then again, I read that asserting facts is a very 'expensive' business, so list handling might be the way to go, even for large lists? Any thoughts? Or is it sometimes better to use one and not the other, depending on (what) situation? eg. for memory optimisation, use method 2, including asserting facts and, for speed use method 1? season(spring). season(summer). season(autumn). season(winter). % Season handling showseason(Season) :- atom_length(Season, LenSeason), write('Season Length is '), write(LenSeason), nl. % ------------------------------------------------------------- % Method 1 - Findall facts/iterate through the list and process each %-------------------------------------------------------------- % Iterate manually through a season list lenseason([]). lenseason([Season|MoreSeasons]) :- showseason(Season), lenseason(MoreSeasons). % Findall to build a list then iterate until all done methodone :- findall(Season, season(Season), AllSeasons), lenseason(AllSeasons), write('Done'). % ------------------------------------------------------------- % Method 2 - Use fail to force recursion %-------------------------------------------------------------- methodtwo :- % Get one season and show it season(Season), showseason(Season), % Force prolog to backtrack to find another season fail. % No more seasons, we have finished methodtwo :- write('Done').

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  • Recursion problem in algorithm

    - by Marthin
    I'm not sure if this is the right place to post this, but the problem actually belongs to a programming assignment. Solve the recursion: T(0) = 2; T(n) = T(n-1) + 2; Solution: T(n) = 2(n+1) Could someone please show me how they got to that solution?

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  • python recursive iteration exceeding limit for tree implementation

    - by user3698027
    I'm implementing a tree dynamically in python. I have defined a class like this... class nodeobject(): def __init__(self,presentnode=None,parent=None): self.currentNode = presentnode self.parentNode = parent self.childs = [] I have a function which gets possible childs for every node from a pool def findchildren(node, childs): # No need to write the whole function on how it gets childs Now I have a recursive function that starts with the head node (no parent) and moves down the chain recursively for every node (base case being the last node having no children) def tree(dad,children): for child in children: childobject = nodeobject(child,dad) dad.childs.append(childobject) newchilds = findchildren(child, children) if len(newchilds) == 0: lastchild = nodeobject(newchilds,childobject) childobject.childs.append(lastchild) loopchild = copy.deepcopy(lastchild) while loopchild.parentNode != None: print "last child" else: tree(childobject,newchilds) The tree formation works for certain number of inputs only. Once the pool gets bigger, it results into "MAXIMUM RECURSION DEPTH EXCEEDED" I have tried setting the recursion limit with set.recursionlimit() and it doesn't work. THe program crashes. I want to implement a stack for recursion, can someone please help, I have gone no where even after trying for a long time ?? Also, is there any other way to fix this other than stack ?

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  • Tail recursion in Erlang

    - by dagda1
    Hi, I am really struggling to understand tail recursion in Erlang. I have the following eunit test: db_write_many_test() -> Db = db:new(), Db1 = db:write(francesco, london, Db), Db2 = db:write(lelle, stockholm, Db1), ?assertEqual([{lelle, stockholm},{francesco, london}], Db2). And here is my implementation: -module(db) . -include_lib("eunit/include/eunit.hrl"). -export([new/0,write/3]). new() -> []. write(Key, Value, Database) -> Record = {Key, Value}, [Record|append(Database)]. append([H|T]) -> [H|append(T)]; append([]) -> []. Is my implementation tail recursive and if not, how can I make it so? Thanks in advance

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  • Recursion in Unity and Dispose pattern implementation

    - by Budda
    My class is inherited from UnityContainer (from Unity 2.0), here is source code: public class UnityManager : UnityContainer { private UnityManager() { _context = new MyDataClassesDataContext(); // ... } protected override void Dispose(bool disposing) { if ( disposing ) { _context.Dispose(); } base.Dispose(disposing); } private readonly CMCoreDataClassesDataContext _context; } When Dispose method is called for the instance of UnityManager class it drop into recursion... Why? As far as I know base.Dispose should call the Dispose method of base class only... isn't it? Who call back the Dispose(bool) of UnityManager? How to prevent that? Thanks.

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  • Detecting Infinite recursion in Python or dynamic languages

    - by drozzy
    Recently I tried compiling program something like this with GCC: int f(int i){ if(i<0){ return 0;} return f(i-1); and it ran just fine. When I inspected the stack frames the compiler optimized the program to use only one frame, by just jumping back to the beginning of the function and only replacing the arguments to f. And - the compiler wasn't even running in optimized mode. Now, when I try the same thing in Python - I hit maximum recursion wall (or stack overflow). Is there way that a dynamic language like python can take advantage of these nice optimizations? Maybe it's possible to use a compiler instead of an interpreter to make this work? Just curious!

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  • javascript too much recursion?

    - by Ken
    Hi, I'm trying to make a script that automatically starts uploading after the data has been enter in the database(I need the autoId that the database makes to upload the file). When I run the javascript the scripts runs the php file but it fails calling the other php to upload the file. too much recursion setTimeout(testIfToegevoegd(),500); the script that gives the error send("/projects/backend/nieuwDeeltaak.php",'deeltaakNaam='+f.deeltaaknaam.value+'&beschrijving='+ f.beschrijving.value+'&startDatum='+f.startDatum.value+'&eindDatum='+f.eindDatum.value +'&deeltaakLeider='+f.leiderID.value+'&projectID='+f.projectID.value,id); function testIfToegevoegd(){ if(document.getElementById('resultaat').innerHTML == "<b>De deeltaak werd toegevoegd</b>"){ //stop met testen + upload file document.getElementById('nieuwDeeltaak').target = 'upload_target'; document.forms["nieuwDeeltaak"].submit() }else{ setTimeout(testIfToegevoegd(),500); } } testIfToegevoegd(); sorry for the dutch names we have to use them it is a school project. when I click the button that calls all this for a second time (after the error) it works fine.

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  • PHP object parent/child recursion

    - by Damien
    I've got a parent-child OO relationship. Parent obejcts has many child objects and every child object knows about it's parent by reference. The parent can be a child too (basically its a tree). When i do a var_dump() on the root object it says ["parent"]=RECURSION many times and the generated description will be really long. I'm wondering if i do something wrong. If yes, i'm interested in the "best practice". Thanks for the help!

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  • .NET Performance: Deep Recursion vs Queue

    - by JeffN825
    I'm writing a component that needs to walk large object graphs, sometimes 20-30 levels deep. What is the most performant way of walking the graph? A. Enqueueing "steps" so as to avoid deep recursion or B. A DFS (depth first search) which may step many levels deep and have a "deep" stack trace at times. I guess the question I'm asking is: Is there a performance hit in .NET for doing a DFS that causes a "deep" stack trace? If so, what is the hit? And would I better better off with some BFS by means of queueing up steps that would have been handled recursively in a DFS? Sorry if I'm being unclear. Thanks.

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  • Infinite recursion trying to check all elements of a TreeCtrl

    - by mavnn
    I have a TreeCtrl in which more than one Item can be assigned the same object as PyData. When the object is updated, I want to update all of the items in the tree which have that object as their PyData. I thought the following code would solve the problem quite neatly, but for some reason the logical test (current != self.GetFirstVisibleItem()) always returns true leading to infinite recursion. Can anyone explain why? def RefreshNodes(self, obj, current=None): print "Entered refresh" current = current or self.GetFirstVisibleItem() if current.IsOk(): print self.GetPyData(current).name if self.GetPyData(current) == obj: self.RefreshNode(current) current = self.GetNextVisible(current) if current != self.GetFirstVisibleItem(): self.RefreshNodes(obj, current) Edit: the above is obviously part of a class based on wx.TreeCtrl

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  • Tree + Recursion..

    - by RBA
    Hi.. I came across an article on Binary Trees Search . It uses intensive Recursive Algorithms.. I am just so confused with these stuff.. Please guide my path so as I understand these problems at ease, or any good website to read about recursion first and then solving these problems.. Please share your experience on it.. Its very urgent, and I want to learn these concepts as soon as possible.. Thankss... Regards.

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  • How to solve recursion relations in mathematica efficiently?

    - by Qiang Li
    I have a recursion to solve for. f(m,n)=Sum[f[m - 1, n - 1 - i] + f[m - 3, n - 5 - i], {i, 2, n - 2*m + 2}] + f[m - 1, n - 3] + f[m - 3, n - 7] f(0,n)=1, f(1,n)=n However, the following mma code is very inefficient f[m_, n_] := Module[{}, If[m < 0, Return[0];]; If[m == 0, Return[1];]; If[m == 1, Return[n];]; Return[Sum[f[m - 1, n - 1 - i] + f[m - 3, n - 5 - i], {i, 2, n - 2*m + 2}] + f[m - 1, n - 3] + f[m - 3, n - 7]];] It takes unbearably long to compute f[40,20]. Could anyone please suggest an efficient way of doing this? Many thanks!

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  • Limited recursion in C?

    - by function
    I ran this program and it output ... 65088 65089 65090 and then it stopped. Windows 7 said a.exe stopped working. Here is the code: #include <stdio.h> void go(void); main() { go(); } void go(void) { static int i = 0; printf("%d\n", i++); go(); } I think this program should keep on printing numbers indefinitely due to recursion, but it stops at 65090! The C code is compiled with gcc. Any ideas?

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  • Prevent recursive CTE visiting nodes multiple times

    - by bacar
    Consider the following simple DAG: 1->2->3->4 And a table, #bar, describing this (I'm using SQL Server 2005): parent_id child_id 1 2 2 3 3 4 //... other edges, not connected to the subgraph above Now imagine that I have some other arbitrary criteria that select the first and last edges, i.e. 1-2 and 3-4. I want to use these to find the rest of my graph. I can write a recursive CTE as follows (I'm using terminology from MSDN): with foo(parent_id,child_id) as ( // anchor member that happens to select first and last edges: select parent_id,child_id from #bar where parent_id in (1,3) union all // recursive member: select #bar.* from #bar join foo on #bar.parent_id = foo.child_id ) select parent_id,child_id from foo However, this results in edge 3-4 being selected twice: parent_id child_id 1 2 3 4 2 3 3 4 // 2nd appearance! How can I prevent the query from recursing into subgraphs that have already been described? I could achieve this if, in my "recursive member" part of the query, I could reference all data that has been retrieved by the recursive CTE so far (and supply a predicate indicating in the recursive member excluding nodes already visited). However, I think I can access data that was returned by the last iteration of the recursive member only. This will not scale well when there is a lot of such repetition. Is there a way of preventing this unnecessary additional recursion? Note that I could use "select distinct" in the last line of my statement to achieve the desired results, but this seems to be applied after all the (repeated) recursion is done, so I don't think this is an ideal solution. Edit - hainstech suggests stopping recursion by adding a predicate to exclude recursing down paths that were explicitly in the starting set, i.e. recurse only where foo.child_id not in (1,3). That works for the case above only because it simple - all the repeated sections begin within the anchor set of nodes. It doesn't solve the general case where they may not be. e.g., consider adding edges 1-4 and 4-5 to the above set. Edge 4-5 will be captured twice, even with the suggested predicate. :(

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  • Marshalling C# com items when using recursion

    - by Kevin
    I am using the SourceSafe COM object (SourceSafeTypeLib) from C# to automate a SourceSafe recursive get (part of a larger build process). The recursive function is shown below. How do I ensure that all the COM objects created in the foreach loop get released correctly? /// <summary> /// Recursively gets files/projects from SourceSafe (this is a recursive function). /// </summary> /// <param name="vssItem">The VSSItem to get</param> private void GetChangedFiles(VSSItem vssItem) { // 'If the object is a file perform the diff, // 'If not, it is a project, so use recursion to go through it if(vssItem.Type == (int)VSSItemType.VSSITEM_FILE) { bool bDifferent = false; //file is different bool bNew = false; //file is new //Surround the diff in a try-catch block. If a file is new(doesn't exist on //the local filesystem) an error will be thrown. Catch this error and record it //as a new file. try { bDifferent = vssItem.get_IsDifferent(vssItem.LocalSpec); } catch { //File doesn't exist bDifferent = true; bNew = true; } //If the File is different(or new), get it and log the message if(bDifferent) { if(bNew) { clsLog.WriteLine("Getting " + vssItem.Spec); } else { clsLog.WriteLine("Replacing " + vssItem.Spec); } string strGetPath = vssItem.LocalSpec; vssItem.Get(ref strGetPath, (int)VSSFlags.VSSFLAG_REPREPLACE); } } else //Item is a project, recurse through its sub items { foreach(VSSItem fileItem in vssItem.get_Items(false)) { GetChangedFiles(fileItem); } } }

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  • Knight's tour / recursion

    - by Layne
    Hey, I'm trying to learn a little bit more about recursion but somehow I can't solve the knight's tour and I'm hoping someone can point out my logic error. public class main { static int fsize = 5; static int board[][] = new int[fsize][fsize]; static int[] sprung_x = {1,2,2,1,-1,-2,-2,-1}; static int[] sprung_y = {-2,-1,1,2,2,1,-1,-2}; static void SucheWeg(int schrittnummer, int x, int y) { board[x][y] = schrittnummer; if( schrittnummer == ((fsize*fsize)-1)) { for(int i = 0; i<fsize; i++) { for(int c=0; c<fsize; c++) { System.out.printf("%3d", board[i][c]); } System.out.println("\n"); } } else { for(int i = 0; i<8; i++) { for(int c = 0; c<8; c++) { if( (x+sprung_x[i]) >= 0 && (x+sprung_x[i]) < fsize && (y+sprung_y[c]) >= 0 && (y+sprung_y[c]) < fsize ) { if(board[x+sprung_x[i]][y+sprung_y[c]] == -1) { System.out.println("Move: "+schrittnummer + "\n"); SucheWeg(schrittnummer+1, (x+sprung_x[i]), (y+sprung_y[c])); } } } } board[x][y] = -1; } } public static void main(String[] args) { System.out.println("Begin: \n"); for(int i = 0; i<fsize; i++) { for(int c = 0; c<fsize; c++) { board[i][c] = -1; } } SucheWeg(0, 0, 0); System.out.println("\nEnd"); } }

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  • I need to stop execution during recursive algorithm

    - by Shaza
    Hey, I have a problem in choosing the right method to accomplish my goal. I'm working on Algorithms teaching system, I'm using C#. I need to divide my algorithm into steps, each step will contain a recursion. I have to stop execution after each step, user can then move to the next step(next recursion) using a button in my GUI. After searching, threads was the right choice, but I found several methods: (Thread.sleep/interrupt): didn't work, my GUI freezed !! (Suspend/Resume): I've read that it's a bad idea to use. (Waithandles): still reading about them. (Monitor wait/resume). I don't have much time to try and read all previous methods, please help me in choosing the best method that fits my system.Any suggestions are extremal welcomed.

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  • jQuery rotator not rotating properly - too much recursion

    - by Matt Nathanson
    I've built a custom jQuery rotator using just basic animation to rotate the 3 Divs (images). I've built the function and then reinitiate the function using it as a call back. Here's the code: function ImageRotate() { var CurrentFeature = "#container" + featureNumber; $(CurrentFeature).stop(false, true).delay(4500).animate({'top' : '330px'}, 3000); var featureNumber2 = featureNumber-1; if ( featureNumber == 1) {featureNumber2=3} var CurrentFeature2 = "#container" + featureNumber2; $(CurrentFeature2).stop(false, true).delay(4500).animate({'top' : '0px'}, 3000); $('#container2').stop(false, true).delay(4500).animate({'top' : '-330px'}, 25); var featureNumber3 = featureNumber+1; if ( featureNumber == 3){featureNumber3=1} var CurrentFeature3 = "#container" + featureNumber3; $(CurrentFeature3).stop(false, true).delay(7500).animate({'top' : '0px'}, 3000); $(CurrentFeature2).stop(false, true).delay(4500).animate({'top' : '330px'}, 3000); $(CurrentFeature).stop(false, true).delay(4500).animate({'top' : '-330px'}, 25); if (featureNumber ==1) {featureNumber=3} else{featureNumber--}; $(CurrentFeature).stop(false, true).delay(7500).animate({'top' : '0px'}, 3000); $(CurrentFeature3).stop(false, true).delay(4500).animate({'top' : '330px'}, 3000); $(CurrentFeature2).stop(false, false).delay(4500).animate({'top' : '-330px'}, 25,ImageRotate()); }; It's worth noting that when calling the function again I also tried making another function called ImageRotate2(); and it did the same thing. It loops, but i get all sorts of funkiness. Edit: I've also tried some answers in the replies and they both leave me with recursion errors each second.

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  • Learning Haskell maps, folds, loops and recursion

    - by Darknight
    I've only just dipped my toe in the world of Haskell as part of my journey of programming enlightenment (moving on from, procedural to OOP to concurrent to now functional). I've been trying an online Haskell Evaluator. However I'm now stuck on a problem: Create a simple function that gives the total sum of an array of numbers. In a procedural language this for me is easy enough (using recursion) (c#) : private int sum(ArrayList x, int i) { if (!(x.Count < i + 1)) { int t = 0; t = x.Item(i); t = sum(x, i + 1) + t; return t; } } All very fine however my failed attempt at Haskell was thus: let sum x = x+sum in map sum [1..10] this resulted in the following error (from that above mentioned website): Occurs check: cannot construct the infinite type: a = a -> t Please bear in mind I've only used Haskell for the last 30 minutes! I'm not looking simply for an answer but a more explanation of it. Thanks in advanced.

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  • too much recursion in javascript with jquery mouseover

    - by Stacia
    I am a JS novice. Using some code I found here to help me with mouseover scale/rotate images but now when I try to apply it to more than one object I'm getting errors saying "too much recursion". Before the function didn't take any arguments, it just was on s_1 and it worked fine. I am tempted to just write different code for each object but that isn't very good programming practice. var over = false; $(function(){ $("#s_1").hover(function(){ over = true; swing_left_anim("#s_1"); }, function(){ over = false; }); $("#np_1").hover(function(){ over = true; swing_left_anim("np_1"); }, function(){ over = false; }); }); function swing_left_anim(obj){ $(obj).animate({ rotate: '0deg' }, { duration: 500 }); if (over) { $(obj).animate({ rotate: '25deg' }, 500, swing_right_anim(obj)); } } function swing_right_anim(obj){ $(obj).animate({ rotate: '-25deg' }, 500, swing_left_anim(obj)); }

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  • Nested loops through recursion with usable iterators

    - by narandzasto
    Help! I need to generate a query with loops,but there must be undefinitly numbers of loop or much as many a client wants. I know that it can be done with recursion,but don't know exectly how.One more thing. Notice,please,that I need to use those k,i,j iterators later in "if" condition and I don't know how to catch them. Thanks class Class1 { public void ListQuery() { for (int k = 0; k < listbox1.Count; k++) { for (int i = 0; i < listbox2.Count; i++) { for (int j = 0; j < listbox3.Count; j++) { if (k == listbox1.count-1 && i == listbox2.count-1 && j == listbox3.count-1) { if (!checkbox1) Query += resultset.element1 + "=" + listbox1[k]; if (!checkbox2) Query += resultset.element2 + "=" + listbox1[i]; if (!checkbox3) Query += resultset.element3 + "=" + listbox1[j]; } } } } } }

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  • C# Recursion SumOfOnlyNeg Elements

    - by Chris
    Hello, A array gets filled up with random elements (negative and positive). Now i want to calculate the sum of ONLY the postive elements. Iterative there is no problem, but in the recursion version i can only get the sum of both negative and postive. How can i "check" in the recursive version that it only sums up the Postive elements? Best Regards. Iterative version: public int IterSomPosElem(int[] tabel, int n) { n = 0; for (int i = 0; i < tabel.Length; i++) { if (tabel[i] >= 0) { n += tabel[i]; } } return n; } Recursive version atm (sums up all the elements insteed, of only the positive) public int RecuSomPosElem(int[] tabel, int n) { if(n == 1) return tabel[0]; //stopCriterium else { return (tabel[n - 1] + RecuSomPosElem(tabel, n - 1)); // how to check, so it only sums up the postive elements and "ignores" the negative elements. } }

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  • Trie Backtracking in Recursion

    - by Darksky
    I am building a tree for a spell checker with suggestions. Each node contains a key (a letter) and a value (array of letters down that path). So assume the following sub-trie in my big trie: W / \ a e | | k k | | is word--> e e | ... This is just a subpath of a sub-trie. W is a node and a and e are two nodes in its value array etc... At each node, I check if the next letter in the word is a value of the node. I am trying to support mistyped vowels for now. So 'weke' will yield 'wake' as a suggestion. Here's my searchWord function in my trie: def searchWord(self, word, path=""): if len(word) > 0: key = word[0] word = word[1:] if self.values.has_key(key): path = path + key nextNode = self.values[key] return nextNode.searchWord(word, path) else: # check here if key is a vowel. If it is, check for other vowel substitutes else: if self.isWord: return path # this is the word found else: return None Given 'weke', at the end when word is of length zero and path is 'weke', my code will hit the second big else block. weke is not marked as a word and so it will return with None. This will return out of searchWord with None. To avoid this, at each stack unwind or recursion backtrack, I need to check if a letter is a vowel and if it is, do the checking again. I changed the if self.values.has_key(key) loop to the following: if self.values.has_key(key): path = path + key nextNode = self.values[key] ret = nextNode.searchWord(word, path) if ret == None: # check if key == vowel and replace path # return nextNode.searchWord(... return ret What am I doing wrong here? What can I do when backtracking to achieve what I'm trying to do?

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