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  • Url shortening algorithm

    - by Bozho
    Now, this is not strictly about URL shortening, but my purpose is such anyway, so let's view it like that. Of course the steps to URL shortening are: Take the full URL Generate a unique short string to be the key for the URL Store the URL and the key in a database (a key-value store would be a perfect match here) Now, about the 2nd point. Here's what I've come up with: ByteArrayOutputStream baos = new ByteArrayOutputStream(); DataOutputStream dos = new DataOutputStream(baos); UUID uuid = UUID.randomUUID(); dos.writeLong(uuid.getMostSignificantBits()); String encoded = new String(Base64.encodeBase64(baos.toByteArray()), "ISO-8859-1"); String shortUrl = StringUtils.left(6); // returns the leftmost 6 characters // check if exists in database, repeat until it does not I wonder if this is good enough. Is it?

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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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  • 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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  • Finding patterns in source code

    - by trex279
    If I wanted to learn about pattern recognition in general what would be a good place to start (recommend a book)? Also, does anybody have any experience/knowledge on how to go about applying these algorithms to find abstraction patterns in programs? (repeated code, chunks of code that do the same thing, but in slightly different ways, etc.) Thanks Edit: I don't mind mathematically intensive books. In fact, that would be a good thing.

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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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  • Random access gzip stream

    - by jkff
    I'd like to be able to do random access into a gzipped file. I can afford to do some preprocessing on it (say, build some kind of index), provided that the result of the preprocessing is much smaller than the file itself. Any advice? My thoughts were: Hack on an existing gzip implementation and serialize its decompressor state every, say, 1 megabyte of compressed data. Then to do random access, deserialize the decompressor state and read from the megabyte boundary. This seems hard, especially since I'm working with Java and I couldn't find a pure-java gzip implementation :( Re-compress the file in chunks of 1Mb and do same as above. This has the disadvantage of doubling the required disk space. Write a simple parser of the gzip format that doesn't do any decompressing and only detects and indexes block boundaries (if there even are any blocks: I haven't yet read the gzip format description)

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  • PHP script to draw a spiral?

    - by Vacilando
    Does anybody have a PHP (or other language) script for drawing a spiral. A simple (Archimedean spiral) would be just fine. Of course the principle is simple but coding it in SVG or GD would take some time, so I wonder if somebody has one ready :-)

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  • how to determine base of a number?

    - by evil.coder
    Given a integer number and its reresentation in some arbitrary number system. The purpose is to find the base of the number system. For example, number is 10 and representation is 000010, then the base should be 10. Another example: number 21 representation is 0010101 then base is 2. One more example is: number is 6 and representation os 10100 then base is sqrt(2). Does anyone have any idea how to solve such problem?

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  • Write a function that returns the longest palindrome in a given string. e.g "ccddcc" in the string "

    - by Learner
    I thought of a solution but it runs in O(n^2) time Algo 1: Steps: Its a brute force method Have 2 for loops for i = 1 to i less than array.length -1 for j=i+1 to j less than array.length This way you can get substring of every possible combination from the array Have a palindrome function which checks if a string is palindrome so for every substring (i,j) call this function, if it is a palindrome store it in a string variable If you find next palindrome substring and if it is greater than the current one, replace it with current one. Finally your string variable will have the answer Issues: 1. This algo runs in O(n^2) time. Algo 2: Reverse the string and store it in diferent array Now find the largest matching substring between both the array But this too runs in O(n^2) time Can you guys think of an algo which runs in a better time. If possible O(n) time

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  • General Address Parser for Freeform Text

    - by Daemonic
    We have a program that displays map data (think Google Maps, but with much more interactivity and custom layers for our clients). We allow navigation via a set of combo boxes that prefill certain fields with a bunch of data (ie: Country: Canada, the Province field is filled in. Select Ontario, and a list of Counties/Regions is filled in. Select a county/region, and a city is filled in, etc...). While this guarantees accurate addresses, it's a pain for the users if they don't know where a street address or a city are located (ie, which county/region is kitchener in?). So we are looking at trying to do an address parser with a freeform text field. The user could enter something like this (similar to Google Maps, Bing Maps, etc...): 22 Main St, Kitchener, On And we could compartmentalize it into sections and do lookups on the data and get to the point they are looking for (or suggest alternatives). The problem with this is that how do we properly compartmentalize information? How do we break up the sections and find possible matches? I'm guessing we wouldn't be guaranteed that the user would enter data in a format we always expected (obviously). A follow up to this would be how to present the data if we don't find an exact match (or find multiple exact matches... two cities with the same street name in different counties, for example). We have a ton of data available in the mapping data (mapinfo tab format mostly). So we can do quick scans of street names, cities, states, etc. But I'm not sure about the best way to go about approaching this problem. Sure, using Google Maps would be nice, bue most of our clients are in closed in networks where outside access is not usually allowed and most aren't willing to rely on google maps (since it doesn't contain as much information as they need, such as custom map layers). They could, obviously, go to google and get the proper location then move to our software, but this would time consuming and speed of the process can be quite important.

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  • All words in a trie data-structure

    - by John Smith
    I'm trying to put all words in a trie in a string, a word is detonated by the eow field being true for a certain character in the trie data structure, hence a trie can could have letters than lead up to no word, for ex "abc" is in the trie but "c"'s eow field is false so "abc" is not a word Here is my Data structure struct Trie { bool eow; //when a Trie field isWord = true, hence there is a word char letter; Trie *letters[27]; }; and here is my attemped print-all function, basically trying to return all words in one string seperated by spaces for words string printAll( string word, Trie& data) { if (data.eow == 1) return word + " "; for (int i = 0; i < 26; i++) { if (data.letters[i] != NULL) printAll( word + data.letters[i]->letter, *(data.letters[i])); } return ""; } Its not outputting what i want, any suggestions?

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  • Diagonal of polygon is inside or outside?

    - by Himadri
    I have three consecutive points of polygon, say p1,p2,p3. Now I wanted to know whether the orthogonal between p1 and p3 is inside the polygon or outside the polygon. I am doing it by taking three vectors v1,v2 and v3. And the point before the point p1 in polygon say p0. v1 = (p0 - p1) v2 = (p2 - p1) v3 = (p3 - p1) With reference to this question, I am using the method shown in the accepted answer of that question. It is only for counterclockwise. What if my points are clockwise. I am also knowing my whole polygon is clockwise or counterclockwise. And accordingly I select the vectors v1 and v2. But still I am getting some problem. I am showing one case where I am getting problem. This polygon is counterclockwise. and It is starting from the origin of v1 and v2.

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  • CODE1 Spoj - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code and it's obviously math problem. The full problem For those who are lazy to go and read, it's like this: a0,a1,a2,...,an - sequence of N numbers B - some number known to us X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0,a1,..an. I don't know how or where to start, because not even N is known, just X and B. Can you help me ?

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  • MATLAB: Reading floating point numbers and strings from a file

    - by xsound
    I am using the following functions for writing and reading 4098 floating point numbers in MATLAB: Writing: fid = fopen(completepath, 'w'); fprintf(fid, '%1.30f\r\n', y) Reading: data = textread(completepath, '%f', 4098); where y contains 4098 numbers. I now want to write and read 3 strings at the end of this data. How do I read two different datatypes? Please help me. Thanks in advance.

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  • maintaing a sorted list that is bigger than memory

    - by tcurdt
    I have a list of tuples. [ "Bob": 3, "Alice: 2, "Jane": 1, ] When incrementing the counts "Alice" += 2 the order should be maintained: [ "Alice: 4, "Bob": 3, "Jane": 1, ] When all is in memory there rather simple ways (some more or some less) to efficiently implement this. (using an index, insert-sort etc) The question though is: What's the most promising approach when the list does not fit into memory. Bonus question: What if not even the index fits into memory? How would you approach this?

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  • Algorithms behind load-balancers?

    - by Vimvq1987
    I need to study about load-balancers, such as Network Load Balancing, Linux Virtual Server, HAProxy,...There're somethings under-the-hood I need to know: What algorithms/technologies are used in these load-balancers? Which is the most popular? most effective? I expect that these algorithms/technologies will not be too complicated. Are there some resources written about them? Thank you very much for your help.

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  • RSA Factorization problem

    - by dada
    At class we found this programming problem, and currently, we have no idea how to solve it. The positive integer n is given. It is known that n = p * q, where p and q are primes, p<=q and |q-k*p|<10^5 for some given positive integer k. You must find p and q. Input: 35 1 121 1 1000730021 9 Output: 5 * 7 11 * 11 10007 * 100003 It's not a homework, we are just trying to solve some interesting problems. If you have some ideas, please post them here so we can try something, thanks.

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  • Find three numbers appeared only once

    - by shilk
    In a sequence of length n, where n=2k+3, that is there are k unique numbers appeared twice and three numbers appeared only once. The question is: how to find the three unique numbers that appeared only once? for example, in sequence 1 1 2 6 3 6 5 7 7 the three unique numbers are 2 3 5. Note: 3<=n<1e6 and the number will range from 1 to 2e9 Memory limits: 1000KB , this implies that we can't store the whole sequence. Method I have tried(Memory limit exceed): I initialize a tree, and when read in one number I try to remove it from the tree, if the remove returns false(not found), I add it to the tree. Finally, the tree has the three numbers. It works, but is Memory limit exceed. I know how to find one or two such number(s) using bit manipulation. So I wonder if we can find three using the same method(or some method similar)? Method to find one/two number(s) appeared only once: If there is one number appeared only once, we can apply XOR to the sequence to find it. If there are two, we can first apply XOR to the sequence, then separate the sequence into 2 parts by one bit of the result that is 1, and again apply XOR to the 2 parts, and we will find the answer.

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  • CODE1 at SPOJ - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code on SPOJ, and it's obviously a math problem. The full problem For those who are lazy to go and read, it's like this: a0, a1, a2, ..., an - sequence of N numbers B - a Complex Number (has both real and imaginary components) X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0, a1, ..an. I don't know how or where to start, because not even N is known, just X and B. The problem is not as easy as expressing a number in a base B, because B is a complex number. How can it be solved?

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  • How Can I Improve This Algorithm (LCS)

    - by superguay
    (define (lcs lst1 lst2) (define (except-last-pair list) (if (pair? (cdr list)) (cons (car list) (except-last-pair (cdr list))) '())) (define (car-last-pair list) (if (pair? (cdr list)) (car-last-pair (cdr list)) (car list))) (if (or (null? lst1) (null? lst2)) null (if (= (car-last-pair lst1) (car-last-pair lst2)) (append (lcs (except-last-pair lst1) (except-last-pair lst2)) (cons (car-last-pair lst1) '())) **(if (> (length (lcs lst1 (except-last-pair lst2))) (length (lcs lst2 (except-last-pair lst1)))) (lcs lst1 (except-last-pair lst2)) (lcs lst2 (except-last-pair lst1)))))) I dont want it to run over and over.. Regards, Superguay

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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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