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Search found 14 results on 1 pages for 'anagram'.

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• Java Anagram Solver

  - by Alex

  I can work out how to create anagrams of a string but I don't know how I can compare them to a dictionary of real words to check if the anagram is a real word. Is there a class in the Java API that contains the entire English dictionary?

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• Anagram of a word

  - by Codenotguru

  How to print anagrams of a word from a dictionary? how to optimize it if we have to do this repeatedly?? is there any java code explanation ??

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• Java anagram recursion List<List<String>> only storing empty lists<Strings>

  - by Riff Rafffer

  Hi In this recursion method i am trying to find all anagrams and add it to a List but what happens when i run this code is it just returns alot of empty Lists. private List<List<String>> findAnagrams(LetterInventory words, ArrayList<String> anagram, int max, Map<String, LetterInventory> smallDict, int level, List<List<String>> result) { ArrayList<String> solvedWord = new ArrayList<String>(); LetterInventory shell; LetterInventory shell2; if (level < max || max == 0) { Iterator<String> it = smallDict.keySet().iterator(); while (it.hasNext()) { String k = it.next(); shell = new LetterInventory(k); shell2 = words; if (shell2.subtract(shell) != null) { anagram.add(k); shell2 = words.subtract(shell); if (shell2.isEmpty()) { //System.out.println(anagram.toString()); it prints off fine here result.add(anagram); // but doesnt add here } else findAnagrams(shell2, anagram, max, smallDict, level + 1, result); anagram.remove(anagram.size()-1); } } } return results; }

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• Word Anagram Hashing Algorithm?

  - by Ahmed Said

  Given set of words, we need to find the anagram words and display each category alone using the best algorithm input: man car kile arc none like output: man car arc kile like none the best solution I am developing now is based on a hashtable, but I am thinking about equation to convert anagram word into integer value exmaple: man = 'm'+'a'+'n' but this will not give unique values any suggestions?

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• Check my anagram code from a job interview in the past.

  - by Michael Dorgan

  Had the following as an interview question a while ago and choked so bad on basic syntax that I failed to advance (once the adrenalin kicks in, coding goes out the window.) Given a list of string, return a list of sets of strings that are anagrams of the input set. i.e. "dog","god", "foo" should return {"dog","god"}. Afterward, I created the code on my own as a sanity check and it's been around now for a bit. I'd welcome input on it to see if I missed anything or if I could have done it much more efficiently. Take it as a chance to improve myself and learn other techniques: void Anagram::doWork(list input, list &output) { typedef list SortType; SortType sortedInput; // sort each string and pair it with the original for(list<string>::iterator i = input.begin(); i != input.end(); ++i) { string tempString(*i); std::sort(tempString.begin(), tempString.end()); sortedInput.push_back(make_pair(*i, tempString)); } // Now step through the new sorted list for(SortType::iterator i = sortedInput.begin(); i != sortedInput.end();) { set<string> newSet; // Assume (hope) we have a match and pre-add the first. newSet.insert(i->first); // Set the secondary iterator one past the outside to prevent // matching the original SortType::iterator j = i; ++j; while(j != sortedInput.end()) { if(i->second == j->second) { // If the string matches, add it to the set and remove it // so that future searches need not worry about it newSet.insert(j->first); j = sortedInput.erase(j); } else { // else, next element ++j; } } // If size is bigger than our original push, we have a match - save it to the output if(newSet.size() > 1) { output.push_back(newSet); } // erase this element and update the iterator i = sortedInput.erase(i); } }

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• anagram!! problem with this code

  - by danielDhobbs

  hello people!! i have a problem with this code can you fix it for me? int anagram(char* word, int cur, int len){ int i, b = cur+1; char temp=0; char arrA[len]; printf("//%d**%d//", b, cur); for (i = 0 ; i < len ; i++) { arrA[i] = word[i]; } for (i = cur ; i < len ; i++) { if (b < len) { printf("%s\n", arrA); temp = arrA[cur]; arrA[cur] = arrA[b]; arrA[b] = temp; b++; } else if (b == len) anagram(arrA, b, len); } return 0; }

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• Code golf: find all anagrams

  - by Charles Ma

  An word is an anagram if the letters in that word can be re-arranged to form a different word. Task: Find all sets of anagrams given a word list Input: a list of words from stdin with each word separated by a new line e.g. A A's AOL AOL's Aachen Aachen's Aaliyah Aaliyah's Aaron Aaron's Abbas Abbasid Abbasid's Output: All sets of anagrams, with each set separated by a separate line Example run: ./anagram < words marcos caroms macros lump's plum's dewar's wader's postman tampons dent tend macho mocha stoker's stroke's hops posh shop chasity scythia ... I have a 149 char perl solution which I'll post as soon as a few more people post :) Have fun!

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• Finding anagaram(s) of dictionary words

  - by Codenotguru

  How can I take an input word (or sequence of letters) and output a word from a dictionary that contains exactly those letters? Does java has an English dictionary class (list of words) that I can use, or are there open source implementations of this? How can I optimize my code if this needs to be done repeatedly?

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• Shortest anagrams finder [C++]

  - by Cente

  Hey, take this as a challange, my informatics professor made it in less than 10 rows, it work perfectly and its very efficient. I'd like to "surprise" him giving a better/shorted version of it. Ofcourse i wouldn't take the credits, hah, u can trust in me. It's a simple anagras finder! If you can.. write it in c++! thanks!

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• "Anagram solver" based on statistics rather than a dictionary/table?

  - by James M.

  My problem is conceptually similar to solving anagrams, except I can't just use a dictionary lookup. I am trying to find plausible words rather than real words. I have created an N-gram model (for now, N=2) based on the letters in a bunch of text. Now, given a random sequence of letters, I would like to permute them into the most likely sequence according to the transition probabilities. I thought I would need the Viterbi algorithm when I started this, but as I look deeper, the Viterbi algorithm optimizes a sequence of hidden random variables based on the observed output. I am trying to optimize the output sequence. Is there a well-known algorithm for this that I can read about? Or am I on the right track with Viterbi and I'm just not seeing how to apply it?

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• Need help with formulating LINQ query

  - by eponymous23

  I'm building a word anagram program that uses a database which contains one simple table: Words --------------------- varchar(15) alphagram varchar(15) anagram (other fields omitted for brevity) An alphagram is the letters of a word arranged in alphabetical order. For example, the alphagram for OVERFLOW would be EFLOORVW. Every Alphagram in my database has one or more Anagrams. Here's a sample data dump of my table: Alphagram Anagram EINORST NORITES EINORST OESTRIN EINORST ORIENTS EINORST STONIER ADEINRT ANTIRED ADEINRT DETRAIN ADEINRT TRAINED I'm trying to build a LINQ query that would return a list of Alphagrams along with their associated Anagrams. Is this possible?

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• links for 2010-03-11

  - by Bob Rhubart

  Andy Mulholland: (Information Technology) + (Business Technology) ÷ Clouds = Infostructure "Internal information technology with its dedicated users, applications, licenses, client-server, data-centric and close coupled integration architecture cannot support externally oriented business technology where almost every condition is different. Internet connectivity and the emergence of people centric services in the web 2.0 world has led business and user expectations to shift dramatically and give rise to the expectation of a new and completely different working environment, based in the cloud, or more correctly, clouds." -- Andy Mulholland, CTO Blog, Capgemini (tags: enterprisearchitecture cloud web2.0 entarch) @myfear: Getting started with (GSW #2): GlassFish v3 "If the application server/container of your choice is a Java EE compliant one, you are on the right track. This list is not too long these days, if you look for Java EE 6 compliant servers. The most prominent and well-known is also the Java EE 6 reference implementation (RI): The Oracle GlassFish v3." -- Oracle ACE Markus "@myfear" Eisele (tags: oracle otn oracleace glassfish java) @oraclenerd: The"Database is a Bucket" Mentality "Could it be that everyone out there believes that the sole purpose of a database is to store data? That it can't do anything else?" -- Chet "@oraclenerd" Justice (tags: otn oracle database dba) The Encyclopedia of SOA "SOA is an anagram for OSA, which means female bear in spanish. It is a well-known fact in the spanish-speaking world that female bears are able to model business processes and optimize reusable IT assets better than any other hibernating animal." -- One of the surprisingly funny nuggets of wisdom available in the Encyclopedia of SOA. (tags: architecture chucknorris humor soa software technology webservices) Marina Fisher: Book Review - Web 2.0 Fundamentals Marina Fisher reviews WEB 2.0 FUNDAMENTALS by Oswald Campesato and Kevin Nilson. (tags: sun web2.0 bookreview socialnetworking)

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• Algorithm to generate all possible letter combinations of given string down to 2 letters

  - by Alan

  Algorithm to generate all possible letter combinations of given string down to 2 letters Trying to create an Anagram solver in AS3, such as this one found here: http://homepage.ntlworld.com/adam.bozon/anagramsolver.htm I'm having a problem wrapping my brain around generating all possible letter combinations for the various lengths of strings. If I was only generating permutations for a fixed length, it wouldn't be such a problem for me... but I'm looking to reduce the length of the string and obtain all the possible permutations from the original set of letters for a string with a max length smaller than the original string. For example, say I want a string length of 2, yet I have a 3 letter string of “abc”, the output would be: ab ac ba bc ca cb. Ideally the algorithm would produce a complete list of possible combinations starting with the original string length, down to the smallest string length of 2. I have a feeling there is probably a small recursive algorithm to do this, but can't wrap my brain around it. I'm working in AS3. Thanks!

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• Accessing parent-level controls from inside a ComboBox's child controls

  - by eponymous23

  I have XAML similar to this: <ListBox ItemsSource="{Binding SearchCriteria, Source={StaticResource model}}" SelectionChanged="cboSearchCriterionType_SelectionChanged"> <ListBox.ItemTemplate> <DataTemplate> <StackPanel Name="spCriterion" Orientation="Horizontal" Height="20"> <ComboBox Name="cboSearchCriterionType" Width="120" SelectionChanged="cboSearchCriterionType_SelectionChanged"> <ComboBox.Items> <ComboBoxItem IsSelected="True" Content="Anagram Match" /> <ComboBoxItem Content="Pattern Match" /> <ComboBoxItem Content="Subanagram Match" /> <ComboBoxItem Content="Length" /> <ComboBoxItem Content="Number of Vowels" /> <ComboBoxItem Content="Number of Anagrams" /> <ComboBoxItem Content="Number of Unique Letters" /> </ComboBox.Items> </ComboBox> <TextBox x:Name="SearchSpec" Text="{Binding SearchSpec}" /> <TextBox x:Name="MinValue" Text="{Binding MinValue}" Visibility="Collapsed" /> <TextBox x:Name="MaxValue" Text="{Binding MaxValue}" Visibility="Collapsed" /> </StackPanel> </DataTemplate> </ListBox.ItemTemplate> As you can tell from the markup, I have a listbox that is bound to a collection of SearchCriterion objects (collectively contained in a SearchCriteria object). The idea is that the user can add/remove criterion items from the criteria, each criterion is represented by a listbox item. Inside the listbox item I have a combobox and three textboxes. What I'm trying to do is change the visibility of the TextBox controls depending on the item that is selected in the ComboBox. For example, if the user selects "Pattern Match" then I want to show only the first textbox and hide the latter two; conversely, if the user selects "Length" or any of the "Number of..." items, then I want to hide the first TextBox and show the latter two. What is the best way to achieve this? I was hoping to do something simple in the SelectionChanged event handler for the listbox but the textbox controls are presumably out of the SelectionChanged event's scope. Do I have to programmatically traverse the control hierarchy and find the controls?

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