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  • Refining data stored in SQLite - how to join several contacts?

    - by Krab
    Problem background Imagine this problem. You have a water molecule which is in contact with other molecules (if the contact is a hydrogen bond, there can be 4 other molecules around my water). Like in the following picture (A, B, C, D are some other atoms and dots mean the contact). A B . . O / \ H H . . C D I have the information about all the dots and I need to eliminate the water in the center and create records describing contacts of A-C, A-D, A-B, B-C, B-D, and C-D. Database structure Currently, I have the following structure in the database: Table atoms: "id" integer PRIMARY KEY, "amino" char(3) NOT NULL, (HOH for water or other value) other columns identifying the atom Table contacts: "acceptor_id" integer NOT NULL, (the atom near to my hydrogen, here C or D) "donor_id" integer NOT NULL, (here A or B) "directness" char(1) NOT NULL, (this should be D for direct and W for water-mediated) other columns about the contact, such as the distance Current solution (insufficient) Now, I'm going through all the contacts which have donor.amino = "HOH". In this sample case, this would select contacts from C and D. For each of these selected contacts, I look up contacts having the same acceptor_id as is the donor_id in the currently selected contact. From this information, I create the new contact. At the end, I delete all contacts to or from HOH. This way, I am obviously unable to create C-D and A-B contacts (the other 4 are OK). If I try a similar approach - trying to find two contacts having the same donor_id, I end up with duplicate contacts (C-D and D-C). Is there a simple way to retrieve all six contacts without duplicates? I'm dreaming about some one page long SQL query which retrievs just these six wanted rows. :-) It is preferable to conserve information about who is donor where possible, but not strictly necessary. Big thanks to all of you who read this question to this point.

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  • Clojure - tail recursive sieve of Eratosthenes

    - by Konrad Garus
    I have this implementation of the sieve of Eratosthenes in Clojure: (defn sieve [n] (loop [last-tried 2 sift (range 2 (inc n))] (if (or (nil? last-tried) (> last-tried n)) sift (let [filtered (filter #(or (= % last-tried) (< 0 (rem % last-tried))) sift)] (let [next-to-try (first (filter #(> % last-tried) filtered))] (recur next-to-try filtered)))))) For larger n (like 20000) it ends with stack overflow. Why doesn't tail call elimination work here? How to fix it?

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  • How does Batcher Merge work at a high level?

    - by Mike
    I'm trying to grasp the concept of a Batcher Sort. However, most resources I've found online focus on proof entirely or on low-level pseudocode. Before I look at proofs, I'd like to understand how Batcher Sort works. Can someone give a high level overview of how Batcher Sort works(particularly the merge) without overly verbose pseudocode(I want to get the idea behind the Batcher Sort, not implement it)? Thanks!

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  • Finding unreachable sections of a 2D map

    - by dada
    I don't want you to solve this problem for me, i just want to ask for some ideas. This is the input below, and it represents a map. The 'x' represents land, and the dots - water. So with the 'x' you can represent 'islands' on the map. xxx.x...xxxxx xxxx....x...x ........x.x.x ..xxxxx.x...x ..x...x.xxx.x ..x.x.x...x.. ..x...x...xxx ...xxxxxx.... x............ As you can see, there are some islands which are closed, i.e. if some boat is inside its territory, it won't be able to get out, for ex: ..xxxxx. ..x...x. ..x.x.x. ..x...x. ..xxxxx. And there are some open islands which is possible to get out of them, ex: .xxxxx .x...x .x.x.x .xxx.x The problem is this: For a given NxM map like those above, calculate howm any of the islands are open, and how many are closed. I repeat: I don't want you to solve it, just need some sugestions, ideas for solving. thanks

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  • B-Tree Revision

    - by stan
    Hi, If we are looking for line intersections (horizontal and vertical lines only) and we have n lines with half of them vertical and no intersections then Sorting the list of line end points on y value will take N log N using mergesort Each insert delete and search of our data structue (assuming its a b-tree) will be < log n so the total search time will be N log N What am i missing here, if the time to sort using mergesort takes a time of N log N and insert and delete takes a time of < log n are we dropping the constant factor to give an overal time of N log N. If not then how comes < log n goes missing in total ONotation run time? Thanks

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  • Algorithm for Negating Sentences

    - by Kevin Dolan
    I was wondering if anyone was familiar with any attempts at algorithmic sentence negation. For example, given a sentence like "This book is good" provide any number of alternative sentences meaning the opposite like "This book is not good" or even "This book is bad". Obviously, accomplishing this with a high degree of accuracy would probably be beyond the scope of current NLP, but I'm sure there has been some work on the subject. If anybody knows of any work, care to point me to some papers?

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  • How to combine elements of a list

    - by Addie
    I'm working in c#. I have a sorted List of structures. The structure has a DateTime object which stores month and year and an integer which stores a value. The list is sorted by date. I need to traverse the list and combine it so that I only have one instance of the structure per date. For example: My initial list would look like this: { (Apr10, 3), (Apr10, 2), (Apr10, -3), (May10, 1), (May10, 1), (May10, -3), (Jun10, 3) } The resulting list should look like this: { (Apr10, 2), (May10, -1), (Jun10, 3) } I'm looking for a simple / efficient solution. The struct is: class CurrentTrade { public DateTime date; public int dwBuy; } The list is: private List<CurrentTrade> FillList

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  • How to find nth element from the end of a singly linked list?

    - by Codenotguru
    The following function is trying to find the nth to last element of a singly linked list. For example: If the elements are 8->10->5->7->2->1->5->4->10->10 then the result is 7th to last node is 7. Can anybody help me on how this code is working or is there a better and simpler approach? LinkedListNode nthToLast(LinkedListNode head, int n) { if (head == null || n < 1) { return null; } LinkedListNode p1 = head; LinkedListNode p2 = head; for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead if (p2 == null) { return null; // not found since list size < n } p2 = p2.next; } while (p2.next != null) { p1 = p1.next; p2 = p2.next; } return p1; }

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  • Drawing N-width lines?

    - by user146780
    Given a series of points, how could I calculate the vector for that line 5 pixels away? Ex: Given: \ \ \ How could I find the vector for \ \ \ \ \ \ The ones on the right. But bear in mind that it may not always be a nice straight line. I'm trying to figure out how programs like Flash can make thick outlines. Thanks

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  • Zoom image to pixel level

    - by zaf
    For an art project, one of the things I'll be doing is zooming in on an image to a particular pixel. I've been rubbing my chin and would love some advice on how to proceed. Here are the input parameters: Screen: sw - screen width sh - screen height Image: iw - image width ih - image height Pixel: px - x position of pixel in image py - y position of pixel in image Zoom: zf - zoom factor (0.0 to 1.0) Background colour: bc - background colour to use when screen and image aspect ratios are different Outputs: The zoomed image (no anti-aliasing) The screen position/dimensions of the pixel we are zooming to. When zf is 0 the image must fit the screen with correct aspect ratio. When zf is 1 the selected pixel fits the screen with correct aspect ratio. One idea I had was to use something like povray and move the camera towards a big image texture or some library (e.g. pygame) to do the zooming. Anyone think of something more clever with simple pseudo code? To keep it more simple you can make the image and screen have the same aspect ratio. I can live with that. I'll update with more info as its required.

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  • Grouping rectangles (getting the bounding boxes of rects)

    - by hyn
    What is a good, fast way to get the "final" bounding boxes of a set of random (up to about 40, not many) rectangles? By final I mean that all bounding boxes don't intersect with any other. Brute force way: in a double for loop, for each rect, test for intersection against every other rect. The intersecting rects become a new rect (replaced), indicating the bounding box. Start over and repeat until no intersection is detected. Because the rects are random every time, and the rect count is relatively small, collision detection using spatial hashing seems like overkill. Is there a way to do this more effectively?

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  • curious ill conditioned numerical problem

    - by aaa
    hello. somebody today showed me this curious ill conditioned problem (apparently pretty famous), which looks relatively simple ƒ = (333.75 - a^2)b^6 + a^2 (11a^2 b^2 - 121b^4 - 2) + 5.5b^8 + a/(2^b) where a = 77617 and b = 33096 can you determine correct answer?

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  • Refresh decorator

    - by Morgoth
    I'm trying to write a decorator that 'refreshes' after being called, but where the refreshing only occurs once after the last function exits. Here is an example: @auto_refresh def a(): print "In a" @auto_refresh def b(): print "In b" a() If a() is called, I want the refresh function to be run after exiting a(). If b() is called, I want the refresh function to be run after exiting b(), but not after a() when called by b(). Here is an example of a class that does this: class auto_refresh(object): def __init__(self, f): print "Initializing decorator" self.f = f def __call__(self, *args, **kwargs): print "Before function" if 'refresh' in kwargs: refresh = kwargs.pop('refresh') else: refresh = False self.f(*args, **kwargs) print "After function" if refresh: print "Refreshing" With this decorator, if I run b() print '---' b(refresh=True) print '---' b(refresh=False) I get the following output: Initializing decorator Initializing decorator Before function In b Before function In a After function After function --- Before function In b Before function In a After function After function Refreshing --- Before function In b Before function In a After function After function So when written this way, not specifying the refresh argument means that refresh is defaulted to False. Can anyone think of a way to change this so that refresh is True when not specified? Changing the refresh = False to refresh = True in the decorator does not work: Initializing decorator Initializing decorator Before function In b Before function In a After function Refreshing After function Refreshing --- Before function In b Before function In a After function Refreshing After function Refreshing --- Before function In b Before function In a After function Refreshing After function because refresh then gets called multiple times in the first and second case, and once in the last case (when it should be once in the first and second case, and not in the last case).

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  • Return a number between 0 and 4

    - by munchine
    How do I return a number between 0 and 4, depending the input number? For example if I pass it number 23 it will return 3. The number set should look like 0 5 10 15 20 .. 1 6 11 16 21 .. 2 7 12 17 22 .. 3 8 13 18 23 .. 4 9 14 19 24 What's the math for this?

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  • Detecting Singularities in a Graph

    - by nasufara
    I am creating a graphing calculator in Java as a project for my programming class. There are two main components to this calculator: the graph itself, which draws the line(s), and the equation evaluator, which takes in an equation as a String and... well, evaluates it. To create the line, I create a Path2D.Double instance, and loop through the points on the line. To do this, I calculate as many points as the graph is wide (e.g. if the graph itself is 500px wide, I calculate 500 points), and then scale it to the window of the graph. Now, this works perfectly for most any line. However, it does not when dealing with singularities. If, when calculating points, the graph encounters a domain error (such as 1/0), the graph closes the shape in the Path2D.Double instance and starts a new line, so that the line looks mathematically correct. Example: However, because of the way it scales, sometimes it is rendered correctly, sometimes it isn't. When it isn't, the actual asymptotic line is shown, because within those 500 points, it skipped over x = 2.0 in the equation 1 / (x-2), and only did x = 1.98 and x = 2.04, which are perfectly valid in that equation. Example: In that case, I increased the window on the left and right one unit each. My question is: Is there a way to deal with singularities using this method of scaling so that the resulting line looks mathematically correct? I myself have thought of implementing a binary search-esque method, where, if it finds that it calculates one point, and then the next point is wildly far away from the last point, it searches in between those points for a domain error. I had trouble figuring out how to make it work in practice, however. Thank you for any help you may give!

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  • KD-Trees and missing values (vector comparison)

    - by labratmatt
    I have a system that stores vectors and allows a user to find the n most similar vectors to the user's query vector. That is, a user submits a vector (I call it a query vector) and my system spits out "here are the n most similar vectors." I generate the similar vectors using a KD-Tree and everything works well, but I want to do more. I want to present a list of the n most similar vectors even if the user doesn't submit a complete vector (a vector with missing values). That is, if a user submits a vector with three dimensions, I still want to find the n nearest vectors (stored vectors are of 11 dimensions) I have stored. I have a couple of obvious solutions, but I'm not sure either one seem very good: Create multiple KD-Trees each built using the most popular subset of dimensions a user will search for. That is, if a user submits a query vector of thee dimensions, x, y, z, I match that query to my already built KD-Tree which only contains vectors of three dimensions, x, y, z. Ignore KD-Trees when a user submits a query vector with missing values and compare the query vector to the vectors (stored in a table in a DB) one by one using something like a dot product. This has to be a common problem, any suggestions? Thanks for the help.

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  • C++ string array binary search

    - by Jose Vega
    string Haystack[] = { "Alabama", "Alaska", "American Samoa", "Arizona", "Arkansas", "California", "Colorado", "Connecticut", "Delaware", "District of Columbia", "Florida", "Georgia", "Guam", "Hawaii", "Idaho", "Illinois", "Indiana", "Iowa", "Kansas", "Kentucky", "Louisiana", "Maine", "Maryland", "Massachusetts", "Michigan", "Minnesota", "Mississippi", "Missouri", "Montana", "Nebraska", "Nevada", "New Hampshire", "New Jersey", "New Mexico", "New York", "North Carolina", "North Dakota", "Northern Mariana Islands", "Ohio", "Oklahoma", "Oregon", "Pennsylvania", "Puerto Rico", "Rhode Island", "South Carolina", "South Dakota", "Tennessee", "Texas", "US Virgin Islands", "Utah", "Vermont", "Virginia", "Washington", "West Virginia", "Wisconsin", "Wyoming"}; string Needle = "Virginia"; if(std::binary_search(Haystack, Haystack+56, Needle)) cout<<"Found"; If I also wanted to find the location of the needle in the string array, is there an "easy" way to find out?

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  • Is there a website to look up common, already written functions?

    - by pinnacler
    I'm sitting here writing a function that I'm positive has been written before, somewhere on earth. It's just too common to have not been attempted, and I'm wondering why I can't just go to a website and search for a function that I can then copy and paste into my project in 2 seconds, instead of wasting my day reinventing the wheel. Sure there are certain libraries you can use, but where do you find these libraries and when they are absent, is there a site like I'm describing? Possibly a wiki of some type that contains free code that anybody can edit and improve? Edit: I can code things fine, I just don't know HOW to do them. So for example, right now, I'm trying to localize a robot/car/point in space. I KNOW there is a way to do it, just based off of range and distance. Triangulation and Trilateration. How to code that is a different story. A site that could have psuedo code, step by step how to do that would be ridiculously helpful. It would also ensure the optimal solution since everybody can edit it. I'm also writing in Matlab, which I hate because it's quirky, adding to my desire for creating a website like I describe.

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  • Reduce number of points in line

    - by culebrón
    I'm searching for algorithms to reduce the LOD of polylines, lines (looped or not) of nodes. In simple words, I want to take hi-resolution coastline data and be able to reduce its LOD hundred- or thousandfold to render it in small-scale. I found polygon reduction algorithms (but they require triangles) and Laplacian smoothing, but that doesn't seem exactly what I need.

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  • red black tree balancing?

    - by Anirudh Kaki
    i am working to generate tango tree, where i need to check whether every sub tree in tango is balanced or not. if its not balanced i need to make it balance? I trying so hard to make entire RB-tree balance but i not getting any proper logic so can any one help me out?? here i am adding code to check how to find my tree is balanced are not but when its not balanced how can i make it balance. static boolean verifyProperty5(rbnode n) { int left = 0, right = 0; if (n != null) { bh++; left = blackHeight(n.left, 0); right = blackHeight(n.right, 0); } if (left == right) { System.out.println("black height is :: " + bh); return true; } else { System.out.println("in balance"); return false; } } public static int blackHeight(rbnode root, int len) { bh = 0; blackHeight(root, path1, len); return bh; } private static void blackHeight(rbnode root, int path1[], int len) { if (root == null) return; if (root.color == "black"){ root.black_count = root.parent.black_count+1; } else{ root.black_count = root.parent.black_count; } if ((root.left == null) && (root.right == null)) { bh = root.black_count; } blackHeight(root.left, path1, len); blackHeight(root.right, path1, len); }

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