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  • Opposite of Bloom filter?

    - by abc
    Hi, I'm trying to optimize a piece of software which is basically running millions of tests. These tests are generated in such a way that there can be some repetitions. Of course, I don't want to spend time running tests which I already ran if I can avoid it efficiently. So, I'm thinking about using a Bloom filter to store the tests which have been already ran. However, the Bloom filter errs on the unsafe side for me. It gives false positives. That is, it may report that I've ran a test which I haven't. Although this could be acceptable in the scenario I'm working on, I was wondering if there's an equivalent to a Bloom filter, but erring on the opposite side, that is, only giving false negatives. I've skimmed through the literature without any luck.

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  • What's the best general programming book to review basic development concepts?

    - by Charles S.
    I'm looking for for a programming book that reviews basic concepts like implementing linked lists, stacks, queues, hash tables, tree traversals, search algorithms, etc. etc. Basically, I'm looking for a review of everything I learned in college but have forgotten. I prefer something written in the last few years that includes at least a decent amount of code in object-oriented languages. This is to study for job interview questions but I already have the "solving interview questions" books. I'm looking for something with a little more depth and explanation. Any good recommendations?

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  • Quickest algorithm for finding sets with high intersection

    - by conradlee
    I have a large number of user IDs (integers), potentially millions. These users all belong to various groups (sets of integers), such that there are on the order of 10 million groups. To simplify my example and get to the essence of it, let's assume that all groups contain 20 user IDs (i.e., all integer sets have a cardinality of 100). I want to find all pairs of integer sets that have an intersection of 15 or greater. Should I compare every pair of sets? (If I keep a data structure that maps userIDs to set membership, this would not be necessary.) What is the quickest way to do this? That is, what should my underlying data structure be for representing the integer sets? Sorted sets, unsorted---can hashing somehow help? And what algorithm should I use to compute set intersection)? I prefer answers that relate C/C++ (especially STL), but also any more general, algorithmic insights are welcome. Update Also, note that I will be running this in parallel in a shared memory environment, so ideas that cleanly extend to a parallel solution are preferred.

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  • Essence of BiMap in Google collections

    - by littleEinstein
    I am still quite puzzled at the BiMap in Google collections/Guava. It was claimed that The two bimaps are backed by the same data; any changes to one will appear in the other. I browsed through the source code, and I found the use of delegate in ForwardingMap. But in any actually subclass of StandardBiMap, I do see the data are put into both the forward and reverse map. So what is the essence, and why it claims to have saved space by keeping only one copy of the data? Is it just the actual objects are one set, but two distinct sets of references to these objects are still needed, one set maintained in forward map, and the other in reverse map? private V putInBothMaps(K key, V value, boolean force) { boolean containedKey = containsKey(key); if (containedKey && Objects.equal(value, get(key))) { return value; } if (force) { inverse().remove(value); } else if (containsValue(value)) { throw new IllegalArgumentException( "value already present: " + value); } V oldValue = super.put(key, value); updateInverseMap(key, containedKey, oldValue, value); return oldValue; }

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  • Deletion procedure for a Binary Search Tree

    - by Metz
    Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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  • Aging Data Structure in C#

    - by thelsdj
    I want a data structure that will allow querying how many items in last X minutes. An item may just be a simple identifier or a more complex data structure, preferably the timestamp of the item will be in the item, rather than stored outside (as a hash or similar, wouldn't want to have problems with multiple items having same timestamp). So far it seems that with LINQ I could easily filter items with timestamp greater than a given time and aggregate a count. Though I'm hesitant to try to work .NET 3.5 specific stuff into my production environment yet. Are there any other suggestions for a similar data structure? The other part that I'm interested in is aging old data out, If I'm only going to be asking for counts of items less than 6 hours ago I would like anything older than that to be removed from my data structure because this may be a long-running program.

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  • Graph representation

    - by Carlucho
    Given graph, how could i represent it using adj matrix?. I have read lots of tutorials, post, slides, etc but i cant get my head round it, i just need that little push.

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  • Behavior Tree Implementations

    - by Hamza Yerlikaya
    I am looking for behavior tree implementations in any language, I would like to learn more about how they are implemented and used so can roll my own but I could only find one Owyl, unfortunately, it does not contain examples of how it is used. Any one know any other open source ones that I can browse through the code see some examples of how they are used etc? EDIT: Behavior tree is the name of the data structure.

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  • A very basic auto-expanding list/array

    - by MainMa
    Hi, I have a method which returns an array of fixed type objects (let's say MyObject). The method creates a new empty Stack<MyObject>. Then, it does some work and pushes some number of MyObjects to the end of the Stack. Finally, it returns the Stack.ToArray(). It does not change already added items or their properties, nor remove them. The number of elements to add will cost performance. There is no need to sort/order the elements. Is Stack a best thing to use? Or must I switch to Collection or List to ensure better performance and/or lower memory cost?

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  • I need help with creating a data structure in PHP

    - by alex
    What I need to do is have a data structure that shows jobs organised into 14 day periods, but only when an id is the same. I've implemented all sorts of stuff, but they have failed miserably. Ideally, maybe a SQL expert could handle all of this in the query. Here is some of my code. You can assume all library stuff works as expected. $query = 'SELECT date, rig_id, comments FROM dor ORDER BY date DESC'; $dors = Db::query(Database::SELECT, $query)->execute()->as_array(); This will return all jobs, but I need to have them organised by 14 day period with the same rig_id value. $hitches = array(); foreach($dors as $dor) { $rigId = $dor['rig_id']; $date = strtotime($dor['date']); if (empty($hitches)) { $hitches[] = array( 'rigId' => $rigId, 'startDate' => $date, 'dors' => array($dor) ); } else { $found = false; foreach($hitches as $key => $hitch) { $hitchStartDate = $hitch['startDate']; $dateDifference = abs($hitchStartDate - $date); $isSameHitchTimeFrame = $dateDifference < (Date::DAY * 14); if ($rigId == $hitch['rigId'] AND $isSameHitchTimeFrame) { $found = true; $hitches[$key]['dors'][] = $dor; } } if ($found === false) { $hitches[] = array( 'rigId' => $rigId, 'startDate' => $date, 'dors' => array($dor) ); } } } This seems to work OK splitting up by rig_id, but not by date. I also think I'm doing it wrong because I need to check the earliest date. Is it possible at all to do any of this in the database query? To recap, here is my problem I have a list of jobs with all have a rig_id (many jobs can have the same) and a date. I need the data to be organised into hitches. That is, the rig_id must be the same per hitch, and they must span a 14 day period, in which the next 14 days with the same rig_id will be a new hitch. Can someone please point me on the right track? Cheers

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  • A[i] * A[j] = k in O(nlog(n))

    - by gleb-pendler
    A is an Array of n positive int numbers k given int Algorithm should find if there is a pair of numbers which product gives the result a. A[i] * A[j] = k b. A[i] = A[j] + k if there is such a couple the algorithm should return thier index. thanks in advance.

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  • check if a tree is a binary search tree

    - by TimeToCodeTheRoad
    I have written the following code to check if a tree is a Binary search tree. Please help me check the code: Pair p{ boolean isTrue; int min; int max; } public boo lean isBst(BNode v){ return isBST1(v).isTrue; } public Pair isBST1(BNode v){ if(v==null) return new Pair(true, INTEGER.MIN,INTEGER.MAX); if(v.left==null && v.right==null) return new Pair(true, v.data, v.data); Pair pLeft=isBST1(v.left); Pair pRight=isBST1(v.right); boolean check=pLeft.max<v.data && v.data<= pRight.min; Pair p=new Pair(); p.isTrue=check&&pLeft.isTrue&&pRight.isTrue; p.min=pLeft.min; p.max=pRight.max; return p; } Note: This function checks if a tree is a binary search tree

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  • Is there a way to generate a short random id, avoiding collisions, without hitting persistent storag

    - by bshacklett
    If you've used GoToMeeting, that's the type of ID I want. I'd like it to be random so that it obfuscates the number of items being tracked and short, so that it's easy to reference manually; UUIDs are way too long. I'd like to avoid hitting persistent storage merely for performance reasons, but I can't think of any other way to avoid collisions. Is 9 digits enough to do something time-based?

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  • Could the cause of recent Toyota computing problems be an interface mismatch ?

    - by Spux
    Any ideas if the recent Toyota computing errors had something to do with the fact they where using an object orintated approach then took a data orientated approach thus causing user interface errors ? Studying programming languages with in interface robotic design and wondered if the car computing glitch Toyota has been having could have something to do with using a different programming approach with out reprogramming the whole system from scratch.

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  • Different Types of Linked Lists!

    - by Jay
    What are the different types of Linked Lists which are commonly used? I know and have used the following: Singly Linked List Doubly Linked List Circular List What are the other kinds of lists that have been used by you or known to you?

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  • Graph visualization code in javascript?

    - by Chris Farmer
    Hi. I have a data structure that represents a directed graph, and I want to render that dynamically on an HTML page. Does anyone know of any javascript code that can do a reasonable job with graph layout? These graphs will usually be just a few nodes, maybe ten at the very upper end, so my guess is that performance isn't going to be a big deal. Ideally, I'd like to be able to hook it in with jQuery so that users can tweak the layout manually by dragging the nodes around. Edit: Google's Visualization API seems to be more "graphs as charts" oriented than "graphs as nodes" oriented. I didn't see any node-oriented visualizations already built there, anyway. Do you know that one exists?

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  • How to move an element in a sorted list and keep the CouchDb write "atomic"

    - by karlthorwald
    I have elements of a list in couchdb documents. Let's say these are 3 elements in 3 documents: { "id" : "783587346", "type" : "aList", "content" : "joey", "sort" : 100.0 } { "id" : "358734ff6", "type" : "aList", "content" : "jill", "sort" : 110.0 } { "id" : "abf587346", "type" : "aList", "content" : "jack", "sort" : 120.0 } A view retrieves all "aList" documents and displays them sorted by "sort". Now I want to move the elements, when I want to move "jack" to the middle, I could do this atomic in one write and change it's sort key to 105.0. The view now returns the documents in the new sort order. After a lot of sorting I could end up with sort keys like 50.99999 and 50.99998 after some years and in extreme situations run out of digits? What can you recommend, is there a better way to do this? I'd rather keep the elements in seperate documents. Different users might edit different elements in parallel (which also can get tricky). Maybe there is a much better way?

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  • What is a data structure for quickly finding non-empty intersections of a list of sets?

    - by Andrey Fedorov
    I have a set of N items, which are sets of integers, let's assume it's ordered and call it I[1..N]. Given a candidate set, I need to find the subset of I which have non-empty intersections with the candidate. So, for example, if: I = [{1,2}, {2,3}, {4,5}] I'm looking to define valid_items(items, candidate), such that: valid_items(I, {1}) == {1} valid_items(I, {2}) == {1, 2} valid_items(I, {3,4}) == {2, 3} I'm trying to optimize for one given set I and a variable candidate sets. Currently I am doing this by caching items_containing[n] = {the sets which contain n}. In the above example, that would be: items_containing = [{}, {1}, {1,2}, {2}, {3}, {3}] That is, 0 is contained in no items, 1 is contained in item 1, 2 is contained in itmes 1 and 2, 2 is contained in item 2, 3 is contained in item 2, and 4 and 5 are contained in item 3. That way, I can define valid_items(I, candidate) = union(items_containing[n] for n in candidate). Is there any more efficient data structure (of a reasonable size) for caching the result of this union? The obvious example of space 2^N is not acceptable, but N or N*log(N) would be.

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